Boiling and vapor pressure

  • #26
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Hmm, So vapor pressure is acting upward?
We know that, at a given location, pressure acts equally in all directions (up, down, left, right). However, pressure can (and does) vary with location.
But it still does pressure on the surface because of it is weight??? ( Shouldn't the pressure on the top of the liquid increase? Should This be calculated as the following Mbefore - Mafter = Mvapor then Mg/A? And when it tells me that the pressure of the vapor is for example 10 Pa does that mean that it is acting upward?)
At a given location, the pressure acts equally in all directions. If half the weight of the water is now above the liquid/vapor interface, there are two ways of calculating the pressure at the interface.

Method 1. Pressure at piston + Mvapor x g/A=10000+5 /0.01 = 10500 Pa

Method 2. Pressure at base of cylinder - Mliquid x g/A = 10500 - 5/0.01=10500 Pa
Solutions to the question:
1) 11000 N/m^2
Correct
2) Piston pressure is 100/0.01 which is 10000 N/M^2 + 0.5*10/0.01 = 10500 N/M^2 Or Pa
Not correct. Pressure at piston is always determined by weight of piston in this problem, and is always equal to 10000 Pa.

Summary:

Gas pressure at piston face = 10000 Pa
Gas and liquid pressure at liquid/vapor interface = 10500 Pa
Liquid pressure at base of cylinder = 11000 Pa

So, under these circumstances, when half the liquid has formed vapor, what is the (uniform) temperature in the system?
 
  • #27
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51
We know that, at a given location, pressure acts equally in all directions (up, down, left, right). However, pressure can (and does) vary with location.

At a given location, the pressure acts equally in all directions. If half the weight of the water is now above the liquid/vapor interface, there are two ways of calculating the pressure at the interface.

Method 1. Pressure at piston + Mvapor x g/A=10000+5 /0.01 = 10500 Pa

Method 2. Pressure at base of cylinder - Mliquid x g/A = 10500 - 5/0.01=10500 Pa

Correct

Not correct. Pressure at piston is always determined by weight of piston in this problem, and is always equal to 10000 Pa.

Summary:

Gas pressure at piston face = 10000 Pa
Gas and liquid pressure at liquid/vapor interface = 10500 Pa
Liquid pressure at base of cylinder = 11000 Pa

So, under these circumstances, when half the liquid has formed vapor, what is the (uniform) temperature in the system?
Why the second solution is wrong
"What will be the pressure on the upper surface of the liquid?"
I answered 10500 Pa
 
  • #28
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Why the second solution is wrong
"What will be the pressure on the upper surface of the liquid?"
I answered 10500 Pa
Sorry. I thought you were saying that was the pressure at the piston face.

So, we are in agreement now, right?

Chet
 
  • #30
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51
I get 45.7 C for the initial levitation temperature at 10000 Pa.
I have used an approximation through a table. However I am going to learn the mathematical way too.

Anyway, Something have bothered me a bit. You said half of the water evaporated right? Well the vapor pressure should be bigger than the piston pressure in order to push it upward then they will equalize (Because?)
And If the vapor pressure acts in every direction why dont we just add its pressure to the water instead of calculating its weight and using the equation (P= F/ A)
(I know it is wrong just asking why)

And Why at the piston face the pressure is 10000? Dont we consider the vapor pressure?
These question are my very first questions in the thread.
 
  • #31
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I have used an approximation through a table. However I am going to learn the mathematical way too.
Yes. I wasn't aware that you hadn't covered linear interpolation in your algebra course yet. So, if you prefer, I will do the interpolations for you.
Anyway, Something have bothered me a bit. You said half of the water evaporated right? Well the vapor pressure should be bigger than the piston pressure in order to push it upward then they will equalize (Because?)
The piston was already at force equilibrium. That is, the forces on the piston were already in balance. So the very slightest incremental increase in gas pressure below would be sufficient to start the piston accelerating upward, albeit at a very low acceleration. So maybe, during the transient period, the pressure would temporarily have to rise to say 10000.001 Pa. But, when the system re-equilibrated thermally, the piston would stop moving, and the pressure at its lower face would again have to be 10000. Pa.
And If the vapor pressure acts in every direction why dont we just add its pressure to the water instead of calculating its weight and using the equation (P= F/ A)
(I know it is wrong just asking why)
At each location within our system, there is only one value of pressure. When we calculate the weight of the water and add it to that of the piston, we are determining the pressure (force per unit area) on the top of the liquid surface. There are not two different values of the pressure present at this interface. Just one. The upward pressure of the liquid at the interface matches the downward pressure of the vapor.

In the atmosphere, we measure a pressure of 100000 at the surface of the earth. This is just the weight of the atmosphere. It weights 100000 N for every square meter of surface of the earth. By Newton's 3rd law, the surface of the earth pushes back on the atmosphere with a pressure of 100000 Pa.
And Why at the piston face the pressure is 10000? Dont we consider the vapor pressure?
That is the pressure of the vapor at the piston face, and it is the pressure that is required to hold the piston in equilibrium. But it is not the "equilibrium vapor pressure" with liquid at the temperature of the system. The pressure at the vapor-liquid interface is 10500 Pa, and this is the "equilibrium vapor pressure" with the liquid at the system temperature. The only place that the "vapor pressure - temperature" relation must be satisfied is at the interface between the vapor and liquid. The vapor at the piston face is "superheated," in that its pressure is less than the equilibrium vapor pressure at the system temperature. The liquid at the bottom of the cylinder is "subcooled," in that, if we evaluated the equilibrium vapor pressure at the system temperature (10500 Pa), the liquid pressure at the bottom of the cylinder (11000 Pa) would be higher than the "equilibrium vapor pressure," so a vapor phase could not form at this location.

