Question

what is the compressibility factor of water vapour at 10 c and 1 atm pressure. it's molar volume is 33.18 dm^3

Answer 1

Answer:

Given,

Density of water vapour=133.2gm/L

Pressure=327.6atm

Temperature=776.4K

Now we know,Ideal gas law

PV=nRT

Molar volume=n/V=P/RT

n/V=327.6/(0.0831*776.4)

=5.0775mol/L

Molar volume of real gas given calculation:

We know,1 mole water gas=18g

133.2g water gas=133.2/18=7.4 mole

Actual volume of given gas=7.4mol/L

Now we know,

Z=actual volume of real gas/volume for ideal gas

Z=7.4/5/0775=1.457