Question

One-litre of oxygen at a pressure of $1 \, \, \text{atm}$ and two-litre of nitrogen at a pressure of $0.5 \, \, \text{atm}$ , are introduced into a vessel of volume one-litre. If there is no change in temperature, the final pressure of the mixture of the gases $\left[\right.$ in $atm\left]\right.$ is

• A. $1.5$
• B. $2.5$
• C. $2$
• D. $4$

Answer 1

Answer: (C)

Ideal gas equation is given by
$pV=nRT$ ...(i)
For oxygen, $p=1$ atm, $V=1$ L, $n=n_{O}_{2}$
Therefore, Eq. (i) becomes
$\therefore \, \, \, 1\times 1=n_{O}_{2}RT$
$\Rightarrow \, \, n_{O}_{2}=\frac{1}{R T}$
For nitrogen $p=$ 0.5 atm, $V=$ 2 L, $n=n_{N_{2}}$
$\therefore \, \, \, 0.5\times 2=n_{N}_{2}RT$
$\Rightarrow \, \, n_{N}_{2}=\frac{1}{R T}$
For mixture of gas
$p_{m i x}V_{m i x}=n_{m i x}RT$
Here, $n_{m i x}=n_{O}_{2}+n_{N}_{2}$
$\therefore \, \frac{p_{m i x} V_{m i x}}{R T}=\frac{1}{R T}+\frac{1}{R T}$
$\Rightarrow \, \, p_{m i x}V_{m i x}=2$ $\left(\right.V_{m i x}=1\left.\right)$