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1. (JEE Main 2023 (Online) 10th April Morning Shift )

Match List I with List II :

List I List II
(A) 3 Translational degrees of freedom (I) Monoatomic gases
(B) 3 Translational, 2 rotational degrees of freedoms (II) Polyatomic gases
(C) 3 Translational, 2 rotational and 1 vibrational degrees of freedom (III) Rigid diatomic gases
(D) 3 Translational, 3 rotational and more than one vibrational degrees of freedom (IV) Nonrigid diatomic gases

Choose the correct answer from the options given below:

(A) (A)-(I), (B)-(III), (C)-(IV), (D)-(II)

(B) (A)-(IV), (B)-(III), (C)-(II), (D)-(I)

(C) (A)-(IV), (B)-(II), (C)-(I), (D)-(III)

(D) (A)-(I), (B)-(IV), (C)-(III), (D)-(II)

Correct answer is (A)

Monoatomic gases possess only translational degrees of freedom. So, option (I) matches with (A).

Polyatomic gases have translational and rotational degrees of freedom. So, option (II) matches with (D).

Rigid diatomic gases have translational, rotational and vibrational degrees of freedom. So, option (III) matches with (B).

Non-rigid diatomic gases possess translational, rotational and vibrational degrees of freedom. So, option (IV) matches with (C).

2. (JEE Main 2023 (Online) 31st January Morning Shift )

The correct relation between γ = c p c v and temperature T is :

(A) γ T

(B) γ 1 T

(C) γ 1 T

(D) γ T

Correct answer is (D)

γ = C P C V

At low temperature ( T ) , γ is independent of T .

3. (JEE Main 2023 (Online) 25th January Evening Shift )

According to law of equipartition of energy the molar specific heat of a diatomic gas at constant volume where the molecule has one additional vibrational mode is :-

(A) 9 2 R

(B) 5 2 R

(C) 3 2 R

(D) 7 2 R

Correct answer is (D)

Diatomic gas molecules have three translational degree of freedom, two rotational degree of freedom \& it is given that it has one vibrational mode so there are two additional degree of freedom corresponding to one vibrational mode, so total degree of freedom = 7

C v = fR 2 = 7 R 2

4. (JEE Main 2023 (Online) 24th January Evening Shift )

Let γ 1 be the ratio of molar specific heat at constant pressure and molar specific heat at constant volume of a monoatomic gas and γ 2 be the similar ratio of diatomic gas. Considering the diatomic gas molecule as a rigid rotator, the ratio, γ 1 γ 2 is :

(A) 35 27

(B) 25 21

(C) 21 25

(D) 27 35

Correct answer is (B)

For monoatomic gas γ 1 = 5 3

For diatomic gas at low temperatures

γ 2 = 7 5 γ 1 γ 2 = 5 3 7 5 = 25 21

5. (JEE Main 2022 (Online) 28th July Evening Shift )

A vessel contains 14   g of nitrogen gas at a temperature of 27 C . The amount of heat to be transferred to the gas to double the r.m.s speed of its molecules will be :

Take R = 8.32   J   mol 1 k 1 .

(A) 2229 J

(B) 5616 J

(C) 9360 J

(D) 13,104 J

Correct answer is (C)

n = 0.5

T = 300

For vrms to be doubled T' = 4 × 300 = 1200

Heat transferred

= ( 0.5 ) ( 5 2 ) ( 8.32 ) ( 900 )

= 9360 J