Internal energy and entropy of a Van der Waals gas
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7.4
Consider n moles of a Van der Waals gas. Show that (dU/dV)_T = n^2a/V^2. Hence show that the internal energy is
U = the integral from zero to T of C_vdT - an^2/V + U0
where U0 is a constant. {Hint: Express U = U(T,V)}.
7.5
As in the previous question, consider n moles of Van der Waals gas. Show that
(a) S = the integral from zero to T of C_v/TdT + nRln(V - nb) + S0
where S0 is a constant. {Hint: Use dS = 1/T(dU + PdV)
(b) The equation for a reversible adiabatic process is
T(V -nb)^(nR/C_v) = a constant
if C_v is assumed to be independent of T.
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Solution Summary
I explain in detail how the desired expressions for the internal energy and entropy can be obtained.
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The fundamental thermodynamic relation expresses dU in terms of dS and dV
dU = T dS - P dV (1)
If we want to consider U as a function of T and V, then all we need to do is rewrite this in terms of dT and dV. We substitute:
dS = (dS/dT)_V dT + (dS/dV)_T dV (2)
Here (dX/dY)_Z denotes the partial derivative of X w.r.t. Y at constant Z. Substituting (2) in (1) gives:
dU = T (dS/dT)_V dT + (T (dS/dV)_T - P) dV
The term T (dS/dT)_V is the heat capacity at constant volume. The term T dS is always the heat supplied to the system in case of a quasistatic change, so the T times the derivative of S w.r.t. temperature is the heat capacity. Because we keep V constant in the derivative, it is the heat capacity at constant volume. So, we can write:
dU = Cv dT + (T (dS/dV)_T - P) dV (3)
We can simplify the term (dS/dV)_T ...
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