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- Thread starter Chestermiller
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I'm now going to provide an analysis to explain how this all plays out. I'm going to model the ideal expansion and recompression of an ideal gas for both the reversible case and the irreversible case. There is an ideal gas in a cylinder with a piston. The gas expands from equilibrium state ##P_1, V_1, T## to equilibrium state ##P_2,V_2, T## during the expansion step; and it goes from equilibrium state ##P_2,V_2, T## to equilibrium state ##P_1, V_1, T## during the compression step. During both of these processes, the cylinder is held in contact with a constant temperature reservoir at temperature T. Because both end states are at temperature T, we must have that $$P_2=\frac{P_1V_1}{V_2}$$

ANALYSIS FOR REVERSIBLE PROCESS PATHS

For reversible process paths, the gas passes through a continuous sequence of thermodynamic equilibrium states, and the pressure of the gas on the surroundings during such a sequence can be determined using the ideal gas law (which, strictly speaking, applies only to gases in thermodynamic equilibrium). So during the entire expansion and the compression, the gas pressure on the surroundings can be calculated from $$P=\frac{P_1V_1}{V}$$where V is the gas volume during the process. From this, it follows that the work done by the gas on the surroundings during the expansion step is $$W_E=\int_{V_1}^{V_2}{PdV}=P_1V_1\ln{(V_2/V_1)}$$And the work done by the gas on the surroundings during the compression is $$W_C=\int_{V_2}^{V_1}{PdV}=P_1V_1\ln{(V_1/V_2)}=-\int_{V_1}^{V_2}{PdV}=P_1V_1\ln{(V_2/V_1)}=-W_E$$

So the net work done by the gas on the surroundings during the combination of the two steps is zero.

It also follows from the 1st law of thermodynamics that since, for each of these isothermal steps, the change in internal energy is zero, the heat transferred from the surroundings to the system during these steps is given by$$Q_E=W_E=-W_C=-Q_C$$So there is not net heat exchange with the surroundings during these two reversible steps.

As a result of all this, after the combination of isothermal reversible expansion and compression is complete, there is no change whatsoever in the surroundings. And, there was no change in the system (the gas) either. So overall, nothing has changed in the universe (system + surroundings).

I think I'll stop here for now and continue with an analysis of the irreversible case later.

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ANALYSIS OF IRREVERSIBLE PROCESS PATHS

Unlike the case of reversible process paths, for the irreversible process paths, the ideal gas law cannot be used to determine the pressure that the gas exerts on the piston to determine the work. This is because, in the irreversible case, except for the initial and final thermodynamic equilibrium states, the system passes through all non-equilibrium states; and the ideal gas law applies only to thermodynamic equilibrium states. So we must adopt a different strategy. This is done in the present development (and most other developments in the thermo literature) by manually controlling the pressure that the surroundings exert on the gas at the inside face of the piston (where the work is being done).

In the irreversible expansion step to be evaluated here, at time zero, we suddenly drop the pressure of the surroundings at the inside face of the piston from ##P_1## to ##P_2## and then hold it at this value while the gas rapidly expands and eventually equilibrates to the new pressure at the new larger volume ##V_2##. And, in the irreversible compression step, we suddenly increase the pressure of the surroundings at the inside face of the piston from ##P_2## to ##P_1## and then hold it at this value while the gas rapidly compresses and eventually re-equilibrates to the pressure ##P_1## and at the original volume ##V_1##.

Based on this description of the irreversible process, the work done by the gas during the expansion step is given by $$W_E=\int_{V_1}^{V_2}{P_{imposed}dV}=P_2(V_2-V_1)=P_2V_2\left(1-\frac{V_1}{V_2}\right)=P_1V_1\left(1-\frac{V_1}{V_2}\right)$$Numerical comparison of this with the work done by the gas during the reversible expansion ##P_1V_1\ln{(V_2/V_1)}## shows that the reversible expansion work is greater than the irreversible expansion work for all values of the volume ratio ##V_2/V_1## greater than unity. This is consistent with what the questioner described in his request for an explanation.

Applying the 1st law of thermodynamics to this irreversible isothermal expansion step shows that the heat absorbed from the surroundings is equal to the work done by the gas:

$$Q_E=W_E=P_1V_1\left(1-\frac{V_1}{V_2}\right)$$If follows from this that the heat absorbed from the surroundings during the reversible expansion step is likewise greater than the heat absorbed from the surroundings during the irreversible expansion step.

In a similar way, we find that the heat and work done by the gas on the surroundings during the irreversible isothermal compression step is given by $$W_C=Q_C=-P_1V_1\left(\frac{V_2}{V_1}-1\right)$$This is the negative of the work and heat done by the surroundings on the gas ##P_1V_1\left(\frac{V_2}{V_1}-1\right)## during the compression. Numerical comparison of this with the work done by the surroundings on the gas during the reversible compression ##P_1V_1\ln{(V_2/V_1)}## shows that the reversible compression work is less than the irreversible compression work for all values of the volume ratio ##V_2/V_1## greater than unity. This again is consistent with what the questioner described in his request for an explanation.

