Question

A vessel has nitrogen gas and water vapour in equilibrium with liquid water at a total pressure of $1$ atm. The partial pressure of water vapours is $0.3$ atm. The volume of this vessel is reduced to one-third of the original volume, at the same temperature, then find total pressure in atm of the system. (Neglect volume occupied by liquid water)

Answer 1

Let us consider the problem :

$P_{N_2}+P_{H_{2}O}=1\text{atm}(Since,P_{H_{2}O}=0.3\text{atm}\text{and}{P}_{N_2}=0.7\text{atm}\text{in}\text{1}\text{vessel)}\text{.}$

Therefore,

New pressure of $N_2$ in another vessel may be calculated as :

$P_1{V}_1=P_2{V}_2$

Implies that :

$0.7\times V_1=P_2\times \frac{V_1}{3}\text{atm}$

Since,

$P_{N_2}=2.1\text{atm}$

Implies that :

$P_{H_{2}O}=0.3\text{atm}$

$P_T=2.4\text{atm}$

$P_{H_{2}O}$ remains constant at constant temperature.