Problem : A vessel has N2 gas and water vapours at a total pressure of 1 atm. The partial pressure of water vapours is 0.3 atm. The contents of this vessel are transferred to another vessel having one third of the capacity of original volume, completely at the same temperature the total pressure of this system in the new vessel is

(A) 3.0 atm

(B) 1 atm

(C) 3.33 atm

(D) 2.4 atm

Ans:(D)

Solution: $\large P’_{N_2} + P’_{H_2 O} = 1 atm$

$\large P’_{H_2 O} = 0.3 atm $

$\large P’_{N_2} = 0.7 atm$

Now new pressure of N2 in another vessel of volume V/3 at same temperature T is given by

$\large P”_{N_2} \times \frac{V}{3} = 0.7 V $

$\large P”_{N_2} = 2.1 atm $

since aqueous tension remains constant, and thus total pressure in new vessel

$\large = P”_{N_2} + P’_{H_2 O} = 2.1 + 0.3 = 2.4 atm $