How does the result compare with the corresponding equation for an ideal gas?
Z = 1 + B?P + C? + D?
+ . . .
Derive an equation for the work of mechanically reversible, isothermal compression of 1 mol of a gas from an initial pressure to a final pressure
when the equation of state is the virial expansion truncated to:
Z = 1 + B?P
We have, in general for a mechanically reversible process, dW = -PdV. If our equation of state is V = RT/P (1 + B’P) = RT /P + RTB’, then dV = -RT/ dP. So, dW = -P(-RT/
) dP = RT/P dP. Integrating this from
to
gives W = RT ln(
/
),which is identical to the result for an ideal gas. The second term in this two-term form of the virial expansion does not affect the work of an isothermal expansion, because when we write the volume as an explicit function of pressure, the term containing the 2nd virial coefficient is independent of pressure.
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