Category: Class 11

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Tamilnadu Samacheer Kalvi 11th English Solutions Poem Chapter 6 The Hollow Crown

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The Hollow Crown Poem Summary Warm up

(a) Work with a partner take this short quiz to find out how well informed you are about history.

Hollow Crown Poem Summary Question 1.
Name a few wars and battles you have read about.
Answer:
World War I, Indo-Pak War. Battle of Panipet War of Roses.

The Hollow Crown Poem Question 2.
What is the difference between a war and a battle?
Answer:
A war is a long drawn affair. The conflict may continue even for years. Battles are small segments of a big war.

The Hollow Crown Summary Question 3.
Why do rulers wage wars and battles?
Answer:
Rulers are greedy. They want to expand their kingdoms. So, they wage wars and battles.

Summary Of The Poem The Hollow Crown Question 4.
Is the outcome of a war always fair?
Answer:
No, the outcome of war is not always fair.

Summary Of The Hollow Crown Question 5.
Do you think rulers understand the true meaning of life – in defeat or in victory?
Answer:
No, rulers involve a large number of people whose lives or deaths don’t matter for them. So, rulers usually don’t understand the true meaning of life.

Hollow Crown Poem Question 6.
Can you name a few kings and leaders who have fallen from glory to disgrace?
Answer:
Chandragupta Maurya / Rajputs and Nelson Mandela

(b) The Historical Background:

The poem is an extract from William Shakespeare’s play King Richard the Second. The play is based on true events that occurred towards the end of the 14th century. Richard II was crowned the King of England in the year 1367. He continued to be the British Monarch until 1399, when he was deposed by his cousin, Henry of Bolingbroke, who crowned himself King Henry the Fourth in the same year. Shakespeare’s play is a dramatic rendition of the last two years of King Richard IPs life.

In this brief span of time, he was ousted from his royal position and sent to prison, where he died in captivity. The following extract is set in the Coast of Wales. King Richard and some of his followers awaited the arrival of the Welsh army [after facing defeat at the hands of his cousin, Bolingbroke], of about 10000 warriors. But to their shock and surprise, they received the message that the army was not coming to their rescue. His followers tried to boost their King’s courage against the news, only in vain. When Richard came face to face with the reality of his terrible fate, he spoke the following verse, famously known as the “Hollow Crown” speech in theatrical circles. In it, King Richard is reminded of the power of Death that overshadows everything else, including the power of rulers, and renders them as powerless as any commoner at a moment’s notice.

Samacheer Kalvi 11th English The Hollow Crown Textual Questions

First, listen to a reading of the complete poem. Then, read silently and try to answer the questions briefly, based on your understanding. You may refer to the glossary given at the end of the monologue to help you.

Let’s talk of graves, of worms, and epitaphs,
Make dust our paper, and with rainy eyes
Write sorrow on the bosom of the earth.
Let’s choose executors and talk of wills.
And yet not so – for what can we bequeath
Save our deposed bodies to the ground?
Our lands, our lives, and all, are
Bolingbroke’s,

And nothing can we call our own but death;
And that small model of the barren earth
Which serves as paste and cover to our bones.
For God’s sake let us sit upon the ground
And tell sad stories of the death of kings:
How some have been depos’d, some slain in war,
Some haunted by the ghosts they have deposed,
Some poisoned by their wives, some sleeping kill’d,

All murdered – for within the hollow crown
That rounds the mortal temples of a king
Keeps Death his court, and there the antic sits,
Scoffing his state and grinning at his pomp,
Allowing him a breath, a little scene,
To monarchize, be fear’d, and kill with looks;
Infusing him with self and vain conceit,
As if this flesh which walls about our life
Were brass impregnable; and, humour’d thus,
Comes at the last, and with a little pin
Bores through his castle wall, and farewell king!
Cover your heads, and mock not flesh and blood
With solemn reverence; throw away respect,
Tradition, form, and ceremonious duty;
For you have but mistook me all this while.
I live with bread like you, feel want,
Taste grief, need friends – subjected thus,
How can you say to me, I am a king?

Hollow Crown Summary Question 1.
Pick out the phrase that suggests that King Riehard was sorrowful.
Answer:
The phrase “Talk of graves of worms and epitaphs” suggest that King Richard was sorrowful.

The Hollow Crown Poem Ppt Question 2.
Why does the King suggest that it is now time for his will to be executed?
Answer:
The King knows pretty well that he will be executed very soon by Bolingbroke. So, he wants his will to be executed.

Hollow Crown Meaning In Tamil Question 3.
What is the only thing we bequeath to our descendants?
Answer:
We bequeath only immovable property to our descendants.

Hollow Crown By Shakespeare Question 4.
What are the vanquished men left with?
Answer:
The vanquished men are left with sorrow and thoughts about death.

The Hollow Crown Poem Summary In Tamil Question 5.
What does the ‘small model’ refer to here?
Answer:
The perishable human body stands as a ‘small model’ of the barren earth.

The Hollow Crown Poem Summary In English Question 6.
What does a monarch’s crown symbolize?
Answer:
Monarch’s crown symbolizes “empty power” because real power is vested with death

The Hollow Crown Poem Figures Of Speech Question 7.
What mocks the ruler’s power and pomp?
Answer:
Death mocks the ruler’s power and pomp.

A. Fill in the blanks using the words given in the box to complete the summary of the poem:

King Richard the second had surrendered to his (a) _______ cousin, Bolingbroke. He experienced deep distress at the horror of his circumstances. In that desperate situation, he speaks of (b) _______ , (c) _______ , (d) and other things connected with death. He spoke of how people leave nothing behind and can call nothing their own, except for the small patch of (e) _______ where they will be buried. King Richard yielded to dejection and talked of all the different ways in which defeated kings suffer and how some had been deposed, (f) _______ in war, (g) _______ by their wives and so forth. He attributed this loss of lives to (h) _______ , who he personified as the jester who watches over the shoulder of every ruler, who mocks kings by allowing them to think their human flesh, was like (i) _______ brass. However, Death penetrates through the castle walls, silentlyand unnoticed like a sharp (j) _______ thus bidding (k) _______ to him and all his pride forever. Finally, Richard appealed to his soldiers not to mock his mere flesh and blood by showing (l) and respect to him. He added that he too needed bread to live, felt want, tasted (m) _______ and needed (n) _______ . He concluded thus, urging his men not to call him a (o) _______ as he was only human, just like the rest of them.
Answer:
(a) rebellious
(b) graves
(c) epitaphs
(d) worms
(e) barren earth
(f) slain
(g) poisoned
(h) death
(i) impregnable
(j) Pin
(k) farewell
(l) reverence
(m) grief
(n) friends
(o) king

B. The words used by Shakespeare find a place in the present day conversations also. Here are a few examples of how these poetic, standardized English words could be used by common people in their regular speech.

(a) Fill in the blanks with appropriate words from the box and complete the statements suitably:

[bequeath, antics, monarchise, impregnable, hollow]

1. Shravan never keeps his promises. His friends know that his words are ______
2. The spectators died laughing at the ______ of the clown.
3. The business woman wished to ______ all her riches to an orphanage, after her death.
4. The fortress was ______ and could not be conquered by the enemies.
5.  Alexander the Great, wished to conquer many lands and ______ the entire world.

Answer:

1. hollow
2. antics
3. bequeath
4. impregnable
5. monarchise

(b) Complete the passage given below, with suitable words from the box:

Lima, a (a) _______ and (b) _______ woman, kept (c) _______ at her colleagues and went on taxing them with hard labour. Though they were (d) _______ to her, she being their head, were offended and filled with (e) _______ It so happened, that Lima was (f) _______ from her high position due to a serious blunder she had committed. Lima, having lost all her (g) _______ and glory, realized how arrogant she had been. She gave up her pride and with (h) _______ sought an apology from everyone. She thus turned over a new Leaf and bid (i) _______ to them.
Answers
(a) vain
(b) conceited
(c) scoffing
(d) ceremonious
(e) sorrow
(f) deposed
(g) pomp
(h) reverence
(z) farewell

C. From your understanding of the poem, answer the following questions briefly in a sentence or two:

Question 1.
What do the three words,‘graves, worms and epitaphs’,refer to?
Answer:
Graves, epitaphs and worms refers to death and what happens to man after its visit.

Question 2.
What does the executor mentioned in the poem do?
Answer:
Executor is one who implements the contents of a will.

Question 3.
Who is Bolingbroke? Is he a friend or foe?
Answer:
Bolingbroke is a foe. He was a cousin of King Richard II. But the power craze turned him into a foe.

Question 4.
Are all deposed kings slain by the deposer?
Answer:
No, some, of the deposed kings are jailed and some are slain.

Question 5.
What does the crown of rulers stand for?
Answer:
Crown of kings stands for power and the right to rule a kingdom.

Question 6.
What hides within the crown and laughs at the king’s grandeur?
Answer:
Death hides within the crown and laughs at the king’s grandeur.

Question 7.
What does ‘flesh’mean here?
Answer:
Flesh means the human body here.

Question 8.
What are the various functions and objects given up by a defeated king?
Answer:
A defeated king abdicates his crown. He parts with his sceptre too. He hands over his right to rule the kingdom to the victorious king. He gives up the right to levy taxes on subjects. Fie also gives up his right and listens to the woes of ordinary subjects and solve them.

Question 9.
How does the king establish that he and his subjects are equal in the end?
Answer:
In the end, King Richard II pathetically explains that he is also an ordinary mortal with desires, need for friends and the compulsion to taste grief. Even a king has a cup of misery in his life.

Question 10.
Bring out King Richard’s feelings when he was defeated.
Answer:
King Richard started feeling distress about his impending death. He uses the words graves, epitaphs and worms. He realizes his possessions will be reduced to a patch of land. He recalls how kings get slain in battle field or poisoned to death by their own spouses. The king feels he is also an ordinary mortal deceived by the jester ‘death’. He also needs to taste grief and needs the support of friends during distress.

D. Explain the following lines with reference to the context in about 5 to 8 lines:

Question (i)
“Our lands, our lives, and all, are Bolingbroke’s,
And nothing can we call our own but death;”
Answer:
Reference: These lines are from the poem “The Hollow Crown” by William Shakespeare. The poem is an excerpt from the play “Richard II”.
Context: Richard II after being defeated by his rebellious cousin Bolingbroke says these words in dejection.
Explanation: Henry II is routed in the war. Some of his loyal nobles try to cheer him up. But Richard II faces the hard reality. He openly admits his failure. He says their lands, lives and all belong to the victor Bolingbroke. They can call nothing but death as their own.
Comment: Death is inevitable.

Question (ii)
“All murdered – for within the hollow crown ‘
That rounds the mortal temples of a king
Keeps Death his court, …”
Answer:
Reference: These lines are from the poem “The Hollow Crown” by William Shakespeare. The poem is an excerpt from the play “Richard II”.
Context: The defeated king thinks about death which is looming large. He remembers how other kings had met with their death. He says these words while sharing his understanding of the power of death who rules men who wear the crowns.
Explanation: A king wears a crown as a symbol of his power over the country he rules. But the empty space within the crown houses death. In the empty space, death conducts his court and gives his verdict when it is time.
Comment: The life of the dead is placed in the memory of the living.

Question (iii)
“Comes at the last, and with a little pin
Bores through his castle wall, and farewell king!”
Answer:
Reference: These lines are from the poem “The Hollow Crown” by William Shakespeare. The poem is an excerpt from the play “Richard II”.
Context: King Richard says these words while discussing the power of death over kings. Explanation: A king when he is ruling a country, looks very strong. He seems to be like an impregnable brass castle. But death with a small pin prick can easily shatter the castle. It can bid farewell easily to the king and send him to heaven.
Comment: Death may be greatest of all human blessings.

Question (iv)
“How can you say to me, I am a king? ”
Reference: This lines is from the poem “The Hollow Crown” by William Shakespeare. The poem is an excerpt from the play “Richard II”.
Context: King Richard says these words to his loyal nobels when he talks about the power of death over monarchs.
Explanation: British subjects usually believe that a king is bom with a divine right to rule. People respect his crown as a symbol of great power. After he is deposed from power, Henry II realizes the bitter truth that he is no way different from ordinary subjects. He also has wants, need for friends and the compulsion to taste grief. Nobody can escape death.
Comment: Death – the only thing inevitable in life.

Speaking Activity

E. Working with your partner, discuss the following adages and share your views with the class. You may need to give your ideas and justify your point of view. Remember to take turns while making your presentation/short speech.

Question (a)
War begets war.
Answer:
Mahatma Gandhi said, “If you are indictive and take an eye for an eye, the whole world will be blind”. Today most lethal weapons of mass destruction are being piled up in China, USA and North Korea, Russia and Iran. The leaders of these countries claim that balance of power is required in North and South. But weapons of mass destruction will not create conditions of peace. Peace has to be created by dialogues between countries. War always begets war.

Question (b)
Uneasy lies the head that wears a crown.
Answer:
Whoever is heading an organization, a team, of players, a country does have heavy responsibility. The leadership may give the person a social recognition but in day to day life, the responsibilities of a leader are really heavy. A captain of the army during Kargil war, found one of his soldiers wounded. The Kargil war was heading to a victory for India. The captain did not allow his junior officers to go and bring the wounded soldier. He went and received the bullets. Yet he pulled the wounded soldier to safety. He brought the wounded soldier to the bunk. While returning also he was shot many times. He dropped down dead. He had saved the wounded soldier and the subordinate officer at the cost of his life. Sometimes, there is a coldwar, people try to usurp power by secret dealings.

Aurangzeb killed many of his brothers to ascend to the throne. While in power, kings are really worried about the conspiracy being cooked by relatives to overthrow him. King’s wife poisons king to death. Kings heading battles get killed too. So, we should never be jealous of people in power. Each post or power carries its own stress and unresolved conflicts, occasionally resulting in depression too. Being the head of an army, or that of a country is not always a matter of pride or glory. The grandeur conceals pain, anxiety and ever fear of impending death.

F. Poetic Devices

(a) Read the poem once again carefully and identify the figure of speech that has been used in each of the following lines from the poem:

1. “Let’s talk of graves, of worms, and epitaphs Make dust our paper, and with rainy eyes Write sorrow on the bosom of the earth”.
2. (“And yet not so – for what can we bequeath Save our deposed bodies to the ground?”
3. “Keeps Death his court, and there the antic sits,..
4. “How can you say to me, I am a king?”
5. “Scoffing his state and grinning at his pomp,…”
6. “Bores through his castle wall, and farewell

Answer:

1. Personification (Earth)
2. Metaphor
3. Personification
4. Interrogation
5. Personification
6. Personification

(b) Pick out the words in alliteration from the following lines:

Question (i)
“Our lands, our lives, and all, are Bolingbroke’s,…”
Answer:
lands, fives

Question (ii)
“And tell sad stories of the death of kings:”
Answer:
sad, stories

Question (iii)
“Comes at the last, and with a little pin…”
Answer:
last, little

G. Based on your reading of King Richard’s speech, answer the following questions in about 100 – 150 words each. You may add your own ideas if required to present and justify your point of view.

Question 1.
What are the causes for King Richard’s grief?
Answer:
King Richard II was a popular king. He had many nobles at the service. His rebellious cousin
Bolingbroke attacks him with 10,000 men on his side. He sends message to the Welsh King for . sending his army to defeat Bolingbroke. But to his shock, Welsh army is not sent. He realizes with alarm the terrible fate he would suffer in the hands of his foe and his most impending death in captivity. King Richard is reminded of the power of death that overshadows everything else. Death scoffs at the power of rulers. Losing the battle, non-receipt of Welsh army and the prospect being jailed and killed worries Richard II.

He realizes that in the hollow crown death had reigned him. Infact, death, a jester had misled him to believe that he was monarchising England. He can now own only a patch of barren land. He is not an impregnable castle of brass anymore. He is an ordinary mortal. He too needs friends and needs to taste grief and face death.

“Life and death are illusions. We are in a constant state of transformation.”

Question 2.
How are the eternal truths and wisdom brought to the reader here?
Answer:
Human’s glorious life gets reduced to graves, epitaphs and worms. Men is left with nothing but his mortal remains to gift to the earth. The earth only serves as a paste and cover to the dead bodies. Great kings too have had inglorious death. Duncan was killed in bed. Hamlet was poisoned to death. Macbeth was slain in the war. The death gives freedom to monarchs from monarchising the country.

The king realizes with a shudder that Death has occupied a prominent position right inside the crown. He scoffs at the pomp and show of the temporal kings. Even the most powerful monarch who feels as strong as a brass castle is brought down by just a pin prick of death. Death is a great leveller who makes kings believe that they are also ordinary mortals with wants, need for friends and the need to taste grief.

“Life is a brief intermission between Birth and Death. Enjoy it.”

Question 3.
Death has been cited to in many ways in this monologue. Identify the poetic devices used in those references.
Answer:
bequeath deposed bodies – Metaphor
small model of barren earth-Metaphor
hollow crown – Metaphor
antics – Personification
Dust our paper – Metaphor
scoffing his state grinning at his pomp – Personification

Question 4.
Who does the future generations remember easily – the victor or the vanquished? Give reasons. Also, cite relevant references from King Richard’s speech.
Answer:
Unusually future generations remember victors. But there are rare instances of just rulers falling due to the conspiracy and greed of an aggressor. On such occasions, future generations remember the vanquished. A Shiva devotee king was very generous. His enemies entered his kingdom under the guise of Shiva devotees in saffron clothes and slew the king and captured his kingdom. Alexander, King Richard was a just ruler. He was loved by his subjects and loyal nobles. He was defeated by his rebellious cousin simply because he wanted to be a king. When Richard was thinking about the welfare of his subjects, Bolingbroke was secretly raising an army to dethrone him.

People who are mad after power resort to unjust means. So, British subjects respected and loved the vanquished but were helpless and defeated Porus who had fought so valiantly and wanted to be treated with respect befitting a king. Alexander himself respected him and returned his kingdom and sealed a life time friendship with him. From King Richard’s speech one understands that he was good at heart but in the strategy of war, he was not good. Like a crooked end of a straight walking stick, a ruler has to have some secret deals with neighbouring countries to be protected during crisis. Bolingbroke turned out to be a more assertive and Shrewd king. But people would remember a just and noble person more even if defeated.

“Nobility of spirit has more to do with Simplicity than Ostentation, Wisdom than Wealth, Commitment rather than Ambition.

The Hollow Crown About the Poet

William Shakespeare (1564 – 1616), an English poet and playwright is widely regarded as the greatest writer in English language and the world’s pre-eminent dramatist. He was born and brought up in Stratford-upon-Avon, Warwickshire. He wrote about 39 plays, 154 sonnets, two long narrative poems, and a few other verses. He was often called England’s National Poet and nicknamed the Bard of Avon. The first publishing of Shakespeare’s works is the ‘The First Folio’. Playwright Ben Johnson wrote a preface to this book including the quote ‘(Shakespeare) is not of an age, but for all time.’ His plays have been translated into every major living language and are constantly studied and performed throughout the world.

The Hollow Crown Summary

King Richard II surrenders to his rebellious cousin Bolingbroke. The King talks to the few loyal friends on the nature of temporal power and how death over takes everything and everybody. Under critical circumstances, King Richard II talk about graves, epitaphs and worms. Shakespeare portrays the fleeting nature of human glory. He explains how even monarchs leave nothing behind to call as their own except a small patch of land into which they will get buried. The dejected King talks on various ways Kings get killed. Some are slain in the battle field.

Some are poisoned to death by their own spouses. The Kings who believed their bodies to be forts or impregnable brass are shattered by just a pinprick. The whole castle wall, the human body, is gone. Death like a jester waits for the King. In fact, he only allows the King to act as if he is ruling and in control of everything. In fact, death is in supreme command. He chides his loyal friends who still believe that he is a monarch. He tells them that he is an ordinary mortal just like them with basic wants and the need to taste grief. He is humbled and realizes he can no more be called a King as he is powerless before the impending death.

The Hollow Crown Glossary

Textual:
antic – someone who attention through silly or funny acts (here a court jester)
bequeath – pass on something to the next generation by means Of a will
ceremonious – being very formal
deposed – removed from office or power
epitaph – short pieces of writing inscribed on tombstones in memory of the dead
executors – persons who put someone’s terms of will into effect
grinning – smiling wildly
impregnable – impossible to pass through
monarchize – rule , carry out the duties functions of a ruler
scoffing – expressing mockery
slain – kill

Additional:
critical – serious
humble – modest
jester – clown
monarch – king
portrays – describes
spouse – wife
temporal – temporary

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You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4

11th Maths Exercise 2.4 Solutions Question 1.
Construct a quadratic equation with roots 7 and -3.
Solution:
The given roots are 7 and -3
Let α = 7 and β = -3
α + β = 7 – 3 = 4
αβ = (7)(-3) = -21
The quadratic equation with roots α and β is x² – (α + β) x + αβ = 0
So the required quadratic equation is
x² – (4) x + (-21) = 0
(i.e.,) x² – 4x – 21 = 0

11th Maths Exercise 2.4 Question 2.
A quadratic polynomial has one of its zeros 1 + \(\sqrt{5}\) and it satisfies p(1) = 2. Find the quadratic polynomial.
Solution:
Given α = 1 + \(\sqrt{5}\) So, β = 1 – \(\sqrt{5}\)

The quadratic polynomial is
p(x) = x² – (α + β)x + αβ
p(x) = k (x² – 2x – 4)
p( 1) = k(1 – 2 – 4) = -5 k
Given p (1) = 2

Exercise 2.4 Class 11 Maths Question 3.
If α and β are the roots of the quadratic equation x² + \(\sqrt{2}\)x + 3 = 0, form a quadratic polynomial with zeroes 1/α, 1/β.
Solution:
α and β are the roots of the equation x² + \(\sqrt{2}\)x + 3 = 0

11th Maths Exercise 2.4 Answers Question 4.
If one root of k(x – 1)² = 5x – 7 is double the other root, show that k = 2 or – 25.
Solution:
k(x – 1)² = 5x – 7
(i.e.,) k(x² – 2x + 1) – 5x + 7 = 0
x² (k) + x(-2k – 5) + k + 1 = 0
kx² – x(2k + 5) + (k + 7) = 0
Here it is given that one root is double the other.
So let the roots to α and 2α

2(4k² + 25 + 20k) = 9k (k + 7)
2(4k² + 25 + 20k) = 9k² + 63k
8k² + 50 + 40k – 9k² – 63k = 0
-k² – 23k + 50 = 0
k² + 23k – 5o = 0
(k + 25)(k – 2) = 0
k = -25 or 2

Exercise 2.4 Class 11 Question 5.
If the difference of the roots of the equation 2x² – (a + 1)x + a – 1 = 0 is equal to their product then prove that a = 2.
Solution:

Samacheer Kalvi 11th Maths Solution Question 6.
Find the condition that one of the roots of ax² + bx + c may be
(i) negative of the other
(ii) thrice the other
(iii) reciprocal of the other.
Solution:
(i) Let the roots be α and -β
Sum of the roots = – b/a = 0 ⇒ b = 0

(ii) Let the roots be α, 3α

Samacheer Kalvi 11 Maths Solutions Question 7.
If the equations x² – ax + b = 0 and x² – ex + f = 0 have one root in common and if the second equation has equal roots that ae = 2(b + f).
Solution:

Samacheer Kalvi Guru 11th Maths Question 8.
Discuss the nature of roots of
(i) -x² + 3x + 1 = 0
(ii) 4x² – x – 2 = 0
(iii) 9x² + 5x = 0
Solution:
(i) -x² + 3x + 1 = 0
⇒ comparing with ax² + bx + c = 0
∆ = b² – 4ac = (3)² – 4(1)(-1) = 9 + 4 = 13 > 0
⇒ The roots are real and distinct

(ii) 4x² – x – 2 = 0
a = 4, b = -1, c = -2
∆ = b² – 4ac = (-1)² – 4(4)(-2) = 1 + 32 = 33 >0
⇒ The roots are real and distinct

(iii) 9x² + 5x = 0
a = 9, b = 5, c = 0
∆ = b² – 4ac = 5² – 4(9)(0) = 25 > 0
⇒ The roots are real and distinct

Samacheer Kalvi 11th Maths Solutions Question 9.
Without sketching the graphs find whether the graphs of the following functions will intersect the x- axis and if so in how many points.
(i) y = x² + x + 2
(ii) y = x² – 3x – 1
(iii) y = x² + 6x + 9
Solution:

Samacheer Kalvi Class 11 Maths Solutions Question 10.
Write f(x) = x² + 5x + 4 in completed square form.
Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 Additional Questions

11th Maths Solutions Samacheer Kalvi Question 1.
Find the values of k so that the equation x² = 2x (1 + 3k) + 7(3 + 2k) = 0 has real and equal roots.
Solution:
The equation is x² – x(2) (1 + 3k) + 7 (3 + 2&) = 0
The roots are real and equal
⇒ ∆ = 0 (i.e.,) b² – 4ac = 0
Here a = 1, b = -2 (1 + 3k), c = 7(3 + 2k)
So b² – 4ac = 0 ⇒ [-2 (1 + 3k)]² – 4(1) (7) (3 + 2k) = 0
(i.e.,) 4 (1 + 3k)² – 28 (3 + 2k) = 0
(÷ by 4) (1 + 3k)² – 7(3 + 2k) = 0
1 + 9k² + 6k – 21 – 14k = 0
9k² – 8k – 20 = 0
(k – 2)(9k + 10) = 0

To solve the quadratic inequalities ax² + bx + c < 0 (or) ax² + bx + c > 0

11th Maths Samacheer Kalvi Solutions Question 2.
If the sum and product of the roots of the quadratic equation ax² – 5x + c = 0 are both equal to 10 then find the values of a and c.
Solution:
The given equation is ax² – 5x + c = 0
Let the roots be α and β Given α + β = 10 and αβ = 10

Samacheer Kalvi 11th Maths Question 3.
If α and β are the roots of the equation 3x² – 4x + 1 = 0, form the equation whose roots are \(\frac{\alpha^{2}}{\beta}\) and \(\frac{\beta^{2}}{\alpha}\)
Solution:

11th Maths Chapter 2 Solution Samacheer Kalvi Question 4.
If one root of the equation 3x² + kx – 81 = 0 is the square of the other then find k.
Solution:

Samacheer Kalvi Guru 11th Maths Solution Question 5.
If one root of the equation 2x² – ax + 64 = 0 is twice that of the other then find the value of a
Solution:

Students can find the most related topics which helps them to analyse the concepts if they practice according to the chapter-wise page. It is necessary for the students to practice more Questions and Answers for Tamilnadu State Board Solutions of 11th Commerce are given in the pdf format in chapter 4 Sole Proprietorship Questions and Answers so that students can prepare in both online and offline modes. So, Download Samacheer Kalvi 11th Commerce Book Solutions Questions and Answers, Notes Pdf, to score good marks.

Samacheer Kalvi 11th Commerce Solutions Chapter 4 Sole Proprietorship

Get the Questions and Answers, in Tamilnadu State Board 11th Commerce Solutions for Chapter 4 Sole Proprietorship. Learn the concepts of 11th Commerce Chapter-Wise by referring to the Tamilnadu State Board Solutions for Chapter 4 Sole Proprietorship Questions and Answers. Hence we suggest the students to Download Samacheer Kalvi 11th Commerce Book Solutions Questions and Answers pdf to enhance your knowledge.

Samacheer Kalvi 11th Commerce Sole Proprietorship Textbook Exercise Questions and Answers

I. Choose the Correct Answer

11th Commerce 4th Lesson Questions And Answers Question 1.
Which is the oldest form of Business organisation?
(a) Sole Proprietorship
(b) Partnership
(c) Co – operative Society
(d) Company
Answer:
(a) Sole Proprietorship

Sole Proprietorship Questions And Answers Question 2.
In which form the owner, establisher and manager is only one?
(a) Joint Enterprise
(b) Government Company
(c) Co – operative Society
(d) Sole Proprietor
Answer:
(d) Sole Proprietor

Samacheer Kalvi Commerce 11th Question 3.
A major disadvantage of sole proprietorship is
(a) Limited liability
(b) Unlimited liability
(c) Easy Formation
(d) Quick decision
Answer:
(b) Unlimited liability

11th Commerce Solutions Samacheer Kalvi Question 4.
From the following which one is Non – corporate form of business?
(a) Joint stock company
(b) Sole trading business
(c) Government company
(d) Co – operatives
Answer:
(b) Sole trading business

II. Very Short Answer Questions

11th Commerce Samacheer Kalvi Question 1.
Who is called Sole Trader?
Answer:
Proprietorship is a form of business organisation in which an individual introduces his own capital, uses his own skill and intelligence in the management of its affairs and is solely responsible for the results of its operations.

Sole Proprietorship Questions And Answers Pdf Question 2.
What are the non – corporate enterprises?
Answer:

1. Sole trading concern
2. partnership firm
3. Joint Hindu family business.

11th Commerce Chapter 4 Book Back Answers Question 3.
What are the corporate enterprises?
Answer:

1. Government – Public Undertakings, Public Utilities.
2. Private – Joint stock companies
3. Co – operative society

Samacheer Kalvi 11th Commerce Question 4.
For which of the following types of business do you think a sole proprietorship form of organisation would be more suitable, and why?

1. Grocery store
2. Medical store
3. Craft centre
4. Legal consultancy
5. Internet cafe

Answer:
Grocery store is more suitable for sole proprietorship business because of limited risk and less educated or uneducated can start the business.

III. Short Answer Questions

Commerce Samacheer Kalvi Question 1.
How is it possible to maintain secrecy in sole proprietorship?
Answer:
In sole proprietorship as the trader is the sole owner of the business, the secrecy can be maintained easily.

Samacheer Kalvi Guru 11th Commerce Question 2.
What is unlimited liability?
Answer:
The liability of the proprietor for the debts of the business is unlimited. The creditors have the right to recover their dues even from the personal property of the proprietor in case the business assets are not sufficient to pay their debts.

11th Samacheer Kalvi Commerce Question 3.
Write any three characteristics of sole proprietorship.
Answer:
1. Ownership by one man:
This is owned by single person. The sole trader contributes the required capital. He is not only the owner of the business but also manages the entire affairs.

2. Freedom of work and Quick Decisions: Since.an individual is himself as a owner, he need not consult anybody else. Hence he can take quick decisions.

3. Unlimited Liability: When his business assets are not sufficient to pay off the business debts he has to pay from his personal property.

Samacheer Kalvi Class 11 Commerce Solutions Question 4.
Give some examples of sole trading business.
Answer:
Saravana Stores, Hotel Saravana Bhavan, Grocery store, Petty shop.

Samacheer Kalvi 11th Commerce Solutions Question 5.
Define Sole trading business.
Answer:
“Sole proprietorship is that form of business organisation which is owned and controlled by a single individual. He receives all the profits and risks of his property in the success or failure of the enterprise” – Wheeler.

IV. Long Answer Questions

Samacheer Kalvi 11th Accountancy Chapter 4 Question 1.
Explain the characteristics of sole trading business.
Answer:
1. Ownership by one man:
This is owned by single person. The sole trader contributes the required capital. He is not only the owner of the business but also manages the entire affairs.

2. Freedom of work and Quick Decisions:
Since an individual is himself as a owner, he need not consult anybody else. Hence he can take quick decisions.

3. Unlimited Liability:
When his business assets are not sufficient to pay off the business debts he has to pay from his personal property.

4. Enjoying Entire Profit:
He strives tirelessly for the improvement and expansion of his business and enjoys all the benefits of his hard work.

5. Absence of Government Regulation:
A sole proprietor concern is free from Government regulations. No legal formalities are to be observed in its formation, management or in its closure.

6. No Separate Entity:
The sole trading concern comes to an end with death, disability, insanity and insolvency of the individual.

7. Maintenance of Secrecy:
Since he/she manages all the affairs of the business, the secrecy can be maintained easily.

Samacheer Kalvi Commerce Question 2.
What are the advantages of sole trading business?
Answer:
1. Easy Formation:
No legal formalities are required to initiate a sole trading concern. Any person capable of entering into a contract can start it, provided he has the necessary resources for it.

