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Students can Download Maths Chapter 3 Algebra Ex 3.3 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths 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 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.3

8th Maths Algebra Exercise 3.3 Question 1.
Expand
(i) (3m + 5)²
(ii) (5p – 1)²
(iii) (2n – 1)(2n + 3)
(iv) 4p² – 25q²
Solution:
(i) (3m + 5)²
Comparing (3m + 5)² with (a + b)² we have a = 3m and b = 5
(a + b)² = a² + 2 ab + b²
(3m + 5)² = (3m)² + 2 (3m) (5) + 5²
= 3²m² + 30m + 25 = 9m² + 30m +25

(ii) (5p – 1)²
Comparing (5p – 1)² with (a – b)² we have a = 5p and b = 1
(a – b)² = a² – 2ab + b²
(5p – 1)² = (5p)² – 2 (5p) (1) + 1²
= 5²p² – 10p + 1 = 25p² – 10p + 1

(iii) (2n – 1)(2n + 3)
Comparing (2n – 1) (2n + 3) with (x + a) (x + b) we have a = -1; b = 3
(x + a) (x + b) = x² + (a + b)x + ab
(2n +(- 1)) (2n + 3) = (2n)² + (-1 + 3)2n + (-1) (3)
= 2²n² + 2 (2n) – 3 = 4n² + 4n – 3

(iv) 4p² – 25q² = (2p)² – (5q)²
Comparing (2p)² – (5q)² with a² – b² we have a = 2p and b = 5q
(a² – b²) = (a + b)(a – b) = (2p + 5q) (2p – 5q)

Maths Chapter 3 Exercise 3.3 Class 8 Question 2.
Expand
(i) (3 + m)³
(ii) (2a + 5)³
(iii) (3p + 4q)³
(iv) (52)³
(v) (104)³
Solution:
(i) (3 + m)³
Comparing (3 + m)³ with (a + b)³ we have a = 3; b = m
(a + b)³ = a² + 3a²b + 3 ab² + b³
(3 + m)³ = 3³ + 3(3)² (m) + 3 (3) m² + m³
= 27 + 27m + 9m² + m³ = m³ + 9 m² + 27m + 27

(ii) (2a + 5)³
Comparing (2a + 5)³ with (a + b)³ we have a = 2a, b = 5
(a + b)³ = a³ + 3a²b + 3ab² + b³ = (2a)³ + 3(2a)² 5 + 3 (2a) 5² + 5³
= 2³a³ + 3(2²a²) 5 + 6a (25) + 125
= 8a³+ 60a² + 150a + 125

(iii) (3p + 4q)³
Comparing (3p + 4q)³ with (a + b)³ we have a = 3p and b = 4q
(a + b) ³ = a³ + 3a²b + 3ab² + b³
(3p + 4q)³ = (3p)³ + 3(3p)² (4q) + 3(3p)(4q)² + (4q)³
= 3³p³ +3 (9p²) (4q) + 9p (16q²) + 43q³
= 27p³ + 108p²q + 144pq² + 64q³

(iv) (52)³ = (50 + 2)³
Comparing (50 + 2)³ with (a + b)³ we have a = 50 and b = 2
(a + b)³ = a³ + 3 a²b + 3 ab² + b³
(50 + 2)³ = 50³ + 3 (50)²2 + 3 (50)(2)² + 2³
52³ = 125000 + 6(2,500) + 150(4) + 8
= 1,25,000 + 15,000 + 600 + 8
52³ = 1,40,608

(v) (104)³ = (100 + 4)³
Comparing (100 + 4)³ with (a + b)³ we have a = 100 and b = 4
(a + b)³ = a³ + 3 a²b + 3 ab² + b³
(100 + 4)³ = (100)³ + 3 (100)² (4) + 3 (100) (4)² + (4)³
= 10,00,000 + 3(10000) 4 + 300 (16) + 64
= 10,00,000 + 1,20,000 + 4,800 + 64 = 11,24,864

Samacheer Kalvi 8 Maths Question 3.
Expand
(i) (5 – x)³
(ii) (2x – 4y)³
(iii) (ab – c)³
(iv) (48)³
(v) (97xy)³
Solution:
(i) (5 – x)³
Comparing (5 – x)³ with (a – b)³ we have a = 5 and b = x
(a – b)³ = a³ – 3a²b + 3ab² – b³
(5 – x)³ = 5³ – 3 (5)² (x) + 3(5)(x²) – x³
= 125 – 3(25)(x) + 15x² – x³ = 125

(ii) (2x – 4y)³
Comparing (2x – 4y)³ with (a – b)³ we have a = 2x and b = 4y
(a – b)³ = a³ – 3a²b + 3ab³ – b³
(2x – 4y)³ = (2x)³ – 3(2x)² (4y) + 3(2x) (4y)² – (4y)³
= 2³x³ – 3(2²x²) (4y) + 3(2x) (4²y²) – (4³y³)
= 8x³ – 48x²y + 96xy² – 64y³

(iii) (ab – c)³
Comparing (ab – c)³ with (a – b)³ we have a = ab and b = c
(a – b)³ = a³ – 3a²b + 3ab² – b³
(ab – c)³ = (ab)³ – 3 (ab)² c + 3 ab (c)² – c³
= a³b³ – 3(a²b²) c + 3abc² – c³
= a³b³ – 3a²b² c + 3abc² – c³

(iv) (48)³ = (50 – 2)³
Comparing (50 – 2)³ with (a – b)³ we have a = 50 and b = 2
(a – b)³ = a³ – 3a²b + 3ab² – b³
(50 – 2)³ = (50)³ – 3(50)²(2) + 3 (50)(2)² – 2³
= 1,25,000 – 15000 + 600 – 8 = 1,10,000 + 592
= 1,10,592

(v) (97xy)³
= 97³ x³ y³ = (100 – 3)³ x³y³
Comparing (100 – 3)³ with (a – b)³ we have a = 100, b = 3
(a – b)³ = a³ – 3a²b + 3ab² – b³
(100 – 3)³ = (100)³ – 3(100)² (3) + 3 (100)(3)² – 3³
97³ = 10,00,000 – 90000 + 2700 – 27
97³ = 910000 + 2673
97³ = 912673
97x³y³ = 912673x³y³

Samacheer Kalvi 8th Maths Book Question 4.
Simplify (i) (5y + 1)(5y + 2)(5y + 3)
(ii) (p – 2)(p + 1)(p – 4)
Solution:
(i) (5y + 1) (5y + 2) (5y + 3)
Comparing (5y + 1) (5y + 2) (5y + 3) with (x + a) (x + b) (x + c) we have x = 5y ; a = 1; b = 2 and c = 3.
(x + a) (x + b) (x + c) = x³ + (a + b + c) x² (ab + bc + ca) x + abc
= (5y)³ + (1 + 2 + 3) (5y)² + [(1) (2) + (2) (3) + (3) (1)] 5y + (1)(2) (3)
= 5³y³ + 6(5²y²) + (2 + 6 + 3)5y + 6
= 125³ + 150y² + 55y + 6

(ii) (p – 2)(p + 1)(p – 4) = (p + (-2))0 +1)(p + (-4))
Comparing (p – 2) (p + 1) (p – 4) with (x + a) (x + b) (x + c) we have x = p ; a = -2; b = 1 ; c = -4.
(x + a) (x + b) (x + c) = x³ + (a + b + c) x² + (ab + be + ca) x + abc
= p³ + (-2 + 1 + (-4))p² + ((-2) (1) + (1) (-4) (-4) (-2)p + (-2) (1) (-4)
= p³ + (-5 )p² + (-2 + (-4) + 8)p + 8
= p³ – 5p² + 2p + 8

Samacheer Kalvi Guru 8th Maths Question 5.
Find the volume of the cube whose side is (x + 1) cm.
Solution:
Given side of the cube = (x + 1) cm
Volume of the cube = (side)³ cubic units = (x + 1)³ cm³
We have (a + b)³ = (a³ + 3a²b + 3ab² + b³) cm³
(x + 1)³ = (x³ + 3x² (1) + 3x (1)² + 1³) cm³
Volume = (x³ + 3x² + 3x + 1) cm³

Samacheerkalvi.Guru 8th Maths Question 6.
Find the volume of the cuboid whose dimensions are (x + 2),(x – 1) and (x – 3).
Solution:
Given the dimensions of the cuboid as (x + 2), (x – 1) and (x – 3)
∴ Volume of the cuboid = (l × b × h) units³
= (x + 2) (x – 1) (x – 3) units³
We have (x + a) (x + b) (x + c) = x³ + (a + b + c) x² + (ab + bc+ ca)x + abc
∴ (x + 2)(x – 1) (x – 3) = x³ + (2 – 1 – 3)x² + (2 (-1) + (-1) (-3) + (-3) (2)) x + (2)(-1) (-3)
x³ – 2x² + (-2 + 3 – 6)x + 6
Volume = x³ – 2x² – 5x + 6 units³

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:

You can Download Samacheer Kalvi 10th 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 10th Maths Solutions Chapter 8 Statistics and Probability Ex 8.2

10th Maths Exercise 8.2 Samacheer Kalvi Question 1.
The standard deviation and coefficient of variation of a data are 1.2 and 25.6 respectively. Find the value of mean.
Solution:
Co-efficient of variation C.V. = \(\mathrm{C.V}=\frac{\sigma}{\overline{x}} \times 100\)

10th Maths Exercise 8.2 Question 2.
The standard deviation and coefficient of variation of a data are 1.2 and 25.6 respectively. Find the value of mean.
Solution:

10th Maths 8.2  Question 3.
If the mean and coefficient of variation of a data are 15 and 48 respectively, then find the value of standard deviation.
Solution:

Samecheer Kalvi.Guru Question 4.
If n = 5 , \(\overline{x}\) = 6, \(\Sigma x^{2}\) = 765, then calculate the coefficient of variation.
Solution:

Ex 8.2 Class 10 Question 5.
Find the coefficient of variation of 24, 26, 33, 37, 29,31.
Solution:

Exercise 8.2 Class 10 Question 6.
The time taken (in minutes) to complete a homework by 8 students in a day are given by 38, 40, 47, 44, 46, 43, 49, 53. Find the coefficient of variation.
Solution:

10th Maths Statistics And Probability Question 7.
The total marks scored by two students Sathya and Vidhya in 5 subjects are 460 and 480 with standard deviation 4.6 and 2.4 respectively. Who is more consistent in performance?
Solution:

Question 8.
The mean and standard deviation of marks obtained by 40 students of a class in three subjects Mathematics, Science and Social Science are given below.

Which of the three subjects shows highest variation and which shows lowest variation in marks?
Solution:

Science subject shows highest variation. Social science shows lowest variation.

Question 9.
The temperature of two cities A and B in a winter season are given below.

Find which city is more consistent in temperature changes?
Solution:



∴ Co-efficient of variation of City A is less than C.V of City B.
∴ City A is more consistent.

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Tamil Book Solutions Guide Pdf Chapter 6.1 வளம் பெருகுக Text Book Back Questions and Answers, Summary, Notes.

