Question

$0.5\text{mole}$ of each $H_2,S{O}_2$ and $C{H}_4$ are kept in a container. A hole was made in the container. After 3 hours, decreasing order of partial pressures of gases in the container will be:

• A. $P_{S{O}_2}>P_{C{H}_4}>P_{H_2}$
• B. $P_{H_2}>P_{S{O}_2}>P_{C{H}_4}$
• C. $P_{C{H}_4}>P_{S{O}_4}>P_{H_2}$
• D. $P_{C{H}_4}>P_{H_2}>P_{S{O}_2}$

Answer 1

The correct option is A $P_{S{O}_2}>P_{C{H}_4}>P_{H_2}$

Answer : D. $P_{S{O}_2}>P_{C{H}_4}>P_{H_2}$

The partial pressure of a gas is directly proportional to mole fraction, therefore directly proportional to the number of moles.

Initially, there are equal number of moles (0.5 moles) of each gases. So their partial pressures are equal.

After 3 hours, some amount of each gases will be effused out through the hole. Gas with larger effusion rate will be effused more through the hole and will have less number of moles remaining in the container, resulting in lower partial pressure compared to others.

We know that,

Molar mass of $H_2$, $M_{H_2}$ = $1\times 2$ = $2$ g

Molar mass of $S{O}_2$, $M_{S{O}_2}$ = $32+(16\times 2)$ = $64$ g

Molar mass of $C{H}_4$, $M_{C{H}_4}$ = $12+(1\times 4)$ = $16$ g

Therefore, the order of molar masses,

$M_{S{O}_2}>M_{C{H}_4}>M_{H_2}$

$\text{Rate of effusion of gas}$ $\alpha$ $\frac{1}{\sqrt{\text{molar mass}}}$

$\Rightarrow$ Order of rates of effusion,

$r_{S{O}_2}<r_{C{H}_4}<r_{H_2}$

$\text{Number of moles remaining in the container}$ $\alpha$ $\frac{1}{\text{Rate of effusion}}$

$\Rightarrow$ Order of number of moles remaining in the container,

$n_{S{O}_2}>n_{C{H}_4}>n_{H_2}$

$\text{Partial pressure}$ $\alpha$ $\text{Number of moles remaining in the container}$

$\Rightarrow$ Order of partial pressures,

$P_{S{O}_2}>P_{C{H}_4}>P_{H_2}$