2. Incentive to Work hard:
There is a direct relationship between effort and reward. The fact that the entire profit can be taken by himself without sharing with anybody else induces him to work ceaselessly.

3. Small Capital:
Small capital is an important as well as specific advantage of sole proprietorship. Sole proprietor can start business with small capital.

4. Credit Standing:
Since his private properties are held liable for satisfying business debts, he can get more financial assistance from others.

5. Personal Contact with the Customers:
Since sole proprietor knows each and every customer individually he can supply goods according to their taste and preferences. Thus he can cultivate personal relationship with the customers.

6. Flexibility:
The sole trader can easily adjust himself to the changing requirements of his business.

Samacheer Kalvi 11 Commerce Question 3.
What are the disadvantages of sole trading business?
Answer:
1. Limited Capital:
Since the capital is contributed by one individual only, business operations have necessarily to be on a limited scale.

2. Limited Managerial Skill:
Single person’s intelligence and experience may not help him beyond a certain stage. Since he has to focus on each and every activity, his managerial ability is bound to be limited.

3. Unlimited Liability:
The creditors have the right to recover their dues even from the personal property of the proprietor in case the business assets are not sufficient to pay their debts.

4. Lack of Specialisation:
Since the business unit is small and the financial resources are limited, experts in different fields cannot be employed to secure maximum advantages.

5. Hasty Decisions:
Sole proprietor is more likely to take hasty decision as he need not consult anybody else.

Samacheer Kalvi 11th Commerce Sole Proprietorship Additional Questions and Answers

I. Choose the Correct Answer:

Question 1.
…………….. is that form of business organisation which is owned and controlled by a single individual.
(a) Sole trading concern
(b) Partnership firm
(c) Joint Hindu family business
(d) Joint stock companies
Answer:
(a) Sole trading concern

Question 2.
…………….. is known as individual entrepreneurship.
(a) Partnership
(b) Sole trader
(c) Joint stock company
(d) Co – operative
Answer:
(b) Sole trader

Question 3.
When his business assets are not sufficient to pay off the business debts, he has to pay from his personal property.
(a) Unlimited Liability
(b) Flexibility
(c) Small capital
(d) Limited Liability
Answer:
(a) Unlimited Liability

Question 4.
“He receives all the profits and risks all of his property in the success or failure of the enterprise”- was said by
(a) Wheeler
(b) J.L. Hansen
(c) H.Haney
(d) O.R. Krishnasamy
Answer:
(a) Wheeler

Question 5.
Which of the following is under non – corporate enterprise?
(a) Government
(b) Co – operative
(c) Company
(d) Sole trading concern
Answer:
(d) Sole trading concern

II. Very Short Answer Questions

Question 1.
How was the decisions taken in sole proprietorship?
Answer:
Since an individual is himself as a owner, he need not consult anybody else. Hence he can take quick decisions.

Question 2.
What should be the Government role in sole proprietorship?
Answer:
A sole proprietor concern is free from Government regulations. No legal formalities are to be observed in its formation, management or in its closure.

III. Short Answer Questions

Question 1.
What is limited capital?
Answer:
Since the capital is contributed by one individual only, business operations have necessarily to be on a limited scale.

Question 2.
What is limited managerial skill?
Answer:
Single person’s intelligence and experience may not help him beyond a certain stage. Since he has to focus on each and every activity, his managerial ability is bound to be limited.

For future Learning

Question a.
Mahesh is a young graduate who has inherited a sum of Rs 1,00,000 by way of family savings. He also has a family house to which he has sole title as the only son of his father. He is thinking of starting a small factory for the manufacture of plastic toys. What form of ownership organisation will you advise him to choose?
Answer:
Sole Trader.

Question b.
Amar started a business on his own. He has his father helping him with the accounts and his brother helps him with looking after customers in the evening. Amar pays a monthly salary to his father and brother. Identify the form of business organisation.
Answer:
Sole Trader.

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Students can Download Accountancy Chapter 10 Depreciation Accounting Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Accountancy Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Accountancy Solutions Chapter 10 Depreciation Accounting

Samacheer Kalvi 11th Accountancy Depreciation Accounting Text Book Back Questions and Answers

I. Multiple Choice Questions
Choose the Correct Answer

Depreciation Accounting Problems And Solutions Pdf Question 1.
Under straight line method, the amount of depreciation is ……………..
(a) Incfeasing every year
(b) Decreasing every year
(c) Constant for all the years
(d) Fluctuating every year
Answer:
(c) Constant for ail the years

Depreciation Problems And Solutions Question 2.
If the total charge of depreciation and maintenance cost are considered, the method that provides a uniform charge is ……………..
(a) Straight line method
(b) Diminishing balance method
(c) Annuity method
(d) Insurance policy method
Answer:
(b) Diminishing balance method

Depreciation Questions And Answers Pdf Question 3.
Under the written down value method of depreciation, the amount of depreciation is ……………..
(a) Uniform in all the years
(b) Decreasing every year
(c) Increasing every year
(d) None of the above
Answer:
(b) Decreasing every year

Depreciation Questions And Answers For Class 11 Pdf Question 4.
Depreciation provided on machinery is debited to ……………..
(a) Depreciation account
(b) Machinery account
(c) Trading account
(d) Provision for depreciation account
Answer:
(a) Depreciation account

Depreciation Accounting Questions And Answers Question 5.
Cash received from sale of fixed asset is credited to ……………..
(a) Profit and loss account
(b) Fixed asset account
(c) Depreciation account
(d) Bank account
Answer:
(b) Fixed asset account

Class 11 Depreciation Questions Question 6.
Depreciation is provided on ……………….
(a) Fixed assets
(b) Current assets
(c) Outstanding charges
(d) All assets
Answer:
(a) Fixed assets

Depreciation Straight Line Method Questions And Answers Pdf Question 7.
Depreciation is caused by ……………..
(a) Lapse of time
(b) Usage
(c) Obsolescence
(d) a, b and c
Answer:
(d) a, b and c

Depreciation Class 11 Practical Problems Question 8.
Depreciation is the process of ……………..
(a) Allocation of cost of the asset to the period of its useful life
(b) Valuation of assets
(c) Maintenance of an asset in a state of efficiency
(d) Adding value to the asset
Answer:
(a) Allocation of cost of the asset to the period of its useful life

Depreciation Problems And Solutions Pdf Question 9.
For which of the following assets, the depletion method is adopted for writing off cost of the asset?
(a) Plant and machinery
(b) Mines and quarries
(c) Buildings
(d) Trademark
Answer:
(b) Mines and quarries

Class 11 Accountancy Chapter 10 Solutions Question 10.
A depreciable asset may suffer obsolescence due to ……………..
(a) Passage of time
(b) Wear and tear
(c) Technological changes
(d) None of the above
Answer:
(c) Technological changes

Depreciation Class 11 Solutions Question 11.
Which method shall be efficient, if repairs and maintenance cost of an asset increases as it grows older?
(a) Straight line method
(b) Reducing balance method
(c) Sinking fund method
(d) Annuity method
Answer:
(b) Reducing balance method

Depreciation Problems With Answers Question 12.
Depreciation is to be calculated from the date when ……………..
(a) Asset is put to use
(b) Purchase order is made
(c) Asset is received at business premises
(d) Invoice of assets is received
Answer:
(a) Asset is put to use

Depreciation Accounting Questions And Answers Pdf Question 13.
If the rate of depreciation is same, then the amount of depreciation under straight line method vis – a – vis written down value method will be
(a) Equal in all years
(b) Equal in the first year but higher in subsequent years
(c) Equal in the first year but lower in subsequent years
(d) Lower in the first year but equal in subsequent years
Answer:
(b) Equal in the first year but higher in subsequent years

Samacheer Kalvi 11th Accountancy Book Pdf Question 14.
Residual value of an asset means the amount that it can fetch on sale at the of its useful life.
(a) Beginning
(b) End
(c) Middle
(d) None
Answer:
(b) End

II. Very Short Answer Questions

Samacheer Kalvi 11th Accountancy Book Question 1.
What is meant by depreciation?
Answer:
The process of allocation of the relevant cost of a fixed asset over its useful life is known as depreciation. It is an allocation of cost against the benefits derived from a fixed asset during an accounting period.

Accounts Samacheer Kalvi Question 2.
List out the various methods of depreciation.
Answer:

1. Straight line method or fixed instalment method or Original cost method.
2. Written down value method or Diminishing balance method or Reducing balance method.
3. Sum of years digits method.
4. Machine hour rate method.
5. Depletion method.
6. Annuity method.
7. Revaluation method.
8. Sinking fund method.
9. Insurance Policy method.

Question 3.
Give the formula to find out the amount and rate of depreciation under straight line method of depreciation.
Answer:
1. Amount of depreciation per year =
2. Rate of depreciation =

Question 4.
What is annuity method?
Answer:
Under this method, not only the original cost of the asset but also the amount of interest on the investment is taken into account while computing depreciation. The idea of considering interest is that if the investment is made in any other asset instead of the relevant fixed asset, it . would have earned a certain rate of interest. To calculate the amount of depreciation, annuity factor is used. Annuity factor can be found out from the annuity table or by using formula. Amount of depreciation is computed as follows:
Amount of depreciation = Annuity factor x original cost of the asset.

Question 5.
What is sinking fund method?
Answer:
This method is adopted especially when it is desired not merely to write off an asset but also to provide enough funds to replace an asset at the end of its working life. Under this method, the amount charged as depreciation is transferred to depreciation fund and invested outside the business. The investment is made in safe securities which offer a certain rate of interest. Interest is received annually and reinvested every year along with the amount of annual depreciation. On the expiry of the life of the asset, the investments are sold and the sale proceeds are used for replacement of the asset. This method of depreciation is suitable for assets of higher value. This method is also known as depreciation fund method.

III. Short Answer Questions

Question 1.
What are the objectives of providing depreciation?
Answer:

1. To find out the true profit or loss
2. To present the true and fair view of financial position
3. To facilitate replacement of fixed assets
4. To avail tax benefits
5. To comply with legal requirements

Question 2.
What are the causes for depreciation?
Answer:

1. Wear and tear
2. Efflux of time
3. Obsolescence
4. Inadequacy for the purpose
5. Lack of maintenance
6. Abnormal factors

Question 3.
State the advantages and limitations of straight line method of depreciation.
Answer:
Advantages:

• Simple and easy to understand
• Equality of depreciation burden
• Assets can be completely written off
• Suitable for the assets having fixed working life

Limitations:

• Ignores the actual use of the asset
• Ignores the interest factor
• Total charge on the assets will be more when the asset becomes older
• Difficulty in the determination of scrap value

Question 4.
State the advantages and limitations of written down value method of depreciation.
Answer:
Advantages:

• Equal charge against income
• Logical method

Limitations:

• Assets cannot be completely written off
• Ignores the interest factor
• Difficulty in determining the rate of depreciation
• Ignores the actual use of the asset

Question 5.
Distinguish between straight line method and written down value method of providing depreciation.
Answer:

IV. Exercises

Straight line method:

Question 1.
A firm purchased a plant for ₹ 40,000. Erection charges amounted to ₹ 2,000. Effective life of the plant is 5 years. Calculate the amount of depreciation per year under straight line method.
Answer:
Calculation of amount of depreciation
Amount of depreciation =
Original cost = Purchase of plant + Erection charges = ₹ 40,000 + ₹ 2,000 = ₹ 42,000
Estimated life = 5 years = \(\frac{₹ 42000-0}{5 \text { years }}\) = ₹ 8,400/-

Question 2.
A company purchased a building for ₹ 50,000. The useful life of the building is 10 years and the residual value is ₹ 2,000. Find out the amount and rate of depreciation under straight line method.
Answer:
(1) Calculation of amount of depreciation:
Amount of depreciation =
Original cost = ₹ 50,000
residual value = ₹ 2,000
Estimated life = 10 years
\(\frac{50,000-2,000}{10 \text { years }}\) = \(\frac { 48,000 }{ 10 }\) = ₹ 4,800/-

(2) Calculation of rate of depreciation:
rate of depreciation = = \(\frac { 48,000 }{ 50,000 }\) x 100 = 9.6%

Question 3.
Furniture was purchased for ₹ 60,000 on 1-7-2016. It is expected to last for 5 years. Estimated scrap at the end of five years is ₹ 4,000. Find out the rate of depreciation under straight line method.
Answer:
(1) Amount of depreciation =
Original cost = ₹ 60,000
Scrap value = ₹ 4,000
Estimated life = 5 years
= \(\frac{60,000-4,000}{5 \text { years }}\) = \(\frac { 56,000 }{ 5 }\) = ₹ 11,200/-

Question 4.
Calculate the rate of depreciation under straight line method from the following information: Purchased a second hand machinery on 1.1.2018 for ₹ 38,000 On 1.1.2018 spent ₹ 12,000 on its repairs
Expected useful life of machine is 4 years
Estimated residual value ₹ 6,000
Answer:
Original cost – residual value
(1) Calculation of amount of depreciation =
Original cost = Purchase of machinery + repair charges = 38,000 + 12,000 = 50,000
Residual value = 6,000
Estimated life = 4 years = \(\frac{50,000-6,000}{4 \text { years }}\) = ₹ 11,000/-
(2) Rate of depreciation =

= \(\frac { 11,000 }{ 50,000 }\) x 100 = 22 %

Question 5.
Calculate the rate of depreciation under straight line method.
Purchase price of a machine ₹ 80,000
Expenses to be capitalised ₹ 20,000
Estimated residual value ₹ 4,000
Expected useful life ₹ 4 years
Answer:
Original cost – residual value =
Original cost = Machine purchased + capitalised expenses
80,000 + 20,000 = 1,00,000
Residual value = 4,000
Estimated life = 4 years
= \(\frac{1,00,000-4,000}{4 \text { years }}\) = \(\frac { 96,000 }{ 4 }\) = ₹ 24,000/-

Question 6.
Machinery was purchased on 1^st January 2015 for ₹ 4,00,000. ₹ 15,000 was spent on its erection and ₹ 10,000 on its freight charges. Depreciation is charged at 10% per annum on straight line method. The books are closed on 31^st March each year. Calculate the amount of depreciation on machinery for the first two years.
Answer:
Calculation of depreciation:
Original cost = Machinery purchased + erection charges + freight charges
= 4,00,000 + 15,000 + 10 ,000
= 4,25,000

First year depreciation = ₹ 10,625
Second year depreciation = ₹ 42,500

Question 7.
An asset is purchased on 1.1.2016 for ₹ 25,000. Depreciation is to be provided annually according to straight line method. The useful life of the asset is 10 years and its residual value is ₹ 1,000. Accounts are closed on 31^st December every year. You are required to find out the rate of depreciation and give journal entries for first two years.
Answer:
Amount of depreciation =
Original cost = ₹ 25,000
Residual value = ₹ 1,000
Estimated life = 10 years
= \(\frac{25,000-1,000}{10 \text { years }}\) = \(\frac { 24,000 }{ 10 }\) = ₹ 24,000/-

(2) Rate of depreciation =
= \(\frac { 2,400 }{ 25,000 }\) x 100 = 9.6%
Journal Entries for first two years

Question 8.
From the following particulars, give journal entries for 2 years and prepare machinery account under straight line method of providing depreciation:
Machinery was purchased on 1.1.2016
Price of the machine ₹ 36,000
Freight charges ₹ 2,500
Installation charges ₹ 1,500
Life of the machine 5 years
Answer:
Calculation of Asset of depreciation:
Amount of depreciation =
Original cost = Price of the machine + Freight charges + Installation charges
= 36,000 + 2,500 + 1,500 = ₹ 40,000
= \(\frac { 40,000 – 0 }{ 5 years }\) = ₹ 8,000
Journal Entries for 2 years

Machinery

Question 9.
A manufacturing company purchased on 1^st April, 2010, a plant and machinery for ₹ 4,50,000 and spent ₹ 50,000 on its installation. After having used it for three years, it was sold for ₹ 3,85,000. Depreciation is to be provided every year at the rate of 15% per annum on the fixed instalment method. Accounts are closed on 31^st March every year. Calculate profit or loss on sale of machinery.
Answer:
Calculation of Profit or Loss on sale of Machinery

Note: If the selling price is more than the book value is called profit.
Selling price – Book value = Profit 3,85,000 – 2,75,000 = 1,10,000
Profit on sale of Machinery is = ₹ 1,10,000.

Question 10.
On 1^st April 2008, Sudha and Company purchased machinery for ₹ 64,000. To instal the machinery expenses incurred was ₹ 28,000. Depreciate machinery 10% p.a. under straight line method. On 30^th June, 2010 the worn out machinery was sold for ₹ 52,000. The books are closed on 31^st December every year. Show machinery account.
Answer:
Workings:

If Book value is more than the selling price it is called loss.
Book value – selling price = loss
71,300 – 52,000 = 19,300
Loss on sale of machinery is = 19,300
Machinery

Question 11.
Ragul purchased machinery on April 1, 2014 for ₹ 2,00,000. On 1^st October 2015, a new machine costing ₹ 1,20,000 was purchased. On 30^th September 2016, the machinery purchased on April 1, 2014 was sold for ₹ 1,20,000. Books of accounts are closed on 31^st March and depreciation is to be provided at 10% p.a. on straight line method. Prepare machinery account and depreciation account for the years 2014 – 15 to 2016 – 17.
Answer:
Workings

Machinery

Written down value method

Question 12.
An asset is purchased for ₹ 50,000. The rate of depreciation is 15% p.a. Calculate the annual depreciation for the first two years under diminishing balance method.
Answer:
Workings: Calculation of depreciation of Machinery

Question 13.
A boiler was purchased on 1st January 2015 from abroad for ₹ 10,000. Shipping and forwarding charges amounted to ₹ 2,000. Import duty ₹ 7,000 and expenses of installation amounted to ₹ 1,000. Calculate depreciation for the first 3 years @10% p.a. on diminishing balance method assuming that the accounts are closed 31^st December each year.
Answer:
Calculation of amount of depreciation on diminishing balance method:

2015 Depreciation ₹ 2,000
2016 Depreciation ₹ 1,800
2017 Depreciation ₹ 1,620

Question 14.
A furniture costing ₹ 5,000 was purchased on 1.1.2016, the installation charges being ₹ 1,000. The furniture is to be depreciated @ 10% p.a. on the diminishing balance method. Pass journal entries for the first two years.
Answer:

Journal Entries

Question 15.
A firm acquired a machine on 1^st April 2015 at a cost of ₹ 50,000. Its life is 6 years. The firm writes off depreciation @ 30% p.a. on the diminishing balance method. The firm closes its books on 31^st December every year. Show the machinery account and depreciation account for three years starting from 1^st April 2015.
Answer:
Workings:

Machinery

Depreciation A/c

Question 16.
A firm purchased a machine for ₹ 1,00,000 on 1-7-2015. Depreciation is written off at 20% on reducing balance method. The firm closes its books on 31^st December each year. Show the machinery account upto 31-12-2017.
Answer:
Workings:

MachineryA/c

Question 17.
On 1^st October 2014, a truck was purchased for ₹ 8,00,000 by Laxmi Transports Ltd. Depreciation was provided @ 15% p.a. under diminishing balance method. On 31^st March 2017, the above truck was sold for ₹ 5,00,000. Accounts are closed on 31^st March every year. Find out the profit or loss made on the sale of the truck.
Answer:
Calculation of Profit or loss on sale of truck:

Note: If Book value is more than the selling price it is called loss:
Book value – selling price = Loss
5,34,650 – 5,00,000 = 34,650
∴ Loss on sale of truck = ₹ 34,650

Question 18.
On 1^st January 2015, a second hand machine was purchased for ₹ 58,000 and ₹ 2,000 was spent on its repairs. On 1^st July 2017, it was sold for ₹ 28,600. Prepare the machinery account for the years 2015 to 2017 under written down value method by assuming the rate of depreciation as 10% p.a. and the accounts are closed on 31^st December every year.
Answer:
Calculation of profit or loss on sale of machinery

Note: If Book value is more than the selling price it is called loss.
Book value – selling price = loss
46,170 – 28,600 = 17,570
∴ Loss on sale of machinery = ₹ 17,570

Question 19.
Raj & Co purchased a machine on 1^st January 2014 for ₹ 90,000. On 1^st July 2014, they purchased another machine for ₹ 60,000. On 1^st January 2015, they sold the machine purchased on 1^st January 2014 for ₹ 40,000. It was decided that the machine be depreciated at 10% per annum on diminishing balance method. Accounts are closed on 31^st December every year. Show the machinery account for the years 2014 and 2015.
Answer:
Workings

Note: If Book value is more than the selling price is called loss.
Book value – selling price = loss
81,000 – 40,000 = 41,000
Machinery

Textbook Case Study Solved

Question a.
Lucky & Co’s income statement shows a loss of ₹ 3,000. The owner thinks that there is no need to provide for depreciation as the company has made a loss. He also suggests his accountant to change the method of depreciation for the next year so as to avoid the loss. But, the accountant is hesitant to make the necessary changes suggested by his owner.
Now, discuss on the following points:

Question 1.
Do you agree on the point that there is no need to charge depreciation when the company has made a loss?
Answer:
No, I don’t agree on the point that there is no need to charge depreciation when the company has made a loss. We have to charge depreciation whether profit or loss, otherwise we cannot find out the actual profit or loss.

Question 2.
Why does the accountant hesitate to make the changes suggested by his owner?
Answer:
The accountant hesitates to make the changes suggested by his owner because it will differ the profit or loss for the business. The depreciation is a necessary one, so it must be deducted every year.

Question 3.
What are the accounting principles not followed if the accountant agrees to his owner’s suggestion?
Answer:
If the accountant agrees to his owner’s suggestion, they will not follow double entry system.

Question 4.
Do you think charging depreciation could be the only reason for the company’s loss?
Answer:
No, charging depreciation could not be the only reason for the company’s loss, because the business activities are subject to change but the depreciation is compulsory when the business is running.

Samacheer Kalvi 11th Accountancy Depreciation Accounting Additional Questions and Answers

I. Multiple Choice Questions
Choose the correct answer

Question 1.
Depreciation is the gradual and permanent decrease in the value of an asset from any cause ……………….
(a) Owen
(b) Wheeler
(c) Spicer and Pegler
(d) R.N. Carter
Answer:
(d) R.N. Carter

Question 2.
Certain assets whether used or not become potentially less useful with the passage of time ……………….
(a) Efflux of time
(b) Lack of maintenance
(c) Abnormal factors
(d) Wear and tear
Answer:
(a) Efflux of time

Question 3.
The normal use of a tangible asset results in physical deterioration which is called ……………….
(a) Wear and tear
(b) Abnormal factors
(c) Obsolescence
(d) Efflux of time
Answer:
(a) Wear and tear

Question 4.
Allocation of acquisition cost of intangible fixed assets such as goodwill is called ……………….
(a) Abnormal factors
(b) Wear and tear
(c) Amortization
(d) Obsolescence
Answer:
(c) Amortization

Question 5.
………………. is also known as residual value.
(a) Book value
(b) Scrap value
(c) Amortization
(d) Wear and tear
Answer:
(b) Scrap value

Question 6.
The following formula is used to complete the rate of depreciation under ……………….
(a) Written down value method
(b) Straight line method
(c) Machine hour rate method
(d) Annuity method
Answer:
(a) Written down value method

II. Very Short Answer Questions

Question 1.
State R.N. Carter’s definition of depreciation.
Answer:
According to R.N. Carter, “Depreciation is the gradual and permanent decrease in the value of an asset from any cause”.

Question 2.
What is wear and tear?
Answer:
The normal use of a tangible asset results in physical deterioration which is called wear and tear. When there is wear and tear, the value of the asset decreases proportionately.

Question 3.
What is obsolescence?
Answer:
It is a reduction in the value of assets as a result of the availability of updated alternative assets. This happens due to new inventions and innovations.

Question 4.
What is Straight line method?
Answer:
Under this method, a fixed percentage on the original cost of the asset is charged every year by way of depreciation. Hence it is called original cost method. As the amount of depreciation remains equal in all years over the useful life of an asset, it is also called as fixed instalment method.

Question 5.
What is written down value method?
Answer:
Under this method, depreciation is charged at a fixed percentage on the written down value of the asset every year. Hence, it is called written down value method.

III. Short Answer Questions

Question 1.
Write a note on sum of years of digits method.
Answer:
This method is similar to the diminishing balance method. The amount of depreciation goes on decreasing year after year in proportion to the unexpired life of the asset. This method is suitable for those assets having more profitability of obsolescence and increased repair charges as the assets grow older. Under this method, amount of depreciation per year is calculated by multiplying the cost of the asset and the number of remaining years of life and dividing it by the sum of the digits of all years of life of the asset.

Question 2.
What is machine hour rate method?
Answer:
Under this method, depreciation per machine hour is calculated. The cost of the machinery after deducting the residual value, if any, is divided by the estimated working hours of the machine to find the depreciation per hour. The actual depreciation for any given period depends upon the working hours during that year. The special feature of this method is that depreciation is found directly in proportion to the actual use of the asset. Under this method life of the asset is estimated in hours and not in years.

Question 3.
What is Depletion method?
Answer:
Depletion means exhaustion of natural resources. That is depletion means quantitative reduction in the content of assets. This is applicable to those assets that get exhausted due to extraction and exploitation. Examples: mines and oil fields, etc. Under this method, depreciation rate is calculated on the basis of the estimated quantities of the output during the whole life of the asset.

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3

11th Maths Exercise 4.3 Question 1.
If ⁿC₁₂ = ⁿC₉ find ²¹C_n.
Solution:
ⁿC_x = ⁿC_y ⇒ x = y or x + y = n
Here ⁿC₁₂ = ⁿC₉ ⇒ 12 ≠ 9 so 12 + 9 = n (i.e) n = 21

Exercise 4.3 Class 11 Maths Question 2.
If ¹⁵C_{2r – 1} = ¹⁵C_{2r + 4}, find r.
Solution:

Exercise 4.3 Maths Class 11 Solutions Question 3.
If ⁿP_r = 720 and ⁿC_r = 120, find n, r.
Solution:

Exercise 4.3 Class 11 Question 4.
Prove that ¹⁵C₃ + 2 × ¹⁵C₄ + ¹⁵C₅ = ¹⁷C₅
Solution:

4.3 Maths Class 11 Question 5.

Solution:

Ex 4.3 Class 11 Question 6.
If ^{(n + 1)}C₈ : ^{(n – 3)}P₄ = 57 : 16, find the value of n.
Solution:

Class 11 Question 7.

Solution:

Mathematical Induction Question 8.
Prove that if 1 ≤ r ≤ n then n × ^{(n – 1)}C_{r – 1} = (n – r + 1)C_{r – 1}.
Solution:


(1) = (2) ⇒ LHS = RHS

Exercise 4.3 Class 11 Pdf Question 9.

(i) A Kabaddi coach has 14 players ready to play. How many different teams of 7 players could the coach put on the court?
Solution:
No. of players in the team = 14
We need 7 players
So selecting 7 from 14 players can be done is ¹⁴C₇ = 3432 ways

(ii) There are 15 persons in a party and if, each 2 of them shakes hands with each other, how many handshakes happen in the party?
Solution:
Total No. of persons = 15
Each two persons shake hands

(iii) How many chords can be drawn through 20 points on a circle?
Solution:
A chord is a line join of 2 points
No. of points given = 20
Selecting 2 from 20 can be done in ²⁰C₂ ways

(iv) In a parking lot one hundred, one year old cars are parked. Out of them five are to be chosen at random for to check its pollution devices. How many different set of five cars are possible?
Solution:
Number of cars =100
Select 5 from 100 cars can be done in ¹⁰⁰C₅ ways

(v) How many ways can a team of 3 boys, 2 girls and 1 transgender be selected from 5 boys, 4 girls and 2 transgenders?
Solution:
We have 5 boys, 4 girls and 2 transgenders. We need 3 boys, 2 girls and 1 transgender The selection can be done as follows Selecting 3 boys from 5 boys can be done is ⁵C₃ ways

Selecting 2 girls from 4 girls can be done in ⁴C₂ ways

Selecting 1 transgender from 2 can be done in ²C₁ = 2 ways
∴ Selecting 3 boys, 2 girls and 1 transgender can be done in 10 × 6 × 2 = 120 ways

Question 10.
Find the total number of subsets of a set with
(i) 4 elements
(ii) 5 elements
(iii) n elements

Solution:
If a set has n elements then the number of its subsets = 2ⁿ

(i) Here n = 4
So number of subsets = 2⁴ = 16

(ii) n = 5
So number of subsets = 2⁵ = 32

(iii) n = n
So number of subsets = 2

Question 11.
A trust has 25 members.

(i) How many ways 3 officers can be selected?
Solution:
Selecting 3 from 25 can be done in ²⁵C₃ ways

(ii) In how many ways can a President, Vice President and a secretary be selected?
Solution:
From the 25 members a president can be selected in 25 ways
After the president is selected, 24 persons are left out.
So a Vice President can be selected in (from 24 persons) 24 ways.
After the selection of Vice President 23 persons are left out
So a secretary can be selected (from the remaining 23 persons) in 23 ways
So a president, Vice president and a secretary can be selected in ²⁵P₃ ways ²⁵P₃ = 25 × 24 × 23 = 13800 ways

Question 12.
How many ways a committee of six persons from 10 persons can be chosen along with a chair person and a secretary?
Solution:
Selecting a chair person from the 10 persons can be done in 10 ways
After the selection of chair person only 9 persons are left out so selecting a secretary (from the remaining a persons) can be done in 9 ways.
The remaining persons = 8
Totally we need to select 6 persons
We have selected 2 persons.
So we have to select 4 persons
Selecting 4 from 8 can be done in ⁸C₄ ways

Question 13.
How many different selections of 5 books can be made from 12 different books if,
Solution:
No. of books given = 12
No. of books to be selected = 5

(i) Two particular books are always selected?
Solution:
So we need to select 3 more books from (12 – 2) 10 books which can be done in ¹⁰C₃ ways

(ii) Two particular books are never selected?
Solution:
Two particular books never to be selected.
So only 10 books are there and we have to select 5 books which can be done in ¹⁰C₅ ways

Question 14.
There are 5 teachers and 20 students. Out of them a committee of 2 teachers and 3 students is to be formed. Find the number of ways in which this can be done. Further find in how many of these committees
(i) a particular teacher is included?
(ii) a particular student is excluded?
Solution:
No. of teachers = 5
No of students = 20
We need to select 2 teachers and 3 students

(i) A particular teacher should be included. So from the remaining 4 teachers one teacher is to be selected which can be done in ⁴C₁ = 4 ways

So selecting 2 teachers and 3 students can be done in 4 × 1140 = 4560 ways

(ii) particular student should be excluded.
So we have to select 3 students from 19 students which can be done in ¹⁹C₃ ways

∴ 2 teachers and 3 students can be selected in 969 × 10 = 9690 ways

Question 15.
In an examination a student has to answer 5 questions, out of 9 questions in which 2 are compulsory. In how many ways a students can answer the questions?
Solution:
No. of questions given = 9
No. of questions to be answered = 5
But 2 questions are compulsory
So the student has to answer the remaining 3 questions (5 – 2 = 3) from the remaining 7 (9 – 2 = 7) questions which can be done in ⁷C₃ ways

Question 16.
Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly three aces in each combination.
Solution:
No. of cards = 52 : In that number of aces = 4
No. of cards needed = 5
In that 5 cards number of aces needed = 3
So the 3 aces can be selected from 4 aces in ⁴C₃ = ⁴C₁ = 4 ways
So the remaining = 5 – 3 = 2
This 2 cards can be selected in ⁴⁸C₂ ways

Question 17.
Find the number of ways of forming a committee of 5 members out of 7 Indians and 5 Americans, so that always Indians will be the majority’ in the committee.
Solution:
We need a majority of Indian’s which is obtained as follows.

