Question

A closed vessel contains a mixture of two diatomic gases A and B. Molar mass of A is 16 times and that of B and mass of gas A, contained in the vessel is 2 times that of B.

• A. Average kinetic energy per molecule of gas A is equal to that of gas B
• B. Root mean square value of translational velocity of gas B is four times that of A
• C. Pressure exerted by gas B is eight times of that exerted by gas A
• D. Number of molecules of gas B in the cylinder is eight times that of gas A.

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A Average kinetic energy per molecule of gas A is equal to that of gas B
B Root mean square value of translational velocity of gas B is four times that of A
C Number of molecules of gas B in the cylinder is eight times that of gas A.
D Pressure exerted by gas B is eight times of that exerted by gas A

Since both the gases are contained in the same vessel, temperature of both the gases is same.

Average KE per molecule of a diatomic gas is 5/2KT. Hence, average KE per molecule of both the gases is same. Therefore. Option (a) is correct.

$v_{mas}=\sqrt{\frac{2RT}{M}}$

Hence,

$\frac{u_{rms}2}{u_{rms}1}$=$\sqrt{\frac{M_1}{M_2}}$ = $\sqrt{1}6=4$

Hence, option (b) is correct.

Let molar mass of B be M, then that of A will be equal to 16M.

Let mass of gass B ini the vessel be m, then that of A will be 2m. The number of moles of a gas, in the vessel will be $n=\frac{m}{M}$ . Hence, number of moles of gases A and B will be

$n_1=\frac{2m}{16M}$ and $n_2=\frac{m}{M}$

Hence $\frac{n_1}{n_2}=\frac{1}{8}$

Hence, option (d) is correct.

Partial pressure exerted by a gas is

$P=\frac{nRT}{V}$

Hence $\frac{P_2}{P_1}$=$\frac{n_2}{n_1}$=8

Therefore, option (c) is also correct.