Entropy after removing partition separating gas into two compartments

In summary, the conversation discusses proving the positive definiteness of entropy change in mixing of gas. The initial answer is shown to be only possible if the initial temperature and pressure of the gas in each compartment are equal. Different methods are suggested and ultimately, it is concluded that the Clausius inequality can prove the positive definiteness of entropy change. Chet Miller's method is also mentioned, which involves establishing the final state of the system at final thermodynamic equilibrium. Further resources are provided for determining the entropy change of a closed system experiencing irreversible processes.
  • #1
burian
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Summary:: Proving that entropy change in mixing of gas is positive definite

>
>An ideal gas is separated by a piston in such a way that the entropy of one prat is## S_1## and that of the other part is ##S_2##. Given that ##S_1>S_2##, if the piston is removed then the total entropy of the system will be..?
>
>Ans: ##S_1 +S_2##

After discussions with Chet Miller on stack exchange, it was concluded that the above answer is only possible if the initial temperature and pressure of the gas in each compartment is equal. Hence the question is illposed.

We can show that by clasius inequality that the entropy change is positive definite, but can we prove it us basic algebra?

I tried to prove it using my method here:
You are most probably right, for the thing about entropy. I tried writing my own proof. Let the compartment with greater temperature be ##T_2## and the other one be ##T_1##, let dQ be transferred from compartment with ##T_2## into one with ##T_1##:
$$−\frac{dQ}{T_2}+\frac{dQ}{T_1}=dS$$

s true for all points in the process. But, $$\frac{dQ}{T_1}>\frac{dQ}{T_2} $$ hence $$dS>0$$

The above method is is apparently wrong due the temperature being continuously at boundary hence ##T_1=T_2##... but couldn't we argue that this is the case for any heat transformed involved and hence there can never be any entropy gain?

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I post Chet Miller's method till the part he had shown me:To get the final temperature and pressure, we use the conditions that the change in internal energy of the combined system is zero, and the final volume is equal to the initial combined volume. This leads to:

$$n_1C_v(T_F-T_1)+n_2C_v(T_F-T_2)=0$$and$$\frac{n_1RT_1}{P_1}+\frac{n_2RT_2}{P_2}=\frac{(n_1+n_2)RT_F}{P_F}$$The solutions to these equations for ##T_F## and ##P_F## are:
$$T_F=\frac{n_1T_1+n_2T_2}{(n_1+n_2)}\tag{1}$$$$\frac{1}{P_F}=\left(\frac{n_1T_1}{n_1T_1+n_2T_2}\right)\frac{1}{P_1}+\left(\frac{n_2T_2}{n_1T_1+n_2T_2}\right)\frac{1}{P_2}\tag{2}$$The entropy change for the system is given by:
$$\Delta S=n_1C_p\ln{\left(\frac{T_F}{T_1}\right)}+n_2C_p\ln{\left(\frac{T_F}{T_2}\right)}$$$$+n_1R\ln{\left(\frac{P_1}{P_F}\right)}+n_2R\ln{\left(\frac{P_2}{P_F}\right)}\tag{3}$$The next step is to show that, unless the initial temperatures and pressures are equal, this entropy change is positive definite.
 
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  • #2
burian said:
Summary:: Proving that entropy change in mixing of gas is positive definite

>
>An ideal gas is separated by a piston in such a way that the entropy of one prat is$ S_1$ and that of the other part is $S_2$. Given that $S_1>S_2$, if the piston is removed then the total entropy of the system will be..?
>
>Ans: $S_1 +S_2$

After discussions with Chet Miller on stack exchange, it was concluded that the above answer is only possible if the initial temperature and pressure of the gas in each compartment is equal. Hence the question is illposed.

We can show that by clasius inequality that the entropy change is positive definite, but can we prove it us basic algebra?

I tried to prove it using my method here:
You are most probably right, for the thing about entropy. I tried writing my own proof. Let the compartment with greater temperature be $T_2$ and the other one be $T_1$, let dQ be transferred from compartment with $T_2$ into one with $T_1$:
$$−\frac{dQ}{T_2}+\frac{dQ}{T_1}=dS$$

s true for all points in the process. But, $$\frac{dQ}{T_1}>\frac{dQ}{T_2} $$ hence $$dS>0$$

The above method is is apparently wrong due the temperature being continuously at boundary hence $T_1=T_2$... but couldn't we argue that this is the case for any heat transformed involved and hence there can never be any entropy gain?
The relationship dS=dQ/T only applies to a reversible change and this process is not reversible. So this argument doesn't work for an irreversible change. Besides, in this case, your dQ is internal to the system as a whole. If you want to apply the Clausius inequality to each of the bodies of gas individually (assuming they each retain their integrity during the process), then you would write $$\Delta S_1>\int{\frac{dQ}{T_B}}$$and $$\Delta S_2>-\int{\frac{dQ}{T_B}}$$where ##T_B## is the temperature at the interface between the two gas masses. Adding these together then yields: $$\Delta S_{Total}=\Delta S_1+\Delta S_2>0$$

So, if you accept the Clausius inequality, then no more work needs to be done.

