Question

Air containing 20.0 mol% water vapor at an initial pressure of 1 atm absolute is cooled in a 1- liter sealed vessel from 200°C to 15°C. (a) What is the pressure in the vessel at the end of the process? (Hint: The partial pressure of air in the system can be determined from the expression pair = nairRT/V and P = pair + pH2O. You may neglect the volume of the liquid water condensed, but you must show that condensation occurs.) (b) What is the mole fraction of water in the gas phase at the end of the process? (c) How much water (grams) condenses?

Answer 1

This solution is quite lengthy

Total system = nRT

n was solved to be 0.02575

nH20 = 0.2x0.02575

= 0.00515

Nair = 0.0206

PH20 = 0.19999

Pair = 1-0.19999

= 0.80001

At 15⁰c

Pair = 0.4786atm

I used antoine's equation to get pressure

The pressure = 0.50

2. Moles of water vapor = 0.0007084

Moles of condensed water = 0.0044416

Grams of condensed water = 0.07994

Please refer to attachment. All solution is in there.