Joule-Thomson Expansion of a van der Waals gas

In summary, the conversation discusses modeling a thermally insulated pipe with a porous plug and a van der Waals gas in thermal equilibrium. The group attempts to find the minimal starting temperature for cooling to occur in a Joule-Thomson expansion. However, the derived equation for enthalpy is incorrect and the correct approach is to set the initial and final temperatures equal to each other. The conversation ends with a discussion about solving for the value of T1 that satisfies this equation.
  • #1
eep
227
0
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 [itex]p_1[/itex] upstream being larger than the pressure [itex]p_2[/itex] 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

[tex]
H = \frac{5}{2}N\tau + N^2(\frac{b\tau}{V - Nb} - \frac{2a}{V})
[/tex]

and I know that enthalphy is conserved in the process.

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

[tex]
\tau_1 = \frac{2Na(\frac{V_1 + V_2}{V_1V_2})}{\frac{5}{2} + \frac{Nb}{V_1 - Nb}}
[/tex]

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 [itex]V_1[/itex] and [itex]V_2[/itex]. Is this correct or should I be approaching this differently?
 
Physics news on Phys.org
  • #2
eep said:
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 [itex]p_1[/itex] upstream being larger than the pressure [itex]p_2[/itex] 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

[tex]
H = \frac{5}{2}N\tau + N^2(\frac{b\tau}{V - Nb} - \frac{2a}{V})
[/tex]

and I know that enthalphy is conserved in the process.

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

[tex]
\tau_1 = \frac{2Na(\frac{V_1 + V_2}{V_1V_2})}{\frac{5}{2} + \frac{Nb}{V_1 - Nb}}
[/tex]

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 [itex]V_1[/itex] and [itex]V_2[/itex]. 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.
 

1. What is the Joule-Thomson Expansion of a van der Waals gas?

The Joule-Thomson Expansion of a van der Waals gas is a process in thermodynamics where a gas expands from a high pressure region to a lower pressure region, causing a decrease in temperature. This phenomenon is described by the Joule-Thomson coefficient, which takes into account the gas's deviation from ideal behavior due to intermolecular interactions.

2. How does the Joule-Thomson Expansion affect the temperature of a gas?

The Joule-Thomson Expansion causes a decrease in temperature of a gas as it expands from a higher pressure to a lower pressure. This is due to the energy required to overcome intermolecular forces, which results in a decrease in the gas's internal energy and temperature.

3. What is the significance of the van der Waals equation in the Joule-Thomson Expansion?

The van der Waals equation is a more accurate representation of real gases compared to the ideal gas law. It takes into account the effects of intermolecular forces and the finite size of gas molecules, which are important factors in the Joule-Thomson Expansion. The van der Waals equation allows for a better understanding of the behavior of gases in this process.

4. How does the Joule-Thomson coefficient relate to the critical point of a gas?

The Joule-Thomson coefficient is directly related to the critical point of a gas. At the critical point, the Joule-Thomson coefficient becomes zero, meaning that there is no change in temperature during expansion. This is because at the critical point, the gas is at its maximum entropy and any further changes in pressure or temperature will not result in a change of state.

5. How is the Joule-Thomson Expansion used in industrial applications?

The Joule-Thomson Expansion is used in various industrial applications, such as refrigeration and liquefaction of gases. By controlling the pressure and temperature during the expansion process, gases can be cooled or condensed for use in various industries, such as food storage, chemical production, and natural gas processing.

Similar threads

  • Advanced Physics Homework Help
Replies
6
Views
2K
  • Advanced Physics Homework Help
Replies
7
Views
2K
  • Advanced Physics Homework Help
Replies
8
Views
2K
  • Advanced Physics Homework Help
Replies
6
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
759
  • Advanced Physics Homework Help
Replies
6
Views
3K
  • Advanced Physics Homework Help
Replies
4
Views
3K
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
2
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
2K
Back
Top