The possible ways are (5I) or (4I and 1A) or (3I and 2A)

Question 18.
A committee of 7 peoples has to be formed from 8 men and 4 women. In how many ways can this be done when the committee consists of
(i) exactly 3 women?
(ii) at least 3 women?
(iii) at most 3 women?
Solution:

We need a committee of 7 people with 3 women and 4 men.
This can be done in (⁴C₃) (⁸C₄) ways

The number of ways = (70) (4) = 280

(ii) Atleast 3 women

So the possible ways are (3W and 4M) or (4W and 3M)

The number of ways (4) (70) + (1) (56) = 280 + 56 = 336

(iii) Atmost 3 women

Question 19.
7 relatives of a man comprises 4 ladies and 3 gentlemen, his wife also has 7 relatives; 3 of them are ladies and 4 gentlemen. In how many ways can they invite a dinner party of 3 ladies and 3 gentlemen so that there are 3 of man’s relative and 3 of the wife’s relatives?
Solution:

We need 3 ladies and 3 gentlemen for the party which consist of 3 Husbands relative and 3 wifes relative.
This can be done as follows

Question 20.
A box contains two white balls, three black balls and four red balls. In how many ways can three balls be drawn from the box, if at least one black ball is to be included in the draw?
Solution:
The box contains 2 white, 3 black and 4 red balls
We have to draw 3 balls in which there should be alteast 1 black ball
The possible draws are as follows
Black balls = 3
Red and White = 2 + 4 = 6

Question 21.
Find the number of strings of 4 letters that can be formed with the letters of the word EXAMINATION.
Solution:
EXAMINATION
(i.e.) A, I, N are repeated twice. So the number of distinct letters = 8
From the 8 letters we have to select and arrange 4 letters to form a 4 letter word which can
be done in ⁸P₄ = 8 × 7 × 6 × 5 = 1680
From the letters A, A, I, I, N, N when any 2 letters are taken as AA, II or AA, NN or II, NN

From AA, II, NN we select one of them and from the remaining we select and arrange 3 which can be done in ways

Total number of 4 letter word = 1680 + 18 + 756 = 2454

Question 22.
How many triangles can be formed by joining 15 points on the plane, in which no line joining any three points?
Solution:
No. of non-collinear points = 15
To draw a Triangle we need 3 points
∴ Selecting 3 from 15 points can be done in ¹⁵C₃ ways.

Question 23.
How many triangles can be formed by 15 points, in which 7 of them lie on one line and the remaining 8 on another parallel line?
Solution:

7 points lie on one line and the other 8 points parallel on another paraller line.
A triangle is obtained by taking one point from one line and second points from the other parallel line which can be done as follows.

∴ Number of triangles = (7) (28) + (21) (8) = 196 + 168 = 364

Question 24.
There are 11 points in a plane. No three of these lies in the same straight line except 4 points, which are collinear. Find,

(i) The number of straight lines that can be obtained from the pairs of these points?
Solution:
4 points are collinear

Total number of points 11.
To get a line we need 2 points

But in that 4 points are collinear

From (1) Joining the 4 points we get 1 line
∴ Number of lines = ¹¹C₂ – ⁴C₂ + 1 = 55 – 6 + 1 = 50

(ii) The number of triangles that can be formed for which the points as their vertices?
A triangle is obtained by joining 3 points.
So selecting 3 from 11 points can be

But of the 11 points 4 points are collinear. So we have to subtract ⁴C₃ = ⁴C₁ = 4
∴ Number of triangles = 165 – 4 = 161

Question 25.
A polygon has 90 diagonals. Find the number of its sides?
Solution:

∴ n = 15

Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 Additional Questions Solved

Question 1.
A group consists of 4 girls and 7 boys. In bow many ways can a team of 5 members be selected, if the team has
(i) no girls
(ii) atleast one boy and one girl.
(iii) at least three girls
Solution:
We have 4 girls and 7 boys and a team of 5 members is to be selected.

(i) If no girl in selected, then all the 5 members are to be selected out of 7 boys

(ii) When at least one boy and one girl are to be selected, then


Hence the required number of ways are (i) 21 ways (ii) 441 ways (iii) 91 ways

Question 2.
A committee of 6 is to be choosen from 10 men and 7 women so as to contain atleast 3 men and 2 women. In how many different ways can this be done it two particular women refuse to serve on the same committee?
Solution:
We have 10 men and 7 women out of which a committee of 6 is to be formed which contain atleast 3 men and 2 women

∴ Total number of committee = 8610 – 810 = 7800
Hence, the value of the filler is 7800

Question 3.
Using the digits 1, 2, 3,4, 5, 6, 7 a number of 4 different digits is formed. Find

Solution:
(a) Total of 4 digit number formed with 1, 2, 3, 4, 5, 6, 7

(b) When a number is divisible by 2 = 4 × 5 × 6 × 3 = 360
(c) Total numbers which are divisible by 25 = 40
(d) Total numbers which are divisible by 4 (last two digits is divisble by 4) = 200
Hence, the required matching is (a) ⟷ (z), (b) ⟷ (iii), (c) ⟷ (iv), (d) ⟷ (ii)

Question 4.
If ²²P_{r + 1} : ²⁰P_{r + 2} = 11 : 52, find r.
Solution:

Question 5.

Solution:

Question 6.
A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of:
(i) exactly 3 girls?
(ii) at least 3 girls?
(iii) almost 3 girls?
Solution:

(ii) We have to select at least 3 girls. So the committee consists of 3 girls and 4 boys or 4 girls and 3 boys.
∴ Number of ways of selection = ⁴C₃ × ⁹C₄ + ⁴C₄ × ⁹C₃

(iii) We have to select at most 3 girls. So the committee consists of no girl and 7 boys or 1 girl and 6 boys or 2 girls and 5 boys or 3 girls and 4 boys.

Question 7.
Determine n if
(i) ²ⁿC₃ : ⁿC₂ = 12 : 1
(ii) ²ⁿC₃ : ⁿC₃ = 11 : 1
Solution:

Question 8.
Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.
Solution:
There are 6 red balls, 5 white balls and 5 blue balls.
We have to select 3 balls of each colour.

Question 9.

Solution:

Question 10.
If ⁿC₄, ⁿC₅ and ⁿC₄ are in A.P. then find n.
[Hint: ²ⁿC₅ = ⁿC₆ + ⁿC₄]
Solution:

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Tamilnadu Samacheer Kalvi 11th English Solutions Supplementary Chapter 2 A Shot In the Dark

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11th English Supplementary A Shot In The Dark Warm up
Can you spot 10 differences between these pictures?
How observant are you? Individually, try to spot as many differences as possible in five minutes. If you have spotted less than five, then you really need to improve your observation skill just to save yourself from being misled.

Answer:

Samacheer Kalvi 11th English Solutions Confessions of A Born Spectator Textual Questions

1. Answer the following questions in about 30-50 words each:

A Shot In The Dark Question Answers Question (a)
Why did Philip Sletherby visit Brill Manor?
Answer:
Philip Sletherby visited Brill Manor because he had been invited by Mrs. Honorio Saltpen Jago for a luncheon at Brill Manor.

Shot In The Dark Questions And Answers Question (b)
How would Sletherby gain from his friendship with Honoria Saltpen-Jago?
Answer:
Mr. Philip Sletherby’s hostess was a person of social importance in London, especially of considerable importance in the Chalkshire. He pinned his hopes on her recommendations to get a ticket to contest the MP election in the eastern division of Chalkshire. One of his club acquaintance assured him that Mrs.Saltpen-Jago had a reputation of being an excellent hostess. ‘

A Shot In The Dark Summary In Tamil Question (c)
Why did his travel companion curse and mutter?
Answer:
Mr. Sletherby’s travel companion was a young man of about twenty-two. He cursed and muttered because he had lost his valet along with the family crest. He claimed that he did so foolishly to seal a letter. He had forgotten to collect the valet.

11th English Unit 2 Supplementary Question (d)
Describe Bertie’s problem.
Answer:
Bertie was returning ffomyachting experience for about six months. He had done a damn awkward thing. He left behind his sovereign purse with four quid in it. It was all his worldly wealth at the moment. He must have left the purse on the table. He was stopping at a little country in a near Brondquay for three days’ fishing. As none knows him there, he needed money for his week-end bill and tips and cab to and from the station. In short, he desperately needed three pounds to manage the crisis.

A Shot In The Dark Supplementary Question (e)
‘There was a tinge of coldness in his voice.’ Why?
Answer:
Bertie claimed that he had not seen his mom Mrs. Saltpen-Jago for about six months , because he had gone for yachting. As he had overheard Mr. Sletherby’s intention of visiting Mrs. Saltpen-Jago at Brill Manor he introduced himself as Bertie, Mrs. Saltpen’s younger son. He had lost his sovereign purse. He needed to borrow three quid. As Mr. Sletherby had some suspicion he asked him what was the crest-like? He immediately replied it was demi-lion holding a cross-crosslet. Disbelieving his words, with a tinge of coldness in his voice, Mr. Sletherby said that Mrs. Saltpen-Jago had a grey hound courant on her note paper.

Question (f)
Compare the two crests.
Answer:
The Saltpen crest was “a demi-lion holding a cross-crosslet in its paw.” The Jago crest was a grey hound. Both were used in the family. Bertie could use either of the crests. Bertie had a feeling that he belonged to Saltpens, who used demi-lion crest always.

Question (g)
What was Sletherby’s response to Bertie’s request?
Answer:
Sletherby initially agreed to lend three pounds. But he wanted to ensure that he was not taken for a ride. He asked him if his mother had a close resemblance with him. Bertie claimed that his mom had dark brown hair like himself. Mr. Sletherby was already biased against lending the money as he believed that Grey hound was the crest found on Mrs. Saltpen-Jago’s note book paper. So, he refused to lend the promised loan to Bertie.

Question (h)
What caught his attention about the car door?
Answer:
The car door had two crests along. One was the demi-lion holding a cross-crosslet in its paw and the other was a greyhound.

Question (i)
Describe Honario Saltpen-Jago.
Answer:
Honario Saltpen-Jago was a brilliant and influential woman. She knew exactly when to take up an individual or drop him. She had very fine looks but she spoiled it. She had k beautiful dark brown hair but applied some dye and changed it into blonde.

Question (j)
Why did K.C.’s words shock Sletherby?
Answer:
Mr. Sletherby had left Mr. Bertie, Mrs. Honario Saltpen-Jago’s younger son on his way to Brill Manor in a railway station penniless and frustrated. He had not believed Bertie’s. description of his mom’s dark brown hair and the family’s double crest. When the footman Mr. KC commented on Mrs. Saltpen’s changed Looks, Mr. Sletherby was alarmed and he realized that he had behaved in a mean way. He had mistaken Bertie to be fraud. Thus K.C.’s words had shocked Mr. Sletherby.

2. Rearrange the sentences given below and write a summary of the story in a paragraph. Begin with:

Philip Sletherby was travelling by train to Brill Manor.
(a) The young man introduced himself as Bertie, the son of Saltpen Jago.
(b) Bertie needed three pounds desperately and asked Sletherby to lend it to him.
(c) All the time Sletherby was gazing at the door panel of the car, on which were the two crests – a demi-lion and a grey hound courant.
(d) He was received by Claude People. K.C.,who had kept on talking about various things.
(e) Bertie had left his purse behind, after sealing an envelope with the crest on the purse.
(f) Bertie stated that it was the Jago crest. He further added that his mother’s hair was dark brown similar to his.
(g) While describing the appearance of Honario Saltpen-Jago, K.C. referred to her altered’ hairstyle.
(h) Sletherby realized that Bertie had not lied to him, but that he had mistaken him for a fraud.
(i) He explained that the Saltpen crest was that of a demi-lion.
(j) Sletherby sat dumbstruck on learning that Mrs. Honario Saltpen Jago had changed her dark brown hair to a blonde, just five weeks ago.
(k) Sletherby pointed out to him that his mother’s letter had a greyhound courant crest.
(l) His companion in the train was a yOungman who was searching for something frantically.
(m) Suspecting foul play, Sletherby did not give him any money, as he knew Mrs.Saltpen- Jago was a blonde.

Answers:

(l) His companion in the train was a young man who was searching for something frantically.
(a) The young man introduced himself as Bertie, the son of Saltpen Jago.
(e) Bertie had left his purse behind, after sealing an envelope with the crest on the purse.
(b) Bertie needed three pounds desperately and asked Sletherby to lend it to him.
(i) He explained that the Saltpen crest was that of a demi-lion.
(k) Sletherby pointed out to him that his mother’s letter had a greyhound courant crest.
(J) Bertie stated that it was the Jago crest. He further added that his mother ’s hair was dark brown similar to his. .
(m) Suspecting foul play, Sletherby did not give him any money, as he knew Mrs. Saltpen- Jago was a blonde.
(d) He was received by Claude People. K.C.,who had kept on talking about various things.
(c) All the time Sletherby was gazing at the door panel of the car, on which were the two crests – a demi-lion and a grey hound courant.
(g) While describing the appearance of Honario Saltpen-Jago, K.C. referred to her altered hairstyle.
(j) Sletherby sat dumbstruck on learning that Mrs. Honario Saltpen Jago had changed her dark brown hair to a blonde, just five weeks ago.
(h) Sletherby realized that Bertie had not lied to him, but that he had mistaken him for a fraud.

Answer (for summary of the story in a paragraph)

Philip Sletherby was travelling by train to Brill Manor. His companion was a young man named Bertie, the son of Saltpen Jago. Bertie had left his purse behind after sealing an envelope with the crest on the purse. He asked Sletherby to lend three pounds to him. He also explained that the Saltpen crest was that of a.demi-lion. Sletherby said that his mother’s letter had a greyhound courant crest. Bertie said that his mother’s hair was dark brown similar to his. Suspecting foul play, Sletherby did not give him any money, as he knew Mrs. Saltpen-Jago was a blonde. He was received by Claude People, K.C., who had kept on talking about various things. While describing the appearance of Honario Saltpen-Jago, K.C. referred to her altered hairstyle and said that she had changed her dark brown hair to a blonde, just five weeks ago. Sletherby realized that Bertie had not lied to him, but that he had mistaken him for a fraud.

Additional Questions

1. Rearrange the sentences given below:

(a) Unable to arrest Bob himself, he sends a plainclothed man to do it.
(b) Bob kept his words and reached the spot before 10 p.m. and waited anxiously.
(c) A patrolman (Jimmy wells) finds a notorious criminal wanted by Chicago police is Bob.
(d) Both promised to meet at the same spot at 10 pm 20 years later.
(e) Jimmy wells and Bob had dinner at Big Joe’ Brady’s restaurant.

Answer:

(e) Jimmy wells and Bob had dinner at Big Joe’ Brady’s restaurant.
(d) Both promised to meet at the same spot at 10 pm 20 years later.
(b) Bob kept his words and reached the spot before 10 p.m. and waited anxiously.
(c) A patrolman (Jimmy wells) finds a notorious criminal wanted by Chicago police is Bob.
(a) Unable to arrest Bob himself, he sends a plainclothes man to do it.

2. Rearrange the sentences given below:

(a) While describing the appearance of Mrs. Saltpen-Jago, K.C. described her altered hairstyle but Sletherby was in a daze after seeing the double crest on the door of the car.
(b) Sletherby realized that Bertie had not lied, but that he had mistaken him for a fraud.
(c) Suspecting foul play, Sletherby did not give him any money, as he knew Mrs. Saltpen- Jago was a blonde.
(d) When Sletherby referred to a greyhound courant, Bertie stated that it was Jago crest. He further added that his mom had dark brown hair similar to his.
(e) Bertie having lost his purse wanted a loan of three pound, from Mr. Sletherby.

Answers:

(e) Bertie having lost his purse wanted a loan of three pound, from Mr. Sletherby.
(d) When Sletherby referred to a greyhound courant, Bertie stated that it was Jago crest. He further added that his mom had dark brown hair similar to his.
(c) Suspecting foul play, Sletherby did not give him any money, as he knew Mrs. Saltpen-Jago was a blonde.
(a) While describing the appearance of Mrs. Saltpen-Jago, K.C. described her altered hairstyle but Sletherby was in a daze after seeing the double crest on the door of the car.
(b) Sletherby realized that Bertie had not lied, but that he had mistaken him for a fraud..

3 Explain the following phrases selected from the story in your own words and work with a partner to make sentences using these phrases:

Question (a)
on approval – on acceptance
Answer:
Seetha went abroad on her dad’s approval.

Question (b)
a rustic holiday – a vacation to be spent in a village
Answer:
The students who went on a rustic holiday to Kurangani forest were caught in a forest fire accident.

Question (c)
a double-distilled idiot – very stupid person
Answer:
Some double – distilled idiots ill-treat their parents in old age.

Question (d)
a tinge of coldness – a shade of hostility lack of friendliness
Answer:
The manager turned down the worker’s plea for a pay hike with a tinge of coldness.

Question (e)
making up for lost time – compensate
Answer:
The boy had to study two hours more everyday to make up for the lost time.

(4) Answer the following questions in a paragraph of about 100-150 words each.

Question (a)
Describe the youth’s strange behaviour when he was in the train.
Answer:
The young man who entered the coach gave out a smothered curse. He was engaged in searching something elusive angrily and uselessly. From time to time, he dug a six penny bit out of a waist coat pocket and stared at it sadly, then resumed his search. He voluntarily broke the silence. He exclaimed that Mr. Sletherby was-going to Bill Manor. He introduced himself as Bertie, the younger son of Mrs. Saltpen-Jago. He admitted that he was away for about six months and had not seen his own mother. Making use of the lucky coincidence that he was going to Brill Manor, he asked for a loan of three pounds as he had lost his sovereign purse and was desperately in need of help. He promised to meet him on the subsequent Monday.
“There is no need to suffer silently and there is no shame in seeking help.”

Question (b)
One has to be cautious and not be carried away by stories. How did Sletherby exhibit caution?
Answer:
Sletherby is a very cautious person. From the beginning he had decided not to be carried away by the ‘stories’ of the young man. He laid the trap for the young man by feigning absolute ignorance about his mom’s looks. He asked him if his mom resembled him. Bertie replied that his mom had dark brown hair like himself. Besides, in his mental eye, Sletherby compared the crest he found in the invitation received from Mrs. Saltpen-Jago “a grey hound” with the young man’s description of the royal crest “demi-lion holding a cross-crosslet in its paw”. The two details contradicted Mr. Sletherby’s knowledge of Mrs. Saltpen-Jago’s royal family. There is something basically inhuman not to be interested in the affairs of a fellow human being “in distress”. .
“The truth is a beautiful and terrible thing and hence should be treated with caution.”

Question (c)
How did Sletherby’s judgement of Bertie turn out to be a wrong one?
Answer:
Sletherby concluded that Bertie was a fraud. He compared his looks with that of his mom. The footman of Mrs.Saltpen-Jago while describing her appearance, said that she had altered her looks only five weeks ago. She had changed her dark brown hair to a blonde. Sletherby also saw the double crest of the royal family on the door panel of the car which displayed a demi-lion and a greyhound courant. The evidences which made Mr. Sletherby believe that Bertie must have been a fraud turned out to be solid evidences ‘ indicating Bertie’s genuineness as a legal heir of Mrs. Saltpen-Jago. Sletherby realized

with alarm that Bertie had not lied to him, but he had mistaken an honest man in trouble to be an impostor. Sletherby’s political future was heading to a disaster simply because he refused to lend three pounds to a young man in trouble. He was penny wise and pound foolish.

“Before you assume, learn the facts. Before you judge, understand why. ”

Question (d)
As Sletherby, would you apologise to Bertie for your rude behaviour? Give reasons.-
Answer:
Of course, yes. I would earnestly apologise for mistaking Bertie to be a fraud. I would explain the circumstantial evidences which really worked against him. The crest of a grey hound I had found in his mom’s letter and the present altered colour of his mom’s hair had clouded my judgement against him. I would admit that I could have offered at least two pounds to help him overcome the problem caused by his carelessness. I would express my regrets for leaving him behind furious and disappointed.
“Life becomes easier when you learn to accept the apology you never got. ”

Question (e)
‘Seeing is believing’. How is this humorously disproved in this story? Bring out the irony in the situation. .
Answer:
“Seeing is believing” is an age old adage which is also contradicted by Plato’s saying Appearance is always deceptive because it appears to appear”. Some times truth is not quite impressive. She needs the support of limping old maid called time to prove herself. Being a cynic and skeptic, Mr. Sletherby takes up a stand right at the beginning of the encounter, not to show any interest in the young man’s fussy outbursts. He carefully lays the trap by telling a lie that he had not seen his mom. He further asks him cleverly if his mom resembled him.

Both Bertie himself and Mr. Sletherby are unaware of the altered looks of Mrs. Saltpen-Jago. There is a dramatic irony when the son himself gives a different version of his mom’s appearance. This influences the decision of Mr. Sletherby in refusing to lend Bertie a loan of three pounds. Thus the adage “Seeing is believing” is humorously disproved in this story. Enquiring thoroughly and compassionately one can find the truth.
“Everything we hear is an opinion, not a fact.
Everything we see is a perspective, not the truth.

Additional Questions

Question (a)
Attempt a character sketch of Mr. Philip Sletherby.
Answer:
Mr. Philip Sletherby was a politician but much known to the people of Chalkshire. He wishes to contest for a MP seat in Chalkshire with the recommendation of a level-headed woman namely Mrs. Saltpen-Jago. Fortunately he had been invited to spend the weekend with her at Brill Manor. He believes that his political pilgrimage would turn out to be a blessing. There is something mean about his unwillingness to show any interest in the affairs of an angry young travelling companion. Mr. Philip Sletherby loves things and uses people to advance his career prospects. At the very beginning, he doubts the credentials of Bertie who introduces himself as the younger son of Bertie. He is quick to judge people. Though the young man wanted only three pounds, he quizzes him with many questions about his royal crest and contradicts him with his ice-cold logic. He is proud of having laid a trap saying he had never seen his mother.

He offers to assist the young man in distress. But when Bertie wants to leave, he refuses to part with money saying that his mom was a pronounced blonde. He prided himself for his cleverness. But he is shocked to see the double crest on the car’s door panel, demi-lion holding a cross-crosslet in his paw and a greyhound. He is petrified to leam from K.C.’ that Mrs. Saltpen-Jago had recently dyed her beautiful dark brown hair blonde. Mr. Sletherby realized that his cleverness has, in reality, turned out to be his meanness. He becomes troubled that his political future may be lost due to his three pound de-buckle. He has proved himself penny wise and pound foolish.
“You must make the choice, to take the chance if you want anything, in life to change. ”

Question (b)
Attempt a character sketch of Mrs. Saltpen-Jago.
Answer:
Mrs. Saltpen is a brilliant woman. She is level headed. She is a clear thinker. She knows exactly when to take up an individual or a social cause. Being a lady of discernment, she may choose to drop a person if she finds him unworthy of her attention. She is an influential woman. But occasionally, she is eccentric. She spoils herself and her chance by being too restless. No wonder Bertie too is restless like his mom. She is never calm and composed. She used to have beautiful dark brown hair which jelled well with her fresh complexion.

Recently she has electrified people by dying it blonde. It has in-fact ruined her looks. K.C. unravels many details about Mrs. Saltpen-Jago. It is obvious that her influence in Chalkshire is more due to her affluence and large estate than her personality traits. Her inconstant nature is obvious in her ability to raise someone from the gutters to the top or drop someone she does not like. The only thing charming about her is that she treats her guests well as a good hostess.

I. Choose the right option.

Question 1.
Philip Sletherby was on a profitable pilgrimage to ______
(a) Mecca
(b) Israel
(c) Velankanni
(d) Brill Manor
Answer:
(d) Brill Manor

Question 2.
Mrs ______ was a person of some social importance in London.
(a) Thatcher
(b) Victoria
(c) Williams
(d) Saltpen-Jago
Answer:
(d) Saltpen-Jago

Question 3.
Sletherby hostess had considerable influence in the country of ______
(a) Cannibals
(b) Yorkshire
(c) Chalkshire
(d) Scotland
Answer:
(c) Chalkshire

Question 4.
______ was the second son of Mrs. Saltpen-Jago.
(a) Bob
(b) Jimmy Wells
(c) Bertie
(d) James
Answer:
(c) Bertie

Question 5.
Mr. ______ was invited to spend the weekend with Mrs. Saltpen-Jago.
(a) Bertie
(b) Claude
(c) KC
(d) Philip Sletherby
Answer:
(d) Philip Sletherby

Question 6.
Bertie claimed that he had not seen his mother for a period of ______ months.
(a) three
(b) six
(c) four
(d) two
Answer:
(b) six

Question 7.
Bertie was ______ when his mom was in town.
(a) fishing
(b) yachting
(c) travelling
(d) skiing
Answer:
(b) yachting

Question 8.
The young man confessed that he had done an ______ thing.
(a) ugly
(b) awkward
(c) embarrassing
(d) efficient
Answer:
(b) awkward

Question 9.
Bertie claims to have lost his sovereign purse with ______ quid in it.
(a) ten
(b) six
(c) four
(d) five
Answer:
(c) four

Question 10.
Bertie compared himself to a double distilled ______ for leaving his purse behind.
(a) genius
(b) fool
(c) idiot
(d) clown
Answer:
(c) idiot

Question 11.
Bertie planned to spend three days at ______
(a) Brazil
(b) Brondquay
(c) Mumbai
(d) Ceylon
Answer:
(b) Brondquay

Question 12.
Bertie wanted to go for ______ for three days.
(a) trekking
(b) fishing
(c) reading
(d) bungee jumping
Answer:
(b) fishing

Question 13.
Initially Sletherby gave Bertie hopes by agreeing to lend him ______ pounds.
(a) one
(b) two
(c) three
(d) four
Answer:
(c) three

Question 14.
As suspicion crept into Sletherby’s mind about the genuineness of the boy’s claim there was a tinge of ______ in his voice.
(a) pity
(b) anger
(c) coldness
(d) love
Answer:
(c) coldness

Question 15.
Bertie said that his mom had dark ______ hair.
(a) red
(b) brown
(c) black
(d) blond
Answer:
(b) brown

Question 16.
Slether by found Bertie’s mother having ______ hair.
(a) brown
(b) red
(c) black
(d) blond
Answer:
(d) blond

Question 17.
Sletherby congratulated himself on his ______
(a) meanness
(b) judgement
(c) astuteness
(d) cautiousness
Answer:
(c) astuteness

Question 18
______ wanted to play a round of golf with Sletherby.
(a) Mrs. Saltpen-Jago
(b) Peter
(c) Claude People
(d) Bertie
Answer:
(c) Claude People

Question 19.
Sletherby’s eyes were fixed on the ______ on which were displayed two crests: a greyhound courant and a demi-lion holding in its paw a cross-crosslet.
(a) tyre
(b) mirror
(c) door panel
(d) bonnet
Answer:
(c) door panel

Question 20.
The idiotic change Mrs. Saltpen-Jago had made five weeks ago was ______
(a) plastic surgery of her nose
(b) dying her dark brown hair blonde
(c) dying her brown hair white
(d) cutting her long curly hair like a school girl
Answer:
(b) dying her dark brown hair blonde

II. Identify the speaker.

Question 1.
“Oh, you’re staying with Mrs. Saltpen-Jago for the week-end.
Answer:
Club acquaintance of Mrs. Sletherby.

Question 2.
“Didn’t I hear you say you were going down to stay with Mrs. Saltpen-Jago at Brill Manor?
Answer:
Bertie to Philip Sletherby.

Question 3.
“I’ve done a damned awkward thing.”
Answer:
Bertie

Question 4.
“What’s your crest by the way?”
Answer:
Sletherby to Bertie.

Question 5.
“I think I can manage that.”
Answer:
Philip Sletherby.

Question 6.
“Thanks awfully. It’s jolly good of you.”
Answer:
Bertie to Sletherby

Question 7.
“The demi-lion is the Saltpen crest.”
Answer:
Bertie to Sletherby

Question 8.
“My station is the next one.”
Answer:
Bertie to Sletherby

Question 9.
“Does she resemble you at all in feature?”
Answer:
Sletherby to Sletherby

Question 10.
“She has the same dark brown hair.”
Answer:
Bertie about his mom to Sletherby

Question 11.
“You’ve forgotten the three quid.”
Answer:
Bertie to Sletherby

Question 12.
“I’ve no intention of lending you three pounds, or three shillings.”
Answer:
Sletherby to Bertie

Question 13.
“She is a pronounced blonde.”
Answer:
Sletherby about Mrs. Saltpen-Jago to Bertie

Question 14.
“Hullo, Sletherby! You spending the week-end at Brill?”
Answer:
Claude People K.C.

Question 15.
“We’ll have a round of golf together.”
Answer:
Claude People K.C,

Question 16.
“Good appearance, until she made that idiotic change.”
Answer:
Claude People K.C.

Question 17.
“Change?…. What change?”
Answer:
Sletherby to K.C.

Question 18.
“She used to have beautiful dark brown hair, which went very well with her fresh complexion” ,
Answer:
Claude People K.C. to Sletherby

Question 19.
“Five weeks ago, she electrified everybody by appearing as a brilliant blonde.”
Answer:
Claude People K.C. to Sletherby

Question 20.
“I say, what’s the matter with you? You look rather ill.”
Answer:
Claude People K.C. to Sletherby.

III. Reading Comprehension.

Question 1.
He had scarcely glanced at a couple of pages, however,, when a smothered curse caused him to glance hastily at the only other occupant of the carriage. His travelling companion was a young man of about two-and-twenty, with dark hair, fresh complexion, and the blend of smartness and disarray that marks the costume of a ‘nut’ who is bound on a rustic holiday. He was engaged in searching furiously and ineffectually for some elusive or non-existent object; from time to time he dug a sixpenny bit out of a waistcoat pocket and stared at it ruefully, then recommenced the futile searching operations. A cigarette-case, matchbox, latchkey, silver pencil case, and railway ticket were turned out on to the seat beside him, but none of these articles seemed to afford him satisfaction; he cursed again, rather louder than before. The vigorous pantomime did not draw forth any remark from Sletherby, who resumed his scrutiny of the magazine.

Question (a)
What made Sletherby look at his fellow traveller?
Answer:
The young travelling companion let out a smothered curse. This made him look at the young travelling companion.

Question (b)
What, did Mr. Sletherby deduce from the looks of the young man?
Answer:
The young man must be around twenty two. He was bound for a rustic holiday.

Question (c)
What was the young man doing inefficiently?
Answer:
The young man was furiously searching for some elusive or non-existent object.

Question (d)
What did the young man’s rueful stare at the six penny imply?
Answer:
The young man’s rueful stare at the six penny implied the young man was in a financial . trouble.

Question (e)
Did the vigorous pantomime get the young man any help? Why?
Answer:
The vigorous pantomime turned out to be useless. Mr. Sletherby did not evince any interest in knowing the young man’s trouble.

2. “Not exactly, but left behind, which is almost as bad; just as inconvenient, anyway. I’ve come away .without my sovereign-purse, with four quid in it, all my worldly wealth for the moment. It was in my pocket all right, just before I was starting, and then I wanted to seal a letter, and
the sovereign-purse happens to have my crest on it, so I whipped it out to stamp the seal with, and, like a double-distilled idiot, I must have left it on the table. I had some silver loose in my pocket, but after I’d paid for a taxi and my ticket I’d only got this forlorn little six pence left. I’m stopping at a little country inn near Brondquay for three days’ fishing; not a soul knows me there, and my week-end bill, and tips, and cab to and from the station, and my ticket on to

Brill, that will mount up to two or three quid, won’t it? If you wouldn’t mind lending me two pound ten, or three for preference, I shall be awfully obliged. It will pull me out of no end of a hole.”

Question (a)
What was left behind?
Answer:
The young man’s sovereign purse was left behind.

Question (b)
What according to the young man, was in the purse?
Answer:
There was four quid in the purse.

Question (c)
What was the need to open his purse?
Answer:
He had to take out his crest to seal a letter.

Question (d)
Why did the young man call himself a distilled idiot?
Answer:
The young man had left his purse on the table while sealing the letter. So, he called himself a double distilled idiot.

Question (e)
“Haste makes waste.” Relate this saying to the young man’s loss.
Answer:
The young man was probably in a hurry to hire a cab and reach the railway station. In a haste he forgot to collect the purse he had kept on the table. Thus, it is true that haste makes waste.

