Thermodynamic process and entropy -- Two chambers with different gasses with interconnecting membranes

In summary, you would need to calculate the work done and the heat transferred in order to find the entropy variation.
  • #1
LCSphysicist
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Homework Statement
In the second case the chamber is divided by two rigid, perfectly selective
membranes, the membrane on the left is perfectly permeable to gas A but
impermeable to gas B. The membrane on the right is just the reverse.
The two membranes are connected by rods to the outside and the whole
chamber is connected to a heat reservoir at temperature T. The gases can
be mixed in this case by pulling left hand membrane to the left and the
right hand one to the right.
Relevant Equations
All below
1596709087732.png

1596709165056.png


We need to find the system's entropy variation.
I don't think i understood pretty well what is happening in this process, can someone help me how to start?
 
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  • #2
LCSphysicist said:
Homework Statement:: In the second case the chamber is divided by two rigid, perfectly selective
membranes, the membrane on the left is perfectly permeable to gas A but
impermeable to gas B. The membrane on the right is just the reverse.
The two membranes are connected by rods to the outside and the whole
chamber is connected to a heat reservoir at temperature T. The gases can
be mixed in this case by pulling left hand membrane to the left and the
right hand one to the right.
Relevant Equations:: All below

View attachment 267350
View attachment 267351

We need to find the system's entropy variation.
I don't think i understood pretty well what is happening in this process, can someone help me how to start?
You might approach this by calculating the work done (assume a quasi-static equilibrium, i.e., if the membranes are moved, they are moved slowly). There will pressure differences on the membranes, and thus forces on the membranes, and you can calculate the work by applying [itex] W = \int \vec F \cdot \vec {dx} [/itex] (work is equal to force times distance).

Keep in mind the chamber is connected to a heat reservoir, keeping everything at a constant temperature. If I'm not missing something, this makes the problem easier (I think).

Here is a recent video from Cody's Lab. It won't solve the problem, but it might give you some insight on what to think about:
[Edit: and in this problem you'll want to work with partial pressures. If an ideal membrane is perfectly permeable to a given gas, then the partial pressure of that gas on one side of the membrane is equal to the partial pressure of that gas on the other side of the membrane, when in quasi-static equilibrium.]
 
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  • #3
LCSphysicist said:
Homework Statement:: In the second case the chamber is divided by two rigid, perfectly selective
membranes, the membrane on the left is perfectly permeable to gas A but
impermeable to gas B. The membrane on the right is just the reverse.
The two membranes are connected by rods to the outside and the whole
chamber is connected to a heat reservoir at temperature T. The gases can
be mixed in this case by pulling left hand membrane to the left and the
right hand one to the right.
Relevant Equations:: All below

View attachment 267350
View attachment 267351

We need to find the system's entropy variation.
I don't think i understood pretty well what is happening in this process, can someone help me how to start?
You need to find the entropy variation as a function of what?
 
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  • #4
Chestermiller said:
You need to find the entropy variation as a function of what?
Just as a function of the moles, the volume and the pression, that is, the variables in the box i posted above.

WIth the help above, i tried to find the change in entropy exactly as would be if the gases were expanding freely.
Anyway, the problem i think became easier because the constant temperature, i found:

R(na*log(Va+Vb)/Va + nb*log(Va+Vb)/Vb)

Actually, the most problematic thing here to me was to see that the membrane in its initial state had no width significance (the problem said nothing about it, so it was tricky hard to understand) , so we could say that V total = Va + Vb, on the other case we would not use the formula above
 
  • #5
LCSphysicist said:
Just as a function of the moles, the volume and the pression, that is, the variables in the box i posted above.

