Derive the energy equation for a van der Waal gas.

[McAfee]

Homework Statement

Derive the energy equation for a van der Waal gas when T and v are the independent variables.

Homework Equations

I'm required to start from: du=\frac{\partial u}{\partial T} dT +\frac{\partial u}{\partial v} dv

The Attempt at a Solution

We'll when I first started this problem. I just started doing random derivatives that made sense to me but after looking it over I had no clue what I was doing.

Does anyone know what the answer I should be getting after I fully derive it?

 

[vanhees71]

I don't know, what's given. So I can't help to find an answer to your question. One way to derive the caloric equation of state, i.e., the internal energy U is to use the canonical partition sum
Z=\frac{z^N}{N!}
with the single-particle partition function
z=\frac{V^*}{\Lambda^3} \exp \left (-\frac{\phi}{2 T} \right),
where V^*=V-2 \pi N d^3/3 is the available volume (geometrical volume minus excluded volume due to the hard-sphere model for the molecules), and \phi is the interaction energy of the particles
\phi=\frac{N}{V} \int_d^\infty \mathrm{d} r \; U_{\text{rel}}(r) 4 \pi r^2,
with the effective two-body potential
U_{\text{rel}}(r)=\begin{cases}<br /> \infty &amp; \text{for} \quad r \leq d ,\\<br /> -\alpha (d/r)^6 &amp; \text{for} \quad r&gt;d.<br /> \end{cases}<br />
The Helmholtz free energy is given by
A(T,V,N)=-T \ln Z,
and from this you can derive all thermodynamical quantities from the usual thermodynamic relations like
p=-\left (\frac{\partial A}{\partial V} \right)_{T,N}
etc. The internal energy is given by the usual Legendre transformation
U=A+T S.

 

[Chestermiller]

dU=TdS-PdV=T\left(\frac{∂S}{∂T}\right)dT+T\left(\frac{∂S}{∂V}\right)dV-PdV=<br /> C_VdT+(T\left(\frac{∂S}{∂V}\right)-P)dV
The next step is to determine ∂S/∂V at constant T. This can be obtained from a Maxwell relation, starting from the equation dA=-SdT-PdV.

S=-∂A/∂T
P=-∂A/∂V

So \frac{∂S}{∂V}=\frac{∂P}{∂T}

Therefore,
dU=C_VdT+(T\left(\frac{∂P}{∂T}\right)-P)dV

The second term in this equation is zero for an ideal gas, but not for a real gas. Just substitute the van der Waals equation into the second term of this equation.

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