Tamilnadu Samacheer Kalvi 8th Tamil Solutions Chapter 6.1 வளம் பெருகுக

கற்பவை கற்றபின்

Question 1.
உமது பகுதியில் நடைபெறும் ஏதேனும் ஒரு தொழிலின் பல செயல்களை வரிசைப்படுத்தி எழுதுக.
Answer:
மண்பாண்டத் தொழில் :
குளங்கள், ஆற்றங்கரைகள், வயல்வெளிகள் ஆகிய இடங்களிலிருந்து களிமண்ணை எடுத்து வருவர், பெரிய பள்ளம் தோண்டி அதில் களிமண்ணை நிரப்பி தண்ணீர் ஊற்றி ஒரு நாள் முழுவதும் ஊற வைப்பர். பிறகு அதனுடன் மெல்லிய மணல் சாம்பல் ஆகியவற்றைக் கலந்து பயன்படுத்துவார்கள். பிறகு பானை செய்யும் சக்கரத்தில் வைத்து வேண்டிய வடிவங்களில் அதை உருவாக்குவார்கள். உரிய வடிவம் வந்ததும் அடிப்பகுதியில் நூல் அல்லது ஊசியால் அறுத்து எடுத்து காய வைப்பர்.

ஓரளவுகாய்ந்ததும், தட்டுப்பலகை கொண்டுதட்டி பானையின் அடிப்பகுதியில் இருக்கும் ஓட்டையை மூடி பானையை முழுமையாக்குகின்றனர். பிறகு உருட்டுக்கல் கொண்டு தேய்த்து பானையைப் பளபளபாக்குகின்றனர். பிறகு வண்ணங்களையும், ஓவியங்களையும் தகுந்தாற்போல வரைகின்றனர்.

பாடநூல் வினாக்கள்

சரியான விடையைத் தேர்ந்தெடுத்து எழுதுக.

Question 1.
தோட்டத்தில் தம்பி ஊன்றிய ……………………. எல்லாம் முளைத்தன.
அ) சத்துகள்
ஆ) பித்துகள்
இ) முத்துகள்
ஈ) வித்துகள்
Answer:
ஈ) வித்துகள்

Question 2.
என் நண்பன் செய்த தொழிலில் அவனுக்கு …………………… பெருகிற்று.
அ) காரி
ஆ) ஓரி
இ) வாரி
ஈ) பாரி
Answer:
இ) வாரி

Question 3.
‘அக்களத்து’ என்ற சொல்லைப் பிரித்து எழுதக் கிடைப்பது …………………….
அ) அ + களத்து
ஆ) அக் + களத்து
இ) அக்க + அளத்து
ஈ) அம் + களத்து
Answer:
அ) அ + களத்து

Question 4.
கதிர் + ஈன என்பதனைச் சேர்த்தெழுதக் கிடைக்கும் சொல் ………………….
அ) கதிரென
ஆ) கதியீன
இ) கதிரீன
ஈ) கதிரின்ன
Answer:
இ) கதிரீன

குறுவினா

Question 1.
பயிர்கள் வாட்டமின்றி கிளைத்து வளரத் தேவையானது யாது?
Answer:
தகுந்த காலத்தில் பெய்யும் மழையே பயிர்கள் வாட்டமின்றி கிளைத்து வளரத் தேவையானது ஆகும்.

Question 2.
உழவர்கள் எப்போது ஆரவார ஒலி எழுப்புவர்?
Answer:
நெற்போரினை அடித்து நெல்லினைக் கொள்ளும் (எடுக்கும்) காலத்தில் உழவர்கள் ஆரவார ஒலி எழுப்புவர்.

சிறு வினா

Question 1.
உழவுத் தொழில் பற்றித் தகடூர் யாத்திரை கூறுவன யாவை?
Answer:

• சேரனின் நாட்டில் பெருகிய மழைநீரால் வருவாய் சிறந்து விளங்குகிறது.
• அகலமான நிலப்பகுதியில் விதைகள் குறைவின்றி முளை விடுகின்றன.
• முளைத்த விதைகள் செழிப்புடன் வளர தட்டுப்பாடின்றி மழை பொழிகின்றது.
• தகுந்த காலத்தில் மழை பொழிவதால் பயிர்கள் வாட்டம் இன்றி கிளைத்து வளர்கிறது. செழித்த பயிர்கள் பால் முற்றிக் கதிர்களைப் பெற்றிருக்கின்றன.
• அக்கதிர்கள் அறுவடை செய்யப் பெற்று ஏரினால் வளம் சிறக்கும் செல்வர்களின் களத்தில் நெற்போர் காவல் இல்லாமலே இருக்கின்றது.
• நெற்போரினை அடித்து நெல்லினைக் கொள்ளும் (எடுக்கும்) காலத்தில் உழவர்கள் எழுப்பும் ஆரவார ஒலியால் நாரை இனங்கள் அஞ்சித் தம் பெண் பறவைகளோடு பிரிந்து செல்லும் சிறப்புடைய, சேர மன்னரின் அகன்ற பெரிய நாடு புது வருவாயுடன் சிறந்து விளங்குகின்றது.

சிந்தனை வினா

Question 1.
உழவுத் தொழில் சிறக்க இன்றியமையாதனவாக நீங்கள் கருதுவன யாவை?
Answer:
உழவுத் தொழில் உயிர் தொழில்

நாகரீகம் என்ற பெயரில் இன்று யாரும் உழவுத் தொழில் செய்ய முன்வருவதில்லை. ஒவ்வொரு வீட்டிலும் உள்ள ஒருவர் கட்டாயம் உழவுத் தொழில் செய்தல் வேண்டும். உழவுத் தொழில், அரசுப் பணிகளில் ஒன்றாகச் சேர்க்கப்பட வேண்டும் உழவுத் தொழிலில் சிறந்து விளங்கும் உழவர்களுக்கு ஆண்டுதோறும் விருதுகளும் பரிசுத் தொகையும் கொடுக்க வேண்டும். இன்றைய இளைஞர்கள் வேலை விருப்பப் பட்டியலில் உழவுத்தொழிலைச் சேர்த்துக் கொள்ள வேண்டும். இவ்வாறு செய்தால் மட்டுமே உழவுத் தொழில் நிச்சயம் சிறக்கும்.

கூடுதல் வினாக்கள்

சரியான விடையைத் தேர்ந்தெடுத்து எழுதுக.

Question 1. ம
ன்பதை காக்கும் மாபெரும் சிறப்பு …………………… க்கு உண்டு .
அ) மழை
ஆ) வயல்
இ) நிலம்
ஈ) உழவர்
Answer:
அ) மழை

Question 2.
தகடூர் யாத்திரை பாடல் பேசும் தொழில் ……………………..
அ) உழவுத் தொழில்
ஆ) நெய்தல் தொழில்
இ) மீன்பிடித் தொழில்
ஈ) மண்பாண்டத் தொழில்
Answer:
அ) உழவுத் தொழில்

Question 3.
புது வருவாய் என்னும் பொருளினைக் குறிக்கும் சொல் …………………
அ) வாரி
ஆ) எஞ்சாமை
இ) ஒட்டாது
ஈ) யாணர்
Answer:
ஈ) யாணர்

Question 4.
வளம் பெருக பாடல் ………………….. மன்னர் பற்றியது.
அ) சோழர்
ஆ) சேரர்
இ) பாண்டியர்
ஈ) பல்ல வர்
Answer:
ஆ) சேரர்

Question 5.
தர்மபுரியின் பழைய பெயர் ……………………..
அ) மாமண்டூர்
ஆ) வடுவூர்
இ) தகடூர்
ஈ) குரும்பூர்
Answer:
இ) தகடூர்

குறுவினா

Question 1.
சேரநாட்டில் வருவாய் சிறந்து விளங்கக் காரணம் யாது?
Answer:
பெருகிய மழை நீரால் சேர நாட்டின் வருவாய் சிறந்து விளங்குகிறது.

Question 2.
அறுவடை செய்யப்பட்ட கதிர்கள் எங்கு நிறைகின்றன?
Answer:
அறுவடை செய்யப்பட்ட கதிர்கள் ஏறினால் வளம் சிறக்கும் செல்வர்களின் களத்தில் வந்து நிறைகின்றன.

Question 3.
நாரை இனங்கள் பெண்பாற் பறவைகளோடு பிரிந்து செல்லக் காரணம் யாது?
Answer:
நெற்போரினை அடித்து நெல்லினைக் கொள்ளும் (எடுக்கும்) காலத்தில் உழவர்கள் எழுப்பும் ஆரவார ஒலியால், நாரை இனங்கள் தன் பெண்பாற் பறவைகளோடு பிரிந்து செல்கின்றன.

சொல்லும் பொருளும்

வாரி – வருவாய்
எஞ்சாமை – குறைவின்றி
முட்டாது – தட்டுப்பாடின்றி
ஒட்டாது – வாட்டம் இன்றி
வைகுக – தங்குக
ஓதை – ஓசை
வெரீஇ – அஞ்சி
யாணர் -புது வருவாய்

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Tamil Book Solutions Guide Pdf Chapter 2.6 திருக்குறள் Text Book Back Questions and Answers, Summary, Notes.

Tamilnadu Samacheer Kalvi 8th Tamil Solutions Chapter 2.6 திருக்குறள்

பாடநூல் வினாக்கள்

சரியான விடையைத் தேர்ந்தெடுத்து எழுதுக.

Question 1.
புகழாலும் பழியாலும் அறியப்படுவது ………………………….
அ) அடக்கமுடைமை
ஆ) நாணுடைமை
இ) நடுவுநிலைமை
ஈ) பொருளுடைமை
Answer:
இ) நடுவுநிலைமை

Question 2.
பயனில்லாத களர்நிலத்திற்கு ஒப்பானவர்கள் …………………….
அ) வலிமையற்றவர்
ஆ) கல்லாதவர்
இ) ஒழுக்கமற்றவர்
ஈ) அன்பில்லாதவர்
Answer:
ஆ) கல்லாதவர்

Question 3.
‘வல்லுருவம்’ என்னும் சொல்லைப் பிரித்து எழுதக் கிடைப்பது ……………………….
அ) வல் + உருவம்
ஆ) வன்மை + உருவம்
இ) வல்ல + உருவம்
ஈ) வ + உருவம்
Answer:
ஆ) வன்மை + உருவம்

Question 4.
நெடுமை + தேர் என்பதனைச் சேர்த்தெழுதக் கிடைக்கும் சொல் ……………………..
அ) நெடுதேர்
ஆ) நெடுத்தேர்
இ) நெடுந்தேர்
ஈ) நெடுமைதேர்
Answer:
இ) நெடுந்தேர்

Question 5.
‘வருமுன்னர்’ எனத் தொடங்கும் குறளில் பயின்று வந்துள்ள அணி ………………………
அ) தற்குறிப்பேற்ற அணி
ஆ) எடுத்துக்காட்டு உவமை அணி
இ) உவமை அணி
ஈ) உருவக அணி
Answer:
இ) உவமை அணி

குறுவினா

Question 1.
சான்றோர்க்கு அழகாவது எது?
Answer:
துலாக்கோல் போல நடுவுநிலைமையுடன் சரியாகச் செயல்படுவதே சான்றோர்க்கு அழகாகும்.

