# Homework Help: Joule-Thomson Expansion of a van der Waals gas

1. Mar 9, 2006

### eep

Hi,
Consider a pipe with thermally insulated walls. A thermally insulating porous plug in the pipe provides a constriction to the flow of gas. We model this as a sudden jump in pressure. A continous stream of gas flows from left to right, with the pressure $p_1$ upstream being larger than the pressure $p_2$ downstream. The gas is in thermal equilibrium on each of the two sides.

For a van der Waals gas, what is the minimal starting temperature for cooling to occur in a Joule-Thomson expansion?

I was able to derive that the enthalphy for a van Der Waals gas is

$$H = \frac{5}{2}N\tau + N^2(\frac{b\tau}{V - Nb} - \frac{2a}{V})$$

and I know that enthalphy is conserved in the process.

I set the enthalphy on both sides equal to one another, and then set $\tau_2$ equal to zero since that is the lowest possible temperature. Solving the equation for $\tau_1$ left me with

$$\tau_1 = \frac{2Na(\frac{V_1 + V_2}{V_1V_2})}{\frac{5}{2} + \frac{Nb}{V_1 - Nb}}$$

I assumed that N would be the same on both sides of the barrier. I don't like this result, however, as it depends on $V_1$ and $V_2$. Is this correct or should I be approaching this differently?