Question

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

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

Answer 1

Answer: (B)

Given, $V_1=4.0L,{\mu }_1=0.75mol$, when $0.25mol$ of gas is added, then ${\mu }_2={\mu }_1+0.25=0.75+0.25=1mol$. 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.)
$\Rightarrow \frac{4}{0.75}=\frac{V_2}{1}$
$\Rightarrow V_2=\frac{4}{0.75}=5.33L$