Question 2.
பழியின்றி வாழும் வழியாகத் திருக்குறள் கூறுவது யாது?
Answer:
தலைவன் முதலில் தன்குற்றத்தைக் கண்டு நீக்கி, அதன்பின் பிறருடைய குற்றத்தை ஆராய்ந்தால், அவனுக்கு எந்தப் பழியும் ஏற்படாது.

Question 3.
‘புலித்தோல் போர்த்திய பசு’ என்னும் உவமையால் திருக்குறள் விளக்கும் கருத்து யாது?
Answer:
மனத்தை அடக்கும் வல்லமை இல்லாதவர் மேற்கொண்ட தவக்கோலம், புலியின் தோலைப் போர்த்திக் கொண்ட பசு பயிரை மேய்ந்ததைப் போன்றது.

திருக்குறளைச் சீர் பிரித்து எழுதுக.

1. தக்கார் தகவிலரென்பது அவரவர் எச்சத்தால் காணப்படும்.
2. தொடங்கற்க எவ்வினையு மெள்ளற்க முற்று மிடங்கண்ட பின் அல்லது.

சீர்பிரித்தது

1. தக்கார் தகவிலர் என்பது அவரவர்
எச்சத்தால் காணப் படும்.

2. தொடங்கற்க எவ்வினையும் எள்ளற்க முற்றும்
இடங்கண்ட பின்அல் லது.

கோடிட்ட இடத்தை நிரப்புக.

1. வலியில் நிலைமையான் வல்லுருவம் ……………………….
புலியின்தோல் ……………………. மேய்ந் தற்று.

2. விலங்கொடு ………………….. அனையர் ……………………….
கற்றாரோடு ஏனை யவர்.
Answer:
1. பெற்றம், போர்த்து
2. மக்கள், இலங்குநூல்

சீர்களை முறைப்படுத்தி எழுதுக.

யாழ்கோடு அன்ன கொளல் கணைகொடிது
வினைபடு பாலால் செவ்விதுஆங்கு.

முறைப்படுத்தியது
கணைகொடிது யாழ்கோடு செவ்விது ஆங்கு அன்ன
வினைபடு பாலால் கொளல்.

படங்களுக்குப் பொருத்தமான திருக்குறள்களை எழுதுக.

1. வலியில் நிலைமையான் வல்லுருவம் பெற்றம்
புலியின் தோல் போர்த்துமேய்ந் தற்று.

2. கடல்ஓடா கால்வல் நெடுந்தேர் கடல் ஓடும்
நாவாயும் ஓடா நிலத்து.

கூடுதல் வினாக்கள்

சரியான விடையைத் தேர்ந்தெடுத்து எழுதுக.

Question 1.
எல்லா மக்களுக்கும் பொருந்தும் அறக்கருத்துகளைக் கொண்ட நூல் ……………………
அ) பரிபாடல்
ஆ) குறுந்தொகை
இ) திருக்குறள்
ஈ) நற்றிணை
Answer:
இ) திருக்குறள்

Question 2.
திருக்குறளின் பெருமையை விளக்கும் நூல் …………………..
அ) ஆத்திசூடி
ஆ) புறநானூறு
இ) கலித்தொகை
ஈ) திருவள்ளுவமாலை
Answer:
ஈ) திருவள்ளுவமாலை

Question 3.
நடுவுநிலைமையுடன் செயல்படுவதே ………………. அழகாகும்.
அ) கயவர்க்கு
ஆ) சான்றோர்க்கு
இ) கல்லாதார்க்கு
ஈ) மூடர்க்கு
Answer:
ஆ) சான்றோர்க்கு

Question 4.
‘வலியில்’ எனத் தொடங்கும் குறளில் பயின்று வந்துள்ள அணி ……………………
அ) உவமையணி
ஆ) எடுத்துக்காட்டு உவமையணி
இ) உருவக அணி
ஈ) இல்பொருள் உவமை அணி
Answer:
ஈ) இல்பொருள் உவமை அணி

Question 5.
மக்களின் பண்புகளை அவரவர் தோற்றத்தால் அல்லாமல் …………………………… வகையால் உணர்ந்து கொள்ள வேண்டும்.
அ) அன்பு
ஆ) செயல்
இ) நட்பு
ஈ) பாசம்
Answer:
ஆ) செயல்

Question 6.
கற்றவர்க்கும் கல்லாதவர்க்கும் இடையே உள்ள வேறுபாடு மக்களுக்கும் ……………………… இடையே உள்ள வேறுபாட்டிற்கு இணையானது.
அ) பறவைகளுக்கும்
ஆ) கயவர்களுக்கும்
இ) விலங்குகளுக்கும்
ஈ) மரங்களுக்கும்
Answer:
இ) விலங்குகளுக்கும்

Question 7.
……………. வருமுன்னே சிந்தித்து நம்மைக் காத்துக் கொள்ள வேண்டும்.
அ) வலி
ஆ) பழி
இ) சளி
ஈ) உளி
Answer:
ஆ) பழி

Question 8.
பழிவருமுன்னே காக்காதவர் வாழ்க்கை, நெருப்பின் அருகில் வைக்கப்பட்ட …………………….. போல அழிந்துவிடும்.
அ) வைக்கோல்போர்
ஆ) பஞ்சு
இ) தாள்
ஈ) மரம்
Answer:
அ) வைக்கோல்போர்

Question 9.
‘கடல் ஓடா’ எனத் தொடங்கும் குறளில் பயின்று வந்துள்ள அணி ………………………..
அ) பிறிதுமொழிதல் அணி
ஆ) தற்குறிப்பேற்ற அணி
இ) உருவக அணி
ஈ) உவமையணி
Answer:
அ) பிறிது மொழிதல் அணி

Question 10.
முதற்பாவலர் என்று அழைக்கப்படுபவர் ……………………
அ) நக்கீரர்
ஆ) கம்ப ர்
இ) வள்ளுவர்
ஈ) கபிலர்
Answer:
இ) வள்ளுவர்

Question 11.
திருவள்ளுவர் ……………………… ஆண்டுகளுக்கு முற்பட்டவர் என்பர்.
அ) 1000
ஆ) 500
இ) 1500
ஈ) 2000
Answer:
ஈ) 2000

Question 12.
………………………. உலகின் பல்வேறு மொழிகளில் மொழி பெயர்க்கப்பட்ட சிறந்த நூல் ஆகும்.
அ) பெரிய புராணம்
ஆ) திருக்குறள்
இ) தேவாரம்
ஈ) நற்றிணை
Answer:
ஆ) திருக்குறள்

Question 13.
திருக்குறள் ……………….. பகுப்புக் கொண்டது.
அ) ஐம்பால்
ஆ) எண்பால்
இ) முப்பால்
ஈ) ஒன்பால்
Answer:
இ) முப்பால்

Question 14.
அறத்துப்பால் ……………………….. இயல்களைக் கொண்டது.
அ) இரண்டு
ஆ) மூன்று
இ) நான்கு
ஈ) ஐந்து
Answer:
இ) நான்கு

Question 15.
பொருட்பால் ………………. இயல்களைக் கொண்டது.
அ) இரண்டு
ஆ) மூன்று
இ) நான்கு
ஈ) ஐந்து
Answer:
ஆ) மூன்று

Question 16.
இன்பத்துப்பால் ………………………. இயல்களைக் கொண்டது.
அ) இரண்டு
ஆ) மூன்று
இ) நான்கு
ஈ) ஐந்து
Answer:
அ) இரண்டு

Question 17.
‘அவரவர்’ என்னும் சொல்லைப் பிரித்து எழுதக் கிடைப்பது ……………………….
அ) அவர + வர்
ஆ) அவர + அவர்
இ) அ + அவர்
ஈ) அவர் + அவர்
Answer:
ஈ) அவர் + அவர்

Question 18.
‘தகவிலர்’ என்னும் சொல்லைப் பிரித்து எழுதக் கிடைப்பது ……………………..
அ) தக + விலர்
ஆ) தக + இலர்
இ) தகவு + இலர்
ஈ) தகவு + பலர்
Answer:
இ) தகவு + இலர்

Question 19.
‘நிலைமையான்’ என்னும் சொல்லைப் பிரித்து எழுதக் கிடைப்பது …………………….
அ) நிலை + யான்
ஆ) நிலைமை + ஆன்
இ) நிலைமை + யான்
ஈ) நிலைமை + ஏன்
Answer:
ஆ) நிலைமை + ஆன்

Question 20.
‘மேய்ந்தற்று’ என்னும் சொல்லைப் பிரித்து எழுதக் கிடைப்பது ……………………..
அ) மேய் + அற்று
ஆ) மேய்ந் + தற்று
இ) மேயும் + அற்று
ஈ) மேய்ந்து + அற்று
Answer:
ஈ) மேய்ந்து+அற்று

Question 21.
‘பாலால்’ என்னும் சொல்லைப் பிரித்து எழுதக் கிடைப்பது …………………..
அ) பா + ஆல்
ஆ) பால் + ஆல்
இ) பாலு + ஆல்
ஈ) பா + லால்
Answer:
ஆ) பால் + ஆல்

Question 22.
‘கல்லாதவர்’ என்னும் சொல்லைப் பிரித்து எழுதக் கிடைப்பது …………………..
அ) க + அவர்
ஆ) கல் + லாதவர்
இ) கல்லாது + அவர்
ஈ) கல்லா + அவர்
Answer:
இ) கல்லாது + அவர்

Question 23.
‘விலங்கொடு’ என்னும் சொல்லைப் பிரித்து எழுதக் கிடைப்பது …………………….
அ) விலங் + கொடு
ஆ) விலங்கு + ஒடு
இ) விலங்க + கொடு
ஈ) விலங்கு + ஓடு
Answer:
ஆ) விலங்கு + ஒடு

Question 24.
‘கற்றாரோடு’ என்னும் சொல்லைப் பிரித்து எழுதக் கிடைப்பது ……………………….
அ) கற்றார் + ஓடு
ஆ) கற்றார் + ஒடு
இ) கற்றா + ரோடு
ஈ) கற் + றாரோடு
Answer:
அ) கற்றார் + ஓடு

Question 25.
‘வைத்து + ஊறு’ என்பதனைச் சேர்த்தெழுதக் கிடைக்கும் சொல் ………………
அ) வைத்து ஊறு
ஆ) வைத்தாறு
இ) வைத்தூறு
ஈ) வைத்துறு
Answer:
இ) வைத்தூறு

Question 26.
‘இடம் + கண்ட’ என்பதனைச் சேர்த்தெழுதக் கிடைக்கும் சொல் ………………….
அ) இடம்கண்ட
ஆ) இடகண்ட
இ) இடத்தைக்கண்ட
ஈ) இடங்கண்ட
Answer:
ஈ) இடங்கண்ட

Question 27.
‘நாவாய் + உம்’ என்பதனைச் சேர்த்தெழுதக் கிடைக்கும் சொல் ………………
அ) நாவாய்உம்
ஆ) நாவாயும்
இ) நாவாஉம்
ஈ) நாவாஊம்
Answer:
ஆ) நாவாயும்

அணிகள்

1. இல்பொருள் உவமை அணி
அணி விளக்கம் :
உலகத்தில் இல்லாத பொருளை உவமையாக்கிக் கூறுவது இல்பொருள் உவமை அணி ஆகும்.