WIth the help above, i tried to find the change in entropy exactly as would be if the gases were expanding freely.
Anyway, the problem i think became easier because the constant temperature, i found:

R(na*log(Va+Vb)/Va + nb*log(Va+Vb)/Vb)

Actually, the most problematic thing here to me was to see that the membrane in its initial state had no width significance (the problem said nothing about it, so it was tricky hard to understand) , so we could say that V total = Va + Vb, on the other case we would not use the formula above
This problem was meant to be solved by designing a reversible process between the initial and final thermodynamic equilibrium states of the system, and determining the work done and the heat transferred. If you want, I can lead you through exactly how to do this.
 
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  • #6
Chestermiller said:
This problem was meant to be solved by designing a reversible process between the initial and final thermodynamic equilibrium states of the system, and determining the work done and the heat transferred. If you want, I can lead you through exactly how to do this.
It would be grateful if you do it.
 
  • #7
LCSphysicist said:
It would be grateful if you do it.
OK. If the membrane that is permeable to A is situated between a volume of pure A and a mixture of A and B, what is the condition for the equilibrium of A across the membrane?
 
  • #8
Chestermiller said:
OK. If the membrane that is permeable to A is situated between a volume of pure A and a mixture of A and B, what is the condition for the equilibrium of A across the membrane?
Maybe the forces (so pressure) need to be equal.

Par = Pmixing
nar/Vaf = (na'+nb')/Vmix
nar = na + na'

where nar is the a's mol remanent
na' is the mol inside membrane
Vaf is the volume of A in the instant
 
  • #9
LCSphysicist said:
Maybe the forces (so pressure) need to be equal.

Par = Pmixing
nar/Vaf = (na'+nb')/Vmix
nar = na + na'

where nar is the a's mol remanent
na' is the mol inside membrane
Vaf is the volume of A in the instant
What I was looking for was that, for equilibrium of A across the semi-permeable membrane, the partial pressure of A in the mixture ##p_A## has to be equal to the total pressure of pure A in the volume of pure A, ##P_A##:
$$p_A=P_A$$
Were you aware of this equilibrium condition for an ideal semi-permeable membrane for an ideal gas?
 
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  • #10
Chestermiller said:
What I was looking for was that, for equilibrium of A across the semi-permeable membrane, the partial pressure of A in the mixture ##p_A## has to be equal to the total pressure of pure A in the volume of pure A, ##P_A##:
$$p_A=P_A$$
Were you aware of this equilibrium condition for an ideal semi-permeable membrane for an ideal gas?
Actually i think i didn't know about this yet, can you say more about it or recommend some reading about?
 
  • #11
LCSphysicist said:
Actually i think i didn't know about this yet, can you say more about it or recommend some reading about?
Knowing this is key to solving this problem. There is no fundamental reference for this. It is just the definition of an ideal semi-permeable membrane.

Are you ready to continue with the rest of the analysis?
 

1. What is a thermodynamic process?

A thermodynamic process is a change in the state of a system, such as a gas or liquid, that occurs due to the transfer of heat or work. It can be described by various parameters, including temperature, pressure, and volume.

2. How do interconnecting membranes affect the behavior of gases in two chambers?

Interconnecting membranes allow for the exchange of particles, such as gas molecules, between two chambers. This can affect the pressure, temperature, and composition of the gases in each chamber, as they try to reach equilibrium.

3. What is the purpose of studying entropy in thermodynamics?

Entropy is a measure of the disorder or randomness in a system. In thermodynamics, it is used to predict the direction and efficiency of energy transfer and the tendency of a system to reach equilibrium. Understanding entropy is crucial for designing efficient processes and systems.

4. How does the Second Law of Thermodynamics apply to this scenario?

The Second Law of Thermodynamics states that the total entropy of a closed system will always increase over time. In the scenario of two chambers with different gases and interconnecting membranes, the gases will eventually mix and reach equilibrium, increasing the overall entropy of the system.

5. Can the entropy of a system ever decrease?

In isolated systems, the entropy can remain constant or decrease, but this is a rare occurrence. In most cases, the Second Law of Thermodynamics applies, and the entropy will increase. However, local decreases in entropy can occur if energy is input into the system to maintain order, such as in living organisms.

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