Question

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

• A. $3.0atm$
• B. $1atm$
• C. $3.33atm$
• D. $2.4atm$

Answer 1

Answer: (D)

$P_{N_2}+P_{H_{2}O(v)}=1atm,$
$\because P_{H_{2}O}=0.3atm$
$\therefore P_{N_2}=0.7atm$
Now new pressure of $N_2$ in another vessel of volume $V/3$ at same $T$ is given by :
$P_{N_2}\times \frac{V_1}{3}=0.70\times V_1$
$\therefore P_{N_2}=2.1atm$
Since aqueous tension remains constant,
and thus total pressure in new vessel.
$=P_{N_2}+P_{H_{2}O}=2.1+0.3=2.4atm.$