எ.கா:
‘வலியில் நிலைமையான் வல்லுருவம் பெற்றம்
புலியின்தோல் போர்த்துமேய்ந் தற்று.’
அணிப்பொருத்தம் : பசுவானது புலியின்தோலைப் போர்த்திக் கொண்டு பயிரை மேய்ந்தல் உலகில் இல்லாத செயலாகும். எனவே அதனை உவமையாக்கிக் கூறியுள்ளமையால், இது இல்பொருள் உவமை அணி ஆயிற்று.

2. உவமை அணி :

அணி விளக்கம் :
உவமை, பொருள், உவம உருபு ஆகியவை வெளிப்படையாக தோன்றுமாறு அமைவது உவமை அணி ஆகும்.

எ.கா – 1:
சமன்செய்து சீர்தூக்கும் கோல்போல் அமைந்து ஒருபால்
கோடாமை சான்றோர்க்கு அணி.

அணிப்பொருத்தம் :
இப்பாடலில் துலாக்கோலை உவமையாகக் கூறி, சான்றோர்களின் நடுவுநிலைமையை உவமிக்கப்படும் பொருளாக்கி, போல என்ற உவம உருபு வெளிப்படையாகத் தோன்றுமாறு அமைக்கப்பட்டுள்ளதால் அது உவமை அணி ஆயிற்று.

எ.கா – 2:
வருமுன்னர்க் காவாதான் வாழ்க்கை எரிமுன்னர்
வைத்தூறு போலக் கெடும்.

அணிப்பொருத்தம் :
நெருப்பின் அருகில் வைக்கப்பட்ட வைக்கோல் போரை உவமையாகக் கூறி, பழிவருமுன்னே சிந்தித்து செயல்படுபவன் வாழ்க்கை என்பதை உவமிக்கப்படும் பொருளாக்கி, போல என்ற உவம உருபு பெற்று, வெளிப்படையாய் அமைந்துள்ளதால் இது உவமை அணி ஆயிற்று.

3. பிறிது மொழிதல் அணி :

அணி விளக்கம் :
உவமையை மட்டும் கூறிப் பொருளினைப் பெற வைப்பது பிறிது மொழிதல் அணி.

எ.கா:
கடல் ஓடா கால்வல் நெடுந்தேர் கடல் ஓடும்
நாவாயும் ஓடா நிலத்து.

அணிப்பொருத்தம் :
அவரவர் தமக்குரிய இடங்களிலேயே சிறப்பாகச் செயல்பட முடியும் என்பதே புலவர் கூற விரும்பிய கருத்து. ஆனால் அதனைக் கூறாமல் பெரிய தேர் கடலில் ஓடாது. கடலில் ஓடும் கப்பல் தரையில் ஓடாது என உவமையாக வேறொன்றைக் கூறியுள்ளார். ஆகவே இது பிறிது மொழிதல் அணி ஆயிற்று.

Students can Download Maths Chapter 1 Numbers Ex 1.4 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths 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 8th Maths Solutions Term 3 Chapter 1 Numbers Ex 1.4

Samacheer Kalvi 8th Maths Book Solutions Exercise 1.4 Question 1.
Fill in the blanks:
(i) (-1 )^{even integer} is…………
(ii) For a ≠ 0, a° is……….
(iii) 4^-3 × 5^-3 =…………
(iv) (-2)⁷ =………….
(v) \((-\frac{1}{3})^{-5}\) =…………
Solution:
(i) 1
(ii) 1
(iii) 20^-3
(iv) \(\frac{-1}{128}\)
(v) -243

Samacheer Kalvi 8th Maths Solutions Term 3 Pdf Question 2.
Say True or False:
(i) If 8^x = \(\frac{1}{64}\), the value of x is -2
(ii) The simplified form of (256)^-1/4 is \(\frac{1}{4}\)
(iii) The standard form of 2 x 10^-4 is 0.0002.
(iv) The scientific form of 123.456 is 1.23456 x 10^-2.
(v) The multiplicative inverse of (3)^-7 is 3⁷.
Solution:
(i) True
(ii) True
(iii) True
(iv) False
(v) True

8th Maths Exercise 1.4 Samacheer Kalvi Question 3.
Evaluate:
(i) \(\left(\frac{1}{2}\right)^{3}\)
Solution:
\(\left(\frac{1}{2}\right)^{3}=\frac{1^{3}}{2^{3}}=\frac{1}{2 \times 2 \times 2}=\frac{1}{8}\)

(ii) \(\left(\frac{1}{2}\right)^{-5}\)
Solution:
\(\left(\frac{1}{2}\right)^{-5}=\frac{1^{-5}}{2^{-5}}=\frac{1}{2^{-5}}=2^{5}\) = 2 × 2 × 2 × 2 × 2 = 32

(iii) (-3)^-3
Solution:
(-3)^-3 = \(\frac{1}{(-3)^{3}}=\frac{1}{-3 \times-3 \times-3}=\frac{1}{-27}=\frac{-1}{27}\)

(iv) (-3)⁴
Solution:
(-3)⁴ = -3 × -3 × -3 × -3 = 81

(v) \(\left(\frac{-5}{6}\right)^{-3}\)
Solution:
\(\left(\frac{-5}{6}\right)^{-3}=\frac{(-5)^{-3}}{6^{-3}}=\frac{6^{3}}{(-5)^{3}}=\frac{6 \times 6 \times 6}{-5 \times-5 \times-5}=-\frac{216}{125}\)

(vi) \(\left( { 2 }^{ -5 }\div { 2 }^{ 7 } \right) \times { 2 }^{ -2 }\)
Solution:
\(\left( { 2 }^{ -5 }\div { 2 }^{ 7 } \right) \times { 2 }^{ -2 }\) = \(\left( { 2 }^{ -5-7 }\right)\times { 2 }^{ -2 }\)
\(2^{-12} \times 2^{-2}=2^{-12+(-2)}=2^{-14}\)

(vii) \(\left( { 2 }^{ -1 }\times { 3 }^{ -1 } \right) \div 6^{ -2 }\)
Solution:
\(\left( { 2 }^{ -1 }\times { 3 }^{ -1 } \right) \div 6^{ -2 }\) = \((2 \times { 3})^{ -1 } \div 6^{ -2 }\)
= \((6)^{-1} \div 6^{-2}=6^{(-1)-(-2)}=6^{1}=6\)

(viii) \(\left(-\frac{1}{3}\right)^{-2}\)
Solution:
\(\left(-\frac{1}{3}\right)^{-2}=\left(-\frac{3}{1}\right)^{2}=\frac{(-3)^{2}}{1^{2}}=\frac{-3 \times-3}{1}\) = 9

Samacheer Kalvi 8th Maths Book Solutions Question 4.
Evaluate:
(i) \(\left(\frac{2}{5}\right)^{4} \times\left(\frac{2}{5}\right)^{2}\)
Solution:
\(\left(\frac{2}{5}\right)^{4} \times\left(\frac{2}{5}\right)^{2}\) = \(\left(\frac{2}{5}\right)^{4+2}=\left(\frac{2}{5}\right)^{6}\)

(ii) \(\left(\frac{4}{5}\right)^{-2} \times\left(\frac{4}{5}\right)^{-3}\)
Solution:
\(\left(\frac{4}{5}\right)^{-2} \times\left(\frac{4}{5}\right)^{-3}\) = \(\left(\frac{4}{5}\right)^{-2+(-3)}=\left(\frac{4}{5}\right)^{-5}\)

(iii) \(\left(\frac{1}{2}\right)^{-3} \times\left(\frac{1}{2}\right)^{7}\)
\(\left(\frac{1}{2}\right)^{-3} \times\left(\frac{1}{2}\right)^{7}\) = \(\left(\frac{1}{2}\right)^{-3+7}=\left(\frac{1}{2}\right)^{4}\)

Samacheer Kalvi Guru 8th Maths Question 5.
Evaluate:
(i) \(\left( { 5 }^{ 0 }+{ 6 }^{ -1 } \right) \times { 3 }^{ 3 }\)
Solution:

(ii) \(\left( 2^{ -1 }\times { 3 }^{ -1 } \right) \div { 6 }^{ -1 }\)
Solution:
\(\left( 2^{ -1 }\times { 3 }^{ -1 } \right) \div { 6 }^{ -1 }\) = \((2 \times 3)^{-1} \div 6^{-1}=6^{-1}+6^{-1}=6^{-1-(-1)}=6^{0}\) = 1

(iii) \(\left( 3^{ -1 }+{ 4 }^{ -2 }+{ 5 }^{ -3 } \right) ^{ 0 }\)
Solution:
\(\left( 3^{ -1 }+{ 4 }^{ -2 }+{ 5 }^{ -3 } \right) ^{ 0 }\) = 1 [ ∴ a° = 1 where a ≠ 0]

Samacheer Kalvi Guru 8th Maths Book Solutions Question 6.
Simplify
(i) \(\left(3^{2}\right)^{3} \times\left(2 \times 3^{5}\right)^{-2} \times(18)^{2}\)
Solution:

Samacheer Kalvi.Guru 8th Maths Question 7.
Solve for x.
(i) \(\frac{10^{x}}{10^{-3}}=10^{9}\)
Solution:
\(\frac{10^{x}}{10^{-3}}=10^{9}\)
\({10^{x+3}}=10^{9}\)
Equating the powers of same base 10
x + 3 = 9
x + 3 – 3 = 9 – 3
x = 6

(ii) \(\frac{2^{2x-1}}{2^{x+2}}\) = 4
Solution:
\(2^{2x-1-1(x+2)}\) = 2²
\(2^{2x-1-x-2}\) = 2²
\(2^{x-3}\) = 2²
Equating the powers of the same base 2.
x – 3 = 2
x – 3 + 3 = 2 + 3
x = 5

(iii) \(\frac{5^{5} \times 5^{-4} \times 5^{x}}{5^{12}}=5^{-5}\)
Solution:
\(\frac{5^{5} \times 5^{-4} \times 5^{x}}{5^{12}}=5^{-5}\) = \(5^{-5} \Rightarrow \frac{5^{5-4+x}}{5^{12}}=5^{-5}\)
\(\Rightarrow \frac{5^{1+x}}{5^{12}}\) = 5^-5
⇒ \(5^{1+x-12}\) = 5^-5
⇒ \(5^{x-11}\) = 5^-5
Equating the powers of same base 5.
x – 11 = -5
x – 11 + 11 = -5 + 11
x = 6

Samacheer Kalvi Guru 8th Maths Guide Question 8.
Expand using exponents:
(i) 6054.321
(ii) 897.14
Solution:

Samacheerkalvi.Guru 8th Maths Question 9.
Find the number is standard form:
(i) 8 x 10⁴ + 7 x 10³ + 6 x 10² + 5 x 10¹+ 2 x 1 + 4 x 10^-2 + 7 x 10^-4
Solution:
8 x 10⁴ + 7 x 10³ + 6 x 10² + 5 x 10¹ + 2 x 1 + 4 x 10^-2 + 7 x 10^-4
= 8 x 10000 + 7 x 1000 + 6 x 100 + 5 x 10 + 2 x 1 + 4 x \(\frac{1}{100}\) + 7 x \(\frac{1}{10000}\)
= 80000 + 7000 + 600 + 50 + 2 + \(\frac{4}{100}\) + \(\frac{7}{10000}\)
= 87652.0407

(ii) 5 x 10³ + 5 x 10¹ + 5 x 10^-1 + 5 x 10^-3
Solution:
5 x 10³ + 5 x 10¹ + 5 x 10^-1 + 5 x 10^-3
= 5 x 1000 + 5 x 10 + 5 x \(\frac{1}{10}\) + 5 x \(\frac{1}{1000}\)
= 5000 + 50 + \(\frac{5}{10}\) + \(\frac{5}{1000}\) = 5050.505

8th Maths Exercise 1.4 Question 10.
The radius of a hydrogen atom is 2.5 x 10 11 m. Express this number in standard notation.
Solution:
Radius of a hydrogen atom = 2.5 x 10¹¹ m
= 2.5 x \(\frac{1}{10^{11}}\) m = \(\frac{2.5}{10^{11}}\) m = 0.000000000025 m

Samacheer Kalvi Guru 8 Maths Question 11.
Write each number in scientific notation:
(i) 467800000000
(ii) 0.000001972
Solution:

Samacheer Kalvi 8th Maths Solutions Question 12.
Write in scientific notation:
(i) Earth’s volume is about 1,083,000,000,000 cubic kilometers.
Solution:

Earth’s volume = 1.083 x 10¹² cubic kilometers.