If you wish to precisely determine the entropy change for this irreversible process and demonstrate that it is positive definite, then you need to approach this differently. The first step must be to establish the final state of the system at final thermodynamic equilibrium (using the 1st law of thermodynamics). Since entropy is a function of state, you cannot determine the entropy change without knowing the final state.

I have written a Physics Forums Insights article giving the recipe for determining the entropy change of a closed system experiencing any irreversible processes. It includes several examples to show how the method is applied to frequently encountered problems. Please check it out:
https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

burian said:
-----

I post Chet Miller's method till the part he had shown me:To get the final temperature and pressure, we use the conditions that the change in internal energy of the combined system is zero, and the final volume is equal to the initial combined volume. This leads to:

$$n_1C_v(T_F-T_1)+n_2C_v(T_F-T_2)=0$$and$$\frac{n_1RT_1}{P_1}+\frac{n_2RT_2}{P_2}=\frac{(n_1+n_2)RT_F}{P_F}$$The solutions to these equations for $T_F$ and $P_F$ are:
$$T_F=\frac{n_1T_1+n_2T_2}{(n_1+n_2)}$$$$\frac{1}{P_F}=\left(\frac{n_1T_1}{n_1T_1+n_2T_2}\right)\frac{1}{P_1}+\left(\frac{n_2T_2}{n_1T_1+n_2T_2}\right)\frac{1}{P_2}$$The entropy change for the system is given by:
$$\Delta S=n_1C_p\ln{\left(\frac{T_F}{T_1}\right)}+n_2C_p\ln{\left(\frac{T_F}{T_2}\right)}+n_1R\ln{\left(\frac{P_1}{P_F}\right)}+n_2R\ln{\left(\frac{P_2}{P_F}\right)}$$The next step is to show that, unless the initial temperatures and pressures are equal, this entropy change is positive definite.
The analysis represents the result of the first step in my recipe.
 
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  • #3
I had gone through the article a long time back. I'll go through it again by around tomorrow. Some real life eventsare keeping me busy.
 
  • #4
burian said:
I had gone through the article a long time back. I'll go through it again by around tomorrow. Some real life eventsare keeping me busy.
I've done a lot more work on this problem with some very interesting quantitative results, but before I post what I've done, I need you to sign off on the analysis I've presented so far.
 
  • #5
Chestermiller said:
The relationship dS=dQ/T only applies to a reversible change and this process is not reversible. So this argument doesn't work for an irreversible change. Besides, in this case, your dQ is internal to the system as a whole. If you want to apply the Clausius inequality to each of the bodies of gas individually (assuming they each retain their integrity during the process), then you would write $$\Delta S_1>\int{\frac{dQ}{T_B}}$$and $$\Delta S_2>-\int{\frac{dQ}{T_B}}$$where ##T_B## is the temperature at the interface between the two gas masses. Adding these together then yields: $$\Delta S_{Total}=\Delta S_1+\Delta S_2>0$$

So, if you accept the Clausius inequality, then no more work needs to be done.

If you wish to precisely determine the entropy change for this irreversible process and demonstrate that it is positive definite, then you need to approach this differently. The first step must be to establish the final state of the system at final thermodynamic equilibrium (using the 1st law of thermodynamics). Since entropy is a function of state, you cannot determine the entropy change without knowing the final state.

I have written a Physics Forums Insights article giving the recipe for determining the entropy change of a closed system experiencing any irreversible processes. It includes several examples to show how the method is applied to frequently encountered problems. Please check it out:
https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/The analysis represents the result of the first step in my recipe.
Had been busy with some college entry related stuff but more or less done now. Reading the article, I realize that we don't really need the step of 'process devisation' here since we have entropy as a state function. We plugged in the variables already. Now, we just need to prove it's positive definite.

Seems a bit more difficult. I suppose maybe we could interpret the equation with a path and process then it could be intuitively concluded.
 
  • #6
burian said:
Had been busy with some college entry related stuff but more or less done now. Reading the article, I realize that we don't really need the step of 'process devisation' here since we have entropy as a state function. We plugged in the variables already. Now, we just need to prove it's positive definite.

