Is the differential of heat in a reversible process in an isolated system equal to zero?

  • #1
zenterix
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Homework Statement
It can be shown (with a simple exactness test) that the differential ##\frac{\delta q_{rev}}{T}=dU+\frac{nRT}{V}{dV}## is exact.

This differential is ##dS##, the differential of entropy.
Relevant Equations
Since this differential is exact, it comes from a state function (ie, a path independent function), ##S## and ##\oint dS=0##.

That is, for any cylic reversible process the entropy doesn't change.
If a process is irreversible, on the other hand, then

$$\oint \frac{\delta q}{T}\leq 0=\oint dS\tag{1}$$

Apparently, from this equation we can conclude that

$$dS \geq \frac{\delta q}{T}\tag{2}$$

How do we mathematically justify the step from (1) to (2)?

Next, consider an isolated system.

We can restate the second law by saying that the entropy increases in a spontaneous process in an isolated system because for an isolated system, ##\delta q=0## and therefore ##\Delta S>0## for a spontaneous process. The entropy of an isolated system (in which the internal energy is constant) can continue to increase as long as spontaneous processes occur.

Consider the expression ##\Delta S>0##.

$$\int dS=\Delta S> \int \frac{\delta q}{T}=0$$

But isn't ##\delta q_{rev}## also zero in an isolated system?
 
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  • #2
zenterix said:
Homework Statement: It can be shown (with a simple exactness test) that the differential ##\frac{\delta q_{rev}}{T}=dU+\frac{nRT}{V}{dV}## is exact.
Entropy on the left and energy on the right?
 
  • #3
Philip Koeck said:
Entropy on the left and energy on the right?
The equation is incorrect as I wrote it.

$$dU=\delta q_{rev}-PdV=\delta q_{rev}-\frac{nRT}{V}dV$$

$$\frac{dU}{T}=dS-\frac{nR}{V}dV$$

$$dS=\frac{dU}{T}+\frac{nR}{V}dV$$
 
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  • #4
zenterix said:
That is, for any cylic reversible process the entropy doesn't change.
I think first you have to be more accurate with your statements.

Since the entropy of the system is a state function its change will be zero for any cyclic process no matter whether reversible or not! This, however, is only true for the entropy of the system, not for that of the surroundings.

In an isolated system (aka as "system + surroundings" or "universe") the total entropy change will be zero for any reversible process, not only for cyclic ones.
 
  • #5
Philip Koeck said:
Since the entropy of the system is a state function its change will be zero for any cyclic process no matter whether reversible or not!
I agree. Entropy is a state function and so doesn't change in any cyclic process.

What does ##\frac{\delta q_{rev}}{T}## mean in an isolated system?

Why can we write ##\int dS=\int\frac{\delta q_{rev}}{T}>\int\frac{\delta q}{T}=0## for an isolated system?

Basically I am asking why ##\delta q## is equal to zero in such a system but ##\delta q_{rev}## apparently not necessarily.
 
  • #6
zenterix said:
What does ##\frac{\delta q_{rev}}{T}## mean in an isolated system?

Why can we write ##\int dS=\int\frac{\delta q_{rev}}{T}>\int\frac{\delta q}{T}=0## for an isolated system?

Basically I am asking why ##\delta q## is equal to zero in such a system but ##\delta q_{rev}## apparently not necessarily.
How is an isolated system defined?
 
  • #7
That is a good question.

Let me preface by saying that I am trying to read a few books, but the one that my course is based on is Silbey, Alberty, Bawendi's "Physical Chemistry" which I frankly hate tremendously.

The question I am asking is based on reading this subpar book. That being said, even with the other resources I am using I don't quite understand the entropy differential very well.

Here are a few snippets that directly generated the question in the OP

1720303009696.png


Does this answer your question about what the definition of isolated system is in this book?

As I understand it, our entire system is inside of rigid walls (no ##PV## work) that are adiabatic (no exchange of heat with surroundings). Thus, ##\delta q=\delta w=dU=0##.

If this is all correct, then I re-ask my question: why is ##\delta q_{rev}## not zero but ##\delta q## is?

Here is a statement of the second law that comes right before the snippet above

1720303210885.png
 
  • #8
zenterix said:
That is a good question.

Let me preface by saying that I am trying to read a few books, but the one that my course is based on is Silbey, Alberty, Bawendi's "Physical Chemistry" which I frankly hate tremendously.

The question I am asking is based on reading this subpar book. That being said, even with the other resources I am using I don't quite understand the entropy differential very well.

Here are a few snippets that directly generated the question in the OP

View attachment 347915

Does this answer your question about what the definition of isolated system is in this book?

As I understand it, our entire system is inside of rigid walls (no ##PV## work) that are adiabatic (no exchange of heat with surroundings). Thus, ##\delta q=\delta w=dU=0##.

If this is all correct, then I re-ask my question: why is ##\delta q_{rev}## not zero but ##\delta q## is?

Here is a statement of the second law that comes right before the snippet above

View attachment 347916
Yes, this books sounds slightly confusing!

Just some quick remarks: just as you say, dq = 0 for an isolated system.

Now in this system there can be processes which are either reversible or irreversible, leading to dS = 0 or dS > 0.