(ii) If you fill a bucket with dirt, the portion of the whole Earth that is in the bucket will be 0.0000000000000000000000016 kg.
Solution:
Portion of earth in the bucket =  kg = 1.6 x 10^-24 kg.

Objective Type Questions

Samacheer Kalvi 8th Maths Book Solution Question 13.
By what number should (-4)^-1 be multiplied so that the product becomes 10^-1?
(a) \(\frac{2}{3}\)
(b) \(\frac{-2}{5}\)
(c) \(\frac{5}{2}\)
(d) \(\frac{-5}{2}\)
Solution:
(b) \(\frac{-2}{5}\)
Hint:

Question 14.
0.0000000002020 in scientific form is
(a) 2.02 x 10⁹
(b) 2.02 x 10^-9
(c) 2.02 x 10^-8
(d) 2.02 x 10^-10
Solution:
(d) 2.02 x 10^-10
Hint:

Question 15.
(-2)^-3 x (-2)^-2 is
(a) \(\frac{-1}{32}\)
(b) \(\frac{1}{32}\)
(c) 32
(d) -32
Solution:
(a) \(\frac{-1}{32}\)

Question 16.
Which is not correct?
(a) \(\left(\frac{-1}{4}\right)^{2}\) = 4^-2
(b) \(\left(\frac{-1}{4}\right)^{2}\) = \(\left(\frac{1}{2}\right)^{4}\)
(c) \(\left(\frac{-1}{4}\right)^{2}\) = 16^-1
(d) \(-\left(\frac{1}{4}\right)^{2}\) = 16^-1
Solution:
(d) \(-\left(\frac{1}{4}\right)^{2}\) = 16^-1
Hint:
\((-2)-3 x(-2)-2=(-2)-3-2=(-2)-5\left(-\frac{1}{2}\right) 5=-\frac{1}{32}\)

Question 17.
If \(\left(\frac{p}{q}\right)^{1-3 x}=\left(\frac{q}{p}\right)^{\frac{1}{2}}\), then x is
(a) 4^-1
(b) 3^-1
(c) 2^-1
(d) 1^-1
Solution:
(c) 2^-1
Hint:

You can Download Samacheer Kalvi 9th 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 9th Maths Solutions Chapter 6 Trigonometry Ex 6.2

9th Maths Exercise 6.2 Samacheer Kalvi Question 1.
Verify the following equalities:
(i) sin² 60° + cos² 60° = 1
(ii) 1 + tan² 30° = sec² 30°
(iii) cos 90° = 1 – 2 sin² 45° = 2 cos² 45° – 1
(iv) sin 30° cos 60° + cos 30° sin 60° = sin 90

Solution:

9th Maths Exercise 6.2 Question 2.
Find the value of the following:

Solution:

9th Maths Trigonometry Exercise 6.2 Question 3.
Verify cos 3A = 4 cos³A – 3 cos A, when A = 30°
Solution:
L.H.S = cos 3A = cos 3(30°) = cos 90° = 0
R.H.S = 4 cos³A – 3 cos A = 4 cos³30° – cos 30°

(1) = (2). Hence it is verified.

9th Maths 6.2 Question 4.
Find the value of 8 sin 2x cos 4x sin 6x, when x = 15°
Solution:
8 sin 2(15°).cos 4(15°).sin 6(15°)

You can Download Samacheer Kalvi 10th 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 10th Maths Solutions Chapter 3 Algebra Ex 3.10

10th Maths Exercise 3.10 Samacheer Kalvi Question 1.
Solve the following quadratic equations by factorization method
(i) 4x² – 7x – 2 = 0
(ii) 3(p² – 6) = p(p + 5)
(iii) \(\sqrt{a(a-7)}=3 \sqrt{2}\)
(iv) \(\sqrt{2} x^{2}+7 x+5 \sqrt{2}=0\)
(v) 2x² – x + \(\frac{1}{8}\) = 0
Solution:

(ii) 3(p² – 6) = p(p + 5)
3p² – 18 = p² + 5p ⇒ 39² – 5p – 18 = 0
⇒ 2p² – 5p – 18 = 0
⇒ (2p – 9) (p + 2) = 0 ⇒ p = \(\frac{9}{2}\), -2

(iii) \(\sqrt{a(a-7)}=3 \sqrt{2}\)
Squaring on both sides
a(a – 7) = 9 × 2
a² – 7a – 18 = 0
a² – 9a + 2a – 18 = 0
a(a – 9) + 2(a – 9) = 0
(a – 9) (a + 2) = 0
⇒ a = 9, a = -2

(v) 2x² – x + \(\frac{1}{8}\) = 0
16x² – 8x + 1 = 0
16x² – 4x – 4x + 1 = 0
4x(4x – 1) – 1(4x – 1) = 0
(4x – 1) (4x – 1) = 0
⇒ x = \(\frac{1}{4}, \frac{1}{4}\)

Ex 3.10 Class 10 Samacheer Question 2.
The number of volleyball games that must be scheduled in a league with n teams is given by G(n) = \(\frac{n^{2}-n}{2}\) where each team plays with every other team exactly once. A league schedules 15 games. How many teams are in the league?
Answer:
Number of games = 15
G(n) = \(\frac{n^{2}-n}{2}\)
\(\frac{n^{2}-n}{2}\) = 15
n² – n = 30 ⇒ n² – n – 30 = 0
⇒ n²– 6n – 5n – 30 = 0
(n – 6) (n + 5) = 0
n – 6 = 0 or n + 5 = 0
[Note: – 5 is neglected because number of team is not negative]
n = 6 or n = -5
∴ Number of teams = 6

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Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 7 Thermodynamics

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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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Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 1 Living World

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Samacheer Kalvi 11th Bio Botany Living World Text Book Back Questions and Answers

Choose the correct answer
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.

Samacheer Kalvi 11th Botany Question 10.
Give a general account on lichens.
Answer:
The symbiotic association between algae and fungi is called lichens. The algal partner is called Phycobiont and the fungal partner is called Mycobiont. Algae provide nutrition for fungal partner and also help in fixing the thallus to the substratum through rhizinae. Asexual reproduction takes place through fragmentation, Soredia and Isidia. Phycobionst reproduce by akinetes, hormogonia, aplanospore, etc. Mycobionts undergo sexual reproduction and produce ascocarps.
Classification:

1. Based on the habitat lichens are classified into following types: Corticolous (on bark) Lignicolous (on wood) Saxicolous (on rocks) Terricolous (on ground) Marine (on siliceous rocks of sea) and Fresh water (on siliceous rock of fresh water).
2. On the basis of morphology of the thallus they are divided into Leprose (a distinct fungal layer is absent) Crustose – crust like; Foliose – leaf like; Fruticose – branched pendulous shrub like.
3. The distribution of algal cells distinguishes lichens into two forms namely Homoiomerous (Algal cells evenly distributed in the thallus) and Fleteromerous (a distinct layer of algae and fungi present).
4. If the fungal partner of lichen belongs to ascomycetes, it is called Ascolichen and if it is basidiomycetes it is called Basidiolichen.

Textbook Activity Solved

Get a button mushroom. Draw diagram of the fruit body. Take a thin longitudinal section passing through the gill and observe the section under a microscope. Record your observations.

Samacheer Kalvi 11th Bio Botany Living World Additional Questions and Answers

I. Multiple Choice Questions
Choose the correct answer
11th Botany Solutions Question 1.
Earth was formed around billion years ago …………… .
(a) 3.3
(b) 5.6
(c) 4.6
(d) 5.9
Answer:
(c) 4.6

11th Bio Botany Samacheer Kalvi Question 2.
According to Mora et al., in 2011, the number of estimated species on Earth is …………… .
(a) 8.7 million
(b) 9.7 million
(c) 7.7 million
(d) 9.7 billion
Answer:
(a) 8.7 million

Question 3.
Which of the following is NOT a prokaryote?
(a) Bacteria
(b) Blue green algae
(c) Oedogonium
(d) Nostoc
Answer:
(c) Oedogonium

Question 4.
Which of the following organism undergoes regeneration?
(a) Spirogyra
(b) Planaria
(c) Yeast
(d) Aspergillus
Answer:
(b) Planaria

Question 5.
Vaccination for small pox was discovered by …………… .
(a) W.M. Stanley
(b) Adolf Mayer
(c) Robert Koch
(d) Edward Jenner
Answer:
(d) Edward Jenner

Question 6.
Who coined the term ‘Bacteriophage’?
(a) F.W. Twort
(b) d’Herelle
(c) Ivanowsky
(d) Robert Gallo
Answer:
(b) d’Herelle

Question 7.
The size of TMV is …………… .
(a) 300 × 20 mm
(b) 300 × 200 µm
(c) 300 × 20 nm
(d) 300 × 20 Å
Answer:
(c) 300 × 20 nm

Question 8.
One nanometer equals to metres …………… .
(a) 10^-9
(b) 10^-6
(c) 10^-5
(d) 10^-12
Answer:
(a) 10^-9

Question 9.
Which is a non – living character of viruses?
(a) Undergoes mutation
(b) Host – specific
(c) Crystallized
(d) Irritability
Answer:
(c) Crystallized

Question 10.
According to David Baltimore, the viruses are classified …………… into classes.
(a) 6
(b) 5
(c) 7
(d) 8
Answer:
(c) 7

Question 11.
Identify the criteria not used in classifying viruses by Baltimore …………… .
(a) ss (or) ds
(b) use of RT
(c) capsid
(d) sense or antisense
Answer:
(c) capsid

Question 12.
Viruses with dsRNA is …………… .
(a) Toga viruses
(b) Retro viruses
(c) Reo viruses
(d) Rhabdo viruses
Answer:
(c) Reo viruses

Question 13.
Which of the plant virus contains DNA as genome?
(a) Tobacco mosaic virus
(b) Cauliflower mosaic virus
(c) Sugarcane mosaic virus
(d) Cucumber mosaic virus
Answer:
(b) Cauliflower mosaic virus

