Question

One liter flask contains air, water vapour and a small amount of liquid water at a pressure of 200 mm Hg. if this is connected to another one-litre evacuated flask, what will be the final pressure of the gas mixture at equilibrium? (Assume the temperature to be ${50}^{o}C$. Vapour pressure at ${50}^{o}C$ = 93 mm Hg).

Answer 1

Given,]

Initial pressure $=200mmHg$

The vapour pressure $=93mmHg$

Thus, the total pressure$=200+93mmHg$

According to Ideal gas equation,

$\frac{P_1\times V_2}{T_1}=\frac{P_2\times V_2}{T_2}$

Here $P_1=293$, $T_1={50}^{o}C=323K$, $V_2=2$ $V_1=1$ and $T_2=323K$

$\therefore \frac{293\times 1}{323}=\frac{P_2\times 2}{323}$

$\therefore P_2=146.5mmHg$