Seems a bit more difficult. I suppose maybe we could interpret the equation with a path and process then it could be intuitively concluded.
The entropy change, as expressed in Eqns. 1-3 of post #1 appears to be a function of the six independent parameters ##T_1##, ##T_2##, ##n_1##, ##n_2##, ##P_1##, and ##P_2## (each of which can vary between 0 and ##\infty##), and evaluating ##\Delta S## over these ranges (to show that it is always positive) would seem to be a daunting task, in reality it is a function only of the following three dimensionless parameters which vary over a very limited range: $$-1<\xi_T=\frac{(T_1-T_2)}{(T_1+T_2)}<1$$$$-1<\xi_P=\frac{(P_1-P_2)}{(P_1+P_2)}<1$$$$-1<\xi_n=\frac{(n_1-n_2)}{(n_1+n_2)}<1$$Evaluating ##\Delta S## over the limited ranges of only these 3 parameters is a much more manageable prospect.

Here is how it plays out:

Write $$T_1=\frac{1}{2}(T_1+T_2)+\frac{1}{2}(T_1-T_2)=\frac{(T_1+T_2)}{2}(1+\xi_T)$$
Similarly,$$T_2=\frac{(T_1+T_2)}{2}(1-\xi_T)$$$$\frac{1}{P_1}=\frac{1}{2}\left(\frac{1}{P_1}+\frac{1}{P_2}\right)(1-\xi_P)$$$$\frac{1}{P_2}=\frac{1}{2}\left(\frac{1}{P_1}+\frac{1}{P_2}\right)(1+\xi_P)$$$$n_1=\frac{(n_1+n_2)}{2}(1+\xi_n)$$$$n_2=\frac{(n_1+n_2)}{2}(1-\xi_n)$$
If we substitute there relationships into Eqns. 1 and 2 of post #1 for ##T_F## and ##P_F## we obtain: $$\frac{T_F}{T_1}=\frac{1+\xi_n\xi_T}{1+\xi_T}$$$$\frac{T_F}{T_2}=\frac{1+\xi_n\xi_T}{1-\xi_T}$$$$\frac{P_1}{P_F}=\frac{1-\lambda \xi_P}{1-\xi_P}$$$$\frac{P_2}{P_F}=\frac{1-\lambda \xi_P}{1+\xi_P}$$where $$-1<\lambda=\frac{n_1T_1-n_2T_2}{n_1T_1+n_2T_2}=\frac{\xi_n+\xi_T}{1+\xi_n\xi_T}<1$$
@burian Please check this over and see if you agree with it so far. Then I will continue with some calculated results.
 
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  • #7
This was a much harder exercise than I thought, I'll go over it carefully
 
  • #8
Chestermiller said:
The entropy change, as expressed in Eqns. 1-3 of post #1 appears to be a function of the six independent parameters ##T_1##, ##T_2##, ##n_1##, ##n_2##, ##P_1##, and ##P_2## (each of which can vary between 0 and ##\infty##), and evaluating ##\Delta S## over these ranges (to show that it is always positive) would seem to be a daunting task, in reality it is a function only of the following three dimensionless parameters which vary over a very limited range: $$-1<\xi_T=\frac{(T_1-T_2)}{(T_1+T_2)}<1$$$$-1<\xi_P=\frac{(P_1-P_2)}{(P_1+P_2)}<1$$$$-1<\xi_n=\frac{(n_1-n_2)}{(n_1+n_2)}<1$$Evaluating ##\Delta S## over the limited ranges of only these 3 parameters is a much more manageable prospect.

Here is how it plays out:

Write $$T_1=\frac{1}{2}(T_1+T_2)+\frac{1}{2}(T_1-T_2)=\frac{(T_1+T_2)}{2}(1+\xi_T)$$
Similarly,$$T_2=\frac{(T_1+T_2)}{2}(1-\xi_T)$$$$\frac{1}{P_1}=\frac{1}{2}\left(\frac{1}{P_1}+\frac{1}{P_2}\right)(1-\xi_P)$$$$\frac{1}{P_2}=\frac{1}{2}\left(\frac{1}{P_1}+\frac{1}{P_2}\right)(1+\xi_P)$$$$n_1=\frac{(n_1+n_2)}{2}(1+\xi_n)$$$$n_2=\frac{(n_1+n_2)}{2}(1-\xi_n)$$
If we substitute there relationships into Eqns. 1 and 2 of post #1 for ##T_F## and ##P_F## we obtain: $$\frac{T_F}{T_1}=\frac{1+\xi_n\xi_T}{1+\xi_T}$$$$\frac{T_F}{T_2}=\frac{1+\xi_n\xi_T}{1-\xi_T}$$$$\frac{P_1}{P_F}=\frac{1-\lambda \xi_P}{1-\xi_P}$$$$\frac{P_2}{P_F}=\frac{1-\lambda \xi_P}{1+\xi_P}$$where $$-1<\lambda=\frac{n_1T_1-n_2T_2}{n_1T_1+n_2T_2}=\frac{\xi_n+\xi_T}{1+\xi_n\xi_T}<1$$
@burian Please check this over and see if you agree with it so far. Then I will continue with some calculated results.
What was your motivation for the substitutions you've made? To a beginner (me), it looks like an arbitary thing to do..
 
