Question

Gas A (Molar mass = 16) and gas B (Molar mass = 4) are confined to a vessel at a partial pressure of 2 atm each at 300 K. The gases are then allowed to effuse out at 300 K through a valve till the pressure inside the vessel becomes 2 atm after which the valve is closed. The gases are again heated at 400 K, identify the correct statements in the following:

• A. The number of moles of A and B diffused out are in the ratio 1 : 2
• B. The number of moles of A and B left after diffusion at 300 K are in the ratio 2 : 1
• C. Partial pressure of B at 400 K is $\frac{8}{9}atm$
• D. Partial pressure of A at 400 K is $\frac{1}{8}atm$

Answer 1

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

The correct options are
A The number of moles of A and B diffused out are in the ratio 1 : 2
B The number of moles of A and B left after diffusion at 300 K are in the ratio 2 : 1
C Partial pressure of B at 400 K is $\frac{8}{9}atm$

At 300 K, total pressure $P_A+P_B=4atm$
$\frac{r_A}{r_B}=\frac{n_A}{n_B}=\frac{P_A}{P_B}\times {\left(\frac{M_B}{M_A}\right)}^{\frac{1}{2}}$
$\frac{n_A}{n_B}=\frac{2}{2}\times {\left(\frac{4}{16}\right)}^{\frac{1}{2}}$
$\frac{n_A}{n_B}=\frac{1}{2}$

The number of moles of A and B diffused out are in the ratio 1 : 2

$A+B\rightleftharpoons C$
Initial moles a a
moles left (a-x) (a-2x)
Total moles after diffusion
= a - x + a - 2x = a (Since, as P drops from 4 atm to 2 atm hence moles should decrease from 2a to a )
$\therefore$ a = 3x

$\therefore$ after diffusion moles of A and B are 3x - x = 2x and 3x - 2x = x
$\therefore$ moles of A and B after diffusion are in the ratio 2 : 1

Partial pressures after diffusion at 300 K,
For A
$\frac{2}{3}\times P=\frac{2}{3}\times 2=\frac{4}{3}$
For B
$\frac{1}{3}\times P=\frac{1}{3}\times 2=\frac{2}{3}$
When the vessel is heated at 400 K
Partial pressure of A = $\frac{4}{3}\times \frac{4}{3}=\frac{16}{9}atm..........(\frac{P_1}{T_1}=\frac{P_2}{T_2})$
Partial pressure of B = $\frac{2}{3}\times \frac{4}{3}=\frac{8}{9}atm$