# Why is the Work Done indeterministic when we know $∆V=0$?

Given: In a process on a closed system in closed container, the initial pressure and volume are equal to the final pressure and volume. Comment about the Net Work Done by the System, Net Heat given to the system.

Given Solution:

As Heat is path function, so here it is not determined if the heat supplied is zero or non-zero.

Since Work done is a path function, so here it is not determined if the Work Done by the system is zero or non-zero.

I have understood the Heat part but confused about the Work part. Here, I am concerned only about the pressure-volume work.

My Solution:

The net work done by the system is zero because the container is closed so, $$∆V = 0$$.

Considering the nature of Work done [Path Function] and we do not know about the path of the process, so we may say that Work done is indeterministic but we know that W = P.∆V and ∆V = 0 so how can there be a case in the given problem where Work is non-zero ?

What should be the correct conclusion about the Work done ?

• Who says the work is the integral of PdV? What about stirring work or electrical work? Commented May 15, 2021 at 12:25
• @Chet Miller OP is talking about pressure-volume work only. Commented May 17, 2021 at 15:52
• @Ritam_Dasgupta How do you know that's what he meant? Are you a mindreader? Commented May 17, 2021 at 16:09
• No, because he's a high school student from my own country unless I'm very mistaken. So I know the thermodynamics curricula and the implicit assumptions that are usually taken in the problems. But yeah, OP should have clarified. Commented May 17, 2021 at 16:17
• @Ritam_Dasgupta Precisely ! Commented May 19, 2021 at 10:01

The answer given in your textbook is correct. The correct expression for work is $$W=\int PdV$$ and not $$W=P\Delta V$$, which is only valid for an isobaric process. Alternatively, you can simply apply the first law of thermodynamics to find out if work was done. Since $$U$$, internal energy, is a state function, $$\Delta U$$ is definitely $$0$$. Also you say you understand that $$\Delta Q$$ may be non-zero. Hence, from the first law, is it not obvious that work can be non-zero too?