Question 14.
Parvo viruses have …………… .
(a) ssDNA
(b) dsDNA
(c) ssRNA
(d) dsRNA
Answer:
(a) ssDNA

Question 15.
Molecular weight of TMV is dalton …………… .
(a) 39 × 10⁶
(b) 39 × 10^-6
(c) 39 × 10⁹
(d) 39 × 10^-9
Answer:
(a) 39 × 10⁶

Question 16.
Approximate number of capsomeres in TMV is …………… .
(a) 3120
(b) 1203
(c) 2130
(d) 3021
Answer:
(c) 2130

Question 17.
The empty protein coat left outside after penetration is …………… .
(a) host
(b) ghost
(c) capsid
(d) capsomeres
Answer:
(b) ghost

Question 18.
The genome of viroid is …………… .
(a) linear ssRNA
(b) dumb – bell shaped ssRNA
(c) circular ssRNA
(d) linear dsRNA
Answer:
(c) circular ssRNA

Question 19.
Viroids were discovered by …………… .
(a) Ivanowsky
(b) Robert Gallo
(c) Diener
(d) d’Herelle
Answer:
(c) Diener

Question 20.
Mad cow disease is caused by …………… .
(a) viroids
(b) virusoids
(c) prions
(d) viruses
Answer:
(c) prions

Question 21.
Match the Following:

(a) 1- d, 2 – b, 3 – a, 4 – c
(b) 1 – c, 2 – d, 3 – a, 4 – b
(c) 1 – c, 2 – a, 3 – b, 4 – d
(d) 1 – d, 2 – a, 3 – b, 4 – c
Answer:
(c) 1- c, 2 – a, 3 – b, 4 – d

Question 22.
Identify the correct sequence regarding lytic cycle of viruses …………… .
(A) Penetration
(B) Adsorption
(C) Assembly
(D) Synthesis

(a) BADC
(b) CABD
(c) BDAC
(d) ADBC
Answer:
(a) BADC

Question 23.
Mycophages infect …………… .
(a) blue green algae
(b) bacteria
(c) fungi
(d) cyanobacteria
Answer:
(c) fungi

Question 24.
Rice tungro is caused by …………… .
(a) fungi
(b) bacteria
(c) mycoplasma
(d) viruses
Answer:
(d) viruses

Question 25.
Father of Botany …………… .
(a) Aristotle
(b) Theophrastus
(c) Lederberg
(d) Whittaker
Answer:
(b) Theophrastus

Question 26.
Three kingdom classification was proposed by …………… .
(a) Copeland
(b) Theophrastus
(c) Linnaeus
(d) Haeckel
Answer:
(d) Haeckel

Question 27.
Which Is not a part of five kindgom classification?
(a) Viruses
(b) Monera
(c) Protista
(d) Mycoplasma
Answer:
(a) Viruses

Question 28.
Six kingdom classification was proposed by …………… .
(a) Haeckel
(b) Copeland
(c) Woese
(d) Cavalier – Smith
Answer:
(d) Cavalier – Smith

Question 29.
Ruggerlo et al., in 2015 proposed …………… kingdom classification.
(a) 5
(b) 6
(c) 7
(d) 8
Answer:
(c) 7

Question 30.
…………… is a new kingdom in seven kingdom classification.
(a) Eubactena
(b) Plantae
(c) Chromista
(d) Archaebacteria
Answer:
(c) Chromista

Question 31.
Actinomycetes comes under …………… kindgom.
(a) fungi
(b) chromista
(c) monera
(d) protista
Answer:
(c) monera

Question 32.
The sourness of curd is due to …………… .
(a) acetic acid
(b) galactic acid
(c) lactic acid
(d) lactone
Answer:
(c) lactic acid

Question 33.
Who is the founder of Modern Bacteriology …………… ?
(a) Aristotle
(b) Robert Koch
(c) Pasteur
(d) Linnaeus
Answer:
(b) Robert Koch

Question 34.
The term bacterium was coined by …………… .
(a) Stanley
(b) Ehrenberg
(c) Gram
(d) Koch
Answer:
(b) Ehrenberg

Question 35.
Plasmids were discovered by …………… .
(a) Ehrenberg
(b) H.Bergy
(c) Joshua Lederberg
(d) Koch
Answer:
(c) Joshua Lederberg

Question 36.
Genophore is seen in …………… .
(a) Amoeba
(b) Cyanobacteria
(c) Chlamydomonas
(d) Euglena
Answer:
(b) Cyanobacteria

Question 37.
Number of domains of life are there according to Carl Woese …………… .
(a) 3
(b) 2
(c) 4
(d) 5
Answer:
(a) 3

Question 38.
Which is not a component of bacterial cell?
(a) Mesosomes
(b) Glycocalyx
(c) Polysomes
(d) Histones
Answer:
(d) Histones

Question 39.
The most abundant polypeptide in bacterial cell wall is …………… .
(a) chitin
(b) amylopectin
(c) porin
(d) pectin
Answer:
(c) porin

Question 40.
Extra chromosomal element in bacterial cells are …………… .
(a) plasmids
(b) mesosomes
(c) histones
(d) genophores
Answer:
(a) plasmids

Question 41.
Bacteriocins are found in …………… .
(a) genophore
(b) plasmids
(c) nucleoids
(d) mesosomes
Answer:
(b) plasmids

Question 42.
Colour revealed by Gram positive bacteria after Gram staining is …………… .
(a) red
(b) indigo
(c) dark violet
(d) blue
Answer:
(c) dark violet

Question 43.
How many number of basal body rings seen in the flagella of Gram negative bacteria?
(a) 2
(b) 9
(c) 4
(d) 1
Answer:
(c) 4

Question 44.
Capnophilic bacteria require for growth …………… .
(a) C₂
(b) CO
(c) CO₂
(d) O₃
Answer:
(c) CO₂

Question 45.
The pigment present in green sulphur bacteria is …………… .
(a) bacterioviridin
(b) bacteriochlorophyll
(c) chlorophyll a
(d) xanthophyll
Answer:
(b) bacteriochlorophyll

Question 46.
The hydrogen donor of purple sulphur bacteria is …………… .
(a) H₂S
(b) thiosulphate
(c) ethanol
(d) acetic acid
Answer:
(b) thiosulphate

Question 47.
Campylobacter is a …………… .
(a) obligate aerobe
(b) obligate anaerobe
(c) capnophilic
(d) aerobe
Answer:
(c) capnophilic

Question 48.
Mycobacterium is a …………… .
(a) parasite
(b) symbiont
(c) saprophyte
(d) free – living
Answer:
(a) parasite

Question 49.
Which is the most common mode of asexual reproduction in bacteria?
(a) Endospore formation
(b) Fission
(c) Budding
(d) Conidia
Answer:
(b) Fission

Question 50.
…………… are thick walled resting spores.
(a) Aplanospores
(b) Endospores
(c) Conidia
(d) Zoospores
Answer:
(b) Endospores

Question 51.
In which of the following method genetic recombination does not occur?
(a) Generalised transduction
(b) Conjugation
(c) Transformation
(d) Fission
Answer:
(d) Fission

Question 52.
During conjugation in bacteria, which of the following is transferred from donor to recipient cell?
(a) R factor
(b) F factor
(c) Ti factor
(d) Ri factor
Answer:
(b) F factor

Question 53.
Griffith used …………… for his experiment.
(a) rat
(b) rabbit
(c) mice
(d) monkey
Answer:
(c) mice

Question 54.
Transformation in bacteria was demonstrated by …………… .
(a) Lederberg
(b) Zinder
(c) Edward
(d) Griffith
Answer:
(d) Griffith

Question 55.
Lederberg studied transduction in bacterium …………… .
(a) Diplococcus pneumoniae
(b) Streptococcus
(c) Salmonella typhi
(d) Escherichia coil
Answer:
(c) Salmonella typhi

Question 56.
Bacteria used in the curing of tea is …………… .
(a) Mycococcus candisans
(b) Eseherichia coli
(c) Acetobacter aceti
(d) Streptococcus lactis
Answer:
(a) Mycococcus candisans

Question 57.
Syphilis k caused by …………… .
(a) Mycococcus candisans
(b) Treponema pallidum
(c) Yersinia pestis
(d) Mycohacterium leprae
Answer:
(b) Treponema pallidum

Question 58.
Methanobacterium is …………… .
(a) Cyanobacteria
(b) Malobacteria
(c) Eubacteria
(d) Archaebacteria
Answer:
(d) Archaebacteria

Question 59.
…………… is NOT a phycobiont in lichens.
(a) Gloeocapsa
(b) Dermacarpa
(c) Scytonema
(d) Nostoc
Answer:
(b) Dermacarpa

Question 60.
Red sea is red colour due to …………… .
(a) Dermacarpa sps.
(b) Trichodesmium sps.
(c) Scytonema sps.
(d) Gloeocapsa sps.
Answer:
(b) Trichodesmium sps.

Question 61.
Filamentous trichome is the plant body of …………… .
(a) Chroococcus
(b) Gloeocapsa
(c) Nostoc
(d) Oscillatoria
Answer:
(c) Nostoc

Question 62.
Stromatolites are the colonies of cyanobacteria bind with …………… .
(a) calcium carbonate
(b) calcium hydroxide
(c) magnesium sulphate
(d) calcium silicate
Answer:
(a) calcium carbonate

Question 63.
…………… sps. is an endophyte in coralloid roots of Cycas.
(a) Gioeocapsa
(b) Scytonerna
(c) Nostoc
(d) Azolla
Answer:
(c) Nosloc

Question 64.
Myxophyceae refers to …………… .
(a) Algae
(b) Fungi
(c) Archaebacteria
(d) Cyanobacteria
Answer:
(d) Cyanobacteria

Question 65.
…………… is used in single cell protein.
(a) Spirulina
(b) Azolla
(c) Dermacarpa
(d) Nostoc
Answer:
(a) Spirulina

Question 66.
…………… is a pleomorphic organism.
(a) Fungi
(b) Mycoplasma
(c) Bacteria
(d) Algae
Answer:
(b) Mycoplasma

Question 67.
Pleuropneumonia is caused by …………… .
(a) bacteria
(b) fungi
(c) mycoplasma
(d) viruses
Answer:
(c) mycoplasma

Question 68.
…………… is also called as Ray fungi.
(a) Basidiomycetes
(b) Ascomycetes
(c) Actinomycetes
(d) Deuteromycetes
Answer:
(c) Actinomycetes

Question 69.
Earthy odour of soil after rain is due to …………… .
(a) Basidiomycetes
(b) Ascomycetes
(c) Actinomycetes
(d) Deuteromycetes
Answer:
(c) Actinomycetes

Question 70.
Viruses that attack blue green algae are called as …………… .
(a) Mycophages
(b) Phycophages
(c) Cyanophages
(d) Bacteriophages
Answer:
(c) Cyanophages

Question 71.
Cell membrane of Archaebacteria has …………… .
(a) glycine and isopropyl ethers
(b) glycerol and isobutyl ethers
(c) glycerol and isopropyl ethers
(d) cellulose and isobutyl ethers
Answer:
(c) glycerol and isopropyl ethers

Question 72.
Which is a true bacteria?
(a) Halobacterium
(b) Thermoplasma
(c) Methanobacteriurn
(d) Azotobacter
Answer:
(d) Azolobacter

