Why is the pressure of a Van der Waals gas less than that of an ideal gas?
Your assertion is in its generality incorrect. Consider the virial expansion of the equation of state corresponding to an interacting gas [1] and compare the terms in this expansion with the similar expansion of the van der Waals equation of state. This shows how the coefficients in the resulting expansion for P/T (P = pressure, T = temperature) vary as a function of the temperature and how their magnitudes and signs depend on the overall shape of the interaction potential [2].
[1] K Huang, Statistical Mechanics, 2nd edition (John Wiley & Sons, New York, 1987).
[2] See in particular Sections 7.6 - 7.7, pp. 168 - 169, in [1]. Note in particular the data in the table on p. 170.
Hello, thanks for your answer.
A professor of my university made that statement. And he asked us the reason.
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