# Thermodynamics problem (ideal gas law, kinetic theory, processes, etc.)

Homework Statement:
Please refer to the picture below.
Relevant Equations:
pV = nRT
Knowledge of vaporisation, condensation, and the 4 thermal processes.
It is a long problem, but it is simple to understand.

I am having trouble with part A. My attempt:
Pressure outside > pressure inside container. pV = constant (isothermal). At equilibrium, all gases are at atmospheric pressure. Because it is quasi-static, the pressures of both compartments are equal at all times, implying that their volumes are equal at all times as well, as pV = const. This means that at equilibrium with the atmosphere, the volume of each compartment is 0.5V0.

However, the answer writes that after equilibrium with the atmosphere is attained, all the water vapour condenses, and throughout this condensation, pressure somehow remains at 1 atm, only that volume of the left chamber decreases to 0. Why would condensation even occur, and if it does how does the pressure remain at 1 atm while the volume goes to 0?

Furthermore, after the complete condensation of water vapour, the N2 gas in the right chamber undergoes yet another isothermal compression, such that its pressure is now 2 atm, despite the outside pressure being only 1 atm. How is this possible?

Chestermiller
Mentor
They said that the starting pressure is 0.5 atm, and that the piston then compresses the gases. So why are you starting it at 1 atm?

You know the number of moles of gas in each container, and the temperature and pressure in each container. From the ideal gas law, what is the value of ##V_0##?

Here is a hint for continuing. Get the pressure vs volume curve for each compartment separately.

phantomvommand
They said that the starting pressure is 0.5 atm, and that the piston then compresses the gases. So why are you starting it at 1 atm?

You know the number of moles of gas in each container, and the temperature and pressure in each container. From the ideal gas law, what is the value of ##V_0##?

Here is a hint for continuing. Get the pressure vs volume curve for each compartment separately.
Yes, it starts at 0.5atm, but the first isothermal compression ends at 1atm, where pressure = atm pressure. Upon reaching 1 atm, the volume of each compartment is 0.5V0. I have obtained this part of the graph; the graph in the answer key must be read from right to left.

what I do not understand is how the water vapour starts to condense at 1 atm.

The graph shown in the answer key shows the total volume against the pressure, instead of plotting graphs for each individually.

Chestermiller
Mentor
Yes, it starts at 0.5atm, but the first isothermal compression ends at 1atm, where pressure = atm pressure. Upon reaching 1 atm, the volume of each compartment is 0.5V0. I have obtained this part of the graph; the graph in the answer key must be read from right to left.

what I do not understand is how the water vapour starts to condense at 1 atm.
When you reach 1 atm at 100C, the water vapor becomes saturated. After that, you continue to remove heat from the contents of that container as the vapor condenses at constant pressure and temperature. What did you expect to happen?

phantomvommand
When you reach 1 atm at 100C, the water vapor becomes saturated. After that, you continue to remove heat from the contents of that container as the vapor condenses at constant pressure and temperature. What did you expect to happen?
May I know how you deduced that after reaching 1atm, heat is still removed from the container? Heat being removed is a result of water vapour condensing. But the question I have is what caused the water vapour to condense in the first place.

Chestermiller
Mentor
If you try to continue compressing the vapor at 1 atm and 100 C, it will try to reach a higher pressure. But it can't do that at 100 C because that will cause it to exceed the equilibrium vapor pressure at 100 C. So the vapor has to start condensing.

phantomvommand
If you try to continue compressing the vapor at 1 atm and 100 C, it will try to reach a higher pressure. But it can't do that at 100 C because that will cause it to exceed the equilibrium vapor pressure at 100 C. So the vapor has to start condensing.
Thanks for this! I get that argument, but in this case, the vapour is compressed from 0.5atm to 1 atm. To me, at 1 atm, the vapour would naturally remain in its vapour form, because equilibrium has been reached, and no further compression beyond 1 atm will happen. Thus, there is no attempt to increase the pressure beyond 1atm, and thus no condensation should occur?

furthermore, the answer claims that all the vapour condenses, so the attempt to increase the vapour’s pressure beyond 1 atm must be persist for enough time for all the vapour to condense. Given that external pressure = vapour pressure = 1 atm, I do not see any sustained attempt to increase the external pressure.

Chestermiller
Mentor
Thanks for this! I get that argument, but in this case, the vapour is compressed from 0.5atm to 1 atm. To me, at 1 atm, the vapour would naturally remain in its vapour form, because equilibrium has been reached, and no further compression beyond 1 atm will happen. Thus, there is no attempt to increase the pressure beyond 1atm, and thus no condensation should occur?
Can 100% liquid water exist at 1 atm and 100 C? And, if you heat this, can you reach 100% vapor without the pressure and temperature changing?

phantomvommand
Can 100% liquid water exist at 1 atm and 100 C?
I don’t think so, but what exactly is the reason why 100% vapour/liquid cannot remain at 100% “purity” at 1 atm and 100C? Furthermore, the answer claims that all the vapour condenses, which does not agree with your suggestion that the vapour would exist in some equilibria with water.

Chestermiller
Mentor
I don’t think so, but what exactly is the reason why 100% vapour/liquid cannot remain at 100% “purity” at 1 atm and 100C? Furthermore, the answer claims that all the vapour condenses, which does not agree with your suggestion that the vapour would exist in some equilibria with water.
Of course water can exist as a pure liquid at 100 C and 1 atm. It will be at equilibrium with any vapor that is present at the same pressure and temperature.

