I have read in multiple places that the virial coefficients in the virial equation of state, Z=1+B/v+C/v^2, are functions of temperature only and are independent of pressure (or equivalently, molar volume). Is this purely an empirical observation or is there some deeper statistical mechanics reason why the virial coefficients should not depend on how close the particles are together (i.e. the molar volume)?

miklintisyt

miklintisyt

Answered question

2022-10-21

I have read in multiple places that the virial coefficients in the virial equation of state, Z = 1 + B v + C v 2 . . ., are functions of temperature only and are independent of pressure (or equivalently, molar volume). Is this purely an empirical observation or is there some deeper statistical mechanics reason why the virial coefficients should not depend on how close the particles are together (i.e. the molar volume)?

Answer & Explanation

Teagan Zamora

Teagan Zamora

Beginner2022-10-22Added 18 answers

The concrete derivation for non-ideal gas is a little tricky and not very illuminating in my opinion. So, let us write the grand canonical partition function
Ω n Z n λ n
where λ = e μ k B T and Z n is n-particle canonical partition function. Now, the virial expansion is nothing else than cluster expansion of Ω. This is just a combinatorial theorem, so that you can write
log ( Ω ) = n B n λ n
with B n = V n 1 f n ( Z 1 , , Z n ) and f n determined by the cluster expansion theorem. Because none of the Z n depends on the pressure, B n can't either. Actually, the pressure enters these formulas only through the thermodynamic identity log ( Ω ) = p V k B T

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