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Suppose we have the reversible isothermal process from state a to b as shown below:

Text

The work done by the system is simply equal to the area under the curve. The fact that this process is isothermal means that the change in internal energy is zero and so the heat transferred to the system during the process is simply the negative of the work done by the system. Now for a general process (reversible or irreversible) between two arbitrary states a and b, we have that $$\Delta S_{sys}=S_{b}-S_{a}=\int^b_a\frac{ \partial Q}{T_{boundary}}+S_{gen}$$

If we apply this formula in the case of the reversible process depicted above we get that $$\Delta S_{sys}=S_b-S_a=\int^b_a\frac{ \partial Q}{T_{boundary}}$$ since $S_{gen}=0$ in a reversible process. But now suppose we once again perform an isothermal process between states a and b however this time we perform it irreversibly. Entropy is a state function and so the change in the systems entropy is the same as in the reversible case however now we have that $S_{gen}>0$. This means that the heat transferred to get from state a to state b must be less in the irreversible case than in the reversible case. But the work done is the same.

I'm pretty sure I have this partly figured out already. The last sentence of the above paragraph is false. When the process is performed irreversibly the work done is actually no longer equal to the area under the curve because no such curve exists for the irreversible path because irreversible implies a lack of contiguous equilibrium states. In actuality, the work done is less in the case of the irreversible process and hence the heat transferred is also able to be less. Is this thinking correct? Also would the internal energy once more be equal to zero in the irreversible case because the reduction in heat transfer is exactly equal to the reduction in the work done?

Any help on this issue or simply confirmation of my thinking would be greatly appreciated!

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  • $\begingroup$ What do you mean by isothermal? Does this include an irreversible process in which you hold the cylinder in contact with a constant temperature reservoir at the initial temperature of the gas throughout the process, but suddenly drop the external pressure from its initial value to a lower value throughout the process? In this case, the pressure internally within the gas could drop during the beginning part of the process, only to return to the initial value at the end. $\endgroup$ – Chet Miller Feb 23 at 15:09
  • $\begingroup$ Initially I wasn't going to include the qualifier isothermal in my original question and instead was going to simply ask the same question but regarding any general expansion because my main issue was whether any irreversible expansion meant less heat transfer to the system. But considering that the work done during irreversible expansion is less, then we still get the same change in internal energy between the two states regardless of reversibilities. Is it correct that the reduction in heat transfer is cancelled exactly by the reduction in work done? $\endgroup$ – SalahTheGoat Feb 24 at 11:03
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    $\begingroup$ Of course. That follows directly from the first law. Irrespective of the process path, the difference between Q and W is the same between the same two end states. $\endgroup$ – Chet Miller Feb 24 at 12:23
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This means that the heat transferred to get from state a to state b must be less in the irreversible case than in the reversible case. But the work done is the same.

Though it depends on the details of the process, you are correct that less heat is transferred to the system in the case of the irreversible expansion process connecting the same two equilibrium states. But the expansion work done is also less.

The diagram below shows a reversible and irreversible isothermal expansion and compression process connecting the same equilibrium states.

For the irreversible expansion process, the external pressure is abruptly dropped to equal the final pressure of state 2 and held constant at that pressure until the system comes to thermal and mechanical equilibrium at state 2. The initial rapid drop in external pressure is such that there is no time for heat transfer. Then heat is transferred at constant external pressure until the final volume is reached. The overall result is less heat transfer and less work for the irreversible process (less area).

For the irreversible compression process going from state 2 back to state 1, the pressure is abruptly increased to the final pressure, then the system allowed to come to equilibrium at state 1. In this case more heat is transferred out of the system for the irreversible process resulting in more work done on the system (more negative work).

It's important to note that the term "isothermal" has a different meaning for an irreversible vs a reversible process. For both processes the system is maintained in contact with a constant temperature thermal reservoir so that the temperature at the boundary between the system and reservoir is constant. But in the case of the reversible process, the system is in internal thermal equilibrium as well, whereas for the irreversible process the temperature is only constant at the boundary while it varies vary within the system, i.e., temperature gradients exist, so that it is not in internal thermal equilibrium. It's the temperature gradient within the system that results in entropy generation, $S_{gen}$.

Hope this helps.

enter image description here

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  • $\begingroup$ (1/2) Thanks for the great answer! Just to check my understanding, you state that the heat transfer is greater in the case of irreversible compression than that of reversible compression. Let us suppose that $\Delta S_{sys}= S_b-S_a= 5 J/K$ where the system is being isothermally expanded from state a to state b like in my original diagram. Then for compression from state b to state a we will have that $S_{sys}=S_a-S_b=-5J/K$. If the compression is irreversible we may have something like $\Delta S_{sys}=\int \frac{dQ}{T}+S_{gen}=-10J/K +5J/K=-5J/K$ where $S_{gen}=5J/K$ but for $\endgroup$ – SalahTheGoat Feb 24 at 10:55
  • $\begingroup$ (2/2) reversible compression we would have that $\Delta S_{sys}=\frac{dQ}{T}+S_{gen}=-5J/K +0J/K=-5J/K$ where we would now have $S_{gen}=0$ . Thus for irreversible compression more heat is transferred to the surroundings which is offset by $S_{gen}$ within the system. Correct? $\endgroup$ – SalahTheGoat Feb 24 at 10:55
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    $\begingroup$ @SalahTheGoat In your example, the $+5J/K$ and $-5J/K$ is not the entropy generated, it's the total entropy change for either the reversible or irreversible process. For the reversible process it equals the entropy transferred to and from the system since no entropy is generated for the reversible process. For the irreversible path the entropy generated equals the total entropy change minus the entropy transfer from the reservoir at the boundary. If that's not clear, I can update my answer to provide more detail than possible here in comment format. $\endgroup$ – Bob D Feb 24 at 12:28
  • $\begingroup$ No its fine, everything is perfectly clear now. Thanks again! $\endgroup$ – SalahTheGoat Feb 24 at 12:49
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But the work done is the same

As you realized, this is not true. Work done is a path dependent process. You are taking a different path and you expect the amount to work done to change. It is hard to say exactly what has happened with the amount of work done, but that is because you have not specified exactly how you made the process irreversible. For example if you added some friction as the system expanded, then work would need to be done overcoming that friction. Since we are counting this towards the entropy of the system, this work is also being done on the system. This means that we have to transfer less heat into the system to arrive at the same final energy.

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You are correct and you can also verify everything by considering an ideal gas.

$$ Q=W$$

and

$$ Q = T \Delta S - TS_g$$

where you can see that the entropy generated always decreases the heat transfer in the algebraic sense.

Consider an ideal gas for the process you've shown.

$$ Q = -mRTln\frac{P_b}{P_a} - TS_g$$

where $P_b < P_a$ therefore the reversible part $ > 0$. The irreversible part decreases the heat transfer to the system, thus decreasing the work output. Makes sense, the 2nd law puts a limit on work we can extract.

It's also useful to walk through this process for $b$ to $a$, where you'll see that entropy generation increases the heat transfer to the system, thus requiring more work to compress.

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