Question 73.
Study of fungus is called as …………… .
(a) phycology
(b) mycology
(c) algology
(d) biology
Answer:
(b) mycology

Question 74.
Who is considered as the founder of mycology?
(a) K.C.Mehta
(b) G.C.Ainsworth
(c) P.A.Micheli
(d) T.S.Sadasivan
Answer:
(c) P.A.Micheli

Question 75.
Asexual phase of fungi is called as …………… .
(a) telomorph
(b) holomorph
(c) metamorph
(d) anamorph
Answer:
(d) anamorph

Question 76.
In which mycelium, the hyphae are arranged loosely?
(a) Prosenchyma
(b) Plectenchyma
(c) Pseudoparenchyrna
(d) Arenchyma
Answer:
(a) Prosenchyma

Question 77.
Number of nuclei in coenocytic mycelium …………… .
(a) 2
(b) many
(c) nil
(d) 9
Answer:
(b) many

Question 78.
Thallospores are produced by …………… .
(a) Aspergillus
(b) Erysiphe
(c) Saccharomyces
(d) Fusarium
Answer:
(b) Erysiphe

Question 79.
In Agaricus, …………… type of sexual reproduction occurs.
(a) spermatization
(b) somatogamy
(c) oogamy
(d) isogamy
Answer:
(b) somatogamy

Question 80.
Albugo belongs to …………… .
(a) oomycetcs
(b) zygomycetes
(c) ascomycetes
(d) deuteromycetes
Answer:
(a) oomycetes

Question 81.
Fungi growing on dung is called as …………… .
(a) Mold fungus
(b) Saprophytes
(c) Capnophilous
(d) Coprophilous
Answer:
(d) Coprophilous

Question 82.
Coprophilous belongs to …………… group.
(a) basidiomycetes
(b) ascomycetes
(c) zygomycetes
(d) oomycetes
Answer:
(c) zygomycetes

Question 83.
Which of the following is a coprophilous fungi?
(a) Albugo
(b) Entomophthora
(c) Rhizopus
(d) Pilobolus
Answer:
(d) Pilobolus

Question 84.
Cup fungus belongs to …………… .
(a) zygomycetes
(b) oomycetes
(c) ascomycetes
(d) actinomycetes
Answer:
(c) ascomycetes

Question 85.
Which group of fungus is called as Sac fungi?
(a) Deuteromycetes
(b) Zygomycetes
(c) Ascomycetes
(d) Oomycetes
Answer:
(c) Ascomycetes

Question 86.
Number of ascospores in an asci is …………… .
(a) 2
(b) 4
(c) 6
(d) 8
Answer:
(d) 8

Question 87.
Shape of perithecium is …………… .
(a) cup shaped
(b) flask shaped
(c) completely closed
(d) open type
Answer:
(b) flask shaped

Question 88.
…………… are called as Club fungi.
(a) Ascomycetes
(b) Zygomycetes
(c) Basidiomycetes
(d) Deuteromycetes
Answer:
(c) Basidiomycetes

Question 89.
Parasexual cycle is observed in …………… .
(a) basidiomycetes
(b) zygomycetes
(c) deuteromycetes
(d) ascomycetes
Answer:
(c) deuteromycetes

Question 90.
Which is called as imperfect fungi?
(a) Basidiomycetes
(b) Zygomycetes
(c) Deuteromycetes
(d) Ascomycetes
Answer:
(c) Deuteromycetes

Question 91.
In basidiomycetes, clamp connections are formed to maintain …………… condition.
(a) monokaryotic
(b) coenocytic
(c) dikaryotic
(d) zygotic
Answer:
(c) dikaryotic

Question 92.
…………… is a single celled fungus used in dairy industry.
(a) Volvariella
(b) Agaricus
(c) Penicillin
(d) Yeast
Answer:
(d) Yeast

Question 93.
Ergot alkaloids are produced by …………… .
(a) Penicillium notatum
(b) Acremonium chrysogenum
(c) Claviceps purpurea
(d) Penicillium griseofulvum
Answer:
(c) Claviceps purpurea

Question 94.
Kojic acid is produced by …………… .
(a) Aspergillus terreus
(b) Aspergillus niger
(c) Aspeigillus oryzae
(d) Agaricus hisporus
Answer:
(c) Aspergillus oryzae

Question 95.
…………… infest dried foods and produce carcinogenic toxin.
(a) Aspergillus flavus
(b) Amanita verna
(c) Amanita phalloides
(d) Rhizopus
Answer:
(a) Aspergillus flavus

Question 96.
Rust of wheat is produced by …………… .
(a) Albugo candida
(b) Puccinia graminis tritici
(c) Candida albicans
(d) Colletotrichum sps
Answer:
(b) Puccinia graminis tritici

Question 97.
VAM is a type of …………… .
(a) Endomycorrhiza
(b) Ectomycorrhiza
(c) Ectendomycorrhiza
(d) Endectomycorrhiza
Answer:
(a) Endomycorrhiza

Question 98.
Algal partner of lichen is …………… .
(a) phycobiont
(b) phytobiont
(c) mycobiont
(d) both (a) & (c)
Answer:
(a) phycobiont

Question 99.
Asexual reproduction by Soredia is seen in …………… .
(a) fungi
(b) lichen
(c) mycorrhiza
(d) algae
Answer:
(b) lichen

Question 100.
Saxicolous lichen grow on …………… .
(a) ground
(b) bark
(c) wood
(d) rock
Answer:
(d) rock

Question 101.
In leprose form of lichen distinct …………… layer is absent.
(a) fungal
(b) algal
(c) both
(d) none
Answer:
(a) fungal

Question 102.
…………… are used as pollution indicators.
(a) Algae
(b) Lichen
(c) Fungi
(d) Mycorrhiza
Answer:
(b) Lichen

Question 103.
…………… acid is obtained from lichen acting as antibiotics.
(a) Alginic
(b) Acetic
(c) Oxalic
(d) Usnic
Answer:
(d) Usnic.

II. Very Short Answer Type Questions

Question 1.
Differentiate plant growth from animal growth.
Answer:
Plant Growth From Animal Growth:

Question 2.
Define Growth.
Answer:
Growth is an intrinsic property of all living organisms through which they can increase cells both in number and mass.

Question 3.
Growth of living thing is an intrinsic property – Justify.
Answer:
Living cells grow by the addition of new protoplasm within the cells. Therefore, growth in living thing is intrinsic.

Question 4.
Define reproduction and Mention its types.
Answer:
Reproduction is the tendency of a living organism to perpetuate its own species. There are two types of reproduction namely asexual and sexual.

Question 5.
What is metabolism? Mention its types.
Answer:
The sum total of all the chemical reactions taking place in a cell of living organism is called metabolism. It is broadly divided into anabolism and catabolism.

Question 6.
What is consciousness and irritability?
Answer:
Animals sense their surroundings by sense organs. This is called consciousness. Respond of plants to the stimuli is called irritability.

Question 7.
List out few attributes of living organisms.
Answer:
The attributes of living organisms are growth, metabolism, movement, reproduction, nutrition, excretion, etc.

Question 8.
Define cyclosis.
Answer:
The movement of cytoplasm inside the cell is called cytoplasmic streaming or cyclosis.

Question 9.
How will you define viruses?
Answer:
Viruses are sub – microscopic, obligate intracellular parasites. They have nucleic acid core surrounded by protein coat.

Question 10.
Mention the size of Bacteriophage and tobacco mosaic virus (TMV).
Answer:
Bacteriophage measures about 10 – 100 nm in size. The size of TMV is 300 × 20 nm.

Question 11.
Classify viruses based on nature of nucleic acid with example.
Answer:
On the basis of nature of nucleic acid viruses are classified into four categories. They are viruses with ssDNA (Parvo viruses), dsDNA (Bacteriophages), ssRNA (TMV) and dsRNA (wound tumour virus).

Question 12.
Distinguish between deoxyviruses and riboviruses.
Answer:
Deoxyviruses and Riboviruses:

1. Deoxyviruses: Viruses having DNA are called deoxyviruses. E.g. Animal viruses except HIV
2. Riboviruses: Viruses having RNA are called riboviruses. E.g. Plant viruses except cauliflower mosaic virus (CMV)

Question 13.
Write the constituents of virions.
Answer:
The virion is made up of two constituents, a protein coat called capsid and a core called nucleic acid.

Question 14.
What are capsomeres?
Answer:
The protein coat of viruses is made up of approximately 2130 identical protein subunits called capsomeres.

Question 15.
Name the two types of phage multiplication.
Answer:
Phages multiply through two different types of life cycle:

1. Lytic or Virulent cycle
2. Lysogenic or Avirulent life cycle

Question 16.
What do you mean by a ‘ghost’ in virology?
Answer:
The empty protein coat left outside by the phage after penetrating the host cell is called as ghost.

Question 17.
What do you understand by “pinning” of phage?
Answer:
Once the contact is established between tail fibres of phase and bacterial cell, tail fibres bend to anchor the pins and base plate to the cell surface. This step is called pinning.

Question 18.
What is prophage?
Answer:
As soon as the phage injects its linear DNA into the host cell, it becomes circular and integrates into the bacterial chromosome by recombination. The integrated phage DNA is now called prophage.

Question 19.
When does a prophage enters lytic cycle?
Answer:
On exposure to UV radiation and chemicals the excision of phage DNA may occur and results in lytic cycle.

Question 20.
Define virion?
Answer:
Virion is an intact infective virus particle which is non – replicating outside a host cell.

Question 21.
What are viroids?
Answer:
Viroid is a circular molecule of ssRNA without a capsid. RNA is of low molecular weight.

Question 22.
Name any two disease caused by viroids.
Answer:
Two Disease:

1. Citrus exocortis
2. Potato spindle tuber disease

Question 23.
What are virusoids?
Answer:
Virusoids are the small circular RNAs which are similar to viroids but they are always linked with larger molecules of the viral RNA.

Question 24.
Who discovered viroids and virusolds?
Answer:
Viroids were discovered by T.O. Diener in 1971. Virusoids were discovered by J.W.Randles in 1981.

Question 25.
Name the causative organism for mad cow disease.
Answer:
Prions are the causative organisms for mad cow disease, Prions are the proteinaceous infectious particles.

Question 26.
What are cyanophages? Who reported it first?
Answer:
Viruses infecting blue green algae are called Cyanophages and are first reported by Safferman and Morris in the year 1963.

Question 27.
Name any two disease caused by Prions.
Answer:
Two Disease:

1. Bovine Spongiform Encephalopathy (BSE) (mad cow disease)
2. Creutzfeldt – Jakob Disease (CJD)

Question 28.
What are mycophages? Who first reported it?
Answer:
Viruses infecting fungi are called mycophages or mycoviruses. Mycophages were first reported by Hollings in 1962.

Question 29.
Expand the following acronyms: (a) SARS and (b) AIDS.
Answer:
Acronyms:
(a) SARS: Severe Acute Respiratory Syndrome
(b) AIDS: Acquired Immuno Deficiency Syndrome

Question 30.
Name the two groups of aninmals according to Aristotle.
Answer:
Two Groups of Aninmals:

1. Enaima – animals with red blood.
2. Anaima – animals without red blood.

Question 31.
Which are the major setbacks of Linnaeus classification?
Answer:
Linnaeus classification faced major setback because prokaryotes and eukaryotes were grouped together. Similarly fungi, heterotrophic organisms were placed along with the photosynthetic plants.

