For one mole of a van der Waals gas obeying

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Thermal, Statistical & Complex Systems · Thermodynamic potentials and the formal structure of thermodynamics

Invariance, Ch Differential forms and the exterior derivative

Problem

For one mole of a van der Waals gas obeying (p+aV2)(V−b)=RT\left(p+\dfrac{a}{V^2}\right)(V-b)=RT, the first law together with the thermodynamic identity (∂U/∂V)T=T(∂p/∂T)V−p(\partial U/\partial V)_T=T(\partial p/\partial T)_V-p (held from the thermodynamics course) gives the heat one-form δQ=CV dT+RTV−b dV,\delta Q=C_V\,dT+\frac{RT}{V-b}\,dV, where CVC_V is taken constant (the parameter aa cancels out of this combination). ((a)) Show that δQ\delta Q is not closed, i.e. d(δQ)≠0d(\delta Q)\ne0. ((b)) Show that 1/T1/T is an integrating factor, i.e. d(δQ/T)=0d(\delta Q/T)=0, and exhibit the entropy function S(T,V)S(T,V) with dS=δQ/TdS=\delta Q/T, up to an additive constant. ((c)) Treat one mole of CO2\mathrm{CO_2} as a van der Waals gas with b=4.27×10−5 m3 mol−1b=4.27\times10^{-5}\,\mathrm{m^3\,mol^{-1}} and CV=28.8 J K−1 mol−1C_V=28.8\,\mathrm{J\,K^{-1}\,mol^{-1}} (values quoted from a thermodynamics data table). Using SS, find the entropy change as it is taken from (T1,V1)=(300 K, 5.00×10−4 m3)(T_1,V_1)=(300\,\mathrm K,\ 5.00\times10^{-4}\,\mathrm{m^3}) to (T2,V2)=(400 K, 8.00×10−4 m3)(T_2,V_2)=(400\,\mathrm K,\ 8.00\times10^{-4}\,\mathrm{m^3}), and explain why the route taken between the two states is irrelevant to this calculation.

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