The vapour pressure of water at `80^(@)C` is `355` torr. A `100 ml` vessel contains water-saturated oxygen at `80^(@)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, what was the total pressure in the final equilibrated state? Neglect the volume of any water which might condense.

Answer

`P_(O_(2))=810mmHg, P_(H_(2))O=355mm Hg, P_("total")=1165mm Hg`

Explanation

In `100ml` 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 `50ml` at the same temperature, pressure of oxygen gets doubled i.e. `P_(O_(2)=810` torr.
But pressure of water vapour will remain constant, as some vapour in this `50ml` vessel, gets condense. So `P