Question

On a particular summer day, the temperature is 30 degrees Celsius and the relative humidity is 80% (which means that the partial pressure of water vapor in the ambient air is 80% of the equilibrium vapor pressure of water). A sample of this air is placed in a 1 Liter flask which is then closed and cooled to 5 degrees Celsius. What is the mass in grams of water that still exists as vapor in the flask? How much liquid water in grams condenses out?

Answer 1

Vapor pressure of water at 80 deg.c =0.4675 atm ( from vapor pressure tables)

given relative humidity =80% =0.8= partial presssure of water vapor/ Vapor pressure of liqud at the given temmperatute

Partial pressure of water vapor= 0.8* vapor pressure of water vapor= 0.8*0.4675 atm=0.374 atm

Partial pressure of water vapor/ partial pressure of dry air = 0.374/(1-0.374)= 0.59744 = moles of water vapor/ moles of dry air

Total moles of wet air at 30 deg.c in 1 L flask can be calculated using gas law equation

PV= nRT

where P= 1atm V= 1L R= 0.08206 Lit.atm/mole.K T= 30 deg.c =30+273.15= 303.15K

n= number of moles of wet air at 30 deg.c =PV/RT= 1*1/ (0.08206*303.15)= 0.040199 moles

Moles of water vapor/moles of dry air = 0.59744

Moles of dry air/ moles of water vapor =1/0.59744=1.674  

Moles of dry air = 1.674 moles of water vapor (1)

but moles of water vapor + moles of dry air = 0.040199 (2)

substitutnng eq.1 in Eq.2

1.674 moles of water vapor + moles of water vapor= 0.040199

2.674* moles of water vapor = 0.040199

Moles of water vapor =0.040199/2.674=0.014671 ( at 30 deg.c)

moles of dry air= 1.674* 0.014671=0.024559

The wet air is cooled to 5deg.c where vapor pressure of water vapor = partial pressure of water vapor = 0.0086 atm

at 5 deg.c

Moles of water vapor/ moles of dry air =partial pressure of water vapor/ parital pressure of dry air

Moles of water vapor/0.024559= 0.0086/(1-0.0086)=0.0086746

Moles of water vapor =0.024559*0.0086746= 0.000213 moles of water vapor = 0.00213*18 gms of water vapor = 0.003835 gms

Water condensed = water vpaor at 30 deg.c - water vapor at 5 deg.c = 0.014671-0.00213= 0.014458 moles= 0.014458*18= 0.2602 gms