- #1

Dazed&Confused

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## Homework Statement

Assume that one mole of an ideal van der Waals fluid is expanded isothermally, at temperature [itex]T_h[/itex] from an initial volume [itex]V_i[/itex] to a final volume [itex]V_f[/itex]. A thermal reselvoir at temperature [itex]T_c[/itex] is available. Apply [tex]

dW_{RWS} = \left ( 1 - \frac{T_{RHS}}{T} \right ) (-dQ) +(-dW) [/tex]

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 [itex]pdV[/itex] to obtain the total work delivered to the reversible work source.

## Homework Equations

Van der Waal equations:

[tex]

u + a/v = cRT

[/tex]

where [itex]u, a, v, c, R, T[/itex] are, respectively, energy per mole, constant, volume per mole, another constant, temperature.

[tex]

p = \frac{RT}{v-b} - \frac{a}{v^2}

[/tex]

The entropy is

[tex]

S = NR\log [ (v-b)(cRT)^c] + Ns_0

[/tex]

where [itex]N[/itex] is the number of moles and [itex]b[/itex] is another constant.

## The Attempt at a Solution

[/B]

Using the first equation with [itex]T=T_h, T_{RHS} = T_c[/itex] and [itex]-dW = pdV[/itex] and integrating we get

[tex]

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}

[/tex]

Also energy conservation gives

[tex]

\Delta u + W + Q =0

[/tex]

and entropy conservation

[tex] R \log \left ( \frac{v_f-b}{v_i - b} \right) + \frac{Q}{T_c} = 0

[/tex]

Finally the energy change is

[tex]

\Delta u = \frac{a}{v_i} - \frac{a}{v_f}

[/tex]

Everything seems to work out except that fraction [itex] T_c/T_h[/itex] is the wrong way round and I see no way of dealing with this. Help would be appreciated.

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