# Partial Pressure of a Vapour

Tags:
1. Aug 12, 2015

### gamz95

1) We know that when both liquid and vapor are present, and system of these is in phase equilibrium; the "partial pressure of the vapor" must be equal to the "vapor pressure" , i.e. :

partial pressure of vapor= vapor pressure.

2) However, what happens if there is no liquid in the system, i.e. only vapor exists.

a) partial pressure of vapor = vapor pressure
b) partial pressure of vapor <vapor pressure

?

2. Aug 12, 2015

### Staff: Mentor

(b)

3. Aug 13, 2015

### gamz95

Thank you. Can you explain why is that?

4. Aug 13, 2015

### Staff: Mentor

If liquid water is not present, then the water vapor in the gas phase cannot be in equilibrium with liquid water (as you indicated). Thus, at the temperature of the system, the partial pressure of water in the gas phase must be less than the equilibrium vapor pressure at that temperature. If it were equal to the vapor pressure of liquid water at that temperature, then liquid water would have to be present, or, at least, water vapor would be on the verge of condensing to liquid water. In this latter case, if even a slight amount of heat were removed from the system, some of the water vapor would condense to form liquid water (without the temperature changing, assuming the partial pressure of water vapor in the gas phase were held constant).

Hope this makes sense.

Chet

5. Aug 13, 2015

### gamz95

Thank you so much Chet. But, if we increase something in amount, wouldn't the pressure of that thing increases? I mean there is no liquid any more in the system; all of them vapor. Why that vapor's pressure doesn't increases? I am missing something bad, but I don't know what:)

Last edited: Aug 13, 2015
6. Aug 13, 2015

### Staff: Mentor

Sorry. I don't quite follow this question. Maybe you can rephrase it, possibly with an example.

Chet

7. Aug 14, 2015

### gamz95

I am sorry. I will try to explain again:

Let's say liquid and vapor are in equilibrium first. Then, let's say all liquid transformed into vapor. In this case, vapor molecules increase in number. Thus, their pressure increases. So that vapor molecules' pressure increases?

8. Aug 14, 2015

### Staff: Mentor

OK, I understand where you are coming from now. I'm recommending that we look at two specific situations for water in a closed container (where water is the only chemical species present, no air).

1. Cylinder with a piston, in which the piston location is controlled so that the pressure in the cylinder is always constant. Container initially contains liquid water in equilibrium with water vapor at 20 C.

2. Rigid closed container (no piston) initially containing liquid water in equilibrium with water vapor at say 20 C.

In both these cases, we add heat to the system and see what happens.

Will looking at these problems be helpful to you? Which one would you prefer looking at first?

Chet

9. Aug 14, 2015

### gamz95

Thank you very much for your effort firstly Chet.

PV=nRT

1. Heat adds, T increases (Q=m*c*DT), Piston goes upward;

a.when there is enough heat energy available liquid starts to transform into vapor,

nvapor increases, Pvapor increases, Vvapor increases
nliquid decreases, Pliquid decreases, Vliquid decreases

b. if not,

nvapor same, Pvapor decreases or same?, Vvapor increases
nliquid same, Pliquid increases or same?, Vliquid same

2. Heat adds, T increases (Q=m*c*DT), Pcontainer increases;

a.when there is enough heat energy available liquid starts to transform into vapor,

nvapor increases, Pvapor increases, Vvapor increases
nliquid decreases, Pliquid decreases, Vliquid decreases

b. if not,

nvapor same, Pvapor ?, Vvapor same
nliquid same, Pliquid ?, Vliquid same

If these assumptions true I would say first one. I am so sorry for my foolishness Chet, I couldn't solve:)

10. Aug 14, 2015

### Staff: Mentor

Case 1 is pretty easy.

If you hold the pressure constant and add heat,
• The temperature and pressure will not change as long as liquid water remains present
• The volume of liquid and vapor in the cylinder will increase
• The amount of liquid water will decrease
• The amount of water vapor will increase
• Once all the liquid water has evaporated, adding more heat will allow the temperature to rise, and the water vapor will follow PV = nRT
Does this make sense to you. If not, which part.

