Van der Waal expansion and delivered work

In summary, the conversation discusses the application of the equation dW_{RWS} = \left ( 1 - \frac{T_{RHS}}{T} \right ) (-dQ) +(-dW) to calculate the work delivered to a reversible work source (RWS) in an isothermal expansion of an ideal van der Waals fluid. The Van der Waal equations, entropy and energy conservation are also mentioned. The final solution shows that everything works except for the fraction T_c/T_h, which can be resolved by considering two separate heat sources.
  • #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.
 
Last edited:
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  • #2
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.
 

Related to Van der Waal expansion and delivered work

1. What is Van der Waal expansion?

Van der Waal expansion is a thermodynamic process that describes the change in volume of a gas due to changes in temperature and pressure. It takes into account the attractive and repulsive forces between gas molecules, which can cause deviations from ideal gas behavior.

2. How does Van der Waal expansion differ from ideal gas expansion?

Unlike ideal gases, which have negligible molecular size and no intermolecular forces, Van der Waal expansion takes into account the size of gas molecules and the attractive and repulsive forces between them. This results in a non-linear relationship between pressure, volume, and temperature, as well as a deviation from the ideal gas law.

3. What is the equation for Van der Waal expansion?

The Van der Waal equation is given by (P + a(n/V)^2)(V - nb) = nRT, where P is the pressure, V is the volume, n is the number of moles, T is the temperature, and a and b are Van der Waal constants specific to each gas.

4. How is delivered work related to Van der Waal expansion?

Delivered work is the work done by or on a system during a thermodynamic process. In the case of Van der Waal expansion, the work done on the gas is equal to the area under the curve on a pressure-volume graph. This is because work is the integral of pressure with respect to volume, and the Van der Waal equation describes the relationship between pressure and volume.

5. What factors affect the Van der Waal constants a and b?

The Van der Waal constants a and b are specific to each gas and are affected by the size of the gas molecules and the strength of intermolecular forces. Generally, larger molecules and stronger intermolecular forces result in larger values for a and b, which lead to a larger deviation from ideal gas behavior.

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