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  • #9
burian said:
What was your motivation for the substitutions you've made? To a beginner (me), it looks like an arbitary thing to do..
Well, there seemed to be too many parameters to do a reasonable number of calculations to test the entropy change. So I was willing to spend some time looking for ways to consolidate the variables to a reasonable number. I was also thinking about expressing the entropy change as a Taylor series in small parameters like I did in the 2nd example of the Insights article. Anyway, I've had lots of decades of experience doing modeling, and I always look for tricks to simplify things (as I was taught to do on the Tilden High School math team in 1958 and 1959).
 
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Here is the continuation of the analysis.

As expressed by Eqn. 3 in post #1, the change in entropy of the system experiencing this irreversible process can be expressed as the sum of a temperature contribution and a pressure contribution: $$\Delta S=\Delta S_T+\Delta S_P$$where
$$\Delta S_T=n_1C_p\ln{\left(\frac{T_F}{T_1}\right)}+n_2C_p\ln{\left(\frac{T_F}{T_2}\right)}$$and $$\Delta S_P=n_1R\ln{\left(\frac{P_1}{P_F}\right)}+n_2R\ln{\left(\frac{P_2}{P_F}\right)}$$Expressed in terms dimensionless parameters, these become:
$$\Delta S_T^*=\frac{2\Delta S_T}{(n_1+n_2)C_p}=(1+\xi_n)\ln{\left(\frac{1+\xi_n\xi_T}{1+\xi_T}\right)}$$$$+(1-\xi_n)\ln{\left(\frac{1+\xi_n\xi_T}{1-\xi_T}\right)}$$
and$$\Delta S_P^*=\frac{2\Delta S_P}{(n_1+n_2)R}=(1+\xi_n)\ln{\left(\frac{1-\lambda\xi_P}{1-\xi_P}\right)}$$$$+(1-\xi_n)\ln{\left(\frac{1-\lambda\xi_P}{1+\xi_P}\right)}$$
Based on these, the dimensionless total entropy change is given by $$\Delta S^*=\Delta S_T^*+\frac{\gamma-1
}{\gamma}\Delta S_P^*$$
Note that the dimensionless temperature contribution to the entropy depends only on ##\xi_n## and ##\xi_T##, while the dimensionless pressure contribution to the entropy depends on all three ##\xi_n##, ##\xi_T##, and ##\xi_P##. I have carried out calculations on an Excel spreadsheet to evaluate the dimensionless temperature contribution as a function of its two parameters, and plotted this contribution as versus ##\xi_T## at a selection of values for ##\xi_n## over its allowable range. The results are presented in this Figure:
1630800824828.png

As can be seen, this dimensionless temperature contribution to the entropy change is positive for all values of its two parameters.
 

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Related to Entropy after removing partition separating gas into two compartments

What is entropy?

Entropy is a measure of the disorder or randomness in a system. It is a thermodynamic quantity that increases over time in closed systems.

How does entropy relate to removing a partition separating gas into two compartments?

When a partition separating gas into two compartments is removed, the gas molecules are free to move and mix in both compartments. This increases the disorder and randomness of the gas system, resulting in an increase in entropy.

What is the formula for calculating entropy?

The formula for calculating entropy is S = k ln W, where S is the entropy, k is the Boltzmann constant, and W is the number of microstates or possible arrangements of a system.

Does removing a partition always increase entropy?

In most cases, removing a partition separating gas into two compartments will increase entropy. However, there are some exceptions, such as when the gas molecules are already evenly distributed in both compartments, or when the gas is at its maximum entropy state.

How does entropy affect the second law of thermodynamics?

The second law of thermodynamics states that the total entropy of a closed system always increases over time. Removing a partition separating gas into two compartments is an example of this, as it leads to an increase in entropy and is therefore in accordance with the second law of thermodynamics.

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