21. (JEE Main 2019 (Online) 12th April Morning Slot )

Two moles of helium gas is mixed with three moles of hydrogen molecules (taken to be rigid). What is the molar specific heat of mixture at constant volume ? (R = 8.3 J/mol K)

(A) 21.6 J/mol K

(B) 17.4 J/mol K

(C) 15.7 J/mol K

(D) 19.7 J/mol K

Correct answer is (B)

$f_{m i x} = \frac{n_{1} f_{1} + n_{2} f_{2}}{n_{1} + n_{2}}$
$\Rightarrow \frac{2 \times 3 + 3 \times 5}{5} = \frac{21}{5}$

$C_{v} = \frac{f R}{5} = \frac{21}{5} \times \frac{R}{2} = 17.4$ J/mol K

22. (JEE Main 2019 (Online) 9th April Evening Slot )

The specific heats, C_P and C_V of a gas of diatomic molecules, A, are given (in units of J mol^–1 K^–1) by 29 and 22, respectively. Another gas of diatomic molecules, B, has the corresponding values 30 and 21. If they are treated as ideal gases, then :-

(A) A is rigid but B has a vibrational mode

(B) A has a vibrational mode but B has none

(C) A has one vibrational mode and B has two

(D) Both A and B have a vibrational mode each

Correct answer is (B)

For A:
$\frac{C_{p}}{C_{v}} = γ = 1 + \frac{2}{f} = \frac{29}{22}$
It gives f = 6.3 $\approx$ 6 (3 translational, 2 rotational and 1 vibrational)

For B:
$\frac{C_{p}}{C_{v}} = γ = 1 + \frac{2}{f} = \frac{30}{21}$
$∴$ f = 4.67 $\approx$ 5 (3 translational, 2 rotational, no vibrational)

23. (JEE Main 2018 (Online) 16th April Morning Slot )

Two moles of helium are mixed with n moles of hydrogen. If $\frac{C p}{C v} = \frac{3}{2}$ for the mixture, then the value of n is :

(A) 1

(B) 3

(C) 2

(D) 3 / 2

Correct answer is (C)

$\frac{C_{p}}{C_{v}} = \frac{f_{m i x} + 2}{f_{m i x}} = \frac{3}{2}$

$\Rightarrow$ f_mix = 4

As, f_mix = $\frac{n_{1} f_{1} + n_{2} f_{2}}{n_{1} + n_{2}}$

$\Rightarrow$ 4 = $\frac{2 \times 3 + n \times 5}{2 + n}$

$\Rightarrow$ n = 2 moles.

24. (JEE Main 2017 (Online) 8th April Morning Slot )

An ideal gas has molecules with 5 degrees of freedom. The ratio of specific heats at constant pressure (C_p ) and at constant volume (C_v) is :

(A) 6

(B) $\frac{7}{2}$

(C) $\frac{5}{2}$

(D) $\frac{7}{5}$

Correct answer is (D)

For ideal gas molecule with 5 degree of freedom,

C_v = $\frac{5}{2}$ R and C_p = $\frac{7}{2}$ R

$∴$ $\frac{C_{p}}{C_{v}}$ = $\frac{\frac{7}{2} R}{\frac{5}{2} R}$ = $\frac{7}{5}$

25. (JEE Main 2017 (Offline) )

C_P and C_v are specific heats at constant pressure and constant volume respectively. It is observed that
C_P – C_v = a for hydrogen gas
C_P – C_v = b for nitrogen gas
The correct relation between a and b is

(A) a = 28 b

(B) a = 1/14 b

(C) a = b

(D) a = 14 b

Correct answer is (D)

As we know, for 1 g mole of a gas,

C_p – C_v = R where C_p and C_v are molar specific heat capacities.

So, when n gram moles are given,

C_p – C_v = $\frac{R}{n}$

For hydrogen (n = 2), C_p – C_v = $\frac{R}{2}$ = $a$

For nitrogen (n = 28), C_p – C_v = $\frac{R}{28}$ = $b$

$∴$ $\frac{a}{b} = 14$

$\Rightarrow$ $a$ = 14 $b$

Kinetic Theory of Gases