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From a video lecture, it is mentioned that "dU≠dW in Joule's free expansion if the process is irreversible and adiabatic" Mentioned in around 36:00-38:00 in the video: https://www.youtube.com/watch?v=RrVq7Yduz2g

What I would like to ask is why in this irreversible adiabatic process, dU≠dW? Is it because the W here doesn't include other sort of work? From first law if dQ is zero dU should be equal to dW.

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  • $\begingroup$ The question clausious inquality is related and the answers to it might help you with this question. $\endgroup$ Commented Jun 4, 2015 at 15:54
  • $\begingroup$ @John Rennie , But if the heat transfer is zero it doesn't matter whether the work is reversible or not, right? $\endgroup$
    – Kelvin S
    Commented Jun 4, 2015 at 16:41

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This is an important and interesting question. It is a tough question, too. It is as simple as it is powerful. I watched the video, but could not get much, to answer this question.

It is easier to answer the question if we invoke the second law. However, we shall answer the question using the first law alone.

To make things simpler, we assume closed system of an ideal gas. When we subject this system to adiabatic expansion, there exist a number of possible final values of V that the system could assume for a given ΔU. In other words, many number of values of delta V correspond to a given value of ΔU (or ΔT).The least value of ΔV corresponds to the reversible process and maximum output of work from the system.This is one extreme. For an irriversible expansion, the output of work would be lower - the greater the irreversibility the lower the output work - maximum irreversibility (free expansion) results in the least out put (zero) work.This is the other extreme.

If the system goes from state A to state B in a reversible adiabatic expansion for a given value of ΔU (or ΔT), it goes from state A to state X (≠ B) in an irreversible expansion for the same value of ΔU . It is impossible to take the system from state X to state B by any adiabatic process - reversible or irreversible. (This is what Caratheodory statement of second law states). The process that takes the system from state X to state B must necessarily involve heat interaction.

Since U is a state function, ΔU from A to B is same for the direct process A to B and the indirect process from A to B via X. However, for the direct (reversible) process ΔU = W, but for the irreversible process via X, ΔU = (W' + Q) = W, Q ≠ 0. Q is the heat rejected to the surroundings in going from X to B.

Thus, we see that ΔU for a process from A to B is equal to W only for a reversible adiabatic process of a closed system, where as, for an irreversible process ΔU ≠ W' for the same end points of the path.

The essential point to note is that the end points of the reversible and irreversible processes must be the same to discuss the question under consideration. With this proviso, we see that ΔU= W for a reversible adiabatic process and ΔU ≠ W' for an irreversible adiabatic process.

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  • $\begingroup$ very good answer. I think what the guy in the video was trying to do is to compare the dU in reversible and irreversible processes, but he got the wrong conclusion. Because the final state that the system achieve by a reversible path is different from that by a irreversible path. One cannot say because dU is not zero in a reversible path then it is also not zero if the gas expand freely. First law of thermodynamics must be obey so if dQ and dW are zero then dU must be zero. $\endgroup$
    – Kelvin S
    Commented Jun 8, 2015 at 14:18
  • $\begingroup$ Did you ever figure out a more detailed answer @Kelvin S $\endgroup$ Commented Oct 1, 2020 at 20:43
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I think that $dW$ and $dQ$ here are both equal to zero, so $dU=0$ in a free expansion as it should.

As for the formula $dW = -PdV$, one needs to be careful with this. The correct equation is $dW= -P_{\rm out}dV$, where $P_{\rm out}$ is the external pressure. This is not the same as the $PdV$ in the equation:

$$dU = TdS - PdV,$$

which is just a statement of partial derivatives. In the latter equation, the partial derivative here is

$$P = -\frac{\partial E}{\partial S}\big|_{S} = \frac{{\partial S}/{\partial V}|_E}{{\partial S}/{\partial E}|_V} = P_{\rm in},$$

which is the internal pressure of the gas. Also, here, $TdS=P_{\rm in}dV$ so the total energy is constant. Note that the first term $TdS$ is nonzero here even though there is no heat. This is because it is the change in entropy due to the expansion, which precisely cancels the $PdV$ term.

The first law in terms of $dQ$ is different and is written

$$dU = dQ + dW = dQ + P_{\rm out} dV,$$

notice the $P_{\rm out}$. What I am writing here may sound surprising but I invite you to check it from as many angles as you like you will see that it holds up.

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