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I saw this post by Chet Miller regarding the work done during an adiabatic reversible and irreversible expansion processes. He had answered to the OP that dU = CvdT = work done during the process. I had a followup question:

Since the dT would be different for each of these processes, the dU would be different as well (assuming constant Cv). Hence, if there is a reversible path and irreversible path between 2 states, the dU for each of these transformations would be different and this would contradict dU being a path independent state function. What am i missing here?

The confusion kind of ties in what one of the MIT lecturers emphasized that dU = (ONLY) reversible work for an adiabatic process.

Thanks

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  • $\begingroup$ No need to worry so much about them. Actually, $dU$ and $dT$ are infinitely small and arbitrary. We are doing so in order to integrate to get the actual change in $U$. $\endgroup$ Jul 12, 2023 at 15:54
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    $\begingroup$ There is no contradiction here. Take another example to understand it. $P$ and $V$ are both state functions but $-PdV=w$ is a path function even though work can be done both reversibly and irreversibly. $\endgroup$ Jul 12, 2023 at 15:56
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    $\begingroup$ There is no adiabatic reversible path between the same two end states as an adiabatic irreversible process . $\endgroup$ Jul 12, 2023 at 17:36
  • $\begingroup$ @ChetMiller Hello Chet Miller. Since you answered my comment in one of my posts, I was wondering if you were still interested in an answer on your question. Just to thanks your comment in my question. $\endgroup$ Jul 12, 2023 at 20:49
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    $\begingroup$ Proscionexium, thanks for replying. The MIT prof. mentioned dU=dW (only for reversible) and said that it is a confusing but important point, He didn't elaborate much. $\endgroup$
    – Guha
    Jul 13, 2023 at 0:27

2 Answers 2

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In an adiabatic irreversible process, entropy is generated within the system even though there is no exchange of heat with the surroundings. In a reversible process, the only way for the entropy of the system to increase in going between the same initial and final state is to exchange heat with the surroundings. This is the only way entropy can change in a reversible process. So there is no adiabatic reversible path between the same two end states as those for an adiabatic irreversible process.

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  • $\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Chemistry Meta, or in Chemistry Chat. Comments continuing discussion may be removed. $\endgroup$
    – andselisk
    Jul 18, 2023 at 15:19
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Let consider reversible and irreversible adiabatic compression of an ideal gas.

If they share the same initial state, they will not reach the same final state unless one of them is combined with some non-adiabatic process.

The irreversible process applies bigger external pressure than system pressure.

\begin{align} p_\text{irrev,ext} &\gt p_\text{rev}\\ \delta W_\text{irrev} = -p_\text{irrev,ext} \mathrm{d}V &\gt -p_\text{rev} \mathrm{d}V = \delta W_\text{rev}\\ \end{align}

Therefore the volume rate of providing work is higher.

\begin{align} \left(\frac{\partial W_\text{irrev}}{\partial V}\right)_Q &\gt \left(\frac{\partial W_\text{rev}}{\partial V}\right)_Q\\ \left(\frac{\partial U_\text{irrev}}{\partial V}\right)_Q &\gt \left(\frac{\partial U_\text{rev}}{\partial V}\right)_Q\\ \left(\frac{\partial T_\text{irrev}}{\partial V}\right)_Q &\gt \left(\frac{\partial T_\text{rev}}{\partial V}\right)_Q \end{align}

The irreversible compression would reach the final temperature and internal energy of the reversible process at bigger volume and lower pressure.

Or, it would reach the final volume of the reversible process at bigger temperature, pressure and internal energy.

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