If an irreversible process happens inside the isolated system we can't calculate dS directly however, because dS = dqrev/T.
What we need to do is find a reversible process which has the same effect on the system as the irreversible one that's actually happening (i.e. same initial and final states).
For this "replacement process" dq > 0. In other words it's not possible to carry out this process if the system is isolated. We would have to add some heat from outside the system to make it happen.
 
  • #9
Consider an isolated system. We go from state 1 to state 2 via an irreversible process (e.g. Joule expansion of a gas). In this process ##q_{irrev}=0##.

Then we take this system in state 2 and reversibly take it back to state 1 (e.g. compress the gas back to initial pressure and volume).

Apparently, this cannot be done with the system in isolation.

One question I have is why we can say this.

If a gas expands against vacuum in an isolated system (Joule expansion) then the internal energy is unchanged. If the gas is an ideal gas then internal energy is a function of temperature only and so temperature is unchanged.

It seems pressure is reduced and volume is increased.

To go back to the initial state in this isolated system we would either need a spontaneous change back to the initial state or we would need to do work. This is all very blurry to me right now.

We could also make the system not isolated anymore and do an isothermal compression, which would require heat.

If we do this latter option, then considering the full cycle we would have

$$\int_1^2\frac{\delta q_{irrev}}{T}+\int_2^1\frac{\delta q_{rev}}{T}\leq 0$$

$$\implies \int_2^1\frac{\delta q_{rev}}{T}=S_1-S_2\leq 0$$

$$\implies S_2-S_1\geq 0$$

which means that the entropy change from state 1 to state 2 is ##\geq 0##.

This is all getting ahead of my original question. The ##\delta q_{rev}/T## from my original question is not zero because of what you said: the reversible part is not done in an isolated system.

So my new question is:
what is the justification for why we can't go from state 2 to state 1 in the isolated system?
 
  • #10
zenterix said:
Consider an isolated system. We go from state 1 to state 2 via an irreversible process (e.g. Joule expansion of a gas). In this process ##q_{irrev}=0##.

Then we take this system in state 2 and reversibly take it back to state 1 (e.g. compress the gas back to initial pressure and volume).

Apparently, this cannot be done with the system in isolation.

One question I have is why we can say this.

If a gas expands against vacuum in an isolated system (Joule expansion) then the internal energy is unchanged. If the gas is an ideal gas then internal energy is a function of temperature only and so temperature is unchanged.

It seems pressure is reduced and volume is increased.

To go back to the initial state in this isolated system we would either need a spontaneous change back to the initial state or we would need to do work. This is all very blurry to me right now.

We could also make the system not isolated anymore and do an isothermal compression, which would require heat.

If we do this latter option, then considering the full cycle we would have

$$\int_1^2\frac{\delta q_{irrev}}{T}+\int_2^1\frac{\delta q_{rev}}{T}\leq 0$$

$$\implies \int_2^1\frac{\delta q_{rev}}{T}=S_1-S_2\leq 0$$

$$\implies S_2-S_1\geq 0$$

which means that the entropy change from state 1 to state 2 is ##\geq 0##.

This is all getting ahead of my original question. The ##\delta q_{rev}/T## from my original question is not zero because of what you said: the reversible part is not done in an isolated system.

So my new question is: what is the justification for why we can't go from state 2 to state 1 in the isolated system?
Because there is no reversible path between the two states in an isolated or adiabatic system for which the actual path is irreversible. The reversible path must allow heat transfer between the system and a new set of surroundings.

The real question is whether you know how to determine the change in entropy for a closed system that has suffered an irreversible process.
 
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  • #11
zenterix said:
Consider an isolated system. We go from state 1 to state 2 via an irreversible process (e.g. Joule expansion of a gas). In this process ##q_{irrev}=0##.

Then we take this system in state 2 and reversibly take it back to state 1 (e.g. compress the gas back to initial pressure and volume).

Apparently, this cannot be done with the system in isolation.

One question I have is why we can say this.

If a gas expands against vacuum in an isolated system (Joule expansion) then the internal energy is unchanged. If the gas is an ideal gas then internal energy is a function of temperature only and so temperature is unchanged.

It seems pressure is reduced and volume is increased.

To go back to the initial state in this isolated system we would either need a spontaneous change back to the initial state or we would need to do work. This is all very blurry to me right now.

We could also make the system not isolated anymore and do an isothermal compression, which would require heat.

If we do this latter option, then considering the full cycle we would have

$$\int_1^2\frac{\delta q_{irrev}}{T}+\int_2^1\frac{\delta q_{rev}}{T}\leq 0$$

$$\implies \int_2^1\frac{\delta q_{rev}}{T}=S_1-S_2\leq 0$$

$$\implies S_2-S_1\geq 0$$

which means that the entropy change from state 1 to state 2 is ##\geq 0##.

This is all getting ahead of my original question. The ##\delta q_{rev}/T## from my original question is not zero because of what you said: the reversible part is not done in an isolated system.

So my new question is: what is the justification for why we can't go from state 2 to state 1 in the isolated system?
Maybe keep it simple and calculate the entropy change for the free expansion in the isolated system and then have a look at any reversible process with the same entropy change.
You should see that this reversible process can't happen in an isolated system.
 
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