3. “Thanks awfully. It’s jolly good of you. What a lucky thing for me that I should have chanced across one of the mater’s friends. It will be a lesson to me not to leave my exchequer lying about anywhere, when it ought to be in my pocket. I suppose the moral of the whole thing is don’t try and convert things to purposes for which they weren’t intended. Still, when a sovereign-purse has your crest on it-”
“What is your crest, by the way?” Sletherby asked, carelessly.

“Not a very common one,” said the youth; “a demi-lion holding a cross-crosslet in its paw.” “When your mother wrote to me, giving me a list of trains, she had, if I remember rightly, a greyhound *courant on her notepaper,” observed Sletherby. There was a tinge of coldness in his voice.

“That is the Jago crest,” responded the youth promptly; “the demi-lion is the Saltpen crest. We have the right to use both, but I always use the demi-lion, because, after all, we are really Saltpens.”

Question (a)
Why did Bertie thank Mr. Sletherby?
Answer:
He thanked Bertie hoping that he would extend him a loan of three pounds.

Question (b)
What was uncommon about the crest Bertie described?
Answer:
It was a demi-lion holding a cross-crosslet in his paw. Thus, it was an uncommon crest.

Question (c)
How did Mr. Sletherby’s observation drive Bertie to defend the genuineness of his crest?
Answer:
Mr. Sletherby observed that in the note paper he had received from Mrs. Saltpen-Jago, he had seen “Greyhound courant” crest printed. This drove Bertie to defend his crest saying his mom used Jago crest whereas he used Saltpen’s crest.

Question (d)
How did Bertie happen to use a different crest?
Answer:
Bertie used Saltpen’s crest. His mom used Jago crest i.e., Greyhound. Children had the option to use either Saltpen’s or Jago crests.

4. “Good-bye,” said Sletherby.

“You’ve forgotten the three quid,” said the young man, opening the carriage-door and pitching his suit-case on to the platform.

“I’ve no intention of lending you three pounds, or three shillings,” said Sletherby severely. “But you said-”
“I know I did. My suspicions hadn’t been roused then, though I hadn’t necessarily swallowed your story. The discrepancy about the crests put me on my guard, not withstanding the really brilliant way in which you accounted for it. Then I laid a trap for you; I told you that I had never met Mrs. Saltpen-Jago. As a matter of fact I met her at lunch on Monday last. She is a pronounced blonde.”

The train moved on, leaving the soi-disant cadet of the Saltpen-Jago family cursing furiously on the platform.

Question (a)
Did Mr, Sletherby really forget three quids. How?
Answer:
No, Sletherby did not forget three quid loan demanded by the young man. He simply took him to be a fraud and refused to lend him even three penny.

Question (b)
What had put Mr. Sletherby on guard?
Answer:
The information the boy disclosed about the family crest of Saltpen-Jago put Mr. Sletherby on guard.

Question (c)
What lie did Mr. Sletherby use to trap the young man?
Answer:
He had never seen Bertie’s mom Mrs. Saltpen-Jago.

Question (d)
Which claim of Bertie about his mom’s looks worked against him?
Answer:
The claim that his mom had dark brown hair similar to him worked against him. This . statement made Mr. Sletherby conclude that Bertie was a fraud.

Question (e)
What did Bertie do to express his feelings on the disappointment caused by Mr. Sletherby?
Answer:
Bertie cursed Sletherby furiously stamping his feet on the platform.

5. Sletherby heard not a single word, noted not one of the details that were being expounded to him. His eyes were fixed on the door panel, on which were displayed two crests: a greyhound courant and a demi-lion holding in its paw a cross-crosslet. “Brilliant woman, level-headed, a clear thinker, knows exactly when to take up an individual or a cause, exactly when to let him or it drop. Influential woman, but spoils herself and her chances by being too restless. No repose. Good appearance, too, till she made that idiotic change.”

“Change?” queried Sletherby, “what change?”

“What change? You don’t mean to say- Oh, of course, you’ve only known her just lately. She used to have beautiful dark brown hair, which went very well with her fresh complexion; then one day, about five weeks ago, she electrified everybody by appearing as a brilliant blonde. Quite ruined her looks. Here we are. I say, what’s the matter with you? You look rather ill.”

Question (a)
Why was Mr. Sletherby speechless?
Answer:
Mr. Sletherby saw two crests on the car’s door panel. One was a grey bound and the other was what Bertie had described, a demi-lion holding in its paw a cross, crosslet. This made him speechless.

Question (b)
Who described Mrs. Saltpen-Jago as a brilliant and influential woman?
Answer:
Mr. K. C. described Mrs. Saltpen-Jago as a brilliant and influential woman.

Question (c)
How did Mrs. Saltpen-Jago spoil her chances?
Answer:
She spoiled her chances by being too restless.

Question (d)
What idiotic change had Mrs. Saltpen-Jago made in her good appearance?
Answer:
She dyed her beautiful dark brown hair into blonde. This idiotic change ruined her looks.

Question (e)
Why did Mr. Sletherby look very ill suddenly?
Answer:
Mr. Sletherby had doubted Bertie’s crest and his claim of brown hair of his mom. But he realized with alarm that Bertie had told the truth about the crest and about his mom’s hair. He looked ill as he had been too mean to Mrs. Salt-pen’s son.

A Shot In the Dark About the Author

Hector Hugh Munro (1870-1916) better known by his pen name Saki was a brilliant British writer. His witty, mischievous and sometimes macabre stories, satirize Edwardian society and culture. He is considered a master of short story and often compared to Q, Henry. His stories also tend to end with a twist like those of O. Henry’s.

A Shot In the Dark Summary

Philip Sletherby travels in a train to meet Mrs. Saltpen Jago. He decided to gain some political mileage by getting her acquaintance. Sletherby wanted to be a candidate for the MP election in the eastern division of Chalkshire. As she had invited him on a friendly luncheon-party at her country house he counted on her for nominating him. The silence in the compartment was spoiled by the smothered cries of a young man who was obviously looking for something. As Mr, Philip Sletherby did not show any interest in his despair, he himself explained his situation. He was glad that the narrator was going to spend his week-end with his mother Mrs. Saltpen – Jago at Brill Manor. The boy said that he hadn’t seen his mom for almost six months. He was Bertie and he had lost his royal seal and the money purse. He needed three quid for his brief stay and ticket fare to his home. He was on a fishing expedition.

Mr. Sletherby agreed to lend but asked him, what was his crest? The young man said it was the demi-lion holding a eross-crosslet in its paw, With a tinge of coldness in his voice, Sletherby said that his mom had written a grey hound courant on her notepaper. The young man said they could use both. Demi-lion was Saltpen crest and Grey Hound was Jago-crest. The young man said that he always used the demi-lion. He also added that he was getting down in the next station. Sletherby asked him if his mom resembled him. The boy said that she had the same dark brown hair like him. When the boy was about to get out, he reminded him of the loan of three quid. Sletherby said that he had met his mom who was a pronounced blonde. The young man stamps his. feet furiously. A tall footman had arrived at the railway station to receive him.

A luxury vehicle was parked. The footman Mr. K.C. explained a lot about the car. But nothing entered his brain as he saw two crests on the car “s door panel-a grey hound courant and a demi-lion holding in its paw a cross-erosslet. The second blow follows when KC said that Mrs. Saltpen-Jago used to have beautiful dark brown hair which agreed well with her complexion. She shocked everybody bv changing her looks as a brilliant blonde. She had ruined her looks. Suddenly Mr. Sletherby turned pale. Mr. K.C. asked if he was ill.

1. Rearrange the sentences given below:

astuteness – cleverness
blonde – pale yellow hair
chaise – horse-drawn carriage
*courant -/ko ra:nt/ – animal represented in the act Of running
crest -a symbol representing a family, reproduced on writing paper
discrepancy – difference
embarked – boarded
espied – caught sight of
exchequer – national treasury
expounded – explained
Honoria /hD ‘ na:na/ – (Latin) honour, a woman of reputation
ineffectually – In vam
mater / ‘ melta/ – (Latin) mother
pantomime – exaggerated behaviour
quid – (British) one pound Sterling
ruefully – regretfully, sorrowfully
sedately – in a dignified manner
soi- disant -self styled
sumptuous – expensive looking
tackle – equipment required for a task or sport
volubility – fluency, talkativeness

Additional:
apologise – seek forgiveness
alarmed – afraid of danger
dumb struck – shocked
exactly – correctly, accurately
frantically – quickly in a disorderly manner

The main aim is to share the knowledge and help the students of 11th English to secure the best score in their final exams. Use the concepts of Samacheer Kalvi 11th English Book Solutions Supplementary Chapter 2 A Shot In the Dark Questions and Answers in Real time to enhance your skills. If you have any doubts you can post your comments in the comment section, We will clarify your doubts as soon as possible without any delay.

Students can Download Computer Science Chapter 7 Composition and Decomposition Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Computer Science Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Computer Science Solutions Chapter 7 Composition and Decomposition

Samacheer Kalvi 11th Computer Science Composition and Decomposition Text Book Back Questions and Answers

PART – 1
I. Choose The Correct Answer

11th Computer Science Chapter 7 Book Back Answers Question 1.
Suppose u, v = 10, 5 before the assignment. What are the values of u and v after the sequence of assignments?
1. u : = v
2. v : = u
(a) u, v = 5, 5
(b) u, v = 10, 5
(c) u, v = 5, 10
(d) u, v = 10, 10
Answer:
(a) u, v = 5, 5

Composition And Decomposition In Computer Science Question 2.
Which of the following properties is true after the assignment (at line 3?
1. – – i + j = 0
2. i, j : = i + 1, j – 1
3. – – ?
(a) i + j > 0
(b) i + j < 0
(c) i + j = 0
(d) i = j
Answer:
(c) i + j = 0

Class 11 Computer Science Chapter 7 Solutions Question 3.
If C1 is false and C2 is true, the compound statement

(a) S1
(b) S2
(c) S3
(d) none
Answer:
(b) S2

Computer Science Samacheer Kalvi Question 4.
If C is false just before the loop, the control flows through

(a) S1 ; S3
(b) S1 ; S2 ; S3
(c) S1 ; S2 ; S2 ; S2 ; S3
(d) S1 ; S2 ; S2 ; S2 ; S3
Answer:
(a) S1 ; S3

Samacheer Kalvi Computer Science Question 5.
If C is true, S1 is executed in both the flowcharts, but S2 is executed in

(a) (1) only
(b) (2) only
(c) both (1) and (2)
(d) neither (1) nor (2)
Answer:
(a) (1) only

Samacheer Kalvi Guru 7th Computer Science Question 6.
How many times the loop is iterated?
i : = 0
while i ≠ 5
i : = i + 1
(a) 4
(b) 5
(c) 6
(d) 0
Answer:
(b) 5

PART – 2
II. Short Answers

Samacheer Kalvi Guru 11th Computer Science Question 1.
Distinguish between a condition and a statement.
Answer:

Samacheer Kalvi Computer Science Book Question 2.
Draw a flowchart for conditional statement.
Answer:

11 Computer Science Samacheer Kalvi Question 3.
Both conditional statement and iterative statement have a condition and a statement. How do they differ?
Answer:
Conditional Statement:

• Statements are executed only once when the condition is true.
• If condition statement.

Iterative Statement:

• Iterative statement repeatedly evaluates a condition and executes a statement until it becomes false.
• While condition statement.

Samacheer Kalvi Guru Computer Science Question 4.
What is the difference between an algorithm and a program?
Answer:
Algorithm:

• Algorithm is for human readers to understand.
• Knowledge of English is needed.
• Easy to understand.

Program:

• Program is for the computers to execute directly.
• Knowledge of programming language is required.
• It is difficult to understand.

Computer Science Textbook Solutions Question 5.
Why is function an abstraction?
Answer:
Once a function is defined, it can be used over and over and over again. Reusability of a single function several times is known as an abstraction.

Computer Chart For 11th Class Question 6.
How do we refine a statement?
Answer:
In refinement, each statement is repeatedly expanded into more detailed statements in the subsequent levels.

PART – 3
Explain in Brief

Computer Charts For Class 11 Question 1.
For the given two flowcharts write the pseudo code.
Answer:

1. Enter A,B
2. Initialise Q = 0, r = A
3. If r > B, then do Q = Q + 1; r = r – B else r, q
4. Exit

Samacheer Kalvi Guru 11 Computer Science Question 2.
If C is false in line 2, trace the control flow in this algorithm.

1. S1
2. – – C is false
3. if C
4. S2
5. else
6. S3
7. S4

Answer:
S1 ; S2 ; S4

Statement Meaning In Computer Question 3.
What is case analysis?
Answer:
Case Analysis statement generalizes the problem into multiple cases. Case Analysis splits the problem into an exhaustive set of disjoint cases.

Question 4.
Draw a flowchart for -3 case analysis using alternative statements.
Answer:

Question 5.
Define a function to double a number in two different ways:

1. n + n
2. 2 x n

Answer:
1. Double (n)
– – inputs: n is a real number or an integer, n > 0
– – Outputs: y is a real number or an integer such that y = n + n

2. Double (n)
– – inputs: n is a real number or an integer, n > 0
– – Outputs: y is a real number or an integer such that y = 2 x n

PART – 4
IV. Explain in Detail

Question 1.
Exchange the contents: Given two glasses marked A and B. Glass A is full of apple drink and glass B is full of grape drink. Write the specification for exchanging the contents of glasses A and B, and write a sequence of assignments to satisfy the specification.
Answer:
Let the variables a, b, c represent Glass A, Glass B and Glass C and a, b, c can store values APPLE, GRAPE or EMPTY.
Specification:

1. exchange (a, b)
2. – – inputs: a, b : = APPLE, GRAPE
3. – – outputs: a, b : = GRAPE, APPLE

Algorithm:

Question 2.
Circulate the contents: Write the specification and construct an algorithm to circulate the contents of the variables A, B and C as shown below: The arrows indicate that B gets the value of A, C gets the value of B and A gets the value of C.

Specifications:

1. circulate
2. – – inputs: a, b, c: = A, B, C
3. – – outputs: a, b, c: = C, A, B

Algorithm:

1. circulate (a, b, c)
2. – – a, b, c: = A, B, C
3. temp : = c
4. c : = b
5. b : = a
6. a : = temp
7. – – a, b, c: = C, A, B

Question 3.
Decanting problem. You are given three bottles of capacities 5, 8, and 3 litres. The 8L bottle is filled with oil, while the other two are empty. Divide the oil in 8L bottle into two equal quantities. Represent the state of the process by appropriate variables. What are the initial and final states of the process? Model the decanting of oil from one bottle to another by assignment. Write a sequence of assignments to achieve the final state.
Answer:
(a)
Let a, b, c be the variables whose maximum value are 8L, 5L and 3L respectively.
Initial State:

a, b, c: = 8, 0, 0

Final State:

a, b, c: = 4, 4, 0

Specifications:

1. decant
2. – – inputs: a, b, c : = 8, 0, 0
3. – – outputs: a, b, c : = 4, 4, 0

Algorithm:
Let us assume that a: = b denote oil in b is poured into a bottle until either “a” is full or “b” becomes empty.

1. decant (a, b, c)
2. – – a, b, c : = 8, 0, 0
3. b : = a
4. – – a, b, c : = 3, 5, 0
5. c : = b
6. – – a, b, c : = 3, 2, 3
7. a : = c
8. – – a, b, c : = 6, 2, 0
9. c : = b
10. – – a, b, c : = 6, 0, 2
11. b : = a
12. – – a, b, c : = 1, 5, 2
13. c : = b
14. – – a, b, c : = 1, 4, 3
15. a : = c
16. – – a, b, c: = 4, 4, 0

Question 4.
Trace the step – by – step execution of the algorithm for factorial(4).
factorial(n)
– – inputs : n is an integer , n ≥ 0
– – outputs : f = n!
f, i : = 1,1
while i ≤ n
f, i : = f x i, i + 1
Answer:

Samacheer kalvi 11th Computer Science Composition and Decomposition Additional Questions and Answers

PART – 1
I. Choose the correct answer

Question 1.
Which one of the following is odd?
(a) Python
(b) C++
(c) C
(d) Ctrl + S
Answer:
(d) Ctrl + S

Question 2.
………………. is a diagrammatic notation for representing algorithms.
(a) Pseudo code
(b) Flowchart
(c) Program
(d) Languages
Answer:
(b) Flowchart

Question 3.
There are important control flow statements.
(a) four
(b) three
(c) two
(d) five
Answer:
(b) three

Question 4.
A ………………. statement is composed of a sequence of statements.
(a) iterative
(b) conditional
(c) sequential
(d) alternative
Answer:
(c) sequential

Question 5.
A ………………. is contained in a rectangular box with a single outgoing arrow, which points to the box to be executed next.
(a) statement
(b) composition
(c) notation
(d) condition
Answer:
(a) statement

Question 6.
The triangle is right – angled, if ……………….
(a) C = a – b
(b) C² = a² + b²
(c) C² = (a + b)²
(d) c² = a² – b²
Answer:
(b) C² = a² + b²

Question 7.
The algorithm can be specified as ……………….
(a) monochromatize (a, b, c)
(b) a = b = 0
(c) C = A + B + C
(d) none
Answer:
(a) monochromatize (a, b, c)

Question 8.
After an algorithmic problem is decomposed into subproblems, we can abstract the subproblems as ……………….
(a) refinement
(b) pseudo – code
(c) decomposition
(d) functions
Answer:
(d) functions

Question 9.
Which one of the following is the elementary problem solving techniques?
(a) Specification
(b) Abstraction
(c) Composition
(d) decomposition
Answer:
(d) decomposition

Question 10.
Which one of the following have only high level details?
(a) Flow chart
(b) Algorithm
(c) Programs
(d) Pseudocode
Answer:
(b) Algorithm

Question 11.
How many different notations are there for representing algorithms?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(b) 3

Question 12.
Which one of the following notations will be executed by computers?
(a) Flow chart
(b) Pseudocode
(c) Programming languages
(d) Compiler
Answer:
(c) Programming languages

Question 13.
Which one of the following algorithmic notation is used for communication among people?
(a) Flow chart
(b) Pseudo code
(c) PL
(d) Interpreter
Answer:
(b) Pseudo code

Question 14.
Which one of the following algorithmic notation is used for giving visual intuition of control flow?
(a) Flow chart
(b) Programming languages
(c) Pseudo code
(d) Compiler
Answer:
(a) Flow chart

Question 15.
The algorithmic notation similar to Programming language is ……………….
(a) Flow chart
(b) Pseudo code
(c) C ++
(d) C
Answer:
(b) Pseudo code

Question 16.
Identify the statement which is wrong ……………….
(a) Programs must obey the grammar of the Programming language exactly.
(b) The punctuations in Programming language must be exactly.
(c) The Programming language is informal.
(d) Translator translates the programs into instructions.
Answer:
(c) The Programming language is informal.

Question 17.
Which one is used for converting programs into computer executable instructions?
(a) Converter
(b) Apps
(c) Translator
(d) exe files
Answer:
(c) Translator

Question 18.
Identify the correct statement from the following?
(a) Pseudo code can be executed by the computer directly.
(b) Pseudo code cannot be executed by the computer directly.
(c) Algorithm cannot be expressed in Pseudo code
(d) Algorithm are written only for python language
Answer:
(b) Pseudo code cannot be executed by the computer directly.

Question 19.
The notation which is not formal nor exact is ……………….
(a) Flow chart
(b) Pseudo code
(c) Compiler
(d) Translator
Answer:
(b) Pseudo code

Question 20.
Identify the correct statements.
(i) Pseudo code uses natural English for statements and conditions..
(ii) Pseudo code notation is formal one.
(iii) There is no need to follow the rules of the Programming language grammer
(iv) It must be rigorous and correct.
(a) (i) (ii) (iii)
(b) (ii) (iii) (iv)
(c) (i) (iii) (iv)
(d) (i) (ii) (iii) (iv)
Answer:
(c) (i) (iii) (iv)

Question 21.
Find the pair which is wrongly matched.
(a) Rectangular boxes – Statements
(b) Diamond boxes – Output
(c) Arrow – Control flow
(d) Parallelogram – Input
Answer:
(b) Diamond boxes – Output

Question 22.
Find the correct flow chart diagram from the following.

Answer:

Question 23.
The inputs and outputs are drawn using ……………….. boxes.
(a) rectangular
(b) diamond
(c) Parallelogram
(d) Oval
Answer:
(c) Parallelogram

Question 24.
The symbol used for representing the box to be executed next is ………………..
(a) ⇒
(b) ⇓
(c) ↓
(d) ++
Answer:
(c) ↓

Question 25.
The flow of control is represented in the flowchart by ………………..
(a) arrow
(b) dot
(c) box
(d) plus
Answer:
(a) arrow

Question 26.
In flow chart, rectangular boxes represents ………………..
(a) statements
(b) condition
(c) Input
(d) End
Answer:
(a) statements

Question 27.
A condition is contained in a diamond shaped box with ……………….. outgoing arrows.
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(a) 2

Question 28.
A collection of boxes containing statements and conditions connected by arrow are ………………..
(a) compiler
(b) Flow chart
(c) Pseudo code
(d) Algorithm
Answer:
(b) Flow chart

Question 29.
Match the following:

(a) (i) -4 (ii) 0 – 1 (iii) – 2 (iv) – 3
(b) (i) – 1 (ii) – 2 (iii) – 3 (iv) – 4
(c) (i) – 4 (ii) – 2 (iii) – 1 (iv) – 3
(d) (i) – 3 (ii) – 2 (iii) – 1 (iv) – 4
Answer:
(a) (i) -4 (ii) 0 – 1 (iii) – 2 (iv) – 3

Question 30.
How many outgoing arrows are needed for rectangular boxes in flow chart?
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(b) 1

Question 31.
Statements composed of other statements are known as:
(a) Simple Statements
(b) Compound Statements
(c) Conditional
(d) Control flow
Answer:
(b) Compound Statements

Question 32.
Which one of the following is not a control flow statements?
(a) Sequential
(b) Assignment
(c) Iterative
(d) Alternative
Answer:
(b) Assignment

Question 33.
Which one of the following statement is used to alter the normal flow of control of the program?
(a) Assignments
(b) Control flow
(c) Compound
(d) both b & c
Answer:
(d) both b & c

Question 34.
Which one of the following statements are executed one after the other as written in the algorithm?
(a) Sequential
(b) Iterative
(c) Conditional
(d) Decisive
Answer:
(a) Sequential

Question 35.
Alternative statements analyses the problem into ……………….. cases.
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(a) 2

Question 36.
Case analysis statement generalizes the statement into ……………….. cases.
(a) 2
(b) 3
(c) 5
(d) multiple
Answer:
(d) multiple

Question 37.
If _s^C is a ……………….. statement.
(a) Conditional
(b) Alternative
(c) Case Analysis
(d) Iterative
Answer:
(a) Conditional

Question 38.
Which one of the following process executes the same action repeatedly?
(a) Conditional
(b) Alternative
(c) Iterative
(d) None of these
Answer:
(c) Iterative

Question 39.
The iterative statement is commonly known as a ………………..
(a) loop
(b) Case Analysis
(c) Alternative
(d) Conditional
Answer:
(a) loop

Question 40.
Testing the loop condition and executing the loop body once is called ………………..
(a) alternative
(b) conditional
(c) Iteration
(d) Decomposition
Answer:
(c) Iteration

PART – 2
II. Short Answers

Question 1.
What is a Programming Language?
Answer:

1. A programming language is a notation for expressing algorithms to be executed by computers.
2. Programs must obey the grammar of the programming language exactly. Ex. C, C++, python.

Question 2.
What is a Pseudo code?
Answer:
Pseudo code is a mix of programming language like constructs and plain English. Algorithms expressed in Pseudo code are not intended to be executed by computers but for human readers to understand.

Question 3.
What is a Flow – chart?
Answer:
Flow chart is a diagrammatic notation for representing algorithms. A flow chart is a collection of boxes containing statements and conditions which are connected by arrows showing the order in which the boxes are to be executed.

Question 4.
What is a control flow statement. Classify it.
Answer:
Control flow statements are compound statements. They are used to alter the control flow of the process depending on the state of the process. They are classified as:

1. Sequential
2. Alternative
3. Iterative

Question 5.
What is meant by Decomposition?
Answer:
Decomposition is one of the elementary problem-solving techniques. An algorithm may be broken into parts, expressing only high level details. Then, each part may be refined into smaller parts, expressing finer details or each part may be abstracted as a function.

Question 6.
When a condition statement will be executed?
Answer:
It will be executed only when the condition statement is true.

Question 7.
Write the algorithm specification to find minimum of 2 numbers.
Answer:
Minimum (a, b)
– – inputs : a, b
– – outputs : result = a >1 b

Question 8.
Draw the flowchart for alternative control flow.
Answer:

PART – 3
III. Explain in Brief

Question 1.
Draw the diagram flowchart for integer division.
Answer:

Question 2.
Explain conditional statement.
Answer:
Conditional statement: Sometimes we need to execute a statement only if a condition is true and do nothing if the condition is false. This is equivalent to the alternative statement in which the else clause is empty. This variant of alternative statement is called a conditional statement. If C is a condition and S is a statement, then

is a statement, called a conditional statement, that describes the following action:

1. Test whether C is true or false.
2. If C is true then do S; otherwise do nothing.

The conditional control flow is depicted in the flowchart.

Question 3.
Write a note on refinement.
Answer:
After decomposing a problem into smaller subproblems, the next step is either to refine the subproblem or to abstract the subproblem.
1. Each subproblem can be expanded into more detailed steps. Each step can be further expanded to still finer steps, and so on. This is known as refinement.

2. We can also abstract the subproblem. We specify each subproblem by its input property and the input – output relation. While solving the main problem, we only need to know the specification of the subproblems. We do not need to know how the subproblems are solved.

Question 4.
Draw various symbols used in flowchart with their usage.
Answer:

Question 5.
Write a Algorithm to find minimum of 2 numbers.
Answer:

Question 6.

Answer:

Question 7.
Construct an iterative algorithm to compute the quotient and remainder after dividing an integer A by another integer B.
Answer:
divide (A, B)
– – inputs: A is an integer and B ≠ 0
– – Outputs: q and r such that A = q x B + r and 0 < r < B
q : = 0, A
While r ≥ B
q, r : = q + 1, r – B

PART – 4
IV. Explain in Detail

Question 1.
Write the disadvantages of flowcharts.
Answer:
Disadvantages of flowcharts:

1. Flowcharts are less compact than representation of algorithms in programming language or pseudo code.
2. They obscure the basic hierarchical structure of the algorithms.
3. Alternative statements and loops are disciplined control flow structures.

Question 2.
Explain the sequential statement in detail with the diagram.
Answer:
A sequential statement is composed of a sequence of statements. The statements in the sequence are executed one after another, in the same order as they are written in the algorithm, and the control flow is said to be sequential. Let S1 and S2 be statements. A sequential statement composed of S1 and S2 is written as
S1
S2
In order to execute the sequential statement, first do S1 and then do S2. The sequential statement given above can be represented in a flowchart. The arrow from S1 to S2 indicates that S1 is

Let the input property be P, and the input-output relation be Q, for a problem. If statement S solves the problem, it is written as

1. – – p
2. S
3. – – Q

If we decompose the problem into two components, we need to compose S as a sequence of two statements S1 and S2 such that the input-output relation of S1, say R, is the input property of S2.

1. – – P
2. S1
3. – – R
4. S2
5. – – Q

Question 3.
Write a note on case analysis.
Answer:
Alternative statement analyses the problem into two cases. Case analysis statement generalizes it to multiple cases. Case analysis splits the problem into an exhaustive set of disjoint cases. For each case, the problem is solved independently. If Cl, C2, and C3 are conditions, and S1, S2, S3 and S4 are statements, a 4 – case analysis statement has the form,

1. case C1
2. S1
3. case C2
4. S2
5. case C3
6. S3
7. else
8. S4

The conditions C1, C2, and C3 are evaluated in turn. For the first condition that evaluates to true, the corresponding sttement is executed, and the case analysis statement ends. If none of the conditions evaluates to true, then the default case S4 is executed.

1. The cases are exhaustive: at least one of the cases is true. If all conditions are false, the default case is true.
2. The cases are disjoint: only one of the cases is true. Though it is possible for more than one condition to be true, the case analysis always executes only one case, the first one that is true. If the three conditions are disjoint, then the four cases are (1) C1, (2) C2, (3) C3, (4) (not C1) and (not C2) and (not C3).

Question 4.
What is a function? Explain in detail.
Answer:
After an algorithmic problem is decomposed into subproblems, we can abstract the subproblems as functions. A function is like a sub-algorithm. Similar to an algorithm, a function is specified by the input property, and the desired input – output relation.

To use a function in the main algorithm, the user need to know only the specification of the function – the function name, the input property, and the input – output relation. The user must ensure that the inputs passed to the function will satisfy the specified property and can assume that the outputs from the function satisfy the input – output relation. Thus, users of the function need only to know what the function does, and not how it is done by the function. The function can be used a “black box” in solving other problems.

Ultimately, someone implements the function using an algorithm. However, users of the function need not know about the algorithm used to implement the function. It is hidden from the users. There is no need for the users to know how the function is implemented in order to use it.

An algorithm used to implement a function may maintain its own variables. These variables are local to the function in the sense that they are not visible to the user of the function. Consequently, the user has fewer variables to maintain in the main algorithm, reducing the clutter of the main algorithm.

Question 5.
Draw the flowchart for Eat breakfast.
Answer:

Question 6.
Explain Alternative statement.
Answer:
Alternative statement:
A condition is a phrase that describes a test of the state. If C is a condition and both S1 and S2 are statements, then

is a statement, called an alternative statement, that describes the following action:

1. Test whether C is true or false.
2. If C is true, then do S1; otherwise do S2.

In pseudo code, the two alternatives S1 and S2 are indicated by indenting them from the keywords if and else, respectively. Alternative control flow is depicted in the flowchart. Condition C has two outgoing arrows, labeled true and false. The true arrow points to the S1 box. The false arrow points to the S2 box. Arrows out of S1 and S2 point to the same box, the box after the alternative statement.

Conditional statement:
Sometimes we need to execute a statement only if a condition is true and do nothing if the condition is false. This is equivalent to the alternative statement in which the else clause is empty. This variant of alternative statement is called a conditional statement.

If C is a condition and S is a statement, then

1. Test whether C is true or false.
2. If C is true then do S; otherwise do nothing.

The conditional control flow is depicted in the flowchart.

Question 7.
Explain various symbols in Flow chart.
Answer:
Flowchart is a diagrammatic notation for representing algorithms. They show the control flow of algorithms using diagrams in a visual manner. In flowcharts, rectangular boxes represent simple statements, diamond-shaped box represent conditions, and arrows describe how the control flows during the execution of the algorithm. A flowchart is a collection of boxes containing statements and conditions which are connected by arrows showing the order in which.the boxes are to be executed.

1. A statement is contained in a rectangular box with a single outgoing arrow,which points to the box to be executed next.
2. A condition is contained in a diamond shaped box with two outgoing arrows, labeled true and false. The true arrow points to the box to be executed next if the condition is true, and the false arrow points to the box to be executed next if the condition is false.
3. Parallelogram boxes represent inputs given and outputs produced.
4. Special boxes marked Start and the End are used to indicate the start and the end of an execution:

The flowchart of an algorithm to compute the quotient and remainder after dividing an integer A by another integer B is shown below, illustrating the different boxes such as input, output, condition, and assignment, and the control flow between the boxes.

Flowcharts also have disadvantages.

1. Flowcharts are less compact than representation of algorithms in programming language or pseudo code.
2. They obscure the basic hierarchical structure of the algorithms.
3. Alternative statements and loops are disciplined control flow structures. Flowcharts do not . restrict us to disciplined control flow structures.

