Question

A certain vessel $X$ has water and nitrogen gas at a total pressure of 2 $atm$ at 300 $K$. All the contents of the vessel are transferred to another vessel $Y$ having half the capacity of the vessel $X$. The pressure of $N_2$ in this vessel was 3.8 $atm$ at 300 $K$. The vessel $Y$ is heated to 320 $K$ and the total pressure observed was 4.32 $atm$. The pressure of water vapor at 320 K is :

[Assume that the volume occupied by the gases in the vessel is equal to the volume of the vessel.]

• A. 0.27
• B. 0.32
• C. 4
• D. 1

Answer 1

Answer: (A)

0.27

Pressure of $N_2$ in Y at $300K=3.8atm$

By Gay Lussac law,

$\therefore \frac{P_1}{T_1}=\frac{P_2}{T_2}$

$\frac{P_{\left(N_2\right)}}{320}=\frac{3.8}{300}$

Pressure of $N_2$ at 320 K $=\frac{3.8}{300}\times 320=4.05atm$.

Total pressure at 320 K $=$ 4.32 atm

$\therefore$ Pressure of water vapour at 320 K $=4.32-4.05=0.27atm$

Hence, option $\left(A\right)$ will be correct.