After you are satisfied with case 1, we can address case 2. Your assignment is to choose a volume for the rigid container, and to specify what fraction is filled with liquid water when we start heating. Because you are interested in the possibility that the liquid water would evaporate, I would suggest specifying a pretty small fraction of liquid water to start with.

Chet

11. Aug 15, 2015

### gamz95

Thank you very and very much Chet.

Case 1:

*Pressure will not change, okay; but, temperature should rise, because we are giving heat into system (Q=m*c*DT ?).
*The volume of vapor increases, okay; but, volume of liquid should be constant as long as liquid remains or decreases as long as there is a transformation.
*If liquid transforms into vapor, okay.
*If liquid transforms into vapor, okay.
*So, we are saying that the thing actually absorbs the heat energy is liquid not the vapor; when liquid exists no more, vapor will absorb the heat energy, right?

Again thank you very much Chet.

12. Aug 15, 2015

### jbriggs444

Like a pot of water on the stove, you can put a flame under it, but the temperature will not rise above boiling while liquid water remains in the pot.

Yes, heat is delivered to the liquid from the external source and removed from the liquid by the evaporation that is taking place.

Edit: Didn't realise that Chet would be online at this hour.

13. Aug 15, 2015

### Staff: Mentor

No. The heat is being used to vaporize the liquid (heat of vaporization)

I didn't write this clearly enough. What I meant was that the sum of the liquid and gas volume (the total volume of water in the cylinder) increases.

Yes. This is one way of thinking about it. I usually just say that the heat flows into the mixture within the cylinder, but, by the time the mixture reaches its final equilibrium state, all the heat has been used to vaporize some liquid.

Chet

14. Aug 15, 2015

### gamz95

Chet and riggs, thank you very much to you. By the way, I am from Turkey, and we share very different time zone:)

One question though; also piston won't go upward before liquid starts to transform, right?

15. Aug 15, 2015

### Staff: Mentor

Suppose

State 1: a fixed amount of water in a cylinder at temperature T and equilibrium vapor pressure P(T)

State 2: same fixed amount of water in cylinder at temperature T and equilibrium vapor pressure P(T), but with less liquid water present, more water vapor present, and larger total volume

It doesn't matter what path you take to get from state 1 to state 2. For example, you can:
• Apply the heat to the liquid exclusively (i.e., at the bottom of the container), and let the piston rise accordingly
• Apply the heat to the vapor exclusively (at the top of the container), and let the piston rise accordingly
• Hold the piston in place while applying the heat and then let the piston rise
• Apply the heat at a slow rate
• Apply the heat at a fast rate
Provided that the location of the piston in state 2 and the total amount of heat added is the same in all these cases, state 2 will be exactly the same.

Chet

16. Aug 15, 2015

### gamz95

I completely understand Chet.

What about case 2? Would you give the same plan like the Case 1:

("If you hold the pressure constant and add heat,
• The temperature and pressure will not change as long as liquid water remains present
• The volume of liquid and vapor in the cylinder will increase
• The amount of liquid water will decrease
• The amount of water vapor will increase
• Once all the liquid water has evaporated, adding more heat will allow the temperature to rise, and the water vapor will follow PV = nRT
Does this make sense to you. If not, which part.")

?

17. Aug 15, 2015

### Staff: Mentor

Case 2 is a little more complicated than case 1. The key feature of case 2 is that the volume of the container is constant. So, in this case, if you add heat, both the temperature and the pressure will have to rise. But they will not be independent. They will rise in tandem, constrained by the condition that the pressure must be at the equilibrium vapor pressure at each temperature. The mass of vapor will increase and the mass of liquid will decrease. Once all the liquid water has evaporated, adding more heat will allow the vapor temperature to rise according to dq = mCvdT, and the pressure will follow PV=nRT.

Incidentally, if you would like to see an actual case 2 problem worked out, see the currently active PF thread https://www.physicsforums.com/threads/thermodynamics-calculate-the-pressure-and-temperature.827818/

Chet

Last edited: Aug 15, 2015
18. Aug 16, 2015

### gamz95

Dear Chet,

Thank you very very very much.

Sincerely,

Gamze.