21. (JEE Main 2020 (Online) 2nd September Morning Slot )

A gas mixture consists of 3 moles of oxygen and 5 moles of argon at temperature T. Assuming the gases to be ideal and the oxygen bond to be rigid, the total internal energy (in units of RT) of the mixture is :

(A) 11

(B) 20

(C) 15

(D) 13

Correct answer is (C)

U $= \frac{f_{1}}{2} n_{1} R T + \frac{f_{2}}{2} n_{2} R T$

$= \frac{5}{2} (3 R T) + \frac{3}{2} \times 5 R T$

$∴$ U $= 15 R T$

22. (JEE Main 2020 (Online) 4th September Morning Slot )

A closed vessel contains 0.1 mole of a monoatomic ideal gas at 200 K. If 0.05 mole of the same gas at 400 K is added to it, the final equilibrium temperature (in K) of the gas in the vessel will be close to _______.

Correct answer is (266)

As work done on gas and heat supplied to the gas are zero,

$∴$ Total internal energy of gases remain same.

u₁ + u₂ = u₁' + u₂'

We know, $Δ$ U = nC_v $Δ$ T

$∴$ $(0.1 \times \frac{3 R}{2} \times 200) + (0.05 \times \frac{3 R}{2} \times 400)$ = $(0.15 \times \frac{3 R}{2} \times T_{f})$

$\Rightarrow$ (20 + 20) = 0.15 T_f

$\Rightarrow$ T_f = 266.67

23. (JEE Main 2019 (Online) 9th April Morning Slot )

An HCl molecule has rotational, translational and vibrational motions. If the rms velocity of HCl molecules in its gaseous phase is $\overset{―}{v}$ , m is its mass and k_B is Boltzmann constant, then its temperature will be :

(A) $\frac{m {\overset{―}{v}}^{2}}{5 k_{B}}$

(B) $\frac{m {\overset{―}{v}}^{2}}{6 k_{B}}$

(C) $\frac{m {\overset{―}{v}}^{2}}{7 k_{B}}$

(D) $\frac{m {\overset{―}{v}}^{2}}{3 k_{B}}$

Correct answer is (C)

An HCl molecule, being diatomic, has:

3 translational degrees of freedom
2 rotational degrees of freedom
2 vibrational degrees of freedom

The total number of degrees of freedom is $3 + 2 + 2 = 7$ .

According to the equipartition theorem, each degree of freedom contributes $\frac{1}{2} k_{B} T$ to the total energy. So the total energy is given by: $\frac{7}{2} k_{B} T$

The translational kinetic energy is related to the root-mean-square (rms) speed $\overset{―}{v}$ by: $\frac{1}{2} m {\overset{―}{v}}^{2} = \frac{7}{2} k_{B} T$

Rearranging to solve for the temperature, we find: $T = \frac{m {\overset{―}{v}}^{2}}{7 k_{B}}$

24. (JEE Main 2019 (Online) 12th January Morning Slot )

An ideal gas occupies a volume of 2m³ at a pressure of 3 $\times$ 10⁶ Pa. The energy of the gas is :

(A) 6 $\times$ 10⁴ J

(B) 9 $\times$ 10⁶ J

(C) 3 $\times$ 10² J

(D) 10⁸ J

Correct answer is (B)

Energy = $\frac{1}{2}$ nRT = $\frac{f}{2}$ PV

= $\frac{f}{2}$ (3 $\times$ 10⁶) (2)

= f $\times$ 3 $\times$ 10⁶

Considering gas is monoatomic i.e. f = 3

E. = 9 $\times$ 10⁶ J

25. (JEE Main 2019 (Online) 11th January Morning Slot )

A gas mixture consists of 3 moles of oxygen and 5 moles of argon at temperature T. considering only translational and rotational modes, the total internal energy of the system is :

(A) 12 RT

(B) 20 RT

(C) 4 RT

(D) 15 RT

Correct answer is (D)

U $= \frac{f_{1}}{2} n_{1} R T + \frac{f_{2}}{2} n_{2} R T$

$= \frac{5}{2} (3 R T) + \frac{3}{2} \times 5 R T$

U $= 15 R T$

Kinetic Theory of Gases