SolutionHard
Question
A vessel has nitrogen gas and water vapours 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 total pressure of the system is :(Neglect volume occupied by liquid water)
Options
A.3.0 atm
B.1 atm
C.3.33 atm
D.2.4 atm
Solution
PN2 + PH2O(v) = 1 atm, ∵ PH2O = 0.3 atm
∴ PN2 = 0.7 atm
Now new pressure of N2 in another vessel of volume V/3 at same T is given by :
PN2 ×
= 0.7 0 × V1
∴ PN2 = 2.1 atm
Since aqueous tension remains constant, and thus total pressure in new vessel.
= PN2 + PH2O = 2.1 + 0.3 = 2.4 atm.
∴ PN2 = 0.7 atm
Now new pressure of N2 in another vessel of volume V/3 at same T is given by :
PN2 ×
= 0.7 0 × V1∴ PN2 = 2.1 atm
Since aqueous tension remains constant, and thus total pressure in new vessel.
= PN2 + PH2O = 2.1 + 0.3 = 2.4 atm.
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