Given: External pressure $\left(\mathrm{P}_{\mathrm{ext}}\right)=1 \mathrm{~atm}$

Number of mole of helium $\left(n_{\mathrm{He}}\right)=0.1 \mathrm{~mol}$

No. of mole of unknown compound

$\left(n_{\text {unknown compound }}\right)=1.0 \mathrm{~mol}$

Vapour pressure of unknown compound

$\left(p_{\text {unknown }}^0\right)=0.68 \mathrm{~atm}$

Temperature of the mixture $0^{\circ} \mathrm{C}=273 \mathrm{~K}$

To Find: The volume of gas (in litre) $=v_{\text {gas }}$

Formula: (i) Vapour pressure of helium $\left(\mathrm{P}_{\mathrm{He}}\right)=$

$P_{\text {ext }}-P_{\text {unknown compound }}$

(ii) $\mathrm{V}_{\mathrm{He}}=\frac{n_{\mathrm{He}} \times \mathrm{R} \times \mathrm{T}}{\mathrm{P}_{\mathrm{He}}}$

Since, the evacuated vessel (with fitted) piston in equilibrium with its surroundings. Hence, external pressure (or pressure outside the vessel) is equal to pressure inside the vessel.

$$ \begin{aligned} & \mathrm{P}_{\text {ext }}=\mathrm{P}_{\text {internal }}=\mathrm{P}_{\mathrm{T}} \\\\ & \mathrm{P}_{\text {ext }}=\mathrm{P}_{\mathrm{He}}+\mathrm{P}_{\text {unknwon compound }} \\\\ & 1 \mathrm{~atm}=\mathrm{P}_{\mathrm{He}}+0.68 \mathrm{~atm} \\\\ & \mathrm{P}_{\mathrm{He}}=0.32 \mathrm{~atm} \\\\ \end{aligned} $$

$\left[\mathrm{P}_{\mathrm{He}} \mathrm{P}_{\text {unknown compound }}\right.$ are partial pressures of helium and unknown gas respectively]

According to ideal gas equation:

$$ \begin{aligned} \mathrm{P}_{\mathrm{He}} \times \mathrm{V}_{\mathrm{He}} & =n_{\mathrm{He}} \times \mathrm{R} \times \mathrm{T} \\\\ \mathrm{V}_{\mathrm{He}} & =\frac{0.1 \times 0.0821 \times 273}{0.32} \\\\ \mathrm{~V}_{\mathrm{He}} & =7.004 \mathrm{~L} \end{aligned} $$