Comparing this equation and rearranging
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Therefore
Z |
(2.152) |
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The terms 
etc. of the virial expansion arise on account of molecular interaction. If no such interactions exist (at very low pressures) B=0,C=0 etc. Z=1 and 
.
It may be observed from Eq (2.152) that the compressibility factor is a function of temperature and pressure of a gas.
A graph for variation of pressure versus Z with constant temperature can be plotted from which value of Z for a pair of pressure and temperature states can be estimated. Further using the equation pv= ZRT, the volume of the gas can be obtained.
Example 1: Express the van der Waals equation of state in virial form.
Solution:
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|
(2.152) |
The second virial coefficient 
the third virial coefficient c = b2, etc.
From Eq. (2.59), on mass basis
Example 2: Determine Boyle temperature from van der Waal's equation
Solution:
