I The relationship between reversible and irreversible processes

[rtareen]

TL;DR Summary

An example of an irreversible free expansion is given and is related to a reversible isothermal expansion of a gas in a piston-cylinder-reservoir system. The changes in entropy of both systems must be the same as they have the same initial and final states.

Hi all. I am referencing the example given in Halliday and Resnick, Chapter 20, Section 1, Subsection "Change in Entropy". The above picture is graph of the free expansion of a gas into a volume that is double its original volume. I n a free expansion there is no heat transfer, the pressure decreases while the volume increases, and there is no work done. But then they go on to say that this has the same change in entropy as a gas in a piston-cylinder where the some mass is removed from the piston so the that gas expands from the same initial state to the same final state, and heat is added to keep it at the same temperature. Here is the plot of the reversible process:

As we can see the temperature remains constant in this second system. But how does this correspond to a free expansion where the temperature obviously decreases? They are saying the initial and final states are the same for both systems, but I don't see how this can be the case. How can the temperature remain the same in a free expansion? The equation $\Delta T = \Delta p \Delta V/nR$ implies a decrease in temperature.
Also, they say that heat is added to keep the temperature constant, and that’s why the second system gains entropy. But there is no heat transfer in a free expansion, so I’m not understanding the relationship between the two examples. Does this mean that entropy of an irreversible process doesn't depend on heat transfer but the corresponding reversible process does? So then what is entropy really a measure of?

 

[kuruman]

Entropy is a state variable that, dependsonly on the end points not how you get from one to the other. Any process that takes the system from point $i$ to point $f$ results in the same entropy change as any other process. A isothermal process is very handy because $\Delta S=\int \dfrac{dQ}{T}$ is easy to calculate.

 

[rtareen]

Entropy is a state variable that, dependsonly on the end points not how you get from one to the other. Any process that takes the system from point $i$ to point $f$ results in the same entropy change as any other process. A isothermal process is very handy because $\Delta S=\int \dfrac{dQ}{T}$ is easy to calculate.

But there is no heat transfer in a free expansion, so how can there be an increase in entropy?

 

[Chestermiller]

The entropy change is not the heat transferred divided by the reservoir (boundary) temperature in an irreversible process. If is only the heat transferred divided by the reservoir temperature between the same two end states for a reversible path.

Also, the pressure does not decrease in free expansion of an ideal gas. If the volume increases by 2X, the pressure decreases by 2X, so PV in the initial and final states is the same. Have you used the 1st law of thermodynamics to determine the final state of the gas after free expansion (within a rigid insulated container)? Let's see your analysis.

For an understanding and a cookbook recipe for how to determine the entropy change for a system that has experienced an irreversible process, see the following Physics Forums Insights article: https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

Also, perhaps this will help:
Unlike energy which is neither created nor destroyed, entropy is an entity which can, in addition to being transferrable between a system and its surroundings can also be generated within the system.

In an irreversible process, entropy is both transferred (by heat flow between system and surroundings) and is also generated (by irreversible viscous heating and significant heat transfer by conduction within the system). But, in a reversible process, no entropy is generated in the system, and only entropy exchange with the surroundings takes place.

In your free expansion example, the irreversible path features substantial entropy generation within the system and no entropy exchange with the surroundings. The reversible path features no entropy generation within the system, but substantial entropy exchange with the surroundings. The entropy generation in the irreversible path exactly matches the entropy transfer in the reversible path.

 

[rtareen]

Also, the pressure does not decrease in free expansion of an ideal gas. If the volume increases by 2X, the pressure decreases by 2X, so PV in the initial and final states is the same. Have you used the 1st law of thermodynamics to determine the final state of the gas after free expansion (within a rigid insulated container)? Let's see your analysis.

This makes sense. I guess a free expansion always maintains a constant temperature. Since there is no work done and no heat added, there is no change in internal energy. But we know that the relationship between P and V is not linear. It is of the form p(V,T) = cT/V. So there is room for a temperature change, and I was under the impression that since there is an increase in volume, it would take more energy to keep it at the same temperature,

Also, perhaps this will help:
Unlike energy which is neither created nor destroyed, entropy is an entity which can, in addition to being transferrable between a system and its surroundings can also be generated within the system.

In an irreversible process, entropy is both transferred (by heat flow between system and surroundings) and is also generated (by irreversible viscous heating and significant heat transfer by conduction within the system). But, in a reversible process, no entropy is generated in the system, and only entropy exchange with the surroundings takes place.

In your free expansion example, the irreversible path features substantial entropy generation within the system and no entropy exchange with the surroundings. The reversible path features no entropy generation within the system, but substantial entropy exchange with the surroundings. The entropy generation in the irreversible path exactly matches the entropy transfer in the reversible path.

This is very explanatory. My book didn't really explain this well. Before this I thought that there needed to be an external heat transfer for a system to gain entropy. You say entropy can be generated within a closed system for an irreversible process, by heat transfer within a system. How does this free expansion generate entropy? Is it because there is heat transfer by the gas into the previously heatless vacuum? Let me know if anything I said is wrong.

 

[Chestermiller]

But we know that the relationship between P and V is not linear. It is of the form p(V,T) = cT/V. So there is room for a temperature change,

Check your math. If temperature is constant and volume doubles, pressure is halved.

I was under the impression that since there is an increase in volume, it would take more energy to keep it at the same temperature,

In a rapid irreversible expansion, there is viscous heating of the gas which offsets the temperature decrease from volume increase. In a reversible expansion, the rate of deformation of the gas is infinitely slow, so no viscous heating occurs (so when expansion work is done, heat has to come from the outside to hold the temperature constant).

This is very explanatory. My book didn't really explain this well. Before this I thought that there needed to be an external heat transfer for a system to gain entropy. You say entropy can be generated within a closed system for an irreversible process, by heat transfer within a system. How does this free expansion generate entropy? Is it because there is heat transfer by the gas into the previously heatless vacuum?

No. As I said above, in a rapidly deforming fluid, viscous heating is what causes the generation of entropy. It is as if there is a heat source within the gas. Viscous heating also occurs when a viscous fluid is sheared between parallel plates.