One litre of oxygen at a pressure of $1 \mathrm{~atm}$ and two litres of nitrogen at a pressure of $0.5…
One litre of oxygen at a pressure of $1 \mathrm{~atm}$ and two litres of nitrogen at a pressure of $0.5 \mathrm{~atm}$, are introduced into a vessel of volume $1 \mathrm{~L}$. If there is no change in temperature, the final pressure of the mixture of gas (in atm) is
$1.5$
$2.5$
$2$
$4$
Solution
Ideal gas equation is given by
$
p V=n R T
$
For oxygen, $p=1 \mathrm{~atm}, V=1 \mathrm{~L}, n=n_{\mathrm{O}}$
Therefore Eq. (i) becomes
$
\begin{aligned}
& \therefore & 1 \times 1 & =n_{\mathrm{O}_2} R T \\
& \Rightarrow & n_{\mathrm{O}_2} & =\frac{1}{R T}
\end{aligned}
$
For nitrogen $p=0.5 \mathrm{~atm}, V=2 \mathrm{~L}, n=n_{\mathrm{N}}$
$
\begin{array}{rlrl}
& \therefore & 0.5 \times 2 & =n_{\mathrm{N}_2} R T \\
\Rightarrow & n_{\mathrm{N}_2} & =\frac{1}{R T}
\end{array}
$
For mixture of gas
$
\begin{array}{rlrl}
& & p_{\text {mix }} V_{\text {mix }} & =n_{\text {mix }} R T \\
\text { Here, } & n_{\text {mix }} & =n_{\mathrm{O}}+n_{\mathrm{N}_2} \\
& \therefore \quad & \frac{p_{\text {mix }} V_{\text {mix }}}{R T} & =\frac{1}{R T}+\frac{1}{R T} \\
& \Rightarrow & p_{\text {mix }} V_{\text {mix }} & =2
\end{array}
$
Here