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Work done in reversible is more than that of irreversible

Suppose we are going from state $(P_1, V_1, T_1)$ to some other state say $(P_2, V_2, T_2)$ for reversible and $(P_2, V_2, T_2')$ for irreversible. Which should be greater, $T_2$ or $T_2'$?

My attempt

Let us say heat supplied is same i.e q for both reversible and irreversible. We already know that $U = q - W$

Reversible as work done is more so $U$ becomes less and in irreversible $W$ is less so $U'$ is more than $U$ i.e $U' > U$

As $U'$ is greater $T_2'$ should be greater. Is this correct explanation?

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For a given number of moles, once $P_1$ and $V_1$ are specified, that determines state 1 and $T_1$. Once $P_2$ and $V_2$ are specified, that determines state 2 and $T_2$. It doesn't matter what process is used to get from state 1 to state 2. Only 2 intensive variables are required to specify the state of the system (assuming no phase change). So $T_2=T_2"$.

Maybe you want to remove either the $P_2$ or the $V_2$ from the problem specification.

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    $\begingroup$ This does not work because one the processes is irreversible which means there is some internal mode of changing the temperature. So if it were an expansion, there would be internal friction which would raise the temperature instantaneously making the overall change smaller. So the OP is right. $\endgroup$ – jheindel Nov 17 '15 at 19:32
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    $\begingroup$ I stand by what I said. The state of gas is established by specifying 2 intensive properties: P and V, P and T, or T and V, where V is the specific volume. If P, V, and T are the final parameter values for an adiabatic reversible expansion, you will not be able to identify an adiabatic irreversible expansion from the same initial state that matches more than one of these final parameter values in the final state. $\endgroup$ – Chet Miller Nov 17 '15 at 20:25
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I believe your answer is correct.

Because we are doing work in this case, we can imagine that this is simply a piston being expanded against the atmosphere.

As you have said, a reversible process corresponds to the maximum amount of work, which in this case also corresponds to the largest change in energy.

Because we are only interested in comparing the temperature $T_2$ and $T_2'$ because the pressures and volumes change by the same amount, we must ask ourselves if the temperature is going to increase or decrease?

Well, without any mathematical proof, I will say that while we are doing work, and thus giving out energy to the atmosphere, the temperature must decrease because our system must lose energy. Thus, while outputting work, the more work we do, the greater the change in temperature.

The above paragraph depends on the process being adiabatic though because if the process were not adiabatic, we would have to consider the relative amounts of heat flow and things could get weird.

But, we have said that $$W_{rev}>W_{irrev}$$thus, $$|\Delta T_{rev}|>|\Delta T_{irrev}|$$but, since temperature is decreasing as we give out energy, $$T_2<T_2'$$

So, you are correct unless we are both making a stupid mistake.

This answer makes sense though, because the for the reverse process, we would be compressing a piston and thus must get back as much energy as possible in the form of a temperature increase. With changes in pressure and volume being the same, we know immediately that the process which begins at the lower temperature and gets back to the original temperature $T_1$ has the largest change in internal energy which must be true for the reversible process.

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    $\begingroup$ I challenge you to offer one single specific example to back up what you are saying here, such that the final pressure and specific volume for the irreversible expansion are exactly the same as the final pressure and specific volume for the reversible expansion (starting from the same initial state). $\endgroup$ – Chet Miller Nov 18 '15 at 0:59
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    $\begingroup$ If the adiabatic reversible and adiabatic irreversible expansions both end at the same (P2, V2) (as specified in the original question), what does the ideal gas law tell you about the final temperatures? $\endgroup$ – Chet Miller Nov 18 '15 at 3:06
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    $\begingroup$ Hmmm.... I'm pretty convinced now that your answer is correct. I've been thinking about it quite a bit over the last few hours and I can't think of any reason why your answer is wrong besides that I feel like the irreversibility must lead to a decrease in work output and hence the temperature shouldn't change as much. Could you point out what is wrong with my mathematical reasoning for my own sake? This is such a straightforward thing I can't believe I'm not understanding this. $\endgroup$ – jheindel Nov 18 '15 at 7:16
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    $\begingroup$ The point I was trying to make is that the original problem statement was over-specified. One cannot say that P2 and V2 are the same for both the adiabatic reversible path and the adiabatic irreversible path. The problem statement would have been fine if the reversible path ended at P2,V2,T2 and the irreversible path ended at P2',V2,T2' (so that only the final volumes were required to be equal). I think that otherwise your analysis would have been fine. $\endgroup$ – Chet Miller Nov 18 '15 at 11:01
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    $\begingroup$ Oh!! Right that makes a lot of sense. I was a little confused by what you meant at the end of your answer but now it makes sense. Ok. Thanks for clarifying! I wasn't trying to be rude I was trying to get to the correct solution. $\endgroup$ – jheindel Nov 18 '15 at 20:43

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