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Samacheer Kalvi 11th Chemistry Thermodynamics Textual Evaluation Solved

I. Choose the correct answer from the following:

11th Chemistry Chapter 7 Book Back Answers Question 1.
The amount of heat exchanged with the surrounding at constant temperature and pressure is called ………
(a) ΔE
(b) ΔH
(c) ΔS
(d) ΔG
Answer:
(b) ΔH

Samacheer Kalvi Guru 11th Chemistry Question 2.
All the naturally occurring processes proceed spontaneously in a direction which leads to
(a) decrease in entropy
(b) increase in enthalpy
(c) increase in free energy
(d) decrease in free energy
Answer:
(d) decrease in free energy

Samacheer Kalvi 11th Chemistry Question 3.
In an adiabatic process, which of the following is true?
(a) q = w
(b) q = 0
(c) ΔE = q
(d) PΔV = 0
Answer:
(a) q = 0

11th Chemistry Samacheer Kalvi Question 4.
In a reversible process, the change in entropy of the universe is ‘
(a) >0
(b) > 0
(c) < 0
(d) = 0
Answer:
(d) = 0

Samacheerkalvi.Guru 11th Chemistry Question 5.
In an adiabatic expansion of an ideal gas
(a) w = -∆u
(b) w = ∆u + ∆H
(c) ∆u = 0
(d) w = 0
Answer:
(a) w = -∆u

Samacheer Kalvi Guru 11 Chemistry Question 6.
The intensive property among the quantities below is
(a) mass
(b) volume
(c) enthalpy
(d) mass/volume
Answer:
(d) mass/volume

Samacheer Kalvi 11th Chemistry Solutions Question 7.
An ideal gas expands from the volume of 1 x 10^-3 m³ to 1 x 10^-2 m³ at 300K against a constant pressure at 1 x 10⁵ Nm^-2. The work done is ………..
(a) – 900 J
(b) 900 kJ
(c) 270 kJ
(d) – 900 kJ
Answer:
(a) – 900 J
Hint:
w = – P∆V
w = – (1 x 105 Nm^-2) (1 x 10^-2 m³ – 1 x 10^-3 m³)
w = – 10⁵ (10^-2– 10^-3)Nm
w = – 10⁵ (10 – 1) 10^3-) J
w = – 10⁵(9x 10^-3) J
w = – 9 x 10² J
w = – 900 J

11 Chemistry Samacheer Kalvi Question 8.
Heat of combustion is always
(a) positive
(b) negative
(c) zero
(d) either positive or negative
Answer:
(b) negative

Samacheer Kalvi.Guru 11th Chemistry Question 9.
The heat of formation of CO and CO₂ are -26.4 kcal and -94 kcal, respectively. Heat of combustion of carbon monoxide will be
(a) +26.4 kcal
(b) -67.6 kcal
(c) -120.6 kcal
(d) +52.8 kcal
Answer:
(b) -67.6 kcal
Hint:
CO_(g) + O_2(g) → CO_2(g)
∆H_C⁰ (CO) = [∆H_f(CO₂) – ∆H_f(CO) + ∆H_f(O₂)]
∆H_C⁰ (CO) = -94 KCal – [- 26.4 KCal + 0]
∆H_C⁰ (CO) = -94 KCal + 26.4 KCal
∆H_C⁰ (CO) = -67.4 KCal

Samacheer Kalvi 11th Chemistry Solution Question 10.
C(diamond) → C(graphite), ∆H = -ve, this indicates that ………..
(a) graphite is more stable than diamond
(b) graphite has more energy than diamond
(c) both are equally stable
(d) stability cannot be predicted
Answer:
(a) graphite is more stable than diamond

Samacheer Kalvi Class 11 Chemistry Solutions Question 11.
The enthalpies of formation of Al₂O₃ and Cr₂O₃ are -1596 kJ and -1134 kJ, respectively. ∆H for reaction 2Al + Cr₂O₃ → 2Cr + Al₂O₃ is ……….
(a) -1365 kJ
(b) 2730 kJ
(c) -2730 kJ
(d) -462 kJ
Answer:
(d) -462 kJ
Answer:
2A1 + Cr₂O₃ → 2Cr + Al₂O₃
∆H_r⁰ = [2 ∆H_f (Cr) + ∆H_f (Al₂O₃)] – [2 ∆H_f (Al) + ∆H_f (Cr₂O₂)]
∆H_r⁰ = [0 + (- 1596 kJ)] – [0 + (- 1134)]
∆H_r⁰ = – 1596 kJ + 1134 kJ
∆H_r⁰ = – 462 kJ

11th Chemistry Solutions Samacheer Kalvi Question 12.
Which of the following is not a thermodynamic function?
(a) internal energy
(b) enthalpy
(c) entropy
(d) frictional energy
Answer:
(d) frictional energy

11th Standard Chemistry Samacheer Kalvi Question 13.
If one mole of ammonia and one mole of hydrogen chloride are mixed in a closed container to form ammonium chloride gas, then …………
(a) ∆H > ∆U
(b) ∆H – ∆U = 0
(c) ∆H + ∆U = 0
(d) ∆H < ∆U
Answer:
(d) ∆H < ∆U

Thermodynamics Chemistry Question 14.
Change in internal energy, when 4 kJ of work is done on the system and 1 kJ of heat is given out by the system is ………….
(a) +1 kJ
(b) -5 kJ
(c) +3 kJ
(d) -3 kJ
Answer:
(c) +3 kJ
Hint:
∆U = q + w
∆U = – 1 kJ + 4 kJ
∆U = + 3 kJc

Class 11 Chemistry Solutions Samacheer Kalvi Question 15.
The work done by the liberated gas when 55.85 g of iron (molar mass 55.85 g mol^-1) reacts with hydrochloric acid in an open beaker at 25°C …………
(a) -2.48 kJ
(b) -2.22 kJ
(c) +2.22 kJ
(d) + 2.48 kJ
Answer:
(a)-2.48 kJ
Hint:
Fe + 2HCl → FeCl₂ + H₂
1 mole of Iron liberates 1 mole of hydrogen gas
55.85 g Iron = 1 mole Iron
∴ n = 1
T = 25°C = 298 K
w = -P\(\left(\frac{\mathrm{nRT}}{\mathrm{P}}\right)\)
w = -nRT
w = -1 x 8314 x 298 J
w = 2477.57 J
w = -2.48 k J

11th Samacheer Kalvi Chemistry Question 16.
The value of AH for cooling 2 moles of an ideal mono atomic gas from 125°C to 25°C at constant pressure will be [given C_P = \(\frac {5}{2}\) R] …………..
(a) -250 R
(b) -500 R
(c) 500 R
(d) +250 R
Answer:
(b) -500 R
Hint:
T_i = 125°C = 398K
T_f = 25°C = 298 K
∆H = nC_p (T_f – T_i)
∆H = 2 x \(\frac {5}{2}\)R (298 – 398)
∆H = -500 R

Samacheer Kalvi Chemistry 11th Question 17.
Given that C_(g) + O_2(g) → CO_2(g) ∆H°= a kJ; 2 CO_(g) + O_2(g) → 2CO_2(g) ∆H° = -b kJ; Calculate the ∆H° for the reaction C_(g) + H₂O_2(g) → CO_(g)
(a) \(\frac {b + 2a}{2}\)
(b) 2a – b
(c) \(\frac {2a – b}{2}\)
(d) \(\frac {b – 2a}{2}\)
Answer:
(d) \(\frac {b – 2a}{2}\)
Hint:
C + O₂ → CO₂  ∆H° = -a kJ ……..(1)
2CO + O₂ → 2CO₂ ∆H° = -b kJ …….(2)
C + ½O₂ → CO ∆H° = ?
(1) x (2)
2C + 2O₂ → 2CO₂ ∆H° = -2a KJ …….(3)
Reverse of equestion (2) will be
2CO₂ → 2CO + O₂ ∆H° = +b KJ …….(4)
(3) + (4)
2C + O₂ → 2CO ∆H° = b – 2a KJ …….(5)
(5) ÷ 2
C + O₂ → CO ∆H° = \(\frac {b – 2a}{2}\) KJ

Class 11 Chemistry Notes Question 18.
When 15.68 litres of a gas mixture of methane and propane are fully combusted at 0°C and 1 atmosphere, 32 litres of oxygen at the same temperature and pressure are consumed. The amount of heat released from this combustion in kJ is (∆H_C (CH₄) = – 890 kJ mol^-1 and ∆H_C (C₃H₈) = -2220 kJ mol^-1)
(a) -889 kJ
(b) -1390 kJ
(c) -3180 kJ
(d) -653.66 kJ
Answer:
(d) -653.66 kJ
Hint:
Given,
∆H_C (CH₄) = -890 kJ mol^-1
∆H_C (C₃H₈) = -2220 kJ mol^-1
Let the mixture contain x lit of and lit of propane.

Volume of oxygen consumed = 2 x + 5 (15.68 – x) = 32 lit
2x + 78.4 – 5 x = 321
78.4 – 3x = 32
3x = 46.41
x = 15.471
Given mixture contains 15.47 liters of methane and 0.2 13 liters of propane. Hence,

∆H_C = [- 614.66 kJ mol^-1] + [- 20.81 kJ mol^-1]
∆H_C = 635.47 kJ mol^-1.

Question 19.
The bond dissociation energy of methane and ethane are 360 kJ mol^-1 and 620 Id mol^-1 respectively. Then, the bond dissociation energy of C-C bond is …………
(a) 170 kJ mol^-1
(b) 50 kJ mol^-1
(c) 80 kJ mol^-1
(d) 220 kJ mol^-1
Answer:
(c) 80 kJ mol^-1
Hint:
4E_C-H = 360 kJ mol^-1
E_C-H = 90 kJ mol^-1
E_C-C + 6E_C-H = 620 KJ mol^-1
E_C-C + 6 x 9O = 62O kJ mol^-1
E_C-C + 540 = 620 kJ mol^-1
E_C-C = 80 kJ mol^-1

Question 20.
The correct thermodynamic conditions for the spontaneous reaction at all temperature is (NEET phase – I)
(a) ∆H<0 and ∆S>0
(b) ∆H<0 and ∆S<0
(c) ∆S>0 and ∆S = 0
(d) ∆H<0 and ∆S>0
Answer:
(a) ∆H<0 and ∆S>0

Question 21.
The temperature of the system, decreases in an …………
(a) isothermal expansion
(b) isothermal compression
(c) adiabatic expansion
(d) adiabatic compression
Answer:
(c) adiabatic expansion

Question 22.
In an isothermal reversible compression of an ideal gas the sign of q, AS and w are respectively
(a) +, -, –
(b) -, +, –
(c) +, -, +
(d) -, -, +
Answer:
(d) -, -, +
Hint:
During compression, energy of the system increases, in isothermal condition, to maintain temperature constant, heat is liberated from the system. Hence q is negative. During compression entropy decreases. During compression work is done on the system, hence w is positive.

Question 23.
Molar heat of vaporization of a liquid is 4.8 kJ mol^-1. If the entropy change is 16 J mol^-1 K^-1. the boiling point of the liquid is ………..
(a) 323 K
(b) 27°C
(c) 164 K
(d) 0.3 K
Answer:
(b) 27°C
Hint:

Question 24.
∆S is expected to be maximum for the reaction
(a) Ca_(S) + O_2(g) → CaO(S)
(b) C_(S) + O_2(g) → CO2(g)
(c) N_2(g) + O_2(g) → 2NO(g)
(d) CaCO_3(S) → CaO_(S) + CO_2(g)
Answer:
(d) CaCO_3(S) → CaO_(S) + CO_2(g)
Hint:
In CaCO_3(S) → CaO_(S) + CO_2(g) entropy change is positive in (a) and (b) entropy change is negative; in (c) entropy change is zero.

Question 25.
The values of H and S for a reaction are respectively 30 kJ mol^-1 and loo kJ mol^-1. Then the temperature above which the reaction will become spontaneous is ………….
(a) 300 K
(b) 30 K
(c) 100 K
(d) 20°C
Answer:
(a) 300 K
Hint:
∆G = ∆H – T∆S
At 300 K.
∆G = 30000 J mol^-1 – 300 K x 100 JK mol^-1
∆G = 0
above 300 K;
∆G will be negative and reaction becomes spontaneous.

II . Answer these questions briefly.

Question 26.
State the first law of thermodynamics.
Answer:
The first law of thermodynamics states that “the total energy of an isolated system remains constant though it may change from one form to another”
(or)
Energy can neither be created nor destroyed, but may be converted from one form to another.

Question 27.
Define Hess’s law of constant heat summation.
Answer:
Hess’s law:
The enthalpy change of a reaction either at constant volume or constant pressure is the same whether it takes place in a single or multiple steps.

Question 28.
Explain intensive properties with two examples.
Answer:
The property that is independent of the mass or size of the system is called as intensive property.
e.g., Refractive index and surface tension.

Question 29.
Define the following terms:
(a) isothermal process
(b) adiabatic process
(c) isobaric process
(d) isochoric process
Answer:
(a) isothermal process:
It is defined as one in which the temperature of the system remains constant, during the change from its initial to final states.

(b) Adiabatic process:
It is def’ined as one in which there is no exchange of heat (q) between the system and surrounding during operations.

(c) Isobaric process:
It is defined as one in which the pressure of the system remains constant during its change from the initial to final state.

(d) Isochoric process:
It is defined as one in which the volume of system remains constant during its change from initial to final state of the process.

Question 30.
What is the usual definition of entropy? What ¡s the unit of entropy?
Answer:

1. Entropy is a thermodynamic state function that is a measure of the randomness or disorderliness of the system.
2. For a reversible change taking place at a constant temperature (T). the change in entropy
3. of the system is equal to heat energy absorbed or evolved (q) by the system divided by the constant temperature (T).
   
4. SI unit of entropy is J K^-1

Question 31.
Predict the feasibility of a reaction when

1. both ∆H and ∆S positive
2. both ∆H and ∆S negative
3. AH decreases but ∆S increases

Answer:

1. When both ∆H and ∆S are positive, the reaction is not feasible.
2. When both ∆H and ∆S are negative, the reaction is not feasible.
3. When ∆H decreases but ∆S increases, the reaction is feasible.

Question 32.
Define Gibb’s free energy.
Answer:
Gibbs free energy is defined as the part of total energy of a system that can be converted (or) available for conversion into work.
G = H -TS,
where G = Gibb’s free energy
H = enthalpy
T = temperature
S = entropy

Question 33.
Define enthalpy of combustion.
Answer:
Enthalpy of combustion of a substance is defined as “the change in enthalpy of a system when one mole of the substance is completely burnt in excess of air or oxygen”. it is denoted as AH.

Question 34.
Define molar heat capacity. Give its unit.
Answer:
Molar heat capacity is defined as “the amount of heat absorbed by one mole of a substance in raising the temperature by I Kelvin”. It is denoted as C_m
Unit of Molar heat capacity: SI unit of C_m is JK^-1 mol^-1.

Question 35.
Define the calorific value of food. What is the unit of calorific value?
Answer:

• The calorific value of food is defined as the amount of heat produced in calories (or Joules) when one gram of food is completely burnt.
• Unit of calorific value (a) Cal g^-1 (b) J Kg^-1

Question 36.
Define enthalpy of neutralization.
Answer:
The enthalpy of neutralization is defined as the change in enthalpy of the system when one gram equivalent of an acid is neutralized by one gram equivalent of a base or vice versa in dilute solution.
H⁺_(aq) + OH^–_(aq) → H₂O_(l) = 57.32 kJ.

Question 37.
What is lattice energy?
Answer:
Lattice energy is defined as the amount of energy required to completely separate one mole of a solid ionic compound into gaseous constituent.

Question 38.
What are state and path functions? Give two examples.
Answer:

• The variables like P. V, T and ‘n’ that are used to describe the state of the system are called as state functions. e.g.. pressure, volume, temperature, internal energy, enthalpy and free energy.
• A path function is a thermodynamic property of the system whose value depends on the path or manner by which the system goes from its initial to final state. e.g., work (w) and heat (q).

Question 39.
Give Kelvin statement of second law of thermodynamics.
Answer:
Kelvin-Planck statement:
It is impossible to take heat from a hotter reservoir and convert a cyclic process heat to a cooler reservoir.

Question 40.
The equilibrium constant of a reaction is 10, what will be the sign of ∆G? Will this reaction be spontaneous?
Answer:
∆G° = -2.303 RT log K_eq
K_eq = 10
∴ ∆G° = -ve value.
So the reaction will be spontaneous.

Question 41.
Enthalpy of neutralization is always a constant when a strong acid is neutralized by a strong base: account for the statement.
Answer:

1. Enthalpy of neutralization of a strong acid by a strong base is always a constant and it is equal to -57.32 kJ irrespective of which acid or base is used.
2. Because strong acid or strong base means it is completely ionized in solution state. For e.g., NaOH (strong base) is neutralized by HCl (strong acid), due to their complete ionization, the net reaction take place is only water formation.

So the enthalpy of neutralization is always constant for strong acid by a strong base.
H⁺Cl^– + Na⁺OH^– → Na⁺Cl^– + H₂O
H⁺NO₃⁺ + K⁺OH^– → K⁺NO₃⁺+ H₂O
(Net reaction) H⁺ + OH^– → H₂O ∆H = -57.32 kJ

Question 42.
State the third law of thermodynamics.
Answer:
It states that the entropy of pure crystalline substance at absolute zero is zero.
(or)
It can be stated as “it is impossible to lower the temperature of an object to absolute zero in a
finite number of steps”. Mathematically

Question 43.
Write down the Born-Haber cycle for the formation of CaCl₂
Answer:
Born – Haber cycle for the formation of CaCl₂
Ca_(S) + Cl_2(l) → CaC_2(S)  ∆H_f°
Sublimation : Ca_(S) → Ca_(S)  ∆H₁°
Ionization : Ca_(g) → Ca²⁺_(g) + 2e^– = ∆H₂°
Vapourisation : Cl_2(l) → Cl_2(g) = ∆H₃°
Dissociation : Cl_2(g) → 2Cl_(g) = ∆H₄°
Electron affinity : 2Cl_2(g) + 2e^– → 2Cl^–2(g)_(g) = ∆H₅°
Lattice enthalpy : Ca²⁺_(g) + 2Cl^–_(g) → CaCl_2(S) = ∆H₆°
∆H_f° = ∆H₁° + ∆H₂° + ∆H₃° + ∆H₄° + ∆H₅° + ∆H₆°

Question 44.
Identify the state and path functions out of the following
(a) Enthalpy
(b) Entropy
(c) Heat
(d) Temperature
(e) Work
(f) Free energy.
Answer:
State function : Enthalpy, entropy, temperature and free energy.
Path function : Heat and work.

Question 45.
State the various statements of second law of thermodynamics.
Answer:
1. Entropy statement:
Whenever a spontaneous process takes place, it is accompanied by an increase in the total entropy of the universe”.

2. Kelvin-Planck statement:
it is impossible to take heat from a hotter reservoir and convert it completely into work by a cyclic process without transferring a part of heat to a cooler reservoir.

3. Efficiency statement:
Even an ideal, frictionless engine cannot convert 100% of its input heat into work.
Efficiency = \(\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{T}_{1}}\)
% Efficiency = \(\left[\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{T}_{1}}\right]\) x 100
% Efficiency = \(\left[\frac{\text { Output }}{\text { Input }}\right]\) x 100
% Efficiency < 100%

4. Clausius statement:
Heat flows spontaneously from hot objects to cold objects and to get it flow in the opposite direction, we have to spend some work.

Question 46.
What are spontaneous reactions? What are the conditions for the spontaneity of a process?
Answer:

1. A reaction that does occur under the given set of conditions is called a spontaneous reaction.
2. The expansion of a gas into a evacuated bulb is a spontaneous process, the reverse process that is gathering of all molecules into one bulb is not spontaneous. This example shows that processes that occur spontaneously in one direction cannot take place in opposite
   direction spontaneously.
3. Increase in randomness favours a spontaneous change.
4. If enthalpy change of a process is negative, then the process is exothermic and occurs spontaneously. Therefore ∆H should be negative.
5. if entropy change of a process is positive, then the process occurs spontaneously, therefore ∆S should be positive.
6. If free energy of a process is negative, then the process occurs spontaneously, therefore ∆G should be negative.
7. For a spontaneous. irreversible process. ∆H <0, ∆S > 0, ∆G < 0. i.e., ∆H = -ve, ∆S = +ve and ∆G = -ve.

Question 47.
List the characteristics of internal energy.
Answer:

• Internal energy of a system is an extensive property. It depends on the amount of the substances present in the system.
• Internal energy of a system is a state function. It depends only upon the state variables (T, P, V. n) of the system.
• The change in internal energy is as ∆U = U₂ – U₁
• In a cyclic process, there is no energy change. ∆U_(cyclic) = 0.
• If the internal energy of the system at final state (U_f) is less than the internal energy of the
  system at its initial state (U_i), then ∆U would be negative.
• if U_f < U_i ∆U = U_f – U_i = -ve
  if U_f > U_i ∆U = U_f – U_i = +ve

Question 48.
Explain how heat absorbed at constant volume is measured using bomb calorimeter with a neat diagram.
Answer:
1. For chemical reactions, heal absorbed at constant volume, is measured in a bomb calorimeter.

2. Description of the apparatus and procedure:
The inner vessel and its cover are made of strong steel. The cover is fitted tightly to the vessel by means of metal lid and screws. A weighed amount of the substance is taken in a platinum cup connected with electrical wires for striking an arc instantly to kindle combustion. The bomb is then tightly closed and pressurizcd with excess

oxygen. The bomb is lowered in water, which is placed inside the calorimeter. A stirreris placed in the bomb to stir the water uniformly. The reaction is started by striking the substance through electrical heating.

3. During burning, the exothermic heat generated inside the bomb raises the temperature of the surrounding water bath. Temperature change can be measured accurately using Beckmann thermometer. Since the bomb calorimeter is sealed, its volume does not change, so the heat measurements in this case corresponds to the heat of reaction at constant volume.

4. In a bomb calorimeter experiment, a weighed sample of benzoic acid (w) is placed in the bomb which is then filled with excess oxygen and sealed. Ignition is brought about electrically. The rise in temperature (AT) is noted. Water equivalent or calorimetry equivalent of the calorimeter is known from the standard value of enthalpy of combustion of benzoic acid.

5. ∆H_C(C₆H₅COOH) = -3227 kJ mol^-1

6. By knowing o value, the enthalpy of combustion of any other substance is determined adopting the similar procedure and using the substance in place of’ benzoic acid. By this experiment, the enthalpy of combustion at constant volume (AU_C°) is known,
∆U_C° = ω_e. ∆T

7. Enthalpy of combustion at constant pressure of the substance is calculated from the equation
∆U°_C(pressure) = ∆U°_C(volume) + ∆n_g RT

Question 49.
Calculate the work involved in expansion and compression process.
Answer:
1. The essential condition of expansion or compression of a system is, there should be difference between external pressure P_ext and internal pressure (P_int).

2. If the volume of the system is increased against the external pressure. the work is done by the system. By convention work done by the system is given a negative sign (-w).

3. If the volume of the system decreased, the work is done on the system. By convention work done on the system is given a positive sign (+w).

4. For understanding pressure-volume work, let us consider a cylinder which contains one mole of an ideal gas fitted with a frictionless piston. Total volume of the gas is V_i and pressure of the gas inside is P_int.

5. If external pressure is P_ext which is greater than P_int piston is moved inward till the pressure inside becomes equal to P_int It is achieved in a single step and the final volume be V_f.

6. During this compression, piston moves a distance x) and is cross-sectional area of the piston is A, then, Change in volume = x A = ∆V = V_f – V_i ………..(1)
P_ext = \(\frac {Force (F)}{Area(A)}\) ………(2)
F = P_ext.A

7. if work is done by the system by pushing out the piston against external pressure (P_ext) then according to the equation,
-w = F.x ………(3)
-w = P_ext . A . x …….(4)
-w = P_ext.∆V ……….(5)
-w = P_ext . (V_f– V_i.) …..(6)
Simply w = – P∆V ………..(7)

8. From the above equation, we can predict the sign of work (w).

9. During expansion, work is done by the system, since V_f>V_i  the sign obtained for work will be negative.

10. During compression, work is done on the system, since V_f<V_i the sign obtained for work will be positive.

Question 50.
Derive the relation between ∆H and ∆U for an ideal gas. Explain each term involved in the equation.
Answer:
1. When the system at constant pressure undergoes changes from an initial state with H₁, U₁, V₁ and P parameters to a final state with H₂, U₂, V₂ and P parameters, the change in enthalpy ∆H, is given by
AH = U + PV

2. At initial state H₁ = U₁ + PV₁ ………(1)
At final state H₁ = U₁ + PV₁ ……..(2)
(2) – (1) ⇒ (H₂ – H₁) = (U₂ – U₁) + P(V₂ – V₁)
∆H = ∆U + P∆V …………(3)
Considering ∆U = q + w ; w = -P∆V
∆H = q + w + P∆V
∆H = q_p – P∆V+ P∆V
∆H = q_p …………(4)
q_p is the heat absorbed at constant pressure and is considered as heat content.

3. Consider a closed system of gases which are chemically reacting to produce product gases at constant temperature and pressure with V. and as the total volumes of the reactant and product gases respectively, and n1 and n_f are the number of moles of gaseous reactants
and products. Then,
For reactants: P V_i = n_i RT
For products: P V_f = n_f RT
Then considering reactants as initial state and products as final state,
P (V_i – V_i) = (n_i – n_i) RT
P∆V = ∆n_g RT
∆H = ∆U + P∆V
∆H = ∆U + ∆n_g RT ……….(5)

Question 51.
Suggest and explain an indirect method to calculate lattice enthalpy of sodium chloride crystal.
Answer:
The Born – Haber cycle is used to determine the lattice enthalpy of NaCl as follows:

Formation of NaCl can be considered in 5 steps. The sum of the enthalpy changes of these steps is equal to the enthalpy change for the overall reaction from which the lattice enthalpy of NaCl is calculated.
Na_(s) + Vi Cl_2(g) → NaCl_(s) ∆H_f = -411.3 kJ mol^-1
Sublimation : Na_(s) → Na_(g) ∆H₁°
Dissociation : ½ Cl_2(g) → Cl_(g) ∆H₂°
Ionisation : Na_(s) → Na⁺_(g) + e^– ∆H₃°
E1etron affinity : Cl_(g) + e^– → Cl^–_(g) ∆H₄°
Lattice enthalpy : Na⁺_(g) + Cl^–_(g) → NaCl_(s) ∆H₅° =?
∆H = ∆H₁° + ∆H₄° + ∆H₄° + ∆H₄° + ∆H₄°
∆H = ∆H_f° – (∆H₁° + ∆H₂°+ ∆H₃°+∆H₄°)
∆H₅°= Lattice enthalpy of NaCl.
By the above method, indirectly lattice enthalpy of NaCl is calculated, if the values of
∆H_f°, ∆H₁°, ∆H₂°- ∆H₃° and ∆H₄° are given.

Question 52.
List the characteristics of Gibbs free energy.
Answer:
Characteristics of Gibbs free energy:
1. Gibbs free energy is defined as the part of total energy of a system that can be converted (or) available for conversion into work.
G = H – TS ………..(1)
Where
H = enthalpy, T = temperature and S = entropy

2. G is a state function and is a single value function.

3. G is an extensive property, whereas ∆G becomes intensive property for a closed system. Both G and ∆G values correspond to the system only.

4. ∆G gives a criteria for spontaneity at constant pressure and temperature.

• If ∆G is negative (∆G < O), the process is spontaneous.
• If ∆G is positive (∆G > O), the process is non-spontaneous.
• If ∆G is zero (AG = O), the process is equilibrium.

5. For any system at constant pressure and temperature,
∆G = ∆H – T∆S ……….. (2)
We know AH = ∆U + P∆V
∆G = ∆U + P∆V – T∆S ………(3)

6. For the first law of thermodynamics, ∆U = q + w
∆G= q+ w+P∆V – T∆S …………(4)

For second law of thermodynamics, ∆S = \(\frac {q}{T}\)
∆G = q + w + P∆V – T\(\frac {q}{T}\)
∆G = w + P∆V …………(5)
∆G = – w – P∆V ……….(6)

7. – P∆V represent the work done due to expansion against a constant external pressure. Therefore, it is clear that the decrease in free energy (-∆G) accompanying a process taking place at constant temperature and pressure is equal to the maximum work obtainable from the system other than the work of expansion.

8. Unit of Gibb’s free energy is J mol^-1

Question 53.
Calculate the work done when 2 moles of an ideal gas expands reversibly and isothermally from a volume of 500 ml to a volume of 2 L at 25°C and normal pressure.
Answer:
Given
n = 2 moles
V_i = 500 ml = 0.5 lit
V_f = 2 lit
T = 25°C = 298 K
w = -2.303 nRT log \(\left(\frac{V}{V_{i}}\right)\)
w = -2.303 x 2 x 8.314 x 298 x log \(\frac {2}{0.5}\)
w = -2.303 x 2 x 8.314 x 298 x log(4)
w = -2.303 x 2 x 8.314 x 298 x 0.6021
w = -6871 J
w =-6.871 kJ.

Question 54.
In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt ¡n excess oxygen at 298 K. The temperature of the calorimeter was found to increase from 298 K to 298.45 K due to the combustion process. Given that the calorimeter constant is 2.5 kJ K^-1. Calculate the enthalpy of combustion of the gas in kJ mol^-1.
Answer:
Given,
T_i = 298 K
T_f = 298.45 K
k = 2.5 kJ K
m = 3.5g
M_m = 28
heat evolved = k∆T
∆H_C = k (T_f – T_i)
∆H_C = 2.5 kJ K’ (298.45 – 298) K^-1
∆H_C = 1.125 kJ
∆H_C = \(\frac {1.125}{3.5}\) x 28 kJ mol^-1
∆H_C = 9 kJ mol^-1

Question 55.
Calculate the entropy change in the system and surroundings, and the total entropy change in the universe during a process in which 245 J of heat flow out of the system at 77°C to the surrounding at 33°C.
Answer:
Given:
T_sys = 77°C = (77 + 273) = 350K
T_sys = 33°C = (33 + 273) = 306K
q = 245 J

∆S_univ = ∆S_sys + ∆S_surr-245 = —07 3K’
∆S_univ = -0.7 JK^-1 + 0.8 JK^-1
∆S_univ = 0.1 JK^-1

Question 56.
1 mole of an ideal gas, maintained at 4.1 atm and at a certain temperature, absorbs heat 3710 J and expands to 2 litres. Calculate the entropy change in expansion process.
Answer:
Given,
n = 1 mole
P = 4.1 atm
V = 2Lit
T = ?
q = 3710 J
∆S = \(\frac {q}{T}\)

∆S = 37.10 JK^-1

Question 57.
30.4 kJ is required to melt one mole of sodium chloride. The entropy change during melting is 28.4 JK^-1 mol^-1. Calculate the melting point of sodium chloride.
Answer:
Given,
∆H_f(NaCl) = 30.4 kJ = 30400 J mol^-1
∆S_f (NaCl) = 28.4 JK^-1 mol^-1
T_f = ?

T_f = 1070.4 K.

Question 58.
Calculate the standard heat of formation of propane, if its heat of combustion is -2220.2 KJ mol^-1, the heats of formation of CO_2(g) and H₂O_(l) are – 393.5 and -285.8 kJ mol^-1 respectively.
Answer:
Given,

Standard heat of formation of propane is ∆H_f⁰(C₃H₈) = -103.5 kJ mol^-1.

Question 59.
You are given normal boiling points and standard enthalpies of vaporization. Calculate the entropy of vaporization of liquids listed below.

