11. (JEE Main 2021 (Online) 25th July Morning Shift )

For a gas C_P $-$ C_V = R in a state P and C_P $-$ C_V = 1.10 R in a state Q, T_P and T_Q are the temperatures in two different states P and Q respectively. Then

(A) T_P = T_Q

(B) T_P < T_Q

(C) T_P = 0.9 T_Q

(D) T_P > T_Q

Correct answer is (D)

C_P $-$ C_V = R for ideal gas and gas behaves as ideal gas at high temperature, so T_P > T_Q

12. (JEE Main 2021 (Online) 25th July Morning Shift )

Two different metal bodies A and B of equal mass are heated at a uniform rate under similar conditions. The variation of temperature of the bodies is graphically represented as shown in the figure. The ratio of specific heat capacities is :

JEE Main 2021 (Online) 25th July Morning Shift Physics - Heat and Thermodynamics Question 139 English

(A) $\frac{8}{3}$

(B) $\frac{3}{8}$

(C) $\frac{3}{4}$

(D) $\frac{4}{3}$

Correct answer is (B)

${(\frac{Δ Q}{Δ t})}_{A} = {(\frac{Δ Q}{Δ t})}_{B}$

$m S_{A} {(\frac{Δ T}{Δ t})}_{A} = m S_{B} {(\frac{Δ T}{Δ t})}_{B}$

$\Rightarrow$ $\frac{S_{A}}{S_{B}} = \frac{{(\frac{Δ T}{Δ t})}_{B}}{{(\frac{Δ T}{Δ t})}_{B}} = \frac{90 / 6}{120 / 3} = \frac{15}{40} = \frac{3}{8}$

13. (JEE Main 2021 (Online) 20th July Evening Shift )

The correct relation between the degrees of freedom f and the ratio of specific heat $γ$ is :

(A) $f = \frac{2}{γ - 1}$

(B) $f = \frac{2}{γ + 1}$

(C) $f = \frac{γ + 1}{2}$

(D) $f = \frac{1}{γ + 1}$

Correct answer is (A)

$γ = 1 + \frac{2}{f}$

$\Rightarrow$ $f = \frac{2}{γ - 1}$

14. (JEE Main 2021 (Online) 17th March Evening Shift )

If one mole of the polyatomic gas is having two vibrational modes and $β$ is the ratio of molar specific heats for polyatomic gas $(β = \frac{C_{P}}{C_{V}})$ then the value of $β$ is :

(A) 1.02

(B) 1.35

(C) 1.2

(D) 1.25

Correct answer is (C)

For polyatomic gas molecule has 3 rotational degrees of freedom, 3 translational degrees of freedom, and 2 vibrational modes.

So, number of vibrational degrees of freedom = 2 $\times$ 2 = 4

Degree of freedom of polyatomic gas

f = T + R + V

f = 3 + 3 + 4 = 10

$β = 1 + \frac{2}{f} = 1 + \frac{2}{10}$

$β = \frac{12}{10} = 1.2$

15. (JEE Main 2021 (Online) 17th March Morning Shift )

A polyatomic ideal gas has 24 vibrational modes. What is the value of $γ$ ?

(A) 1.37

(B) 1.30

(C) 1.03

(D) 10.3

Correct answer is (C)

f = 3T + 3R + 24V

= 30

$γ$ = 1 + $\frac{2}{f}$

$γ$ = 1 + $\frac{2}{30}$

= 1.066

Nearest Ans. = 1.03

Kinetic Theory of Gases