Question

A 4 liter sealed container contains water vapour at 0.4 atm and 90°C.

a) What is the dew point of the vapour?

b) Determine the percentage of the vapour that will condense if the system is cooled to 50°C.

Answer 1

Ans-

Pvap = Ptotal and just looked up that corresponding pressure in the charts to find the dew point as 76.4 degree celsius. .

Pvap_50C = Ptotal_50C

Assuming that the volume of the condensate is negligible compared to the 4L total volume, you now have 4L vapor at the newly found pressure Ptotal_50C = Pvap_50C.
Assuming we can treat the water vapor as an ideal gas (n = PV/RT), the moles of vapor are:
n_initial = (0.4atm)(4L)/[R(273+90)]
n_final = (Pvap_50C)(4L)/[R(273+50)]
and the fraction of vapor remaining as vapor would be
n_final/n_initial = (Pvap_50C)(273+90)/[(0.4atm)(273+50)]

So the fraction condensed would be 1 minus that.