# Going from 1st state to 2nd through irreversible isothermal expansion

• Chestermiller

#### Chestermiller

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My teacher was teaching me that how work done in isothermal reversible expansion is greater than irreversible expansion and also work done in isothermal irreversible compression is greater than that for reversible compression. He then said if someone tries to go from 1st state to the 2nd step ( let it be expansion ) and then come back to the first state ( compression) through isothermal irreversible process for both 1st to 2nd and 2nd to 1st, he will not be able to get back to the first state. The reason he to told that work done in both the cases were different and so would be the final state ( i.e. the first state ). Please explain.

• Lnewqban
In my judgment, what the teacher said is not quite correct. It is obviously possible in an isothermal process to return the system to its original state in the 2nd step by recompressing the gas back to its original volume. In that case, the temperature and volume will be the same as at the beginning of the expansion,, and so, by the ideal gas law, so must the pressure. However, in the irreversible case, the surroundings will have received a net amount of heat, and so, the surroundings will not be in the same state as at the beginning of expansion. So even though the system (the gas) has been returned to its original state, the surroundings have not.

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.

• 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.

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• 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?

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

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?
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.

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

Very interesting . I'm still a little confused about that last bit on entropy. If the (magnitude of) heat rejected to the surroundings is ##\frac{P_1}{V_2}(V_2-V_1)^2##, then the change of entropy of the surroundings is$$\Delta S_{surr} \geq \frac{P_1}{V_2 T}(V_2-V_1)^2$$Similarly the entropy change of the system is, I assume, $$\Delta S_{sys} \geq -\frac{P_1}{V_2 T}(V_2-V_1)^2$$And we have that ##\Delta S_{surr} + \Delta S_{sys} \geq 0##, which is to be expected for the entropy change of the universe due to an irreversible process.

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 universe is indeed not the 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 .

Very interesting . I'm still a little confused about that last bit on entropy. If the (magnitude of) heat rejected to the surroundings is ##\frac{P_1}{V_2}(V_2-V_1)^2##, then the change of entropy of the surroundings is$$\Delta S_{surr} \geq \frac{P_1}{V_2 T}(V_2-V_1)^2$$Similarly the entropy change of the system is, I assume, $$\Delta S_{sys} \geq -\frac{P_1}{V_2 T}(V_2-V_1)^2$$And we have that ##\Delta S_{surr} + \Delta S_{sys} \geq 0##, which is to be expected for the entropy change of the universe due to an irreversible process.

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 universe is indeed not the 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 .
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}## always. This is the key characteristic of an ideal reservoir, which has an infinite capacity to absorb heat without its temperature changing and an in infinite thermal conductivity, so that there are no temperature gradients within the reservoir. This all means that the entire irreversibility and entropy generation are confined to the system, but, because the initial and final states of the system are the same, all the entropy generated within the system during the process is transferred to the surroundings. So the entropy change of the system is zero: $$\Delta S_{sys}=0 > -\frac{P_1}{V_2 T}(V_2-V_1)^2$$This is the proper representation of the Clausius inequality for the system. So we have for the universe, $$\Delta S_{universe}=\Delta S_{surr} + \Delta S_{sys}= \frac{P_1}{V_2 T}(V_2-V_1)^2\gt 0$$

• etotheipi
@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 Last edited by a moderator:
• Chestermiller
There is one more point I wanted to make about this. If, rather than being isothermal (i.e., system in contact with a constant temperature reservoir), the irreversible expansion and compression were carried out adiabatically, there is no way that the system could be returned to its original state. This may have been what the questioner was referring to when he told us that his teacher had said "through isothermal irreversible process for both 1st to 2nd and 2nd to 1st, he will not be able to get back to the first state." It would seem that his teacher got confused between the case of an isothermal process (which actually can be returned to the original state) and the case of an adiabatic process (which can not be returned to the original state).

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

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.

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• etotheipi and Chestermiller