Answer:
For ethanol:
Given:
T_b = 78.4°C = (78.4 + 273) = 351.4 K
∆H_v(ethanol) = + 42.4 kJ mol^-1

∆H_v = + 91.76 J K^-1 mol^-1
For Toluene:
Given:
T_b = 110.6°C = (110.6 + 273) = 383.6 K
∆S_V (toluene) = + 35.2 KJ mol^-1

∆S_V = + 91.76 J KJ mol^-1

Question 60.
For the reaction Ag₂O_(s) → 2Ag_(s) + ½ O_2(g): ∆H = 30.56 kJ mol^-1 and ∆S = 6.66 JK^-1 mol^-1 (at 1 atm). Calculate the temperature at which AG is equal to zero. Also predict the direction of the reaction (i) at this temperature and (ii) below this temperature.
Answer:
Given,
∆H = 30.56 kJ mol^-1
∆H = 30560 J mol^-1
∆S = 6.66 x 10^-3 kJ K^-1 mol^-1
T = ? at which ∆G = 0
∆G = ∆H – T∆S
0 = ∆H – T∆S
T = \(\frac {∆H}{∆S}\)

T = 4589K

(i) At 4589K ; ∆G = 0, the reaction is in equilibrium.
(ii) At temperature below 4598 K, ∆H > T ∆ S
∆G = ∆H – T∆S > 0, the reaction in the forward direction, is non-spontaneous. In other words the reaction occurs in the backward direction.

Question 61.
What is the equilibrium constant K_eq for the following reaction at 400K.
2NOCl_(g) ⇌ 2NO_(g) + Cl_2(g) given that AH° = 77.2 kJ mol^-1 and ∆S° = 122 JK^-1 mol^-1
Answer:
Given,
T = 400 K ; ∆H° = 77.2 kJ mol^-1 = 77200 J mol^-1;
∆S° = 122 JK^-1 mol^-1
∆G° = -2.303 RT log K_eq

log K_eq = -3.7080
K_eq = anti log(-3.7080)
K_eq = 1.95 x 10^-4

Question 62.
Cyan-amide (NH₂CN) ¡s completely burnt in excess oxygen in a bomb calorimeter, ∆U was found to be -742.4 kJ mol^-1 calculate the enthalpy change of the reaction at 298K.
NH₂CN_(s) + 3/2 O_2(g) → N_2(g) + CO_2(g) + H₂O_(l) ∆H = ?
Answer:
Given,
T = 298K ; ∆U = -742.4 kJ mol^-1
∆H = ?
∆H = ∆U + ∆n_(g)RT
∆H = ∆U + (n_p – n_r) RT
∆H = – 742.4 +[2 – \(\frac {3}{2}\)] x 8.314 x 10^-3 x 298
∆H = -742.4 + (0.5 x 8.314 x 10^-3 x 298)
∆H = -742.4 + 1.24 .
∆H = -741.16 kJ mol^-1

Question 63.
Calculate the enthalpy of hydrogenation of ethylene from the following data. Bond energies of C H, C – C, C = C and H – H are 414, 347, 618 and 435 kJ mol^-1.
Answer:
Given,
E_C-H = 414 kJ mol^-1
E_C-H = 347 kJ mol^-1
E_C-H = 618 kJ mol^-1
E_H-H = 435 kJ mol^-1

∆H_r = ∑(Bond energy)_r – ∑(Bond energy)_P
∆H_r = (E_C=C + 4E_C-H + E_H-H) – (E_C-C + 6E_C-H)
∆H_r = (618 + (4 x 414) + 435) – (347 + (6 x 414))
∆H_r = 2709 – 2831
∆H_r = -122 KJ mol^-1

Question 64.
Calculate the lattice enegry of CaCl₂ from the given data
Ca_(s) + Cl_2(g) → CaCl_2(s)  ∆H_f⁰ = -795 KJ mol^-1
Atomisation : Ca_(s) → Ca_(g)  ∆H₁° = + 121 KJ mol^-1
Ionization : Ca_(g) → Ca²⁺_(g) + 2e^–  ∆H₂° = + 242.8 KJ mol^-1
Dissociation : Cl_2(g) → 2 Cl_(g)  ∆H₃° = +242.8 KJ mol^-1
Electron affinity : Cl_(g) + e^– → Cl^–_(g) ∆H₃° = -355 KJ mol^-1
Answer:

∆H_f = ∆H₁ + ∆H₂ + ∆H₃ + 2∆H₄ + u
-795 = 121 + 2422 + 242.8 + (2 x—355) + u
-795 = 2785.8 – 710 + u
-795 = 2075.8 + u
u = -795 – 2075.8
u = -2870.8 KJ mol^-1

Question 65.
Calculate the enthalpy change for the reaction Fe₂O₃ + 3CO → 2Fe + 3CO₂ from the following data.
2Fe + \(\frac{3}{2}\) O₂ → Fe₂O₃; ∆H = -741 kJ
C + \(\frac{1}{2}\) O₂ → CO; ∆H = -137 KJ
C + O₂ → CO₂ ∆H = -394.5 KJ
Answer:
Given,
∆H_f (Fe₂O₃) = -741 kJ mol^-1
∆H_f(CO) = -137 kJ mol^-1
∆H_f(CO₂) = -394.5 kJ mol^-1
Fe₂O₃ + 3CO → 2Fe + 3CO₂ ∆Hr = ?
∆Hr = ∑(∆H_f)_products – ∑(∆H_f)_reactants
∆Hr = [2 ∆H_f(Fe) + 3 ∆H_f(CO₂)] – [∆H_f(Fe₂O₃) + 3∆H_f(CO)]
∆Hr = [- 1183.5 ] – [-1152]
∆Hr = – 1183.5 + 1152
∆Hr = -31.5 KJ mol^-1

Question 66.
When 1-pentyne (A) is treated with 4N alcoholic KOH at 175°C, it is converted slowly into an equilibrium mixture of 13% 1-pentyne(A), 95.2% 2-pentyne(B) and 3.5% of 1,2 pentadiene (C) the equilibrium was maintained at 175°C, calculate AG° for the following equilibria.
B ⇌ A ∆G₁° = ?
B ⇌ C ∆G₂° = ?
Answer:
T = 175°C = 175 + 273 = 448 K
Concentration of 1 – pentyne [A] = 1.3%
Concentration of 2 – pentyne [B] = 95.2%
Concentration of 1, 2 – pentadiene [C] = 3.5%
At equiLibrium
B ⇌ A
95.2% 1.3% ⇒ K₁ = \(\frac{3.5}{95.2}\) = 0.0 136
B ⇌ C
95.2% 3.5% ⇒ K₁ = \(\frac{1.3}{95.2}\) = 0.0367
⇒ ∆G₁° = -2.303 RT log K₁
∆G₁° = – 2.303 x 8.3 14 x 448 x log 0.0136
∆G₁° = + 16010 J
∆G₁° = + 16 kJ
⇒ ∆G₂°= – 2.303 RT log K₂
∆G₂° = -2.303 x 8.314 x 448 x log 0.0367
∆G₂° = + 12312 J
∆G₂° = +12.312 kJ.

Question 67.
At 33K. N₂H₄ is fifty percent dissociated. Calculate the standard free energy change at this temperature and at one atmosphere.
Answer:
T = 33K
N₂O₄ ⇌ 2NO₂
Initial concentration 100%
Concentration dissociated 50%
Concentration remaining at equilibrium 50% – 100%
K_eq = \(\frac{100}{50}\) = 2
∆G° = -2.303 RT log K_eq
∆G° = -2.303 x 8.31 x 33 x 1og 2
∆G° = -190.18 J mol^-1

Question 68.
The standard enthaipies of formation, of SO₂ and SO₃ are -297, kJ. rnol^-1 and -396 kJ mol^-1 respectively. Calculate the standard enthalpy of reaction for the reaction:
SO₂ + \(\frac{1}{2}\) O₂ → SO₃
Answer:
Given,
∆G_f°(SO₂) = – 297 KJ mol^-1
∆G_f°(SO₂) = – 297 KJ mol^-1
SO₂ + \(\frac{1}{2}\)O₂ → SO₃ ∆H_r° = ?
∆H_r° = (∆H_f°)_compound – ∑(∆H_f°)_elements
∆H_r° = ∆H_f° (SO₃) – [∆H_f° (SO₂) + \(\frac{1}{2}\) ∆H_f° (O₂)]
∆H_r° = – 396 kJ mol^-1 – (- 297 kJ mol^-1 + 0)
∆H_r° = – 396 kJ mol^-1+297
∆H_r° = – 99 kJmol^-1

Question 69.
For the reaction at 298 K : 2A + B → C
∆H = 400 J mol^-1 ∆S = 0.2 JK^-1 mol^-1
Determine the temperature at which the reaction would be spontaneous.
Answer:
Given,
T = 298K
∆H = 400 J mol^-1
∆S = 0.2 J K^-1 mol^-1
∆G = ∆H – ThS
if T = 2000K
∆G = 400 – (0.2 x 2000) = 0
if T > 2000 K
∆G will be negative.
The reaction would be spontaneous only beyond 2000 K.

Question 70.
Find out the Value ofequilibrium constant for the following reaction at 298K,
2 NH_3(g) + CO_2(g) ⇌ NH₂CONH_2(aq) + H₂O_(l)
Standard Gibbs energy change, AGr° at the given temperature is – 13.6 kJ mol^-1.
Answer:
Given,
T = 298 K
∆G_r° = – 13.6 kJ mol^-1
= – 13600 J mol^-1
∆G° = – 2.303 RT log K_eq
log K_eq = \(\frac{-∆G°}{2.303 RT}\)
log K_eq
log K_eq = 2.38
K_eq = anti log(2.38)
K_eq = 239.88.

Question 71.
A gas mixture of 3.67 lit of ethylene and methane on complete combustion at 25°C and at I atm pressure produce 6.11 lit of carbon dioxide. Find out the amount of heat evolved in kJ, during this combustion. (∆H_C(CH₄) = – 890 kJ mol^-1 and (∆H_C(C2H₄)= -1423 kJ mol^-1.
Answer:
Given,
∆H_C (CH₄) = – 890 kJ mol^-1
∆H_C (C₂H₄) = -1423 kJ mol^-1
Let the mixture contain x lit of CH₄ and (3.67 – x) lit of ethylene.

Volume of carbon dioxide formed x + 2 (3.67 – x) 6.11 lit
x + 7.34 – 2x = 6.11
x = 1.23 lit
Given mixture contains 1.23 lit of methane and 2.44 lit of ethylene, hence

∆H_C = [-48.87 kJ mol^-1] + [-155 kJ mol^-1]
∆H_C = -203.87 kJ mol^-1

In-Text Questions – Evaluate Yourself

Question 1.
Calculate AH for the reaction CO₂(g) + H₂(g) → CO(g) + H₂O(g) given that ∆H_r° for CO₂(g), CO(g) and H₂O(g) are – 393.5, – 111.31 and – 242 kJ mol^-1 respectively.
Given,
∆H_f° CO₂ = -393.5 kJ mol^-1
∆H_f° CO = -111.31 kJ mol^-1
∆H_f° (H₂O) = – 242 kJ mol^-1
CO₂(g) + H₂(g) → CO(g) + H₂O(g)
∆H_r° = ?
∆H_r° = ∑(∆H_f°)_products – ∑(∆H_f°)_products
∆H_r° = [∆H_f° (CO) + ∆H_f° (H₂O)] – [∆H_f° (CO₂) + ∆H_r° (H₂)]
∆H_r° = [- 111.31 + (-242)] – [- 393.5 +(0)]
∆H_r° = [- 353.31] + 3935
∆H_r° = 40. 19
∆H_r° = 40.19 KJ mol^-1

Question 2.
Calculate the amount of heat necessary to raise 180 g of water from 25°C to 100°C. Molar heat capacity of water is 75.3 J mol^-1 K^-1
Answer:
Given:
Number of moles of water n = \(\frac{180 \mathrm{g}}{18 \mathrm{g} \mathrm{mol}^{-1}}\) = 10 mol molar heat capacity of water
C_p = 75.3 J K^-1 mol^-1
T₂ = 100°C = 373K
T₁ = 25°C = 298K
∆H = ?
∆H = nC_p(T₂ – T₁)
∆H = 10 mol x 75.3 J mol^-1 K^-1 x (373 – 298) K
∆H = 56475 J
∆H = 56.475 kJ

Question 3.
From the following data at constant volume for combustion of benzene, calculate the heat of this reaction at constant pressure condition.
C₆H_6(l) + \(\frac {7}{2}\) O_2(g) → 6 CO_6(g)+ 3 H₂O_(l)
∆U at 25°C = -3268.12 kJ
Answer:
Given,
T = 25°C = 298 K;
∆U = -3268.12 KJ mol^-1
∆H = ?
∆H = ∆U + ∆n_g RT
∆H = ∆U + (n_p – n_r)RT
∆H = 3268.12 + [6 – \(\frac{7}{2}\) x 8.314 x 10^-3 x 298
∆U = -3268.12 + (2.5 x 8.314 x 10^-3 x 298)
∆U = – 3268.12+6.L9
∆U = -3261.93 kJ mol^-1

Question 4.
When a mole of magnesium bromide is prepared from 1 mole of magnesium and 1 mole of liquid bromine, 524 kJ of energy is released. The heat of sublimation of Mg metal is 148 kJ mol^-1. The heat of dissociation of bromine gas into atoms is 193 kJ mol^-1. The heat of vapourisation of liquid bromine is 31 kJ mol^-1. The ionisation energy of magnesium is 2187 kJ mol^-1 and the electron affinity of bromine is – 662 kJ mol^-1. Calculate the lattice energy of magnesium bromide.
Answer:
Given,
Mg(s) + Br₂(l) MgBr₂(s) – ∆H_f° = – 524 KJ mol^-1

Sublimation:
Mg(s) → Mg(g) – ∆H_f° = +148 KJ mol^-1

Ionisation:
Mg(g) → Mg²⁺(g) + 2e^– 2187 – ∆H_f° = 2187 KJ mol^-1

Vapourisation:
Br₂(l) → Br₂(g) – ∆H_f° = + 31 KJ mol^-1

Dissociation:
Br₂(g) → 2Br(g) – ∆H_f = + 193 KJ mol^-1

Electron affinity:
Br(g) + e ^– → Br^– (g) – ∆H_f = -331 KJ mol^-1

∆H_f = ∆H₁ + ∆H₂ + ∆H₃ + ∆H₄ + 2AH₅ + u
-524 = 148 + 2187 + 31 + 193 + (2 x – 331) + u
-524 = 1897 + u
u = -524 -1897
u = – 2421 kJ mol^-1

Question 5.
An engine operating between 127°C and 47°C takes some specified amount of heat from a high temperature reservoir. Assuming that there are no frictional losses, calculate the percentage efficiency of an engine.
Answer:
Given,
T₁ = 127°C = 127 + 273 = 400 K
T₂ = 47°C = 47 + 273 = 320 K
% efficiency η = ?

η = 20%

Question 6.
Urea on hydrolysis produces ammonia and carbon dioxide. The standard entropies of urea, H₂O. CO₂, NH₃ are 173.8, 70, 213.5 and 192.5 J mole^-1K^-1 respectively. Calculate the entropy change for this reaction.
Answer:
Given,
S°(urea) = 173.8 J mol^-1K^-1
S° (H₂O) = 70 J mol^-1 K^-1
S° (CO₂) = 213.5 J mol^-1 K^-1
S° (NH₃) = 192.5 J mol^-1 K^-1
NH₂ – CO – NH₂ + H₂O → 2NH₃ + CO₂
∆S_r° = ∑(S°)_product – ∑(S°)_reactants
∆S_r° = [2 S°(NH₃) + S°(CO₂)] — [S°(urea) + S°(H₂O)]
∆S_r° = [2 x 192.5 + 213.5] – [173.8 + 70]
∆S_r° = [598.5] – [243.8]
∆S_r° = 354.7 J mol^-1 K^-1

Question 7.
Calculate the entropy change when I mole of ethanol is evaporated at 351 K. The molar heat of vapourisation of ethanol is 39.84 kJ mol^-1
Answer:
Given,
T_b = 351 K
∆H_vap = 39840 J mol^-1
∆H_V = ?
∆H_V = \(\frac{\Delta \mathrm{H}_{\mathrm{vap}}}{\mathrm{T}_{\mathrm{b}}}\)
∆H_V = 113.5 JK^-1mol^-1

Question 8.
For a chemical reaction the values of ∆H and ∆S at 300 K are – 10 kJ mol^-1 and -20 J mol^-1 respectively. What is the value of AG of the reaction? Calculate the ∆G of a reaction at,600K assuming ∆H and AS values are constant. Predict the nature of the reaction.
Answer:
Given,
∆H =- 10 kJ mol = -10000 J mol^-1
∆S = -20 JK^-1 mol^-1
T = 300 K
∆G = ?
∆G = ∆H – T∆S
∆G = – lo kJ mol^-1 -300 K x (-20 x 10^-3)kJ K^-1 mol^-1
∆G = (-10 + 6) kJ mol^-1
∆G =-4 kJ mol^-1
At 600 K,
∆G = – 10 kJ mol^-1 -600 K x (-20 x 10^-3)kJ K^-1 mol^-1
∆G =(-10+12) kJ mol^-1
∆G = +2 kJ mol^-1
The value of ∆G is negative at 300K and the reaction is spontaneous, but at 600K the value ∆G becomes positive and the reaction is non-spontaneous.

In-Text Example Problems

Question 1.
A gas contained in a cylinder fitted with a frictionless piston expands against a constant external pressure of 1 atm from a volume of 5 litres to a volume of 10 litres. In doing solt absorbs 400 J of thermal energy from its surroundings. Determine the change in internal energy of system.
Answer:
Given data q = 400 J; V₁ = 5L; V₂ = 10L
Au = q – w (heat is given to the system (±q); work is done by the system(-w)
Au = q – PdV
= 400 J – 1 atm (10 – 5)L
= 400 J – 5 atm L [∴ 1L atm = 101.33 J]
= 400 J – 5 x 101.33 J
= 4003 – 506.65 J
= – 106,65 J

Question 2.
The standard enthalpies of formation of C₂H₅OH_(l), CO_2(g) and H₂O_(l) are – 277, -393.5 and -285.5 kJ mol^-1 respectively. Calculate the standard enthalpy change for the reaction. C₂H₅OH_(l) + 3O_2(g) → 2CO_2(g) + 3H₂O_(l). The enthalpy of formation of O_2(g) in the standard state is zero by definition.
Answer:
The standard enthalpy change for the combustion of ethanol can be calculated from the strndard enthalpies of formation of C₂H₅OH_(l), CO_2(g) and H₂O_(l). The enthalpies of formation are – 277, – 393.5 and – 285.5 kJ mol^-1 respectively.
C₂H₅OH_(l) + 3O_2(g) → 2CO_2(g) + 3H₂O_(l).

= [-787 – 856.5] – [-277]
= -1643.5 + 277
∆H_r° = -1366.5 KJ.

Question 3.
Calculate the value of AU and AH on heating 128 g of oxygen from 0°C to 100°C. C. and C, on an average are 21 and 29 J mol^-1 K^-1. (The difference is 8 J mol^-1 K^-1 which is approximately equal to R)
Answer:
We know
∆U = n C_V (T₂ – T₁)
∆H = n C_P (T₂ – T₁)
Here n = \(\frac {128}{32}\)4 moles;
T₂ = 100°C = 373 K
T₁ = O°C = 273 K
∆U = n C_V(T₂ – T₁)
∆U = 4 x 21 x (373 – 273)
∆U = 8400 J
∆U = 8.4 kJ
∆H = nC_P (T₂ – T₁ )
∆H = 4 x 29 x (373 – 273)
∆H = 11600 J
∆H = 11.6 KJ.

Question 4.
Calculate the enthalpy of combustion of ethylene at 300 K at constant pressure, If Its heat of combustion at constant volume (∆U) is – 1406 KJ.
Answer:
The complete ethylene combustion reaction can be written as,
C₂H_4(g) + 3O_2(g) → 2CO_2(g) + 2H₂O_(l)
∆U = -1406 kJ
∆n = n_p(g) – n_r(g)
∆n = 2 – 4 = -2
∆H = U + RT ∆n_(g)
∆H = -1406 + (8.314 x 10^-3 x 300 x (-2))
∆H = – 1410.9 kJ

Question 5.
Calculate the standard enthalpy of formation ∆H°_f of CH₄ from the values of enthalpy of combustion for H₂, C (graphite) and CH₄ which are -285.8, -393.5, and -890.4 KJ mol^-1 respectively.
Answer:
Let us interpret the information about enthalpy of formation by writing out the equations. It is important to note that the standard enthalpy of formation of pure elemental gases and elements is assumed to be zero under standard conditions. Thermochemical equation for the formation of methane from its constituent elements is,
C(graphite) + 2H_2(g) → CH_4(g)
∆H_f⁰ = X kJ mol^-1 …………(i)
Thermo  chemical equations for the combustion of given substances are,
2H_2(g) + \(\frac {1}{2}\) O₂ → H₂O_(l)
∆H⁰ = – 285.8 KJ mol^-1 …………..(ii)
C(graphite) + O₂ → CO₂
∆H⁰ = -393.5 KJ mol^-1 …………….(iii)
CH_4(g) + 2O₂ → CO_2(g) + 2H₂O_(l)
∆H⁰ = -890.4 KJ mol^-1 ……………(iv)
Since methane is in the product side of the required equation (i), we have to reverse the equation (iv)
CO_4(g) + 2H₂O_(l) → CH_4(g) + 2O₂
∆H° = + 890.4 kJ mol^-1 …………….(v)
In order to get equation (i) from the remaining, (i) = [(ii) x 2] + (iii) + (v)
X = [(-285.8) x 2] + [-393.5] + [+ 890.4] = -74.7kJ
Hence, the amount of energy required for the formation of I mole of methane is -74.7 kJ The heat of formation methane -74.7 kJ mol^-1

Question 6.
Enthalpy for the oxidation of graphite to CO₂ and CO to CO₂ can easily be measured. For these conversions, the heat of combustion values are -393.5 kJ and -283.5 kJ respectively.
Answer:
From these data the enthalpy of combustion of graphite to CO can be calculated by applying Hess’s law. The reactions involved in this process can be expressed as follows:

According to Hess law,
∆H₁ = ∆H₂ + ∆H₃
-393.5 kJ = X – 283.5 kJ
X = – 110.5 kJ

Question 7.
Calcu late the lattice energy of sodium chloride using Born-Haber cycle.
Answer:

∆H_f = heat of formation of sodium chloride = 411.3 kJ mol^-1
∆H₁ = heat of sublimation of Na(s) = 108.7 kJ mol^-1
∆H₂ = ionisation energy of Na(s) = 495 kJ mol^-1
∆H₃ = dissociation energy of Cl₂(s) = 244 kJ mol^-1
∆H₄ = Electron affinity of Cl(s) = – 349 kJ mol^-1
U = lattice energy of NaCl
∆H_f = ∆H₁ + ∆H₂ + ½ ∆H₃ + ∆H₄ + U

∴U = (∆H_f) – (∆H₁ +∆H₂ + ½ ∆H₃ + ∆H₄)
=>U = (-411.3) – (108.7 + 495 + 122 – 349)
U = (-41 1.3) – (376.7)
∴U = -788 kJ mol^-1
This negative sign in lattice energy indicates that the energy is released when sodium is tormea from its constituent gaseous ions Na⁺ and Cl^–

Question 8.
If an automobile engine burns petrol at a temperature of 8 16°C and if the surrounding temperature is 21°C, calculate its maximum possible efficiency.
Answer:

Here
T_h = 816 + 273 = 1098 K
T_c = 21 + 273 = 294 K
Samacheer Kalvi 11th Chemistry Solutions Chapter 7 Thermodynamics
% Efficiency = 73%.

Question 9.
Calculate the standard entropy change for the following reaction (∆H_f°), given the standard entropies of CO_f (g), C(s), O₂(g) as 213.6, 5.740 and 205 JK^-1 respectively.
C(g) + O₂(g) → CO₂(g)
∆S_r° = ∑ S_products° – S_reactants°‘
∆S_r° = {S_CO2°°} – {S_c° + S_O2°°}
∆S_r° = 213.6 – [5.74 + 205]
∆S_r° = 213.6 – [210.74]
∆S_r° = 2.86 JK^-1.

Question 10.
Calculate the entropy change during the melting of one mole of ice into water at 0°C and I atm pressure. Enthalpy of fusion of ice is 6008 J mol^-1.
Answer:
Given,
∆S_fusion = 6008 J mol^-1
T_f = 0°C = 273 K

∆S_fusion = 22.007 JK^-1 mol^-1

Question 11.
Show that the reaction CO +½ O₂ → CO₂ at 300 K is spontaneous. The standard Gibbs free energies of formation of CO₂ and CO are -394.4 and -137.2 kJ mole respectively.
Answer:
CO + ½ O₂ → CO₂
∆G_(reaction)° = ∑G_f(products)° – ∑G_f(reactants)°

∆G_(reaction)° = -394.4 + [137.2 + 0]
∆G_(reaction)° = -257.2 KJ mol^-1
∆G_(reaction)° of a reaction at a given temperature is negative hence the reaction is spontaneous.

Question 12.
Calculate G° for conversion of oxygen to ozone 3/2 O₂ ⇌ O_3(g) at 298 K, if K_p for this conversion is 2.47 x 10^-29 in standard pressure units.
Answer:
∆G° = -2.303 RT log K_p
Where
R = 8.314 JK^-1mol^-1
K = 2.47 x 10^-29
T = 298 K
∆G° = – 2.303(8.314)(298) log (2.47 x 10^-29)
∆G°= 16300 J mol^-1
∆G° = 16.3 KJ mol^-1

Additional:

Question 1.
Calculate the maximum % efficiency of thermal engine operating between 110°C and 25°C.
Answer:
% Efficiency = \(\left[\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{T}_{1}}\right]\) x 100
T₁ = 110°C + 273 = 383 K.
T₂ = 25°C – 273 = 298 K.
% Efficiency = \(\left[\frac{383 – 298}{383}\right]\) x 100
% Efficiency = \(\left[\frac{85 × 100}{383}\right]\) = \(\left[\frac{8500}{383}\right]\) % Efficiency = 22.2%

Question 2.
Calculate the entropy change in the system. and in the surroundings and the total entropy change in the universe when during a process 75 J of heat flow out of the system at 55°C to the surrounding at 20°C.
Answer:
Heat flow (q) = 75 J
Entropy change = ∆S = ?
Temperature of the system = 55°C + 273 = 328 K
Temperature of the surroundings = 20°C + 273 = 293 K

= 0.2286 + 0.2559 = ∆S_universe = 0.4845 JK^-1

Question 3.
1 mole of an ideal gas is maintained at 4.1 atm and at a certain temperature absorbs 3710 J heat and expands to 2 litres. Calculate the entropy change in expansion process.
Answer:
Pressure of an ideal gas P_i = 4.1 atm.
Expansion in volume = ∆V = 2 litres
Heat absorbed = q = 3710 J
Entropy change = ∆S = ?
For an ideal gas PV = RT for one mole.
T = \(\frac {PV}{R}\) = \(\frac {4.1 x 2}{0.0830}\) = 100°C
T = 100 + 273 = 373 K.
∆S = \(\frac{q}{\mathrm{T}_{(\mathrm{K})}}\) = \(\frac{3710}{373}\)
Entropy change = 9.946 JK^-1.

Question 4.
Calculate the entropy change of a process H₂O_(l) → H₂O_(g) at 373K. Enthalpy of vaporization of water is 40850 J Mole^-1.
Answer:
H₂O_(l) → H₂O_(g)
Temperature T = 373 K
Enthalpy of vapourisation of water = ∆V_vap = 40580 J mol^-1
∆S = entropy change
∆S(entropy change) = 109.51 J mol^-1 K^-1

Question 5.
30.4 KJ is required to melt one mole of sodium chloride. The entropy change during melting is 28.4 JK^-1 mol^-1 Calculate the melting point of sodium chloride.
Answer:
Heat required for I mole of NaCl for melting (q) = 30.4 K J
= 30.4 x 1000 J
∆S – entropy change = 28.4 J K^-1 mol^-1
Melting point = T_m = ?
∆S = \(\frac{q}{\Gamma_{m}}\)
∴ T_m = \(\frac{q}{∆S}\)
T_m = \(\frac{30.4 x 1000}{28.4}\) = 10704 K
Melting point of NaCl = 1070.4 K.

Question 6.
Calculate the standard heat of formation of propane, if its heat of combustion is 0-2220.2 KJ mol^-1 [he heats of formation of CO_2(g) and H_2(g)O_(l) are -393.5 and -285.8 kJ mol^-1 respectively.
Solution:
Standard heat of formation of propane,
3C_(g) + 4H_2(g) → C₃H_8(g) ∆H_f° = ?
Data given:
C₃H_8(g) + 5O_2(g) → 3CO₂ + 4H₂O_(l)  ∆H = -2220.2 KJ mol^-1
C_(s) + O_2(g) → CO_2(g)  ∆H = -393.5 KJ mol^-1
H_2(g)(g) + ½ O_2(g) → H₂ O_(l)   ∆H = -285.8 KJ mol^-1
According to Hes&s law, equation
Equation (1) is reversed.
Equation (2) is x 3
Equation (3) is x 4
The add all the equations.
3CO_2(g) + 4H₂O_(l) → C₃H_8(g) + 5O_2(g)  ∆H₁ = + 2220.2 KJ mol^-1
3C_(s) + 3O_2(g) → 3CO2_2(g)  ∆H₂ = -1180.5 KJ mol^-1
4H_2(g) + 2O_2(g) → 4H₂O_(l)  ∆H₃ = -1143.2 KJ mol^-1
3C_(S) + 4H_2(g) → C₃H_8(g)   ∆H_f° = -103.5 KJ mol^-1
Standard enthalpy of formation of propane  ∆H_f° = -103.5 K.J mol^-1

Question 7.
The boiling point of water at a pressure of 50 atm is 265°C. Compare the theoretical efficiencies of a steam engine operating between the boiling point of water at
1. 1 atm pressure
2. 50 atm pressure, assuming the temperature of the sink to be 35°C in each case.

Answer:
Boiling point of water at 50 atm pressure (T_b) = 265°C = 265 + 273 = 538K
1. Boiling point of water at 1 atm pressure (T_b) = 100°C = 100 + 273 = 373K
% Efficiency of steam engine = \(\left[\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{T}_{1}}\right]\) x 100 = \(\left[\frac{538-373}{538}\right]\) x 100 = 0.3066 x 100 = 30.66%

2. BoiLing point of water at 50 atm pressure (T_b) = 35°C = 35 + 273 = 328 K
% Efficiency of steam engine = \(\left[\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{T}_{1}}\right]\) x 100 \(\left[\frac{538-328}{538}\right]\) x 100 = \(\left[\frac{210 x 100}{538}\right]\) = 39.03 %

Question 8.
The standard enthalpies of formation of SO₂ and SO₃ are -297 kJ mol^-1 and -396 kJ mol^-1 respectively.Calculate the standard enthalpy of reaction for the reaction: SO₂ + ½ O₂ → 4 SO₃
Solution:
Data given,
(1) ⇒ S + O₂ → SO₂ ∆H_f° = -297 KJ mol^-1
(2) ⇒ S + 1½O₂ → SO₃ ∆H_f° = -396 KJ mol^-1
SO₂ + ½ O₂ →SO₃ ∆H_r = ?
Equation (1) is reversed and added with equestion (2)

Question 9.
For the reaction at 298 K : 2A + B → C
∆H = 400 KJ mol^-1: ∆S = 0.2 JK^-1 mol^-1
Determine the temperature at which the reaction would be spontaneous.
Solution:
Data given,
2A + B → C at 298 K
∆H = 400 KJ mol^-1
∆S = 0.2 JK^-1 mol^-1
T = 298 K
[∆G = ∆H – T∆S]
if ∆G = 0
∆H – T∆S = 0
∆S = \(\frac {∆H}{T}\)
For ∆G = 0, ∆S = \(\frac {∆H}{T}\) = \(\frac {400}{0.2}\) = \(\frac {4000}{2}\) = 2000K
∴ T = 2000K
At 2000 K. the rcaction is in equilibrium. So. above 2000 K, the reaction will be spontaneous.

Question 10.
Calculate the heat of glucose and its calorific value
1. C_(graphite) + O_2(g) → CO_2(g) ∆H = -395 KJ
2. H_2(g) + hO₂ → H₂O_(l) ; ∆H = -269.4 KJ
3. C + 6H_2(g) + 3O_2(g) → C₆H₁₂O_6(S) ∆H = -1169.8 KJ

Solution:
Calorific value of glucose ∆H_C = ?

