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I get confused in problems where it is necessary to evaluate the work done by a gas that someway moves a mass. I'll make an example.

Consider the tube containing mercury and an ideal gas in A (Picture $(1)$). If the gas expands and increases its volume, what is the work done by the gas?

enter image description here

My reasoning, illustrated in $(2)$ and $(3)$ would be: since the only effect of the expansion is to "move" the red volume of Hg from the initial position to an higher position in the thin tube, then the work done by the gas equals the amount of change in potential energy of the red volume of Hg, that's all, there is no other work done by the gas.

It seems right but also a bit contractictory. Suppose for istance that the section of the tube was the same in the two parts. Then there would be still change in the potential energy of the red volume of Hg, but it seems to me that in that case nothing happens, as the Hg "goes down" and "goes up" of the same height, and the work should be zero.

If all the previous is right, then my question is: is it correct to think about such situations in this way? I mean to look at small volumes of the fluid that overall change their position, whitout caring about the rest of the fluid, to evaluate the change in potential energy?

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  • $\begingroup$ "as the Hg "goes down" and "goes up" of the same height, and the work should be zero." I don't think it is zero, if there is difference in height $d$ is not zero. $\endgroup$ – philip_0008 Jun 18 '16 at 19:14
  • $\begingroup$ The question is actually trickier than I thought. Since the gas is expanding, wouldn't the work done by the gas be always positive? $\endgroup$ – philip_0008 Jun 18 '16 at 19:26
  • $\begingroup$ Hg "goes down" and "goes up" of the same height, and the work should be zero. Why do you think so? It seems to me that you are having problem in calculating the difference in potential energy. To understand things better try to calculate the change in potential energy assuming that there is no atmospheric pressure. Note that the pressure P in PdV is variable(why?) and remember that the expansion is reversible(why?) $\endgroup$ – Jolie Jun 25 '16 at 20:51
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Your intuition that the same amount of fluid goes down and then up by the same amount is incomplete, you are forgetting what happens inside the fluid. It is easier to see using solid blocks as in the figure below:

enter image description here

Here you can see that the effect of moving block 1 down is to shift block 2 to the right, and moving block 3 back up the same amount that block 1 went down, then in addition to that , you move block 4 up. You can compute the change in energy block by block (easy for blocks but difficult for fluids), or by the net change in configuration: imagine that it was block one that moved on top of block 4 and the rest did not move. Both ways of computing it will result in the same positive change in potential energy.

To illustrate the idea, the work done by the gas will be equal to the energy transferred to the environment, if there are no heat losses, this can be calculated as the change in potential energy of the mercury plus the work done on the atmosphere:

$W=P_{atm}\Delta V-m_Ag\Delta h_A/2+m_Bg(d+\Delta h_B/2)$

using $m=\rho_{Hg} \Delta V$ and $\Delta h_i=\Delta V /S_i$ we get:

$W=P_{atm}\Delta V+ \frac{1}{2}\rho_{Hg} g(\Delta V)^2 (\frac{1}{S_B}-\frac{1}{S_A})+\rho_{Hg} g d\Delta V $

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Work is always force times displacement in the direction of the force. The only place where the gas is doing work is at the bottom surface that is moving downward. The force it is exerting there is $PA$, where $P$ is the gas pressure and $A$ is the cross sectional area of the tube. If the lower surface moves downward a differential distance dx, the work done by the gas is $dW=PA \ dx$. But $A\ dx$ is the change in volume $dV$ of the gas. Therefore, the differential amount of work that the gas does is $P\ dV$. You just integrate this to get the total amount of work done.

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Your approach is almost correct!

To see what's wrong, consider the case $d = 0$. Then for a small change in volume, the change in potential energy of the water is zero (since you're just moving water from the left side to water at the same height on the right side). But the gas has definitely done $p \, dV$ work.

The mistake is that you've neglected the work done on the atmosphere, i.e. on the gas at the open part of the right end of the tube. When $d = 0$, the atmospheric pressure is equal to the gas pressure $p$, so the work done on the air is $p\, dV$. Energy simply goes from one gas to another.

In general, your approach will work, but you have to take care to include all contributions. The atmosphere is easy to forget about!

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  • $\begingroup$ Any explanation for the downvotes? $\endgroup$ – knzhou Jun 22 '16 at 21:04
  • $\begingroup$ I voted down because you seem to agree that there is no change in potential energy! and what if the experiment is done in the moon? $\endgroup$ – Wolphram jonny Jun 22 '16 at 21:16
  • $\begingroup$ @Wolphramjonny We're not talking about the same situation. I am taking $d = 0$ and infinitesimal $dV$, in which case the change in potential energy is zero. You are taking finite $dV$. $\endgroup$ – knzhou Jun 22 '16 at 21:18
  • $\begingroup$ In both cases, you must account for atmospheric pressure or else you get the wrong answer. This is the main piece that OP is missing, which is why I focused on that. $\endgroup$ – knzhou Jun 22 '16 at 21:19
  • $\begingroup$ if you do not consider a finite change in height on the liquid, then it does not answer the question. But I will edit you questio to undo my downvote $\endgroup$ – Wolphram jonny Jun 22 '16 at 21:24

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