# Derive the energy equation for a van der Waal gas.

1. Mar 21, 2014

### McAfee

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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?

2. Mar 21, 2014

### 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
$$U_{\text{rel}}(r)=\begin{cases} \infty & \text{for} \quad r \leq d ,\\ -\alpha (d/r)^6 & \text{for} \quad r>d. \end{cases}$$
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
$$U=A+T S.$$

3. Mar 22, 2014

### Staff: Mentor

$$dU=TdS-PdV=T\left(\frac{∂S}{∂T}\right)dT+T\left(\frac{∂S}{∂V}\right)dV-PdV= C_VdT+(T\left(\frac{∂S}{∂V}\right)-P)dV$$
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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