Data given are.
1. C_(graphite) + O_2(g) → CO_2(g) ∆H = -395 KJ ………….(1)
2. H_2(g) + 1/2 O₂ → H₂O_(l) ∆H_C ∆H = -269. 4 KJ ………….(2)
3. C + 6H_2(g) + 3O_2(g) → C₆H₁₂O_6(S) ∆H = – 1169.8 KJ …………(3)
Equation (1) x 6 6C + 6O₂ → 6CO₂ ∆H = – 2370 KJ
Equation (2) x 6 6H₂ + 3O₂ → 6H₂O ∆H = – 1616.4 KJ
Equation (3) is reversed,
C₆H₁₂O₆ → 6C + 6H₂ +3O₂ ∆H = + 1169.8 KJ
Add all equations,
C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O
∆H = – 2370 – 1616.4 + 1169.8
∆H = – 3986.4 + 1169.8
∆H = – 2816.6 KJ
Calorific value of C₆H₁₂O₆ = -2816.6 KJ mol^-1

Question 11.
Calculate the entropy change when 1 mole of ethanol is evaporated at 351 K. The molar heat of vapourisation of ethanol is 39.84 kJ mol^-1
Answer:
Given,
∆H_vap = 39840 J mol^-1
∆H_v = \(\frac{\Delta \mathrm{H}_{\mathrm{vap}}}{\mathrm{T}_{b}}\)
T_b = 351 K
∆H_v = \(\frac{39840}{351}\)
∆H_v = 113.5 J k^-1 mol^-1

Question 12.
Calculate the entropy change of a process possessing ∆H_t = 2090 J mole^-1
Answer:
Given,
S_{n(α 13°C)} – S_{n(β 13°C)}
∆H_t = 2090 J mol^-1
∆S_t = \(\frac{\Delta \mathrm{H}_{t}}{\mathrm{T}_{t}}\)
∆S_t = \(\frac{2090}{286}\)
∆S_t = 7.307 J K^-1 mol^-1

Question 13.
Calculate the standard enthalpy of formation of CH₃OH_(l) from the following data:
(i) CH₃OH_(l) + 3/2O_2(g) → CO_2(g) + 2H₃O_(l) ∆_r\({ H }^{ \ominus }\) = -726 kJ mol^-1
(ii) C_(S) + O_2(g) → CO_2(g); ∆_C \({ H }^{ \ominus }\) = – 393 kJ mol^-1
(iii) H_2(g) + ½ O_2(g) → H₂O_(l) ∆_f \({ H }^{ \ominus }\) = – 286 kJ mol^-1
Answer:
The equation we aim at;
C_(S) + 2H_2(g) + O_2(g) → CH₃OH_(l) ∆_f \({ H }^{ \ominus }\) = ± ? ………..(iv)
Multiply equation (iii) by 2 and add to equation (ii)
C_(S) + 2H_2(g) + 2O_2(g) → CO_2(g) + 2H₂O_(l)  ∆H = – (393 + 522) = – 965 kJ mol^-1
Subtract equation (iv) from equation (i)
CH₃OH_(l) + 3/2 O_2(g) → CO_2(g) + 2H₂O_(l)  ∆H = -726 kJ mol^-1

Subtract:
C_(S) + 2H_2(g) + ½ O_2(g) → CH₃OH_(l)  ∆_f\({ H }^{ \ominus }\)= -239 kJ mol^-1

Question 14.
The equilibrium constant for the reaction is 10. Calculate the value of ∆G-;
Given R = 8.3 14 JK^-1 mol^-1; T = 300 K.
Answer:
∆Ge = -RT in K = -2.303 RT log K
R = 8.314 JK^-1 mol^-1; T = 300 K; K = 10
∆Ge = – 2.303 x 8.314 JK^-1 mol^-1 x (300 K) x log 10
= -5527 J mol^-1 = -5.527 kJ mol^-1

Question 15.
Calculate the entropy change in surroundings when 1 mol^-1 of H₂O_(l) is formed under standard conditions. Given ∆H = – 286 kJ mol^-1
Answer:
q_rev = (-Δ_f\({ H }^{ \ominus }\)) = -286 KJ mol = 286000  J mol

Question 16.
The enthalpy of formation of methane at constant pressure and 300 K is – 78.84 kJ.What will be the enthalpy of formation at constant volume?
Answer:
The equation representing the enthalpy of formation of methane is:
C_(S) + 2H_2(g) – CH_2(g);
∆H = – 78.84 kJ
∆U = 78.84 kJ;
∆^ng = 1 – 2 = -1 mol
R = 8.314 x 10^-13 KJ K^-1 mol^-1; T = 300 K.

According to the relation,
∆H = ∆U + ∆^ng RT – 78.84 kJ
∆U = ∆H – ∆^ng RT
∆U = (- 78.84 kJ) – (1 mol) x (8.314 x 10^-3 KJ K^-1 mol^-1 ) x 300K.
= – 78.84 – 2.49 = -8.314 KJ.

Question 17.
Calculate ∆_rGe for conversion of oxygen to ozone 3/2 O_2(g) → O_3(g) at 298 K_p If for this conversion is 2.47 x 10^-29.
Solution:
We know
∆_r\({ G }^{ \ominus }\) = -2.303 RT log K_p
R = 8.314 JK^-1 mol^-1
∆_r\({ G }^{ \ominus }\)= -2.303 (8.314 J K^-1 mol^-1) x (298 K) (log 2.47 x 10^-29)
= 163000 J mol^-1 = 163 mol^-1

Question 18.
(a) Under what condition, the heat evolved or absorbed in a reaction ¡s equal to its free energy change?
(b) Calculate the entropy change for the following reversible process.
H₂O_(s) ⇌ H₂O_(l)  ∆_fusH is 6 kJ mol^-1
(a) ∆G = ∆H – T∆S
When the reaction is carried out at 0°K
or ∆S = 0
∆G = ∆H
H₂O_(s) ⇌ H₂O_(l)
∆_fusH = 6 kJ mol^-1 = 6000Jmol^-1
∆_fusH = 6 kJ mol^-1
6000 J mol^-1

Samacheer Kalvi 11th Chemistry Thermodynamics Additional Questions Solved

I. Choose the correct answer from the following:

Question 1.
When a liquid boils, there is ……….
(a) increase in entropy
(b) a decrease in entropy
(c) increase in heat of vaporization
(d) an increase in free energy
Answer:
(a) Increase in entropy

Question 2.
if ∆G for a reaction is negative, the change is ……….
(a) spontaneous
(b) non-spontaneous
(c) reversible
(d) equilibrium
Answer:
(a) spontaneous

Question 3.
In which of the following process. the process is always non-feasible?
(a) ∆H >O, ∆S>O
(b) ∆H<O, ∆S>O
(c) ∆H >O, ∆S<O
(d) ∆H<O, ∆S>O
Answer:
(c) ∆H >O, ∆S<O

Question 4.
Change in Gibbs free energy is given by
(a) ∆G = ∆H +T∆S
(b) ∆G = ∆H – T∆S
(c) ∆G = H x T∆S
(d) none of these
Answer:
(b) ∆G = ∆H – T∆S

Question 5.
Which of the following process is feasible at all ternperatures?
(a) ∆H>O.∆S>O
(b) ∆H>O.∆S<O
(c) ∆H<O.∆S>O
(d) ∆H<O.∆S<O
Answer:
(c) ∆H<O.∆S>O

Question 6.
Calculate the entropy change during the melting of one mole of ice into water at 0°C and 1 atm pressure. Enthalpy of fusion of ice is 6008 J Mole^-1.
(a) 22.007 J K^-1Mole^-1
(b) 22.007 J K^-1 Mole^-1
(c) 220.07 J K^-1 Mole^-1
(d) 2.2007 J K^-1 Mole^-1
Answer:
(a) 22.007 J K^-1 Mole^-1
Hint:
Enthalpy of fusion of ice 6008 J mol^-1 = ∆H
∆S = \(\frac{\Delta \mathrm{H}}{\mathrm{T}_{\mathrm{m}}}\) T_m = 0°C = 273 K

∆S = \(\frac{6008}{273}\) = 22.007 J K^-1 mol^-1

Question 7.
Calculate the entropy change of a process H₂O_(l) → H₂O_(g) at 373K. Enthalpy of vaporization of water is 40850 J Mole^-1.
(a) 120 J K^-1 mol^-1
(b) 9.1 x 10 J K^-1 mol^-1
(c) 9.1 x 10 J K^-1 mol^-1
(d) 109.52 J K^-1 mol^-1
Answer:
(d) 109.52 J K^-1 mol^-1
Solution:
EnthaLpy of vaporization of water is = ∆H = 40850 J mol^-1
Boiling point = 373 K = T_b
∆S = \(\frac{\Delta \mathrm{H}}{\mathrm{T}_{\mathrm{b}}}\) = \(\frac{40850}{373}\) = 109.517 = 109.52 JK^-1  mol^-1

Question 8.
The final temperature of an engine whose initial temperature is 400K and having efficiency 25%.
(a) 200K
(b) 400K
(c) 300K
(d) 450K
Answer:
(c) 300K
Solution:
Initial temperature = T₁ = 400 K
Final temperature = T₂ = ?
Efficiency = 25%
η = \(\left[\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{T}_{1}}\right]\) x 100
25 = \(\left[\frac{400-\mathrm{T}_{2}}{400}\right]\) x 100
\(\frac{400-T_{2}}{4}\) = 25
400 – T₂ = 100
– T₂ = 100 – 400
∴ -T₂ = -300
T₂ = 300 K

Question 9.
When solid melts there is ………
(a) an increase of entropy
(b) a decrease in entropy
(c) an increase in free energy
(d) an increase of heat of fusion
Answer:
(a) an increase of entropy

Question 10.
The unit of entropy is ………
(a) J K^-1 mol^-1
(b) J mol^-1
(c) J K mol^-1
(d) J^-1 K^-1
Answer:
(a) J K^-1 mol^-1

Question 11.
If G = 0. then the process is …………
(a) equilibrium
(b) spontaneous
(c) non-spontaneous
(d) none of these
Answer:
(a) equilibrium

Question 12.
The standard conditions for G° are ………..
(a) 1 mm Hg / 25°C
(b) 1 atm /25 K
(c) 1 attn / 298 K
(d) 1 atm / 0K
Answer:
(c) 1 atm / 298 K

Question 13.
The efficiency of engine working between 100 to 400 K
(a) 25%
(b) 75%
(c) 100%
(d) 50%
Answer:
(b) 75%
Solution:
η = \(\left[\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{T}_{1}}\right]\) x 100
T₁ = 400K
T₂ = 100 K
∴ η = \(\left[\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{T}_{1}}\right]\) x 100

= 75%

Question 14.
Entropy is a ………. function.
(a) state
(b) path
(c) defined
(d) undefined
Answer:
(a) state

Question 15.
An efficiency of an engine is always ………..
(a) = 0%
(b) > 100%
(c) < 100%
(d) = 100%
Answer:
(c) < 100%

Question 16.
if system moves from ordered state to disordered state, its entropy ………..
(a) decreases
(b) increases
(c) become zero
(d) increases then decreases
Answer:
(b) increases

Question 17.
In a reversible process, the entropy of Universe is ………..
(a) greater than zero
(b) less than zero
(c) equal to zero
(d) remains constant
Answer:
(c) equal to zero

Question 18.
In which of the following entropy increases?
(a) Condensation of water vapour
(b) Liquid freezes to solid
(c) Sublimation
(d) Gas freezes to a solid
Answer:
(c) Sublimation

Question 19.
Which of the following is a state function?
(a) q
(b) ∆q
(c) w
(d) ∆S
Answer:
(d) ∆S

Question 20.
Which of the following will have highest ∆H_vap Value?
(a) Acetone
(b) Ethanol
(c) Carbon tetrachloride
(d) Chloroform
Answer:
(b) Ethanol

Question 21.
Which of the following is not a state function?
(a) S
(b) H
(c) G
(d) q
Answer:
(d) q

Question 22.
The net work done by the system ………..
(a) w – P∆V
(b) w +P∆V
(c) -w + P∆V
(d) -w – P∆V
Answer:
(d) – w – P∆V

Question 23.
– ∆G is the net work done by the system except ………..
(a) electrical work
(b) expansion work
(c) chemical work
(d) photo chemical work
Answer:
(b) expansion work

Question 24.
The enthalpy of vapourisation of a liquid is 30 kJ^-1 mol^-1 and the entropy of vaporization is 75 JK mol^-1 The boiling point of the liquid at 1 atm is ……….
(a) 250 K
(b) 400 K
(c) 450 K
(d) 600 K
Answer:
(b) 400 K
Solution:
∆H_vap = 30 kJ mol^-1 x 1000 = 30000 J mol^-1
∆S_vap = 75 J mol^-1
T_b = Boiling point = ?
∆S_vap = \(\frac{\Delta \mathrm{H}_{\mathrm{vap}}}{\mathrm{T}_{\mathrm{b}}}\)
∴ T_b = \(\frac{\Delta \mathrm{H}_{\mathrm{vap}}}{\Delta \mathrm{S}_{\mathrm{vap}}}\) = \(\frac {30000}{75}\) = 400K
T_b = 400 K
Tb = 400 K

Question 25.
In a reversible process ∆S_sys + ∆S_surr is ………..
(a) >0
(b)<0
(c) ≥0
(d) = 0
Answer:
(d) = 0

Question 26.
Which of the following does not result in an increase in the entropy?
(a) Crystallization of sucrose from solution
(b) Rusting of Iron
(c) Conversion of ice to water
(d) Vaporization of camphor
Answer:
Crystallization of sucrose from solution

Question 27.
The standard free energy change (∆G°) is related to equilibrium constant (K) as …………
(a) ∆G° = – 1303 RT in K
(b) ∆G° = 2.303 RT log K
(c) ∆G° = RT in K
(d) ∆G° = -2.303 RT log K
Answer:
(d) ∆G° = -2.303 RT log K

Question 28.
enropy change involved in the conversion of I mole of liquid water at 373K to vapour at the same temperature will be (∆H_vap = 2.257 kJ g^-1)
(a) 0.119 kJ
(b) 0.109 kJ
(c) 0.129 k
(d) 0. 120 kJ
Answer:
(b) 0.109 kJ
Solution:
∆H_vap = 2.257 kJ g^-1
1 mole of H₂O = 18 g
∴∆H_vap for 1 mole = 2.257 x 18 = 40.626 k J g^-1
T_b = 373K
∆H_vap = \(\frac{40.626}{373}\) = 0.1089 KJ g^-1
= 0.109 kJ mol^-1

Question 29.
Which of the following units represent largest amount of energy?
(a) calories
(b) Joule
(c) erg
(d) eV
Answer:
(a) calories

Question 30.
The intensive property among these quantities is
(a) mass
(b) volume
(c) enthalpy
(d) mass / volume
Answer:
(d) mass /volume

Question 31.
System in which there is no exchange of matter, work or energy from surrounding is
(a) closed
(b) isolated
(c) adiabatic
(d) isothermal
Answer:
(b) isolated

Question 32.
Which of the following is not an intensive property?
(a) Pressure
(b) Density
(c) Volume
(d) Surface tension
Answer:
(c) Volume

Question 33.
A gas can expand from loo ml to 250 ml under a constant pressure of 2 atm. The work done by the gas is ………..
(a) -30.39 J
(b) 25 J
(c) 5 kJ
(d) 16 J
Answer:
(a) -30.39 J
Solution:
∆V = expansion in volume = 100 to 250
∆V = V₂ – V₂ = 250 – 100
∆V = 150 ml = 0.15 litre
Work done = ?
Pressure = 2 atm
w = -P∆V
= -2 x 0.15 litre x 101.3 JL^-1 atm^-1
= -30.39 J.

Question 34.
An ideal gas expands in volume from 1 x 10^-3 m³ to 1 x 10^-2 m³ at 300K against a constant pressure at 1 x 10⁵ Nm⁵. The work done is ………..
(a) -900 J
(b) 900 kJ
(c) 270 kJ
(d) -900 kJ
Answer:
(a) -900 J

Question 35.
Identify the state quantity among the following
(a) q
(b) q – w
(c) q + w
(d) q/w
Answer:
(b) q – w

Question 36.
In general, for an exothermic reaction to be spontaneous
(a) temperature should be high
(b) temperature should be zero
(c) temperature should be low
(d) temperature has no effect
Answer:
(c) temperature should be low

Question 37.
Heat of neutralization of a strong acid by a strong base is a constant value because
(a) only OH⁺ and OH^– ions react in every case
(b) the strong base and strong acid react completely
(c) the strong base and strong acid react in aqueous solution
(d) salt formed does not hydrolyse
Answer:
(a) only OH⁺ and OH^– ions react in every case

Question 38.
The heat absorbed at constant volume is equal to the system’s change in
(a) enthalpy
(b) entropy
(c) internal energy
(d) free energy
Answer:
Answer:
(c) internal energy

Question 39.
Heat of neutralization is always
(a) positive
(b) negative
(c) zero
(d) positive or negative
Answer:
(b) negative

Question 40.
The heat of formation CO and CO₂ are -26.4 Kcal and -94 Kcal respectively. Heat of combustion of carbon monoxide will be …………
(a) + 26.4 KCal
(b) – 67.6 KCal
(c) – 120.6 K Cal
(d) + 52.8 K Cal
Answer:
(b) -67.6 KCal
Solution:
C + ½ O₂ → CO ∆H = -26.4 K.cal ……….(1)
C + O₂ → CO₂ ∆H = -94 K.cal ……..(2)
Heat of combustion of CO is
C + ½O₂ → CO₂ ∆H = ?
Equation (1) is reversed.

ΔH= -67.6 K. cal

Question 41.
For the reaction H₂ + I₂ ⇌ 2HI, ∆H = 12.40 Kcal the heat of formation of HI is ……….
(a) 12.4 Kcal mol^-1
(b) -12.4 Kcal mol^-1
(c) -6.20 Kcal mol^-1
(d) 6.20 Kcal mol^-1
Answer:
(d) 6.20 Kcal mol^-1
Solution:
H₂ + I₂ → 2HI ∆H = 12.40K.cal
½ H₂ + ½ I₂ → HI = ∆H/2
∴ \(\frac {12.40}{2}\) = 6.20 K.cal mol^-1.

Question 42.
Heat capacity is ……….
(a) \(\frac {dq}{dT}\)
(b) dq.dT
(c) ∑q. \(\frac {1}{dT}\)
(d) none of these
Answer:
(a) \(\frac {dq}{dT}\)

Question 43.
The relation between C_p and C_v is…………
(a) C_p – C_v = R
(b) C_p + C_v = R
(c) – 285 KJ
(d) R – C_v = C_p
Answer:
(a) C_p – C_v = R

Question 44.
The heat required to raise the temperature of a body by I K is called ……….
(a) specific heat
(b) thermal capacity
(c) water equivalent
(d) none of these
Answer:
(b) thermal capacity

Question 45.
Heat liberated when 100 ml of IN NaOH is neutralized by 300 ml of in HCl ………..
(a) 22.92 Ici
(b) 17.19 kJ
(c) 11.46 kJ
(d) 5.73 Id
Answer:
Base = V₁ = 100ml
N₁ = 1N
Acid = V₂ = 300ml
N₂ = 1N
Enthalpy of neutralization of 1000 ml = 57.3 kJ.
∴ Enthalpy of neutralization of 100 ml x 100 = \(\frac {5.73 kJ}{1000}\) x 100 = 5.73 kJ.

Question 46.
In order to decompose 9g of water, 142.5 kJ of heat is required. Hence enthalpy of formation of water is ………..
(a) -142.5 kJ
(b) 142.5 kJ
(c) -285 kJ
(d) 285 kJ
Answer:
(c) -285 kJ
Solution:
H₂O → H₂ + ½ O₂
18(g)
9g H₂O is decomposed by – 142.5 kJ amount of heat.
:. 18g H₂O will be decomposed by = +285 kJ.
:. 18g of H₂O is formed by – 285 kJ amount of heat. (Evolution of heat = – ve sign)

Question 47.
Assertion (A) : Combustion of all organic compounds is an exotherinic reaction ………….
Reason (R) : The enthalpies of all elements in their standard state are zero.
Which of the above statement isare not correct?
(a) both A and R are true and R is the correct explanation of A
(b) both A and R are true and R is not correct explanation of A
(c) both A and R are false
(d) A is false but R is true
Answer:
(b) both A and R are true and R is not correct explanation of A

Question 48.
Assertion (A) : Spontaneous process is an irreversible process and may be reversed by same external agency.
Reason (R) : Decrease in enthalpy is a contributory factor for spontaneity.
(a) both A and R are true and R is the correct explanation of A
(b) both A and R are true and R is not correct explanation of A
(c) both A and R are false
(d) A is false but R is true
Answer:
(b) both A and R are true and R is not correct explanation of A

Question 49.
Assertion (A) : A liquid crystallizes into a solid and accompanied by decrease in entropy.
Reason (R) : In crystals molecules are organised in an ordered manner.
(a) both A and R are true and R is the correct explanation of A
(b) both A and R are true and R is not correct explanation of A
(c) both A and R are false
(d) A is false but R is true
Answer:
(a) both A and R are true and R is the correct explanation of A

Question 50.
Thermodynamics is applicable to ………….
(a) macroscopic system only
(b) microscopic system only Thermodynamics
(c) homogeneous system only
(d) heterogeneous system only
Answer:
(a) macroscopic system only

Question 51
An isochoric process takes place at constant …………
(a) temperature
(b) pressure
(c) volume
(d) concentration
Answer:
(c) volume

Question 52.
For a cyclic process, the change in internal energy of the system is …………
(a) always + ve
(b) equal to zero
(c) always – ve
(d) none of the above
Answer:
(b) equal to zero

Question 53.
Which of the following properties is not a fùnction of state?
(a) Concentration
(b) Internal energy
(c) Enthalpy
(d) Entropy
Answer:
(u) Concentration

Question 54.
Which of the following relation is true?
(a) C_p >C_v
(b)C_v > C_p
(c) C_p = C_v
(d) C_p = C_v = 0
Answer:
(a) C_p >C_v

Question 55.
Which of the following always has a negative value?
(a) heat of reaction
(b) heat of solution
(c) heat of combustion
(d) heat of formation
Answer:
(c) beat of combustion

Question 56.
The bond energy depends upon ………..
(a) size of the atom
(b) electronegativity
(c) bond length
(d) all of the above
Answer:
(d) all of the above

Question 57.
For an endothermic reaction ………..
(a) ∆H is -ve
(b) ∆H is +ve
(e) ∆H is zero
(d) none of these
Answer:
(b) AH is +ve

Question 58.
The process depicted by the equation.
H₂O_(s) → H₂O_(l)
∆H = + 1.43 kcal represents
(a) fusion
(b) melting
(c) evaporation
(d) boiling
Answer:
(a) fusion

Question 59
Which one is the correct unit for entropy?
(a) KJ mol
(b) JK^-1 mol
(c) JK^-1 mol^-1
(d) KJ mol^-1
Answer:
(c) JK^-1 mol^-1

Question 60.
A thermodynamic state function is a quantity ………….
(a) used to determine heat changes
(b) whose value is independent of path
(c) used to determine pressure volume work
(d) whose value depends on temperature only
Answer:
(b) whose value is independent of path

Question 61.
For the process to occur under adiabatic conditions, the correct condition is ………..
(a) ∆T = 0
(b) ∆_p = 0
(c) q = 0
(d) w = 0
Answer:
(c) q = 0

Question 62.
The enthalpies of all elements in their standard states are …………
(a) unity
(b) zero
(c) <0
(d) different for each element
Answer:
(b) zero

Question 63.
∆\({ U }^{ \ominus }\) of combustion of methane is -X kJ mol^-1. The value of ∆He is ………..
(a) = ∆\({ U }^{ \ominus }\)
(b) > ∆\({ U }^{ \ominus }\)
(c) < ∆\({ U }^{ \ominus }\)
(d) 0
Answer:
(c) < ∆U\({ U }^{ \ominus }\)

Question 64.
The enthalpy of combustion of methane, graphite and dihydrogen at 298K are -890.3 kJ mol^-1 -393.5 kJ mol^-1 and 285.8 kJ mol^-1 respectively. Enthalpy of formation of CH₄ will be ………..
(a) – 74.8 kJ mol^-1
(b) – 52.27 k mol^-1
(c) 74.8 kJ mol^-1
(d) + 52.26 kJ mol^-1
Answer:
(a) – 74.8 kJ mol^-1
Solution:
CH₄ + 3O₄ → CO + 2H₂O ∆H₁ = – 89O.3 kJ mol^-1 ……….(1)
C + O₂ → CO₂ = -393.5 kJ mol^-1 ……….(2)
H₅ + 1/2 O₂ → H₂O = – 285.8 kJ mol^-1 ……….(3)
Equation (1) is reversed.
Equation (3) x 2

∆H_f = +890.3 – 965.1 = -74.8 KJ mol^-1

Question 65.
A reaction, A + B → C + D + q is found to have a positive entropy change. The reaction will be ………..
(a) possible at high temperature
(b) possible only at low temperature
(c) not possible at any temperature
(d) possible at any temperature
Answer:
(d) possible at any temperature

Question 66.
Consider the following statements.
(i) Thermodynamics is independent of atomic and molecular structure.
(ii) It includes whether a particular reaction is feasible or not under a given set of temperature and concentration of reactants and products.
(iii) It can determine the rate at which the reaction take place.
Which of the above statements is/are not correct’?
(a) (i) only
(b) (ii) only
(c) (iii) only
(d) (i) and (ii) only
Answer:
(c) (iii) only

Question 67.
A fundamental goal of thermodynamics is the …………
(a) prediction of spontaneity of the reaction.
(b) determination of rate of the chemical reaction.
(c) evaluation of the microscopic properties.
(d) both (b) and (c)
Answer:
(a) prediction of spontaneity of the reaction.

Question 68.
The first law of thermodynamics states that …………
(a) ∆U = q – w
(b) ∆U = q + w
(c) ∆U + q = w
(d) ∆U = w – q
Answer:
(b) ∆U = q + w

Question 69.
Anything which separates the system from its surroundings is called …………..
(a) Boundary
(b) Partition
(c) Universe
(d) Outer layer
Answer:
(a) Boundary

Question 70.
Hot water in a thermos flask is an example of ………….
(a) closed system
(b) open system
(c) isolated system
(d) isochoric system
Answer:
(c) isolated system

Question 71.
Which one of the following is an example for closed system?
(a) Hot water contained in a thermos flask
(b) A gas contained in a cylinder fitted with a piston
(c) All living things
(d) Hot water contained in a open beaker
Answer:
(b) A gas contained in a cylinder fitted with a piston

Question 72.
Statement-I: All living things are open systems.
Statement-II: Because they continuously exchange matter and energy with the surroundings
(a) Statement-I and II arc correct and Statement-II is the correct explanation of Statement-I
(b) Statement-I and II are correct but Statement-II is not the correct explanation of Statement-I
(c) Statement-I is correct and Statement-II is wrong.
(d) Statement-I is wrong but Statement-II is correct.
Answer:
(a) Statement-I and II are correct and Statement-II is the correct explanation of Statement-I.

Question 73.
Which one of the following is an extensive property?
(a) Molar volume
(b) Density
(c) Molaritv
(d) Entropy
Answer:
(d) Entropy

Question 74.
Which one of the following is ail intensive property?
(a) Specific heat capacity
(b) Mass
(c) Enthalpy
(d) Heat capacity
Answer:
(a) Specific heat capacity

Question 75.
Which one of the following is not an extensive property?
(a) Mole
(b) Energy
(c) Molar mass
(d) Free energy
Answer:
(c) Molar mass

Question 76.
Which one of the following is not an intensive property?
(a) Density
(b) Molaritv
(c) Molality
(d) Mole
Answer:
(d) Mole

Question 77.
Which one of the following depend on the mass of the system?
(a) Density
(b) Mole fraction
(c) Mass
(d) Molar mass
Answer:
(c) Mass

Question 78.
Which one of the following is independent to the mass of the system?
(a) Volume
(b) Enthalpy
(c) Entropy
(d) Density
Answer:
(d) Densily

Question 79.
The process in which there is no exchange of heat between thc system and surrounding is called ……….
(a) Adiabatic process
(b) Isothermal process
(c) Isobaric process
(d) Isochoric process
Answer:
(a) Adiabatic process.

Question 80.
For an adiabatic process …………..
(a) dU = 0
(b) dT = 0
(c) q = 0
(d) q = w
Answer:
(c) q = 0

Question 81.
For an isothermal process …………..
(a) dH = 0
(b) dP = 0
(c) dT = 0
(d) dV = 0
Answer:
(c) dT = 0

Question 82.
The process in which the volume of the system remains constant is called ……….
(a) Adiabatic process
(b) Isothermal process
(c) Isobaric process
(d) Isochoric process
Answer:
(d) Isochoric process

Question 83.
For an isochoric process ……………
(a) q = 0
(b) dp = O
(c) dv = O
(d) ∆U = 0
Answer:
(c) dv = O

Question 84.
Combustion of fuel in a bomb calorimeter is an example of …………
(a) adiabatic process
(b) isochoric process
(c) isobaric process
(d) isothemal process
Answer:
(b) isochoric process

Question 85.
Which one of the following is not a path function?
(a) Work
(b) Heat
(c) Pressure
(d) Either (a) or (b)
Answer:
(c) Pressure

Question 86.
Which one of the following is a path function?
(a) Pressure
(b) Volume
(c) Temperature
(d) Heat
Answer:
(d) Heat

Question 87.
Which one of the following is a state function?
(a) Internal energy
(b) Enthalpy
(c) Free energy
(d) All the above
Answer:
(d) All the above

Question 88.
Statement-I: Internal energy of a system is an extensive property.
Statement-II: Internal energy depends on the amount of the substances present in the system.
(a) Statement-I and II are correct and Statement-II is the correct explanation of Statement-I.
(b) Statement-I and II arc correct hut Statement-II is not the correct explanation of Statement-I.
(c) Statement-I is correct but Statement-II is wrong.
(d) Statement-I is wrong but Statement-II is correct.
Answer:
(a) Statement-I and II are correct and Statement-II is the correct explanation of Statement-I.

Question 89.
The SI unit of heat is ………..
(a) Joule
(b) Kelvin
(c) Kg
(d) Kg mol^-1
Answer:
(a) Joule

Question 90.
Which one of the following is the quantity of heat required to raise the temperature of I gm of water by 1 °C?
(a) 1 Joule
(b) 1 Calorie
(c) 1 Kelvin
(d) 1 Kilo joule
Answer:
(b) 1 Calorie

Question 91.
Which one of the following is equal to 1 Joule?
(a) Nm^-1
(b) Nm²
(c) Nm
(d) Kg ms^-2
Answer:
(c) Nm

Question 92.
Consider the following statements.
(i) Work is a state function.
(ii) Work brings a temporary effect in the surroundings.
(iii) Work appears only at the boundary of the system.
Which of the above statements is/are not correct?
(a) (iii) only
(b) (i) and (ii)
(c) (i) and (iii)
(d) (ii) and (iii)
Answer:
(b) (i) and (ii)

Question 93.
Which of the following represents the gravitational work ?
(a) Q_V
(b) F. x
(c) mgh
(d) -P∆V
Answer:
(c) mgh

Question 94.
Match the List-I and List-II using the correct code given below the list.

Answer:

Question 95.
Match the List-I and List-II using the correct code given below the list.