Applying the 1st law of thermodynamics to this irreversible isothermal compression step shows that the heat absorbed from the surroundings is equal to the work done by the gas:$$Q_C=W_C=Q_C=-P_1V_1\left(\frac{V_2}{V_1}-1\right)$$The minus sign indicates that, during the isothermal compression step, heat is rejected to the surroundings. It follows from the previous discussion that the amount of heat rejected to the surroundings during the irreversible compression is greater than the amount of heat rejected to the surroundings during the reversible compression.

For the overall combined two-step irreversible process, the net amount of work done by the system and the net amount of heat absorbed from the surrounding are given by $$W_{net}=Q_{net}=P_1V_1\left(2-\frac{V_2}{V_1}-\frac{V_1}{V_2}\right)=-\frac{P_1}{V_2}(V_2-V_1)^2$$Numerical evaluation of this reveals that this net work and heat are negative for all values of the volumes other than ##V_2=V_1##. This indicates that, for the overall combined expansion-compression process, the system receives a finite amount of work form the surroundings and rejects an equal amount of heat to the surroundings. Thus, although the system does not change over the combined process, the surroundings do change; and the net effect of this change is the conversion of potentially usable mechanical work from the surroundings into heat.

Unlike the case of reversible process paths, for the irreversible process paths, the ideal gas law cannot be used to determine the pressure that the gas exerts on the piston to determine the work. This is because, in the irreversible case, except for the initial and final thermodynamic equilibrium states, the system passes through all non-equilibrium states; and the ideal gas law applies only to thermodynamic equilibrium states. So we must adopt a different strategy. This is done in the present development (and most other developments in the thermo literature) by manually controlling the pressure that the surroundings exert on the gas at the inside face of the piston (where the work is being done).

In the irreversible expansion step to be evaluated here, at time zero, we suddenly drop the pressure of the surroundings at the inside face of the piston from ##P_1## to ##P_2## and then hold it at this value while the gas rapidly expands and eventually equilibrates to the new pressure at the new larger volume ##V_2##. And, in the irreversible compression step, we suddenly increase the pressure of the surroundings at the inside face of the piston from ##P_2## to ##P_1## and then hold it at this value while the gas rapidly compresses and eventually re-equilibrates to the pressure ##P_1## and at the original volume ##V_1##.

Based on this description of the irreversible process, the work done by the gas during the expansion step is given by $$W_E=\int_{V_1}^{V_2}{P_{imposed}dV}=P_2(V_2-V_1)=P_2V_2\left(1-\frac{V_1}{V_2}\right)=P_1V_1\left(1-\frac{V_1}{V_2}\right)$$Numerical comparison of this with the work done by the gas during the reversible expansion ##P_1V_1\ln{(V_2/V_1)}## shows that the reversible expansion work is greater than the irreversible expansion work for all values of the volume ratio ##V_2/V_1## greater than unity. This is consistent with what the questioner described in his request for an explanation.

Applying the 1st law of thermodynamics to this irreversible isothermal expansion step shows that the heat absorbed from the surroundings is equal to the work done by the gas:

$$Q_E=W_E=P_1V_1\left(1-\frac{V_1}{V_2}\right)$$If follows from this that the heat absorbed from the surroundings during the reversible expansion step is likewise greater than the heat absorbed from the surroundings during the irreversible expansion step.

In a similar way, we find that the heat and work done by the gas on the surroundings during the irreversible isothermal compression step is given by $$W_C=Q_C=-P_1V_1\left(\frac{V_2}{V_1}-1\right)$$This is the negative of the work and heat done by the surroundings on the gas ##P_1V_1\left(\frac{V_2}{V_1}-1\right)## during the compression. Numerical comparison of this with the work done by the surroundings on the gas during the reversible compression ##P_1V_1\ln{(V_2/V_1)}## shows that the reversible compression work is less than the irreversible compression work for all values of the volume ratio ##V_2/V_1## greater than unity. This again is consistent with what the questioner described in his request for an explanation.

Applying the 1st law of thermodynamics to this irreversible isothermal compression step shows that the heat absorbed from the surroundings is equal to the work done by the gas:$$Q_C=W_C=Q_C=-P_1V_1\left(\frac{V_2}{V_1}-1\right)$$The minus sign indicates that, during the isothermal compression step, heat is rejected to the surroundings. It follows from the previous discussion that the amount of heat rejected to the surroundings during the irreversible compression is greater than the amount of heat rejected to the surroundings during the reversible compression.

For the overall combined two-step irreversible process, the net amount of work done by the system and the net amount of heat absorbed from the surrounding are given by $$W_{net}=Q_{net}=P_1V_1\left(2-\frac{V_2}{V_1}-\frac{V_1}{V_2}\right)=-\frac{P_1}{V_2}(V_2-V_1)^2$$Numerical evaluation of this reveals that this net work and heat are negative for all values of the volumes other than ##V_2=V_1##. This indicates that, for the overall combined expansion-compression process, the system receives a finite amount of work form the surroundings and rejects an equal amount of heat to the surroundings. Thus, although the system does not change over the combined process, the surroundings do change; and the net effect of this change is the conversion of potentially usable mechanical work from the surroundings into heat.

