Is the Entropy in a Free Expansion Affected by the Gas Type?

[Dario SLC]

Homework Statement

The question is:
What happened with the entropy in a free expansion? The system is isolated and the state equation is:
$$p=AT/V+B/V^2$$

Homework Equations

$$dU=TdS-pdV$$

The Attempt at a Solution

My attempt is:
Because the system is isolated and corresponding to an free expansion then the energy is constant since there aren't interchange of heat and work, but there is a work due to expansion of the gas, that is $pdV$. Then the first and second principle is:
$$TdS-pdV=0$$
then the entropy it can be calculated for integration and using the state equation, and because the initial volume is than minor final volumen, the entropy must increase like must be.

$$S(T)=A\ln\left(\frac{V_2}{V_1}\right)+\frac{B}{T}\left(\frac{V_2-V_1}{V_2V_1}\right)>0$$

My doubt is if this argument valid for any gas or only must be ideal gas? The state equation isn't of the ideal gas, is similar to van der Waals gases.

Thanks a lot!

 

[Chestermiller]

This is not done correctly, because the temperature can change. What is the equation for dU in terms of dT and dV? What is the equation for dS in terms of dT and dV?

 

[Dario SLC]

This is not done correctly, because the temperature can change. What is the equation for dU in terms of dT and dV? What is the equation for dS in terms of dT and dV?

Ok, $$\frac{\partial U}{\partial V}_T=\left(T\left(\frac{ \partial p}{ \partial T}\right)_V-p\right)dV=-\frac{B}{V^2}$$ and
$$\frac{\partial U}{\partial T}_V=T\left(\frac{ \partial S}{ \partial T}\right)_V=c_V(T)$$
now because it is a free expansion and isolated $dU=0$ then $c_V(T)dT=\frac{B}{V^2}dV$ and $C_V$ in a van der Waals gases only depend of T because for definition $C_V(T)=T\left(\frac{ \partial S}{ \partial T}\right)_V$ and derivation respect to $V$:
$$\frac{\partial c_V}{\partial V}=T\left(\frac{ \partial}{ \partial V}\left(\frac{\partial S}{\partial T}\right)_V\right)_T=T\left(\left(\frac{\partial}{\partial T}\frac{\partial p}{\partial T}\right)_V\right)_T=0$$
but I don't know the form for $c_V(T)$, only I learn that:
$$\int C_V(T)dT=B(\frac{1}{V_1}-\frac{1}{V_2})>0$$
if $C_V(T)=C_V$, then $T_2-T_1=\frac{B}{C_V}\left(\frac{1}{V_1}-\frac{1}{V_2}\right)$ but I can't see nothing. I understand that the $T_2=T_1$ but not necessary in the intermediates dots

For dS I have:
$$dS=\left(\frac{ \partial S}{ \partial T}\right)_VdT+\left(\frac{ \partial p}{ \partial T}\right)_VdV$$
in advance it is clear that the entropy must be greater for the Clausius inequality in a irreversible process.
But I see no nothing, the entropy is so dark...I still thinking

 

[Chestermiller]

You've done a pretty good job so far. Assuming that Cv doesn't depend on T, you've shown that:
$$T_2=T_1+\frac{B}{C_v}\left(\frac{1}{V_1}-\frac{1}{V_2}\right)$$
You've also shown that $$dS=\left(\frac{ \partial S}{ \partial T}\right)_VdT+\left(\frac{ \partial p}{ \partial T}\right)_VdV=C_v\frac{dT}{T}+\frac{A}{V}dV$$Like the case of dU, this is also an exact differential. What do you get when you integrate this to get $\Delta S$? Then you can substitute for $T_2$ and you will have $\Delta S$ as a function of $T_1$, $V_1$, and $V_2$.

 

[Dario SLC]

Like the case of dU, this is also an exact differential. What do you get when you integrate this to get $\Delta S$? Then you can substitute for $T_2$ and you will have $\Delta S$ as a function of $T_1$, $V_1$, and $V_2$.

Sorry, but I don't understand this issue. If I write $dU=TdS-pdV$ and using the expression for $dS$:
$$dU=T\left(\frac{c_V}{T}dT+\frac{A}{V}dV\right)-pdV$$
$$dU=c_VdT+\frac{AT}{V}-pdV=c_VdT-\frac{B}{V^2}dV$$
$$dU=c_VdT-\frac{B}{V^2}dV$$
then, and it is the part with that I don't sure:
like
$$dU=TdS-pdV$$ and
$$dU=c_VdT-\frac{B}{V^2}dV$$
$$dS=\frac{c_V}{T}dT+\frac{A}{V}dV$$
therefore and using that $T_2=T_1+\frac{B}{c_V}\left(\frac1{V_1}-\frac1{V_2}\right)$
$$\Delta S=c_V\ln\frac{T_2}{T_1}+A\ln\frac{V_2}{V_1}$$
$$\Delta S=c_V\ln\left(1+\frac{B}{c_VT_1}\left(\frac1{V_1}-\frac1{V_2}\right)\right)+A\ln\left(\frac{V_2}{V_1}\right)>0$$
the first natural logarithm is positive because all terms are positives.

Question: in the intermediate dot for entropy, the temperature is diferent, but in the initial and final state are equal, right? why then the $\Delta S$ not intervenes $T_2=T_1$?
Maybe I write, in place of $\Delta S$
$$S(T_1,V_1,V_2)$$
Yet i have doubts

 

[Chestermiller]

Question: in the intermediate dot for entropy, the temperature is diferent, but in the initial and final state are equal, right? why then the $\Delta S$ not intervenes $T_2=T_1$?

No. The initial and final temperatures are not equal. Only for an ideal gas (B=0) are they equal.

 

[Dario SLC]

No. The initial and final temperatures are not equal. Only for an ideal gas (B=0) are they equal.

Ok! yes now I see the difference. Then, the entropy found it is fine?

 

[Chestermiller]

Ok! yes now I see the difference. Then, the entropy found it is fine?

Yes.

 

[Dario SLC]

Yes.

Really thank you a lot Chet!