If you try to compress the water vapor at 100 C and 1 atm, you will have to remove heat, and you will begin to form liquid water. The liquid water will exist in equilibrium with the water vapor at the same temperature and pressure. As you compress it, the volume (and number of moles) of water vapor decreases, and the volume (and number of moles) of the liquid water increases, so that the total volume is what you have compressed it to. Eventually, if you decrease the volume sufficiently, you will reach 100 % liquid water.

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phantomvommand
Of course water can exist as a pure liquid at 100 C and 1 atm. It will be at equilibrium with any vapor that is present at the same pressure and temperature.

If you try to compress the water vapor at 100 C and 1 atm, you will have to remove heat, and you will begin to form liquid water. The liquid water will exist in equilibrium with the water vapor at the same temperature and pressure. As you compress it, the volume (and number of moles) of water vapor decreases, and the volume (and number of moles) of the liquid water increases, so that the total volume is what you have compressed it to. Eventually, if you decrease the volume sufficiently, you will reach 100 % liquid water.
Please bear with me for a while more... I am trying to understand this.
You write that “if you try to compress vapour, you have to remove heat ...”. I agree with this, but doesn’t this make everything tentative? There must be a reason that vapour is being further compressed, despite already having the same pressure as the surroundings, and what is that reason?

Chestermiller
Mentor
Please bear with me for a while more... I am trying to understand this.
You write that “if you try to compress vapour, you have to remove heat ...”. I agree with this, but doesn’t this make everything tentative? There must be a reason that vapour is being further compressed, despite already having the same pressure as the surroundings, and what is that reason?
The density of the vapor doesn't change when you are decreasing the volume during the compression at 100 C and 1 atm. Instead, the number of moles of vapor is gradually decreasing and the number of moles of liquid water is correspondingly increasing. But the amount of volume that a given mass of liquid occupies is much less than the amount of volume that the same mass of vapor occupies. So this is how the combined volume of liquid water and water vapor decreases during the horizontal section on your graph. The middle of the horizontal section corresponds to when you have converted half the vapor to liquid water, and the left side of the horizontal section corresponds to when you have converted all the vapor to liquid water.

phantomvommand
The density of the vapor doesn't change when you are decreasing the volume during the compression at 100 C and 1 atm. Instead, the number of moles of vapor is gradually decreasing and the number of moles of liquid water is correspondingly increasing. But the amount of volume that a given mass of liquid occupies is much less than the amount of volume that the same mass of vapor occupies. So this is how the combined volume of liquid water and water vapor decreases during the horizontal section on your graph. The middle of the horizontal section corresponds to when you have converted half the vapor to liquid water, and the left side of the horizontal section corresponds to when you have converted all the vapor to liquid water.
Yes I understand the graph. However, I would like to know why the vapour is even being compressed, because when it is at 1 atm, it’s pressure is equal to the external pressure, and there is no compression of the vapour to begin with. Since the vapour does not even compress, as the external pressure is not greater than the vapour pressure, it will not condense.

Chestermiller
Mentor
Yes I understand the graph. However, I would like to know why the vapour is even being compressed, because when it is at 1 atm, it’s pressure is equal to the external pressure, and there is no compression of the vapour to begin with. Since the vapour does not even compress, as the external pressure is not greater than the vapour pressure, it will not condense.
The piston is moving to the left to compress the nitrogen, and the nitrogen pressure is increasing, so this causes the movable partition between the steam vapor and the nitrogen to move to the left and compress the vapor.

phantomvommand
The piston is moving to the left to compress the nitrogen, and the nitrogen pressure is increasing, so this causes the movable partition between the steam vapor and the nitrogen to move to the left and compress the vapor.
But wouldn’t the piston stop moving the moment pressure in each chamber= 1atm, at which point there is no compression?

Chestermiller
Mentor
But wouldn’t the piston stop moving the moment pressure in each chamber= 1atm, at which point there is no compression?
This has nothing to do with the outside atmosphere. We push the piston in manually.

phantomvommand
This has nothing to do with the outside atmosphere. We push the piston in manually.
I don’t think pushing the piston in manually is clear from the question. To me, the question suggests that the piston is pushed inwards at the start as the external pressure = 1atm > gas pressure = 0.5 atm. (The external pressure is driving the entire system) I am quite new to this forum; would you be able to get a third person to check this question out to verify the interpretation?

Chestermiller
Mentor
I don’t think pushing the piston in manually is clear from the question. To me, the question suggests that the piston is pushed inwards at the start as the external pressure = 1atm > gas pressure = 0.5 atm. (The external pressure is driving the entire system) I am quite new to this forum; would you be able to get a third person to check this question out to verify the interpretation?
That's not necessary. I'm the thermo expert on Physics Forums.

It is true that the problem statement is a little ambiguous. But pushing it in manually is the only way the final volume is going to get to ##V_0/4##.

phantomvommand
That's not necessary. I'm the thermo expert on Physics Forums.

It is true that the problem statement is a little ambiguous. But pushing it in manually is the only way the final volume is going to get to ##V_0/4##.
I was not aware of this fact! Great to hear from someone so experienced.
Yes, that's also the only way the pressure could increase beyond 1atm. Thank you so much for helping me throughout this. :)