A vessel has nitrogen gas and water vapor in equilibrium with liquid water at a total pressure of 1 atm. The partial pressure of water vapors 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)

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3.33 atm

1 atm

2.4 atm

3.0 atm

answer is D.

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Detailed Solution

$P_{N_{2}} + P_{H_{2} O}$ = 1 atm (Since, $P_{H_{2} O}$ =0.3 atm and $P_{N_{2}}$ = 0.7 atm in1vessel).
Therefore, new pressure of N₂ in another vessel may be calculated as :
P₁V₁= P₂V₂
Implies that : 0.7 × V₁=P₂× $\frac{v_{1}}{3} atm$
Since, $P_{N_{2}}$ = 2.1 atm
Implies that : $P_{H_{2} O}$ = 0.3 atm
P_T= 2.4 atm
$P_{H_{2} O}$ remains constant at constant temperature.

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