Question

A mole of air is sampled from the atmosphere when the atmospheric pressure is 765 mmHg, the temperature is 25 C, an relative humidity is 75%. The sample of air is placed inside a closed container and heated to 135 C and then compressed to 2 atm. What are the relative humidity, the humidity, and the mole fraction of water in the compressed air?

Answer 1

1)   AH = m(H2O)/V   ,V= of the air and water vapor mixture

Relative humidity=RH=p(H2O)/P*(H2O)

P(H2O)= partial pressure of water vapor in the mixture

P*(H2O)= equilibrium vapor pressure of water at a given temperature

Given RH =75% at when the atmospheric pressure is 765 mmHg, the temperature is 25 C, an relative humidity is 75%.

The saturated vapor pressure of water may be calculated from the relation, Antoine’s equation.

P (mmHg)= exp(20.438 – 5044/T(K)

T=298K

P*(H2O)= exp(20.438 – 5044/298K)=23.78 mmhg

P(H2O)= P(H2O)*0.75=23.78 mmhg*0.75=17.83 mmhg

Atm P=total pressure=1047.38mmhg

Also P(H2O)=mole fraction of water(x(H2O))* total pressure of (air+water=atmospheric)[Dalton’s law of partial pressure]

X(H2O)=p(H2O)/P =17.83 mmhg/765 mmhg=0.0233 (mole fraction of water vapour)

After compression ,to p2=1520 mmhg

T2=408K

X(H2O)=p(H2O)/P (same mole fraction in container)

0.0233=p(H2O)/ 1520 mmhg

p(H2O)=0.0233*1520 mmhg=35.42 mmhg

RH2= p(H2O)/P*(H2O)=35.42/23.78=1.489

RH2%=1.489*100%=148.9%

2) AH=mass/Vnet

Mole of air mix=1

Mole of water=nw=0.0233 mol

Mass of water=0.0233 mol*18g/mol=0.4194 g

PV=nRT

Vnet =nRT/P=1 mol*0.0821 L atm/K mol*408K/2 atm=16.75L

AH=0.4194g/16.75L=0.0250

AH=0.0250g/L