Answer:

Question 96.
Enthalpy is defined as ………..
(a) q + w
(b) q – P∆V
(c) U + PV
(d) w
Answer:
(c) U + PV

Question 97.
Which one of the following always he negative?
(a) Enthalpy of combustion
(b) Enthalpy of fusion
(c) Enthalpy of vapourisation
(d) Enthalpy of sublimation
Answer:
(a) Enthalpy of combustion

Question 98.
For an ideal gas …………
(a) C_p – C_V = O
(b) C_p – C_V = R
(c) C_V – C_p = R
(d) C_V – C_p > R
Answer:
(b) C_p – C_V = R

Question 99.
The standard substance used in the enthalpy of combustion of a substance in bomb calorimeter is …………
(a) methane
(b) acetic acid
(c) propane
(d) benzoic acid
Answer:
(d) benzoic acid

Question 100.
The standard value of enthalpy of combustion of benzoic acid is …………
(a) – 3227 kJ mol^-1
(b) + 3227 kJ mol^-1
(c) – 32.27 Ici mol^-1
(d) + 32.27 kJ mol^-1
Answer:
(a) -3227 kJ mol^-1

Question 101.
The heat of neutralization of a strong acid and strong base is around …………
(a) + 57.32 kJ
(b) – 57.32 kJ
(c) – 3227 kJ mol^-1
(d) + 3227 kJ mol^-1
Answer:
(b) – 57.32 kJ

Question 102.
Which of the following is not a spontaneous process?
(a) All water fall runs down hill.
(b) A lump of sugar dissolves in cup of coffee.
(c) Heat flow from hotter object to colder one.
(d) A water flow from a well to upper reservoir.
Answer:
(d) A water flow from a well to upper reservoir.

Question 103.
Which one of the following is an endothermic process?
(a) CH₄ + 2O₂ → CO₂ + 2H₂O
(b) H⁺ + OH^– → H₂O

(d) C + O₂ → CO₂
Answer:

Question 104.
The SI unit of entropy is …………
(a) Nm
(b) Cal mol^-1
(c) KJ mol^-1
(d) JK^-1
Answer:
(d) JK^-1

Question 105.
In which of the following entropy decreases?
(a) melting of ice
(b) evaporation of water
(c) crystallization of sugar
(d) dissolution of salt
Answer:
(c) crystallization of sugar

Question 106.
Gibbs’s free energy is defined as ………….
(a) G = H+ TS
(b) G = H x TS
(c) G = H – TS
(d) G = H/TS
Answer:
(c) G = H – TS

Question 107.
Match the List-I and List-II using the correct code given below the list.

Answer:

Question 108.
Van’t Hoff equation is ………….
(u) ∆G° = AH° – T∆S°
(b) G = H – TS
(c) ∆G° = -2.303 RT log K_eq
(d) ∆S = ∆H / T
Answer:
(c) ∆G° = -2.303 RT log K_eq

Question 109.
Mathematically, the third law of thermodynamics is expressed as ………..
(a)
(b)
(c) AU = q + w
(d) G = H – TS
Answer:
(a)

II. Answer briefly (2 or 3 marks)

Question 1.
What is the aim of the study of chemical thermodynamics?
Answer:
The main aim of the study of chemical thermodynamics is to learn

• Transformation of energy from one form into another form.
• Utilization of various forms of energies.
• Change in the properties of system produced by chemical or physical effects.

Question 2.
What are the scope of thermodynamics?
Answer:
The scope of thermodynamics:

• To derive feasibility of a given process.
• It also helps in predicting how far a physical (or) chemical change can proceed. until the equilibrium conditions are established.

Question 3.
What are the limitations of the thermodnamics?
Answer:

• Thermodynamics suggests feasibility of reaction but fails to suggest rate of reaction. It is concerned only with the initial and the final states of the system. It is not concerned with the path by which the change occurs.
• It does not reveal the mechanism of a process.

Question 4.
Define
1. System
2. Surroundings.
Answer:
1. System:
A system is defined as any portion of matter under thermodynamic consideration. which is separated from the rest of the universe by real or imaginary boundaries. e.g.. Water taken in a beaker, balloon filled with air, seed, plant, flower and bird.
2. Surroundings:
Everything in the universe that is not the part of system and can interact with system is called as surroundings.

Question 5.
What is meant by isolated system? Give example.
Answer:

• A system which can exchange neither matter nor energy with its surroundings is called an isolated system.
• Here boundary is scaled and insulated.
• Hot water contained in a thermos flask, is an example for an isolated system.

Question 6.
Explain a closed system with an example.
Answer:

• A system which can exchange only energy but not matter with its surroundings is called a closed system.
• Here the boundary is sealed but not insulated.
• Hot water contained in a closed beaker is an example for a closed system.
• In this system heat is transferred to the surroundings but no water vapour can escape from this system.
• A gas contained in a cylinder fitted with a piston constitutes a closed system.

Question 7.
What is meant by open system? Give example.
Answer:

• A system which can exchange both matter and energy with its surroundings is called an open system.
• Hot water contained in an open beaker is an example for open system.
• In this system, both water vapour and heat is transferred to the surroundings through the imaginary boundary.
• All living things are open systems because they continuously exchange matter and energy with the surroundings.

Question 8.
What are extensive properties?
Answer:

• The property that is depend on the mass or size of the system is called an extensive property.
• e.g., Volume, number of moles, mass and internal energy.

Question 9.
What is reversible process? Give an example.
Answer:
The process in which the system and surroundings can be restored to the initial state from the final state without producing any changes in the thermodynamics properties of the Universe is called a reversible process. e.g.,
H₂ + I₂ ⇌ 2HI

Question 10.
What is an irreversible process? Give an example.
Answer:

• The process in which the system and surroundings cannot be restored to the initial state from the final state is called a reversible process.
• e.g.. All the processes occurring in nature are irreversible processes.

Question 11.
Define cyclic process. Give example.
Answer:
When a system returns to its original state after completing a series of changes, then it is said that a cycle is completed. This process is known as a cyclic process. For a cyclic process dU = 0, dH= 0, dP = 0, dV= 0 and dT= 0.

Question 12.
What is meant by internal energy?
Answer:

• Internal energy (U) of a system is equal to the energy possessed by all its constituents namely atoms, ions and molecules.
• The energy of a molecule is equal to the sum of its translational energy (U_v), vibrational energy (U_b), rotational energy (U_e), bond energy (jh’ electronic energy and energy due to molecular interactions (U_i).
• U = U_t + U_v + U_r + U_b + U_e + U_i
• The total energy of all the molecules of the system is called internal energy.

Question 13.
Define Heat. Give its unit.
Answer:

• Heat (q) is regarded as energy in transit across the boundary separating a system from its
  surroundings.
• Heat changes result in temperature differences between system and surroundings.
• Heat is a path function.
• Units of heat: SI unit of heat is the joule (J).

Question 14.
Write a note about the sign convention of heat.
Answer:

• The symbol of heat is q.
• if the heat flows into the system from the surroundings, the energy of a system increases. Hence it is taken to be positive (+ q).
• if heat flows out of the system into the surroundings. energy of the system decreases. Hence it is taken to be negative (- q).

Question 15.
What is meant by work? Give its unit.
Answer:

• Work is defined as the force (F) multiplied by the displacement (x).
• – w = F. x
• Minus (-) sign indicates the work done by the system
• Unit of work: The SI unit of work is Joule (J).

Question 16.
Explain about gravitational work. Give its unit.
Answer:
1. When an object is raised to a certain height against the gravitational field, gravitational work is done on the object.
2. For example, if an object of mass ‘m’ is raised through a height ‘h’ against acceleration due to gravity ‘g’, then the gravitational work carried out is ‘mgh’.
w = m.g.h
w = Kg .ms^-2.m
w = Kg m² s^-2
w = Joule

Question 17.
Define electrical work. Give its unit.
Answer:

• When a charged body moves from one potential region to another electrical work is done.
• If the electrical work done is QV. Where V is the potential difference and Q is the quantity of electricity.
  w = QV
  w = Coulomb . Volts
  w = Joule

Question 18.
Write a note about mechanical work. Give its unit.
Answer:

1. Mechanical work is defined as force multiplied by the displacement through which the force acts.
2. Whenever a force (F) acts on an object and the object undergoes a displacement (x) in the direction of the force, then the mechanical work is said to be done.
3. Mathematically w = F . x
   w = F. x = N.m
   w = Joule

Question 19.
Define Zeroth law of thermodynamics (or) Law of thermal equilibrium.
Answer:
Zeroth law of thermodynamics states that ‘If two systems at different temperatures are separately in thermal equilibrium with a third one, then they tend to be in thermal equilibrium with themselves’.

Question 20.
Define enthalpy of a system. Give its unit.
Answer:
Enthalpy is a thermodynamic property of a system. Enthalpy (H) is defined as sum of the internal energy (U) of a system and the product of pressure and volume of the system.
H = U + PV
Unit of enthalpy: KJ mol^-1.

Question 21.
Define standard heat of formation.
Answer:
The standard heat of formation of a compound is defined as “The change in enthalpy that takes place when one mole of a compound is formed from its elements, all substances being in their standard states (298 K and 1 atm pressure).

Question 22.
Define specific heat capacity of a system.
Answer:
Specific heat capacity of a system is defined as the heat absorbed by one gram of a substance in raising its temperature by one Kelvin at a specified temperature.
Cm = \(\frac{q}{\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right)}\)
Where C specific heat capacity, q = amount otheat absorbed. m = mass and T₁,T₂ temperatures.

Question 23.
Derive the value of molar heat capacity as constant volume.
Answer:
According to the first law of themrndynamics,
dq = dU + PdV
Dividing both sides by dT, we have
\(\frac {dq}{dT}\) = \(\frac {(dU + PdV)}{dT}\)
At constant volume dV = 0,then
\(\frac {dq}{dT}\) = (\(\frac {dU}{dT}\))_V
C_V = (\(\frac {dU}{dT}\))_V
Thus the heat capacity ai constant volume (C_V ) is defined as the rate of change of internal energy with respect to temperature at constant volume.

Question 24.
Derive the value of molar heat capacity at constant pressure.
Answer:
We know.
H = U + PV
Differentiating the above equation with respect to temperature at constant pressure we get,
(\(\frac {dH}{dT}\))_P = ?(\(\frac {dU}{dT}\))_P + P(\(\frac {dV}{dT}\))_P
We know,
C = \(\frac {dq}{dT}\)
But at constant pressure dq = dH
Hence we get,
C_P = (\(\frac {dH}{dT}\))_P
Thus heat capacity at constant pressure (C_P) is defined as the rate of change of enthalpy with respect to temperature at constant pressure.

Question 25.
Prove that for an ideal gas, C_P is greater than C_V.
Answer:
1. It is clear that two heat capacities are not equal and C_P is greater than C_V by a factor which is related to the work done.

2. At a constant pressure, a part of heat absorbed by the system is used up in increasing the internal energy of the system and the other for doing work by the system.

3.  At constant volume, the whole of heat absorbed is utilized in increasing the temperature of the system as there is no work done by the system. Thus C_P is greater than C_V.
C_P = \(\frac {dH}{dT}\) ; C_V = \(\frac {dU}{dT}\)

4. By definition, H = U + PV for 1 mole of an ideal gas.
H = U + RT
By differentiating this equation with respect to temperature T. we get,
\(\frac {dH}{dT}\) = \(\frac {dU}{dT}\) + R
C_P = C_V + R
C_P – C_V = R
Thus for an ideal gas, C_P is greater than C_V by the gas constant R.

Question 26.
What are the applications of Bomb Calorimeter?
Answer:

• Bomb calorimeter is used to determine the amount of heat released in combustion reaction.
• It is used to determine the calorific value of food.
• Bomb calorimeter is used in many industries such as metabolic study. food processing and
  explosive testing.

Question 27.
Define heat of solution.
Answer:
The heat of solution is defined as “the change in enthalpy of the system when one mole of a substance is dissolved in a specified quantity of solvent at a given temperature”.

Question 28.
Define molar heat of fusion.
Answer:
The molar heat of fusion is defined as “the change in enthalpy when one mole of a solid substance is converted into the liquid state at its melting points’.

Question 29.
What is meant by molar heat of vapourisation?
Answer:
The molar heat of vapourisation is defined as the change in enthalpy when one mole of liquid is converted into vapour or gaseous state at its boiling point.

Question 30.

1. What is sublimation?
2. Define molar heat of sublimation.

Answer:

1. Sublimation is a process when a solid changes directly into gaseous state without changing into liquid state.
2. Molar heat of sublimation is defined as the change in enthalpy when one mole of a solid is directly converted into the gaseous state at its sublimation temperature.

Question 31.
Define heat of transition?
Answer:
The heat of transition is defined as the change in enthalpy when one mole of an element changes from one allotropic form to another.

Question 32.
How do you measure the enthalpy of formation of carbon monoxide?
Answer:
1. Hess’s law can be applied to calculate the enthalpy of formation of carbon monoxide. It is very difficult to control the oxidation of graphite to give pure CO. However, enthalpy for the oxidation of graphite to CO₂ can be easily measured and enthalpy of oxidation of CO to CO₂ is also measurable.

2. The application of Hess’s law enables us to estimate the enthalpy of formation of CO.
C + O₂ → CO₂ ∆H° = 393.5 kJ ……….. (1)
CO + 1/2 O₂ → CO₂ ∆H° = -283 kJ ……….. (2)
on inverting equation (2), we get
CO₂ → CO + 1/2 O₂ ∆H° = + 283 kJ ………… (3)
on adding equations (2) and (3), we get
C + 1/2 O₂ → CO ∆H°= 393 5 + 283 = 110.5kJ

Question 33.
What are the important features of lattice enthalpy?
Answer:
1. Higher lattice energy shows greater electrostatic attraction and therefore a stronger bond in the solid.
2. The lattice enthalpy is greater for ions of higher charge and smaller radii.

Question 34.
Why there is a need for second law of thermodynamics? Give its importance.
Answer:

• Thermodynamics first law tells that there is an exact equivalence between various forms of energy and that heat gained is equal to heat loss.
• Practically it is not possible to convert the heat energy into an equivalent amount of work.
• To explain this, another law is needed which is known as second law of thermodynamics.
• The second law of thermodynamics helps us to predict whether the reaction is feasible or not and also tell the direction of the flow of heat.
• It also tells that energy cannot be completely converted into equivalent work.

Question 35.
Write the entropy statement of second law of thermodynamics.
Answer:
Whenever a spontaneous process takes place, it is accompanied by an increase in the total entropy of the Universe.
∆S_universe > ∆S_system + ∆S_surroundings

Question 36.
Write the Clausius statement of second law of thermodynamics.
Answer:
Clausius statement: Heat flows spontaneously from hot objects to cold objects and to get it flow in the opposite direction, we have to spend some work.

Question 37.
What are spontaneous reaction? Give three examples for spontaneous reaction.
Answer:
reaction that does occur under the given set of conditions is called a spontaneous reaction.
Example:

• A waterfall runs downhill, but never up, spontaneously.
• Heat flows from hotter object to a colder one.
• Ageing process.

Question 38.
Define standard entropy of formation.
Answer:
Standard entropy of formation is defined as “the entropy of formation of 1 mole of a compound from the elements under standard conditions”. It is denoted as ∆S_f°. We can calculate the value of entropy of a given compound from the values of S° of elements.
∆S_f° = ∑∆S_products° – ∑∆S_reactants°

Question 39.
What is entropy of fusion?
Answer:
When one mole of the solid melts at its melting point reversibly the heat absorbed is called molar heat of fusion. The entropy change is given by
∆S_f = \(\frac{\Delta \mathrm{H}_{f}}{\mathrm{T}_{f}}\)
Where ∆H_f = molar heat of fusion, T_f is melting point.

Question 40.
What is entropy of Vapourisation?
Answer:
When one mole of liquid is boiled at its boiling point reversibly, the heat absorbed is called as molar heat of vaporization. The entropy change is given by
∆S_v = \(\frac{\Delta \mathrm{H}_{v}}{\mathrm{T}_{b}}\)
where ∆H_v is molar heat of vapourisation. T_b is boiling point.

Question 41.
Define entropy of transition.
Answer:
When one mole of a solid changes reversibly from one allotropic form to another at its transition temperature. The entropy change is given
∆S_t = \(\frac{\Delta \mathrm{H}_{t}}{\mathrm{T}_{t}}\)
Where ∆H_t = molar heat of transition and T_t = transition temperature.

Question 42.
Explain the following:

1. Out of diamond and graphite, which has greater entropy? Why?
2. From thermodynamic point of view, in which system the animals and plants belong?

Answer:

1. Graphite has greater entropy, because it is loosely packed.
2. Animals and plants belong to open system.

Question 43.
What is the condition spontaneity in terms of free energy change?
Answer:

1. If ∆G is negative, process is spontaneous.
2. if ∆G is positive, process is non-spontaneous.
3. if ∆G = 0, the process is in equilibrium.

Question 44.
Why standard entropy of an elementary substance is not zero whereas standard enthalpy of formation is taken as zero?
Answer:
A substance has a perfectly ordered arrangement only at absolute zero. Hence, entropy is zero only at absolute zero. Enthalpy of formation is the heat change involved in the formation of one mole of the substance from its elements. An element formed from its constituents means no heat change.

Question 45.
The equilibrium constant for a reaction is one or more if ∆G ° for it is less than zero. Explain.
Answer:
\({ G }^{ \ominus }\) = – RT in K, thus if \({ G }^{ \ominus }\) is less than zero Le., it is negative, then In K will be positive and hence K will be greater than one.

Question 46.
Many thermo dynamically feasible reactions do not occur under ordinary conditions. Why?
Answer:
Under ordinary conditions, the average energy of the reactants may be less than threshold energy. They require some activation energy to initiate the reaction.

Question 47.
Predict in which of the following, entropy increases or decreases.
Answer:

1.  A liquid crystallizes into a solid
2. Temperature of a crystallized solid is raised from 0 K to 115 K
3. 2NaHCO_3(s) → NaCO₂CO_3(s) + CO_2(s) + H₂O_(s)
4. H_2(s) → 2H_(g)

Answer:

1. After freezing, the molecules attain an ordered state and therefore, entropy decreases.
2. At O K the constituent particles are in static form therefore, entropy is minimum. If the temperature is raised to 115 K particles begin to move and entropy increases.
3. Reactant, NaHCO₃ is solid. Thus, its entropy is less in comparison to product which has high entropy.
4. Here, one molecule gives two atoms. Thus, number of particles increases and this leads to more disordered form.

Samacheer Kalvi 11th Chemistry Thermodynamics 5 – Mark Questions

Question 1.
Explain how heat absorbed at constant pressure is measured using coffee cup calorimeter with neat diagrani.
Answer:

1. Measurement of heat change at constant pressure can be done in a coffee cup calorimeter.
2. We know that ∆H = q_p (at constant P) and therefore, heat absorbed or evolved, q_p at constant pressure is also called the heat of reaction or enthalpy of reaction, ∆H_r
3. in an exoihermic reaction, heat is evolved, and system loses heat to the surroundings. Therefore, q_p will be negative and ∆H_r will also be negative.
4. Similarly in an endothermic reaction, heat is absorbed, q_p is positive and ∆H_r will also be positive.

Question 2.
List the characteristics of entropy.
Characteristics of entropy:
Answer:

1. Entropy is a thermodynamic state function that is a measure of the randomness or disorderliness of the system.
2. In general, the entropy of gaseous system is greater than liquids and greater than solids. The symbol of entropy is S.
3. Entropy is defined as for a reversible change taking place at a constant temperature (T), the change in entropy (∆S) of the system is equal to heat energy absorbed or evolved (q) by the system divided by the constant temperature (T).
   \(\Delta S_{\text {sys }}=\frac{q_{\text {rev }}}{T}\)
4. If heat is absorbed, then AS is positive and there will be increase in entropy. If heat is evolved, ∆S is negative and there is a decrease in entropy.
5. The change in entropy of a process represented by AS and is given by the equation,
   ∆H_sys = S_f – S_i
6. If S_f> S_i, ∆S is positive, the reaction is spontaneous and reversible. If S_f < S_i, ∆S is negative, the reaction is non-spontaneous and irreversible.
7. Unit of entropy: SI unit of entropy is J K ‘.

Question 3.
Explain about the characteristics of work.
Characteristics of work:
Answer:

1. Work is defined as the force (F) multiplied by the displacemcnt (x).
   – w = F.x ……… (1)
   The – ve sign is introduced to indicate that the work has been done by the system by spending a part of its internal energy.
2. Work is a path function.
3. Work appears only at the boundary of the system.
4. Work appears during the change ¡n the state of the system.
5. Work brings a permanent effect in the surroundings.
6. Units of work: The SI unit of work is the joule (J) or Kilojoule (KJ).
7. If work done by the system, the energy of the system decreases, hence by convention work is taken to be negative (- w).
8. If work done by the system, the energy of the system increases, hence by convention work is taken to be positive (+ w).

Question 4.
Derive the relationship between work for a reversible reaction and the charge in volume during comoression and expansion.
Answer:
1. During expansion, work is done by the system; since V1 > V1, the sign obtained for work will be negative.
2. During compression, work is done on the system; since V< V., the sign obtained for work will be positive.
3. If the pressure is not constant, but changes during the process such that it is always infinitesimally greater than the pressure of the gas, then, at each stage of compression, the volume decreases by an infinitesimal amount, dV.
4. We can calculate the work done on the gas by the relation,
w = –\(\int_{V_{i}}^{V_{f}} P d V\) …………(1)
5. In a compression process, P_ext the external pressure is always greater than the pressure of the system. i.e., P_ext = (P_int + dp). In an expansion process, the external pressure is always less than the pressure of the system i.e., P_ext = (P_int – dp).
6. In general case, we can write, P_ext = (P_int ± dp). Such processes are called reversible processes. For a compression process, work can be related to internal pressure of the system under reversible conditions by writing equation.

Since dp . dv is very small, we can write,

For a given system,
P_int V = nRT
P_int V = \(\frac {nRT}{V}\)

7. If V_f > V_i (expansion), the sign to work done by the process is negative.
If V_f < V_i (compression), the sign to work done by the process is positive.

Question 5.
Write the various definition of first law of thermodynamics.
First law of thermodynamics:
Answer:

• The total energy of an isolated system remains constant though it may change from one form to another.
• Whenever energy of a particular type disappears equivalent amount of another type must be produced.
• Total energy of a system and surroundings remains constant.
• Energy can neither be created nor destroyed, but may be converted from one form to another.
• The change in the internal energy of a closed system is equal to the energy that passes through its boundary as heat or work.
• Heat and work are equivalent ways of changing a systems internal energy.

Question 6.
Derive the various mathematical statements of the first law.
Answer:
Mathematical statement of the First law of Thermodynamics is
∆U = q + w
Case 1:
For a cyclic process involving isothermal expansion of an ideal gas
∆U = 0
∴ q = -w
In other words, during a cyclic process, the amount of heat absorbed by the system is equal to work done by the system.

Case 2:
For an isochoric process (no change in volume) there is no work of expansion.
∆V = 0
w = 0
∆U = 0
In other words, during isochoric process, the amount of heat supplied to the system is converted to its internal energy.

Case 3:
For an adiabatic process there is no change in heat .i.e., q O. Hence
q = 0
∆U = w
In other words, in an adiabatic process, the decrease in internal energy is exactly equal to the work done by the system on its surroundings.

Case 4:
For an isobaric process. There is no change in the pressure. P remains constant. Hence
∆U = q + w
∆U = q – P∆V
In other words, in an isobaric process a part of heat absorbed by the system is used for PV expansion work and the remaining is added to the internal energy of the system.

Question 7.
What are the characteristics of enthalpy?
Characteristics of enthalpy:
Answer:
1. Enthalpy is a thermodynamic property of a system. Enthalpy H is defined as sum of the internal energy (U) of a system and the product of pressure and volume of the system. That is,
H = U + PV
2. Enthalpy is a state function which depends entirely on the state functions T, P and U.
3. Enthalpy is usually expressed as the change in enthalpy (∆H) for a process between initial and final states.
∆H = ∆U + P∆V
4. At constant pressure, the heat flow (q) for the process is equal to the change in enthalpy which is defined by the equation.
∆H = q_p
5. In an endothermic reaction heat is absorbed by the system from the surroundings that is q >0 (positive). Therefore, at constant Tand P, by the equation above, if q is positive then ∆H is also positive.
6. In an exothermic reaction heat is evolved by the system to the surroundings that is, q<0 (negative). If q is negative, then ∆H will also be negative.
7. Unit of enthalpy is KJ mol^-1.

Question 8.
What are thermochemical equation? What are the conventions adopted In writing thermochemical equation?
Answer:
A thermochemical equation is a balanced stoichiometric chemical equation that includes the enthalpy change (∆H). Conventions adopted ¡n thermochemical equations:

1. The coefficients in a balanced thermochemical equation refer to number of moles of reactants and products involved in the reaction.
2. The enthalpy change of the reaction ∆H has unit kJ.
3. When the chemical reaction is reversed, the value of AR is reversed in sign with the same magnitude.
4. Physical states (gas, liquid, aqueous and solid) of all species is important and must be specified in a thermochemical reaction since AH depends on the phases of reactants and products.
5. if the thermochemical equation is multiplied throughout by a number, the enthalpy change is also be multiplied by the same number value.
6. The negative sign of ∆H indicates the reaction to be an exothermic and the positive sign of ∆H indicates an endothermic type of reaction.

Question 9.
Calculate the values of ∆U and ∆H for an ideal gas in terms of C_P and C_V
Calculation of ∆U and ∆H:
Answer:
For one mole of an ideal gas, we have
C_V = \(\frac {dU}{dT}\)
dU = C_V .dT
For a finite change, we have,
∆U = (U₂ – U₁) = C_V (T₂ – T₁)
and for n moles of an ideal gas we get
∆U = nC_V (T₂ – T₁) ………….(1)
We know,
∆H = ∆(U + PV)
∆H = ∆U + ∆(PV)
∆H = ∆U + ∆RT [: PV= RT]
∆H = ∆U + R∆T
∆H = C_V (T₂ – T₁) + R (T₂ – T₁)
∆H = (C_V + R) (T₂ – T₁)
∆H = C_P (T₂ – T₁) [∴C_P – C_V = R]
For n moles of an ideal gas we get
∆H = n C_P (T₂ – T₁) …………..(2)

Question 10.
Suggest and explain indirect method to calculate lattice enthalpy of magnesium bromide.
Answer:
Born Haber’s cycle method:
Mg(s) + Br₂ (l) → MgBr₂ ∆H_f°
Sublimation: Mg_(s) → Mg_(g) ∆H₁°
Ionisation: Mg_(g) → Mg²⁺_(g) + 2e ∆H₂°
Vapourisation: Br₂(l) → Br_2(g) ∆H₃°

Dissociation: Br_2(g) → 2Br_(g)∆H₄°∆H₄°
Electron affinity: 2Br_(g) + 2e^– → 2Br^–_(g)∆H₅°
Lattice enthalpy: Mg²⁺_(g) + 2Br^–_(g) → MgBr_2(s) ∆H₆° = ?
∆H_f° = ∆H₁° + ∆H₁° + ∆H₂° + ∆H₃° + ∆H₄° + ∆H₅° + ∆H₆°
∆H₆° = ∆H_f° – (∆H₁° + ∆H₁° + ∆H₂° + ∆H₃° + ∆H₄° + ∆H₅°)
If we know the values of ∆H_f° , ∆H₁°, ∆H₂° , ∆H₃° , ∆H₄° and ∆H₅°, we can calculate the value of ∆H₆° by indirect method.

Question 11.
Derive the relationship between standard free energy (∆G°) and equilibrium constant (K_eq).
Answer:
1. In a reversible process, system is at all times in perfect equilibrium with its surroundings.
2. A reversible chemical reaction can proceed in either direction simultaneously, so that a dynamic equilibrium is set up.
3. This means that the reactions in both the directions should proceed with decrease in free energy, which is impossible.
4. It is possible only ¡fat equilibrium, the free energy of a system is minimum.
5. Lets consider a general equilibrium reaction,
A + B ⇌ C + D
The free energy change of the above reaction in any state (∆G) is related to the standard free energy change of the reaction (∆G°) according to the following equation.
∆G = ∆G° + RT in Q …………(1)
where Q is reaction quotient and is defined as the ratio of concentration of the products to the concentration of the reactants under non-equilibrium condition.
6. When equilibrium is attained, there is no further free energy change i.e. ∆G = 0 and Q becomes equal to equilibrium constant. Hence the above equation becomes,
∆G° = RT in K_eq …………(2)
This equation is known as Van’t Hoff equation.
∆G° = – 2.303 RT log K_eq ……….(3)
We also know that,
∆G° = ∆H°- T∆S° = – RT in K_eq ……….(4)

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Samacheer Kalvi 11th Bio Botany Living World Text Book Back Questions and Answers

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Samacheer Kalvi Guru 11th Bio Botany Question 1.
Which one of the following statement about virus is correct?
(a) Possess their own metabolic system
(b) They are facultative parasites
(c) They contain DNA or RNA
(d) Enzymes are present
Answer:
(c) They contain DNA or RNA

Samacheer Kalvi Guru 11th Botany Question 2.
Identify the incorrect statement about the Gram positive bacteria.
(a) Teichoic acid absent
(b) High percentage of peptidoglycan is found in cell wall
(c) Cell wall is single layered
(d) Lipopolysaccharide is present in cell wall
Answer:
(a) Teichoic acid absent

11th Botany Samacheer Kalvi Question 3.
Identify the Archaebacterium.
(a) Acetobacter
(b) Erwinia
(c) Treponema
(d) Methanobacterium
Answer:
(d) Methanobacterium

Samacheer Kalvi Guru 11th Biology Question 4.
The correct statement regarding Blue green algae is
(a) lack of motile structures
(b) presence of cellulose in cell wall
(c) absence of mucilage around the thallus
(d) presence of floridean starch
Answer:
(a) lack of motile structures

Samacheer Kalvi 11th Botany Book Question 5.
Identify the correctly matched pair
(a) Actinomycete – 1. Late blight
(b) Mycoplasma – 2. Lumpy jaw
(c) Bacteria – 3. Crown gall
(d) Fungi – 4. Sandal spike
Answer:
(c) Bacteria – 3. Crown gall

Samacheer Kalvi 11th Bio Botany Solutions Question 6.
Differentiate Homoiomerous and Heteromerous lichens.
Answer:
Homoiomerous and Heteromerous Lichens:

Samacheer Kalvi 11th Bio Botany Question 7.
Write the distinguishing features of Monera.
Answer:
Distinguishing Features of Monera:

1. This kingdom includes all prokaryotic organisms. Example: Mycoplasma, bacteria, actinomycetes and cyanobacteria.
2. These are microscopic. They do not have a true nucleus and membrane bound organelles.
   
3. Many other bacteria like Rhizobium, Azotobacter and Clostridium can fix atmospheric nitrogen into ammonia.
   
4. Some bacteria are parasites and others live as symbionts.

Samacheer Kalvi 11th Bio Botany Book Back Answers Question 8.
Why do farmers plant leguminous crops in crop rotations/mixed cropping?
Answer:
Rotations / Mixed Cropping:

1. Legumes have bacteria on nodules which are on the roots of the plants. The bacteria on the nodules takes nitrogen from the air and fixes it into the soil, so that other plants that require nitrogen can use it as well.
2. Rotation of crops improves the fertility of the soil and hence brings about an increase in the production of food grains.
3. Rotation of crops helps in saving on nitrogenous fertilizers, because leguminous plants grown during the rotation of crops can fix atmospheric nitrogen in the soil with the help of nitrogen fixing bacteria.
4. Crop rotation adds diversity to an operation.

Samacheer Kalvi 11th Botany Solutions Question 9.
Briefly discuss on five kingdom classification. Add a note on merits and demerits.
Answer:
R.H. Whittaker, an American taxonomist proposed five kingdom classification in the year 1969. The kingdoms include Monera, Protista, Fungi, Plantae and Animalia. The criteria adopted for the classification include cell structure, thallus organization, mode of nutrition, reproduction and phylogenetic relationship. A comparative account of the salient features of each kingdom is given in table.

Merits:

• The classification is based on the complexity of cell structure and organization of thallus.
• It is based on the mode of nutrition.
• Separation of fungi from plants.
• It shows the phylogeny of the organisms

Demerits:

• The kingdom monera and protista accommodate both autotrophic and heterotrophic organisms, cell wall lacking and cell wall bearing organisms thus making these two groups more heterogeneous.
• Viruses were not included in the system.