Question 32.
Name the viruses which are employed as potential insecticides?
Answer:
Cytoplasmic polyhedrosis Granulo viruses and Entomopox virus were employed as potential insecticides.

Question 33.
Who proposed five kingdom classification? Mention the five kingdoms.
Answer:
R.H. Whittaker proposed the five kingdom classification. It includes Monera, Protista, Fungi, Plantae and Animalia.

Question 34.
List out the criteria undertaken for Whittaker’s classification.
Answer:
The criteria adopted for the classification include cell structure, thallus organization, mode of nutrition, reproduction and phylogenetic relationship.

Question 35.
Point out the demerits of five kingdom classification.
Answer:
(a) 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.
(b) Viruses were not included in the system.

Question 36.
Who proposed six kingdom classification? Mention the kingdoms.
Answer:
Thomas Cavalier – Smith proposed six kingdom classification.
The kingdom includes:

1. Archaebacteria
2. Eubacteria
3. Protista
4. Fungi
5. Plantae and
6. Animalia.

Question 37.
How milk is changed into curd, if a few drops of curd is added to it? What is the reason for its sourness?
Answer:
The change is brought by Lactobacillus lactis, a bacterium present in the curd. The sourness is due to the formation of lactic acid.

Question 38.
Define bacteria and bacteriology.
Answer:
Bacteria are prokaryotic, unicellular, ubiquitous, microscopic organisms. The study of bacteria is called bacteriology.

Question 39.
What is Porin? How it helps the bacteria?
Answer:
Porin is an abundant polypeptide present in bacterial cell walls. It helps in the diffusion of solutes.

Question 40.
List out the cytoplasmic inclusions of bacterial cell.
Answer:
Glycogen, poly-β-hydroxybutyrate granules, sulphur granules and gas vesicles.

Question 41.
Define Genophore.
Answer:
The bacterial chromosome is a single circular DNA molecule, tightly coiled and is not enclosed in a membrane as in Eukaryotes. This genetic material is called nucleoid or genophore.

Question 42.
Write the chemical composition of bacterial cell wall.
Answer:
The chemical composition of cell wall is rather complex and is made up of peptidoglycan or mucopeptide (N – acetyl glucosamine, N – acetyl muramic acid and peptide chain of 4 or 5 aminoacids).

Question 43.
What are polysomes?
Answer:
During protein synthesis, the ribosomes are held together by mRNA and form the polysomes.

Question 44.
What are Pili?
Answer:
Pili or fimbriae are hair like appendages found on surface of cell wall of gram – negative bacteria.

Question 45.
What are capnophilic bacteria? Give an example.
Answer:
Bacteria which require CO₂ for their growth are called as capnophilic bacteria.
Example: Campylobacter.

Question 46.
Distinguish between Photolithotrophs and Photoorganotrophs.
Answer:
Between Photolithotrophs and Photoorganotrophs:

1. Photolithotrophs: In photolithotrophs, the hydrogen donor is an inorganic substance. E.g. Chlorobium
2. Photoorganotrophs: In Photoorganotrophs, the hydrogen donor is an organic acid or alcohol. E.g. Rhodospirillum

Question 47.
Name the hydrogen donor of green sulphur bacteria and purple sulphur bacteria.
Answer:
Hydrogen donor of green sulphur bacteria is H₂S. Hydrogen donor of purple sulphur bacteria is thiosulphate.

Question 48.
Name the bacterial pigment of green sulphur bacteria and purple sulphur bacteria.
Answer:
Bacteria’s:

1. Green sulphur bacteria – Bacterioviridin
2. Purple sulphur bacteria – Bacteriochlorophyll

Question 49.
What are endospores?
Answer:
Endospores are thick walled resting spores developed by bacteria during unfavourable condition.
E.g. Clostridium tetani produces endospores.

Question 50.
Mention the various ways by which genetic recombination occurs.
Answer:
Genetic recombination in bacteria occurs by conjugation, transduction and transformation.

Question 51.
Name the eminent persons who demonstrated the conjugation process.
Answer:
J. Lederberg and Edward L. Tatum.

Question 52.
What is transformation? Name the bacteriologist who described it.
Answer:
The Bacteriologist Who Described it:

1. Transfer of DNA from one bacterium to another is called transformation.
2. Fredrick Griffith demonstrated the transformation process.

Question 53.
Which organism and bacterial species was used in Griffith’s transformation experiment?
Answer:
Mice and Diplococcus pneumoniae.

Question 54.
List out the asexual modes of reproduction of bacteria.
Answer:
Asexual reproduction in bacteria includes binary fission, conidia formation and endospore formation.

Question 55.
Who discovered transduction? Define it.
Answer:
Zinder and Lederberg (1952) discovered transduction in Salmonella typhimurum. Phage mediated DNA transfer is called transduction.

Question 56.
Name any two bacterial species and the antibiotic produced by them.
Answer:

Question 57.
How bacteria helps in vinegar production?
Answer:
Acetobacter aceti bacteria oxidises ethanol obtained from molasses by fermentation to form vinegar.

Question 58.
Name any two ammonifying bacteria.
Answer:
Two Ammonifying Bacteria:

1. Bacillus ramosus and
2. Bacillus mycoides.

Question 59.
What do you mean by retting of fibres?
Answer:
The fibres from the fibre yielding plants are separated by the action of Closiridium is called retting of fibres.

Question 60.
Name any two plant disease caused by the bacteria and mention the host.
Answer:

Question 61.
Name any four animal disease caused by bacteria.
Answer:
Four Animal Disease:

1. Anthrax
2. Brucellosis
3. Bovine tuberculosis and
4. black leg.

Question 62.
Name any four human disease caused by bacteria.
Answer:
Four Human Disease:

1. Cholera
2. Typhoid
3. Tuberculosis and
4. Leprosy.

Question 63.
What are Archaebacteria?
Answer:
Archaebacteria are primitive prokaryotes and are adapted to thrive in extreme environments like hot springs, high salinity and low pH.
E.g., Thermoplasma.

Question 64.
How stromatolites are formed?
Answer:
Stromatolites are deposits formed when colonies of cyanobacteria bind with calcium carbonate.

Question 65.
What is the reason for the colour of Red Sea?
Answer:
A cyanobacteria called Trichodesmium erythraeum imparts red colour to sea.

Question 66.
Mention the cyanobacteria leading endophytic relation with Cycas roots.
Answer:
Nostoc and Anabaena.

Question 67.
Define Cyanobacteria.
Answer:
Cyanobacteria are popularly called as ‘Blue green algae’ or ‘Cyanophyceae’. They are photosynthetic, prokaryotic organisms. Cyanobacteria are primitive forms and are found in different habitats.

Question 68.
Blue green algae can also be called as Myxophyceae. How?
Answer:
The presence of mucilage around the thallus is characteristic feature of cyanobacteria group. Therefore, this group is also called Myxophyceae.

Question 69.
Define mycoplasma?
Answer:
The mycoplasma are very small (0.1 – 0.5 μm), pleomorphic gram negative microorganisms.

Question 70.
Name few plant disease caused by mycoplasma.
Answer:
Little leaf of brinjal, witches broom of legumes, phyllody of cloves and sandal spike are some plant diseases caused by mycoplasma.

Question 71.
Draw and label the structure of mycoplasma.
Answer:
The Structure of Mycoplasma:

Question 72.
What is the reason behind the earthy odour after raining?
Answer:
Streptomyces is a mycelial forming Actinobacteria which lives in soil, they impart “earthy odour” to soil after rain which is due to the presence of geosmines (volatile organic compound).

Question 73.
When and by whom the penicillin was discovered?
Answer:
Penicillin was discovered by Alexander Flemming in 1928.

Question 74.
Define Fungi.
Answer:
Fungi are ubiquitous, eukaryotic, achlorophyllous heterotrophic organisms. They exist in unicellular or multicellular forms.

Question 75.
Define mycology. Who is the founder of mycology?
Answer:
Study of fungi is called mycology. P.A. Micheli is considered as the founder of mycology.

Question 76.
Name few eminent Mycologists.
Answer:
John Webster, G.C. Ainsworth, T.S. Sadasivan and C.V. Subramanian.

Question 77.
With example define coenocytic mycelium.
Answer:
In lower fungi the hypha is aseptate, multinucleate and is known as coenocytic mycelium (Example: Albugo).

Question 78.
What is plectenchyma? Mention its types.
Answer:
The mycelium is organised into loosely or compactly interwoven fungal tissues called plectenchyma. It is further divided into two types: prosenchyma and pseudoparenchyma.

Question 79.
Distinguish between Anamorph and Telomorph.
Answer:
Between Anamorph and Telomorph:

Question 80.
What is holomorph?
Answer:
Fungi showing both sexual and asexual phases are called holomorph.

Question 81.
What is planogametic copulation? Mention its types.
Answer:
Fusion of motile gamete is called planogametic copulation.
Types:

1. Isogamy
2. Anisogamy and
3. Oogamy.

Question 82.
List out the asexual spores produced by fungus.
Answer:
Zoospores, conidia, oidia and chlamydospores.

Question 83.
What are coprophilous fungi? Give an example.
Answer:
Fungi growing on dung are called coprophilous fungi.
Example: Pilobolus.

Question 84.
Ascomycetes are called sac fungi. Give reason.
Answer:
In ascomycetes the ascospores are found inside a bag like structure called ascus. Due to the presence of ascus, this group is popularly called “Sac fungi”.

Question 85.
Name the four types of ascocarps produced by ascomycetes.
Answer:
Four Types Of Ascocarps Produced By Ascomycetes:

1. Cleistothecium
2. Perithecium
3. Apothecium and
4. Pseudothecium.

Question 86.
Basidiomycetes are called club fungi. Give reason.
Answer:
In basidiomycetes the basidium is club shaped with four basidiospores, thus this group of fungi is popularly called “Club fungi”. The fruit body formed is called Basidiocarp.

Question 87.
Name the special structures in deuteromycetes that produces conidia.
Answer:
Pycnidium, acervulus, sporodochium and synnemata.

Question 88.
Deuteromycetes are imperfect fungi – Justify.
Answer:
The fungi belonging to deuteromycetes lack sexual reproduction and are called imperfect fungi.

Question 89.
List out the antibiotics produced by fungi.
Answer:
Penicillin, cephalosporins and griseofiilvin.

Question 90.
Name some toxins produced by Fungus.
Answer:
Alfatoxin, Patulin and Ochratoxin – A.

Question 91.
Name 2 fungal species employed as Biopesticides.
Answer:
Two Fungal Species Employed As Biopesticides:

1. Beauveria bassiana and
2. Metarhizium anisopliae.

Question 92.
Name few fungal diseases in plants.
Answer:
Blast of paddy, rust of wheat, red rot of sugarcane and white rust of crucifers.

Question 93.
Name few fungal diseases in Humans?
Answer:
Fungal Diseases in Humans:

Question 94.
What is mycorrhiza?
Answer:
The symbiotic association between fungal mycelium and roots of plants is called as mycorrhiza.