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The best thermodynamics book, using this approach to the 2nd Law (closely following the now classical considerations by (Carnot), Clausius, Planck, Boltzmann, et al.) by postulating the impossibility of a perpetuum mobile of the 2nd kind, is

Becker, Theory of Heat, Springer

- #5

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If, by a closed system, you mean one that can exchange work and heat with its surroundings, but not mass, then no. This inequality is true for the combination of system plus surroundings, not for the system. For the system, the equation that applies is $$\Delta S \geq \int{\frac{dq}{T_{surr}}}$$which is satisfied for the present irreversible process.Very nice, but don't you argue with the 2nd Law (or rather Boltzmann's H-theorem) that ##\Delta S \geq 0## for a closed system with the equal sign only for the reversible processes?

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Of course Boltzmann's H-theorem is for system+surroundings (i.e., a closed system).

- #7

Entropy is a (path independent) state function and we also have, potentially, that ##\Delta S_{sys} \neq 0## for the irreversible analysis. Is this because of what you were talking about above, that the final state of the

Sorry if it is a silly question .

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The change in entropy of the surroundings is exactly $$\Delta S_{surr} =\frac{P_1}{V_2 T}(V_2-V_1)^2$$ not greater. This is because the surroundings are comprised of an ideal reservoir at temperature T, for which ##\Delta S=\frac{Q}{T}##

Entropy is a (path independent) state function and we also have, potentially, that ##\Delta S_{sys} \neq 0## for the irreversible analysis. Is this because of what you were talking about above, that the final state of theuniverseis indeednotthe same (because the surroundings are not in its original state), and as such the system can undergo a change in entropy?

Sorry if it is a silly question .

- #9

@Chestermiller thanks, that makes a lot more sense to me now! That is reassuring, the possibility of the system returning to its exact original state but with a non-zero change in system entropy seemed weird (and slightly oxymoronic...), so I'm very glad that's not the case!

I think I learned a lot from this analysis

I think I learned a lot from this analysis

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I wish that the questioner would join PF so that he could participate in our discussion.

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I thought I would chime into say that there is a perhaps easier way to see the reason for the outcomes questioned on stack exchange.

Say we have an ideal gas as the working substance. So, we know that the internal energy is a function only of temperature, and thus in an isothermal process the internal energy doesn't change. This means that by the first law we have:

$$ \cancelto{0}{\Delta U } =Q-W \rightarrow Q=W$$

Now consider two processes that go from state 1 to state 2 isothermally. Process A is reversible and process B is irreversible. Entropy balances give:

$$\Delta S_A = \oint_{boundary} \,\frac{\delta Q_A}{T} + \cancelto{0}{\sigma_{genA} }$$

and

$$\Delta S_B = \oint_{boundary} \,\frac{\delta Q_B}{T} + \sigma_{genB} $$

Now since both processes occur between the same states, the change in entropy for each must be the same. This gives us

$$\oint_{boundary} \,\frac{\delta Q_A}{T} = \oint_{boundary} \,\frac{\delta Q_B}{T} + \sigma_{genB} $$

Or, upon rearrangement (and since T is constant and the entropy generation term is strictly positive):

$$\oint_{boundary} \,\delta Q_A - \oint_{boundary} \,\delta Q_B = T \sigma_{genB} >0 \rightarrow Q_A - Q_B > 0$$

Recognizing that we can use the first law result above, we get:

$$W_A - W_B>0 \rightarrow W_A>W_B$$

And we are done.

Note that for a compression, work will be negative and so we see that less work input is required for a reversible compression. For an expansion, the work will be positive and so we see that more work output is obtained in a reversible expansion.

Just another way of looking at it.

Say we have an ideal gas as the working substance. So, we know that the internal energy is a function only of temperature, and thus in an isothermal process the internal energy doesn't change. This means that by the first law we have:

$$ \cancelto{0}{\Delta U } =Q-W \rightarrow Q=W$$

Now consider two processes that go from state 1 to state 2 isothermally. Process A is reversible and process B is irreversible. Entropy balances give:

$$\Delta S_A = \oint_{boundary} \,\frac{\delta Q_A}{T} + \cancelto{0}{\sigma_{genA} }$$

and

$$\Delta S_B = \oint_{boundary} \,\frac{\delta Q_B}{T} + \sigma_{genB} $$

Now since both processes occur between the same states, the change in entropy for each must be the same. This gives us

$$\oint_{boundary} \,\frac{\delta Q_A}{T} = \oint_{boundary} \,\frac{\delta Q_B}{T} + \sigma_{genB} $$

Or, upon rearrangement (and since T is constant and the entropy generation term is strictly positive):

$$\oint_{boundary} \,\delta Q_A - \oint_{boundary} \,\delta Q_B = T \sigma_{genB} >0 \rightarrow Q_A - Q_B > 0$$

Recognizing that we can use the first law result above, we get:

$$W_A - W_B>0 \rightarrow W_A>W_B$$

And we are done.

Note that for a compression, work will be negative and so we see that less work input is required for a reversible compression. For an expansion, the work will be positive and so we see that more work output is obtained in a reversible expansion.

Just another way of looking at it.

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