• Astrocyte
In summary: I used the Maxwell relation to calculate the partial derivative of entropy with respect to volume. This allowed me to write the change in entropy in terms of the change in temperature and volume, as shown in the final result. In summary, the conversation discusses the change in temperature and entropy for a van der Waals gas in free expansion. The speakers share their calculations and results, leading to a discussion about the decrease in entropy for this process. The expert summarizes the equations and explains the use of the Maxwell relation in their analysis. They conclude that the overall change in entropy must be positive, despite the negative term in the final result.
Astrocyte
Homework Statement
What is the change in entropy of van der Waals gas in free expansion?
Free expansion from V to 2V
Relevant Equations
(p+aN^2/V^2)(V-Nb)=NkT
Previous of this problem, there was another problem. that is "What is the change in Temperature of van der Waals gas in free expansion?".
I got them.
It was
C_V dT= -aN^2/V^2 dV
Then, I got
T=T0-aN^2/2VC_V
So i knew that the Temperature is decreased by free expansion in adiabatic process.
Then I calculated some about the change of entropy.

And I notice that this result says "entropy can be decreased for Free expansion of Waals gas.
What did i do wrong for this calculation?

Last edited:
I get something similar, solving it a very different way, and working in terms of specific volume v and specific entropy s: $$\Delta s=C_v\ln{\left(1-\frac{a}{2vT_0C_v}\right)}+R\ln{\left(\frac{2v-b}{v-b}\right)}$$where Cv is independent of v and T, and thus equal to the heat capacity in the ideal gas limit.

Astrocyte and etotheipi

And I notice that this result says "entropy can be decreased for Free expansion of Waals gas.
What did i do wrong for this calculation?
I don't think you did anything wrong. From ##\Delta S = \int_V^{2V}\frac{P}{T}dV##, you can see that ##\Delta S## must be positive since ##P##, ##T##, and ##V## are positive. The fact that ##P## is positive, means that in the equation ##P = \frac{NkT}{V-Nb} - \frac{aN^2}{V^2}##, the first term on the right "outweighs" the second term on the right. These two terms lead to the two terms in your final result $$\Delta S = Nk \log \left (1+ \frac{V}{V-Nb} \right)+C_V\log \left (1- \frac{aN^2}{2VT_0C_V} \right)$$

So, although the second term in the result is negative, the overall expression must be positive.

Astrocyte and etotheipi
My analysis was based on the following equations for an arbitrary fluid:
$$ds=\left(\frac{\partial s}{\partial T}\right)_vdT+\left(\frac{\partial s}{\partial v}\right)_Tdv$$with $$\left(\frac{\partial s}{\partial T}\right)_v=\frac{C_v}{T}$$and$$\left(\frac{\partial s}{\partial v}\right)_T=\left(\frac{\partial P}{\partial T}\right)_v$$For a van der Waals gas, the latter equation reduces to:$$\left(\frac{\partial s}{\partial v}\right)_T=\frac{R}{(v-b)}$$These equations imply that Cv depends only on T.

Astrocyte
TSny said:
So, although the second term in the result is negative, the overall expression must be positive.
I understood it. Thank you.
Temperature T is always positive and P is also positive. So, the integration must be positive.
In my result, the second term will be negative. But, the size of negative term is not big enough to exceed the first term.
I should check with graph.

Chestermiller said:
My analysis was based on the following equations for an arbitrary fluid:
$$ds=\left(\frac{\partial s}{\partial T}\right)_vdT+\left(\frac{\partial s}{\partial v}\right)_Tdv$$with $$\left(\frac{\partial s}{\partial T}\right)_v=\frac{C_v}{T}$$and$$\left(\frac{\partial s}{\partial v}\right)_T=\left(\frac{\partial P}{\partial T}\right)_v$$For a van der Waals gas, the latter equation reduces to:$$\left(\frac{\partial s}{\partial v}\right)_T=\frac{R}{(v-b)}$$These equations imply that Cv depends only on T.
Did you use The Maxwell relation? I know about the name. but I'm not learn about that yet.
But, I think the relation makes other problem very easy.

Astrocyte said:
Did you use The Maxwell relation?
Yes

Astrocyte

## 1. What is entropy?

Entropy is a thermodynamic quantity that measures the amount of disorder or randomness in a system. It is a measure of the number of possible arrangements of a system's particles that are consistent with its observed macroscopic properties.

## 2. How is entropy related to the free expansion of a Waals gas?

In the context of the free expansion of a Waals gas, entropy is related to the change in the number of microstates of the gas. As the gas expands, the number of microstates increases, leading to an increase in entropy.

## 3. Is entropy always increased during free expansion of a Waals gas?

Yes, entropy is always increased during free expansion of a Waals gas. This is because the gas is expanding into a larger volume, resulting in an increase in the number of possible microstates and therefore an increase in entropy.

## 4. Can entropy be decreased during free expansion of a Waals gas?

No, entropy cannot be decreased during free expansion of a Waals gas. This is due to the Second Law of Thermodynamics, which states that the total entropy of a closed system always increases over time.

## 5. How does the Second Law of Thermodynamics apply to the free expansion of a Waals gas?

The Second Law of Thermodynamics applies to the free expansion of a Waals gas by stating that the total entropy of the gas and its surroundings must increase during the expansion process. This is because the gas is expanding into a larger volume, resulting in an increase in the number of possible microstates and an overall increase in entropy.

• Introductory Physics Homework Help
Replies
10
Views
20K
• Introductory Physics Homework Help
Replies
2
Views
947
• Introductory Physics Homework Help
Replies
2
Views
257
• Introductory Physics Homework Help
Replies
4
Views
978
• Introductory Physics Homework Help
Replies
8
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
365
• Introductory Physics Homework Help
Replies
8
Views
1K
• Introductory Physics Homework Help
Replies
5
Views
251
• Introductory Physics Homework Help
Replies
3
Views
4K
• Introductory Physics Homework Help
Replies
4
Views
820