Van der Waal expansion and delivered work

[Dazed&Confused]

Homework Statement

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 <br /> 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.

Homework Equations

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

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

The Attempt at a Solution

[/B]
Using the first equation with T=T_h, T_{RHS} = T_c and -dW = pdV and integrating we get
<br /> 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}<br />
Also energy conservation gives
<br /> \Delta u + W + Q =0<br />
and entropy conservation
R \log \left ( \frac{v_f-b}{v_i - b} \right) + \frac{Q}{T_c} = 0<br />
Finally the energy change is
<br /> \Delta u = \frac{a}{v_i} - \frac{a}{v_f}<br />
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.

 

[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.