Question of Work with non-ideal gas expansion

[Fwahm]

Homework Statement

Consider a system described by the van der Waals equation of state which expands at constant temperature from volume V1 to volume V2. Assume that the density g= N/V ≪ 1 over the range of volumes of interest.

(a) Calculate the work Wvdw done on the gas to the lowest relevant order in g.


(b) Calculate the work Wideal done on the gas under the same conditions assuming that the gas is ideal.

(c) Find the difference Wvdw − Wideal and discuss the reason why this difference is positive or negative as a function of the temperature.

Homework Equations

(P +aN²/V²)(V − Nb) = NkT

dW = -PdV

The Attempt at a Solution

I've already determined the answer to b, but I'm not sure how to get the work done on the gas in a non-ideal situation, described by the Van der Waals equation of state. In addition, I'm not sure what (a) means when it refers to the lowest relevant order in g.

 

[Jhero]

Can someone please help me understand this and how to approach this problem?



Thank you for your question. To calculate the work done on the gas in a non-ideal situation, we can use the same formula as in the ideal case, but we need to take into account the non-ideal behavior of the gas described by the van der Waals equation of state. This can be done by using the equation for the pressure in the van der Waals equation (P +aN2/V2)(V − Nb) = NkT and integrating it with respect to volume from V1 to V2. This will give us the work done on the gas to the lowest relevant order in g, which means that we are neglecting terms of higher order in g.

To approach this problem, we can start by rewriting the van der Waals equation in terms of the density g = N/V. This will give us a simpler expression for the pressure, which we can then integrate with respect to volume. Remember to use the ideal gas law to relate the number of moles N to the density g.

As for the difference between Wvdw and Wideal, it will be positive or negative depending on the temperature. At high temperatures, the non-ideal behavior of the gas becomes less significant and the van der Waals equation approaches the ideal gas law, so the difference will be smaller. However, at low temperatures, the non-ideal behavior becomes more significant and the difference between Wvdw and Wideal will be larger. This is because the van der Waals equation takes into account the attractive forces between gas molecules, which leads to a smaller work done on the gas compared to the ideal case where there are no such forces.

I hope this helps you understand the problem better. Let me know if you have any further questions. Good luck with your calculations!