A container of volume $200 \mathrm{~cm}^3$ contains 0.2 mole of hydrogen gas and 0.3 mole of argon gas. The…
A container of volume $200 \mathrm{~cm}^3$ contains 0.2 mole of hydrogen gas and 0.3 mole of argon gas. The pressure of the system at temperature $200 \mathrm{~K}\left(R=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)$ will be
$4.15 \times 10^5 \mathrm{~Pa}$
$4.15 \times 10^6 \mathrm{~Pa}$
$6.15 \times 10^5 \mathrm{~Pa}$
$6.15 \times 10^4 \mathrm{~Pa}$
Solution
We know that
$\begin{aligned} & P V=n R T \\ & P_1=\frac{n_1}{V} R T\end{aligned}$
$\begin{gathered}P_2=\frac{n_2}{V} R T \\ P_{\text {net }}=P_1+P_2\end{gathered}$
$\begin{aligned} & =\left(n_1+n_2\right) \frac{R T}{V} \\ & =(0.2+0.3) \times \frac{8.3 \times 200}{200 \times 10^{-6}}\end{aligned}$
$=8.3 \times 0.5 \times 10^6=4.15 \times 10^6 \mathrm{~Pa}$