Question

The density of water vapour at $327.6atm$ and $776.4K$ is $133.2gm/d{m}^3$. Determine the molar volume, $V_m$ of water and the compression factor.

• A. $Vm$=5.0775mol/L $Z=1.457$
• B. $Vm$=6.0775mol/L $Z=13.457$
• C. $Vm$=5.0775mol/L $Z=4.457$
• D. $Vm$=6.0775mol/L $Z=1.457$

Answer 1

The correct option is A $Vm$=5.0775mol/L $Z=1.457$

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