Question

A mixture of two gases $A$ and $B$ in the mole ratio $2:3$ is kept in a $2$ litres vessel. A second $3$ litres vessel has the same gases in the mole ratio $3:5$. Both gas mixtures have the same temperature and same total pressure. They are allowed to intermix and the final temperature and total pressure are the same as the initial values, the final volume being $5$ litres. Given that the molar masses are $M_A$ and $M_B$ then what is the mean molar mass of the final mixture?

• A. $\frac{77M_A+123M_B}{200}$
• B. $\frac{123M_A+77M_B}{200}$
• C. $\frac{77M_A+123M_B}{250}$
• D. $\frac{123M_A+77M_B}{250}$

Answer 1

The correct option is A $\frac{77M_A+123M_B}{200}$

Since temperature and total pressure are constant, the ratio $\frac{n}{V}$ is the same for vessel A and vessel B.

Let vessel A contains 2a moles of A and 3a moles of B.

Let vessel B contains 3b moles of A and 5b moles of B.

Thus $\frac{2a+3a}{2}=\frac{3b+5b}{3}$ or 1$5a=16b$.

For total pressure, $\frac{5a+8b}{5}=5a$ or $8b=20a$.

When $b=1$, $a=\frac{4}{5}$.

Thus vessel A contains $\frac{8}{5}$ moles of A and $\frac{12}{5}$ moles of B.

Vessel B contains 3 moles of A and 5 moles of B.

When two gases are mixed, the mole fraction of A becomes $\frac{23}{60}$ or $\frac{77}{200}$ and the mole fraction of B becomes $\frac{37}{60}$ or $\frac{123}{200}$.

The mean molar mass of final mixture is $\frac{\frac{n_A{M}_A+n_B{M}_B}{n_A+n_B}}{200}$ or $\frac{77M_A+123M_B}{200}$.

Hence, option $A$ is correct.