Question

A closed vessel contains a mixture of two diatomic gases $A$ and $B$. Molar mass of $A$ is 16 times that of $B$ and mass of gas $A$ contained in the vessel is 2 times that of $B$. Which of the following statements are correct?

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

Answer 1

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

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

Since both the gases are contained in the same vessel, temperature of both the gases is same.
Average K.E 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.
$v_{rms}=\sqrt{\frac{3RT}{M}}$
Hence,
$\frac{(v_{rms}{)}_2}{(v_{rms}{)}_1}=\sqrt{\frac{M_1}{M_2}}=\sqrt{16}=4$
Hence, option (b) is correct.
Let molar mass of $B$ be $M$, then that of $A$ will be equal to $16M$.
Let mass of gas $B$ in the vessel be $m$; then that of $A$ will be $2m$. The number of moles of a gas, in the vessel will be $n=m/M$. 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 ia
$P=\frac{nRT}{V}$
Hence,
$\frac{P_2}{P_1}=\frac{n_2}{n_1}=8$
Therefore, option (c) is also correct.