Joule-Thomson Expansion of a van der Waals gas

[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 continuous 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?

 

[Chestermiller]

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 continuous 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?

This was not done correctly. In order to be at the boundary between heating and cooling, we must have $T_2=T_1$.

For a VDW gas, we find that $$du=C_vdT+a\frac{dv}{v^2}$$The form of this equation implies that Cv is independent of v, and so is equal to the ideal gas Cv. So integrating this equation from the initial to the final state yields $$\Delta u=a\left(\frac{1}{v(T_1,P_1)}-\frac{1}{v(T_1,P_2)}\right)$$So we have $$\Delta h=a\left(\frac{1}{v(T_1,P_1)}-\frac{1}{v(T_1,P_2)}\right)+P_2v(T_1,P_2)-P_1v(T_1,P_1)$$$$=0$$
One needs to solve for the value of T1 that makes good on this equation.