Question

A $4.0 L$ sample at $27^{\circ} C$ and $1.2$ atm of pressure contains $0.75\, mol$ of a gas. If an additional $0.25\, mol$ of gas at the same temperature and pressure are added, then total volume of the gas is

• A. $1.5\, L$
• B. $5.33\, L$
• C. $3\, L$
• D. $4.5\, L$

Answer 1

Answer: (B)

Given, $V_{1}=4.0 L , \mu_{1}=0.75 mol$, when $0.25 mol$ of gas is added, then $\mu_{2}=\mu_{1}+0.25=0.75+0.25=1 mol$. Since, gas constant is same for all gases, i.e. $\frac{p_{1} V_{1}}{T_{1} \mu_{1}}=\frac{p_{2} V_{2}}{T_{2} \mu_{2}} \Rightarrow \frac{V_{1}}{\mu_{1}}=\frac{V_{2}}{\mu_{2}} (\because p$ and $T$ are same.)
$\Rightarrow \frac{4}{0.75}=\frac{V_{2}}{1} $
$\Rightarrow V_{2}=\frac{4}{0.75}=5.33 L$