# Van der Waal expansion and delivered work

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1. Aug 3, 2017

### Dazed&Confused

1. The problem statement, all variables and given/known data
Assume that one mole of an ideal van der Waals fluid is expanded isothermally, at temperature $T_h$ from an initial volume $V_i$ to a final volume $V_f$. A thermal reselvoir at temperature $T_c$ is available. Apply $$dW_{RWS} = \left ( 1 - \frac{T_{RHS}}{T} \right ) (-dQ) +(-dW)$$
to a differential process and integrate to calculate the work delivered to a reversible work source (RWS). RHS is reversible heat source. Corroborate by overall energy and entropy conservation.

Hint: remember to add the direct work transfer $pdV$ to obtain the total work delivered to the reversible work source.

2. Relevant equations
Van der Waal equations:
$$u + a/v = cRT$$
where $u, a, v, c, R, T$ are, respectively, energy per mole, constant, volume per mole, another constant, temperature.
$$p = \frac{RT}{v-b} - \frac{a}{v^2}$$

The entropy is
$$S = NR\log [ (v-b)(cRT)^c] + Ns_0$$
where $N$ is the number of moles and $b$ is another constant.

3. The attempt at a solution

Using the first equation with $T=T_h, T_{RHS} = T_c$ and $-dW = pdV$ and integrating we get
$$W_{RHS} = - \left ( 1- \frac{T_c}{T_h} \right) Q + RT_h \log \left ( \frac{v_f-b}{v_i - b} \right) + \frac{a}{v_f} - \frac{a}{v_i}$$
Also energy conservation gives
$$\Delta u + W + Q =0$$
and entropy conservation
$$R \log \left ( \frac{v_f-b}{v_i - b} \right) + \frac{Q}{T_c} = 0$$
Finally the energy change is
$$\Delta u = \frac{a}{v_i} - \frac{a}{v_f}$$
Everything seems to work out except that fraction $T_c/T_h$ is the wrong way round and I see no way of dealing with this. Help would be appreciated.

Last edited: Aug 3, 2017
2. Aug 3, 2017

### Dazed&Confused

I see my problem: there are two Qs. The first equation was the heat from the subsystem and the energy and entropy the heat RHS.