The vapour pressure of water at 80ºC is 355 torr. A 100 ml vessel contained water − saturated oxygen at 80º C, the total gas pressure being 760 torr. The contents of the vessel were pumped into a 50.0 ml, vessel at the same temperature. What were the partial pressures of oxygen and of water vapour and the total pressure in the final equilibrium state? Neglect the volume of any water which might condense
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
= 810 mm Hg, 
= 355 mm Hg, P total = 1165 mm Hg
In 100 ml vessel which contained water - saturated oxygen, the pressure of O 2 gas = 760 – 355 = 405 torr.
When the contents of this vessel were pumped into 50 ml vessel, at the same temperature, the pressure of oxygen gets doubled i.e. 
= 810 torr.
But pressure of water vapour will remain constant, as some vapour in this 50 ml vessel, gets condensed.
So 
= 355 torr & Total pressure = 810 + 355 = 1165 torr.
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