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

In summary, a system undergoing a free expansion in an isolated state will experience no change in energy due to the absence of heat and work exchange. However, the work done by the gas during expansion will lead to an increase in entropy, which can be calculated through integration using the state equation. This argument is valid for any gas, not just ideal gases. The resulting formula for entropy shows that it will always increase in a free expansion due to the Clausius inequality.
  • #1
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!
 
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  • #2
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?
 
  • #3
Chestermiller said:
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
 
  • #4
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##.
 
  • #5
Chestermiller said:
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
 
  • #6
Dario SLC said:
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.
 
  • #7
Chestermiller said:
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?
 
  • #8
Dario SLC said:
Ok! yes now I see the difference. Then, the entropy found it is fine?
Yes.
 
  • #9
Chestermiller said:
Yes.
Really thank you a lot Chet!
 

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

What is entropy in a free expansion?

Entropy in a free expansion refers to the measure of disorder or randomness in a system that undergoes a spontaneous expansion without any external work being done on it.

Why does entropy increase in a free expansion?

In a free expansion, the volume of a system increases while the energy remains constant. This leads to an increase in the number of microstates or possible arrangements of particles, resulting in an increase in disorder or entropy.

How is entropy related to the second law of thermodynamics?

The second law of thermodynamics states that the total entropy of a closed system always increases over time. In a free expansion, the entropy of the system increases as the gas particles become more dispersed, in accordance with the second law.

Can entropy ever decrease in a free expansion?

No, entropy can never decrease in a free expansion as it is a spontaneous process that leads to an increase in disorder and the second law of thermodynamics dictates that entropy must always increase.

How does the temperature of a gas change during a free expansion?

During a free expansion, the gas particles do not exchange energy with their surroundings, so the temperature remains constant. However, the increase in volume can lead to a decrease in temperature due to the decrease in particle density.

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