To an evacuated vessel with movable piston under external pressure of $1 \mathrm{~atm}$, $0.1$ mole of He…
To an evacuated vessel with movable piston under external pressure of $1 \mathrm{~atm}$, $0.1$ mole of He and $1.0$ mole of an unknown compound (vapour pressure $0.68 mathrm{~atm}$ at $0^{\circ} \mathrm{C}$ ) are introduced. Considering the ideal gas behaviour, the total volume (in litre) of the gases at $0^{\circ} \mathrm{C}$ is close to
Solution
Since, the external pressure is $1.0 \mathrm{~atm}$, the gas pressure is also $1.0 \mathrm{~atm}$ as piston is movable. Out of this $1.0 \mathrm{~atm}$ partial pressure due to unknown compound is $0.68 \mathrm{~atm}$. Therefore, partial pressure of $\mathrm{He}=1.00-0.68=0.32 \mathrm{~atm}$.
$
\begin{aligned}
& \Rightarrow \quad \text { Volume }=\frac{n(\mathrm{He}) \mathrm{RT}}{r(\mathrm{He})}=\frac{0.1 \times 0.082 \times 273}{0.32}=7 \mathrm{~L} \\
& \Rightarrow \text { Volume of container }=\text { Volume of } \mathrm{He} .
\end{aligned}
$