I hope that this makes sense.
 
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  • #32
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I get a temperature of 46.6 degrees when half the water has evaporated and the pressure at the liquid-vapor interface is 10500 Pa.

Now, please describe the situation that prevails when just enough heat has been added to exactly evaporate all the water. What is the pressure at the piston face? What is the pressure at the base of the cylinder? (The temperature is now 47.6 C).

Chet
 
  • #33
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51
I get a temperature of 46.6 degrees when half the water has evaporated and the pressure at the liquid-vapor interface is 10500 Pa.

Now, please describe the situation that prevails when just enough heat has been added to exactly evaporate all the water. What is the pressure at the piston face? What is the pressure at the base of the cylinder? (The temperature is now 47.6 C).

Chet
ًWhen you increase the temperature the pressure will increase thus it will push the piston even further until it balances out again
Assuming that the piston is still then the pressure should equal 10000 Pa
The pressure at the base of the cylinder is the same as in previous calculations
 
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  • #34
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ًWhen you increase the temperature the pressure will increase thus it will push the piston even further until it balances out again
Assuming that the piston is still then the pressure should equal 10000 Pa
The pressure at the base of the cylinder is the same as in previous calculations
Excellent. Now, one final question before we move on to focus problem #2. If, in this problem, the piston had negligible mass, and, instead of the cylinder being in a room with vacuum, there was air pressure in the room at a constant value of 10000 Pa, would anything change?

Chet
 
  • #35
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Excellent. Now, one final question before we move on to focus problem #2. If, in this problem, the piston had negligible mass, and, instead of the cylinder being in a room with vacuum, there was air pressure in the room at a constant value of 10000 Pa, would anything change?

Chet
Inside the cylinder nothing will change.
 
  • #36
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Problem 2:
You have a cylinder with a massless, frictionless piston (axis of cylinder is vertical). There is 1 liter of liquid water in the cylinder at 20 C, and 1 liter of head space containing a mixture of water vapor and air. The cylinder and piston are enclosed in a room containing air at a pressure of 100 kPa. The system is in thermodynamic equilibrium. In this problem, we are going to be neglecting the effects of gravity on the contents of the cylinder, so that the variation of pressure within the cylinder is going to be considered negligible compared to the total pressure.

What is the pressure exerted by the piston on the gas mixture in the head space? What is the pressure throughout the cylinder? What is the equilibrium vapor pressure of water at 20 C? What is the partial pressure of water vapor in the head space? What is the partial pressure of the air in the head space? What is the pressure throughout the liquid water? From the ideal gas law, how many moles of gas mixture are in the head space?
 
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  • #37
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51
Problem 2:
You have a cylinder with a massless, frictionless piston (axis of cylinder is vertical). There is 1 liter of liquid water in the cylinder at 20 C, and 1 liter of head space containing a mixture of water vapor and air. The cylinder and piston are enclosed in a room containing air at a pressure of 100 mPa. The system is in thermodynamic equilibrium. In this problem, we are going to be neglecting the effects of gravity on the contents of the cylinder, so that the variation of pressure within the cylinder is going to be considered negligible compared to the total pressure.

What is the pressure exerted by the piston on the gas mixture in the head space? What is the pressure throughout the cylinder? What is the equilibrium vapor pressure of water at 20 C? What is the partial pressure of water vapor in the head space? What is the partial pressure of the air in the head space? What is the pressure throughout the liquid water? From the ideal gas law, how many moles of gas mixture are in the head space?
1) It should be 100 KPa which is 10^5Pa
2) As you said we are going to neglect gravity here. So it is will be the same
3) 3158.1885 Pa
4)
A bit of problem here.. I could just do 3158.1885/10^5 and get it but where does this get place in this situation?
he only place that the "vapor pressure - temperature" relation must be satisfied is at the interface between the vapor and liquid. The vapor at the piston face is "superheated," in that its pressure is less than the equilibrium vapor pressure at the system temperature.
5) ^^
6)Hmm think it should be 100 KPa
7)
10^5 * 0.01 = n 8.3 * (273+25)
n = 0.404 moles
 
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  • #38
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1) It should be 100mPa which is 10^8 Pa
2) As you said we are going to neglect gravity here. So it is will be the same
3) 3158.1885 Pa
4)
A bit of problem here.. I could just do 3158.1885/10^8 and get it but where does this get place in this situation?
he only place that the "vapor pressure - temperature" relation must be satisfied is at the interface between the vapor and liquid. The vapor at the piston face is "superheated," in that its pressure is less than the equilibrium vapor pressure at the system temperature.
5) ^^
6)Hmm think it should be 100 mPa
7)
10^8 * 0.01 = n 8.3 * (273+25)
n = 404.3 moles
Which is a lot XD
Uh oh. I meant for the external pressure to be only 100 kPa, not 100 mPa. Can you please edit your answer, and I will edit my question. Sorry for the error.

Chet
 
  • #39
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51
Uh oh. I meant for the external pressure to be only 100 kPa, not 100 mPa. Can you please edit your answer, and I will edit my question. Sorry for the error.

Chet
Done.
 
  • #40
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1) It should be 100 KPa which is 10^5Pa
2) As you said we are going to neglect gravity here. So it is will be the same
3) 3158.1885 Pa
In post #15, you gave the correct value of 2338.8 Pa for the equilibrium vapor pressure of water at 20 C.
4) and 5) ^^
A bit of problem here.. I could just do 3158.1885/10^5 and get it but where does this get place in this situation?
the only place that the "vapor pressure - temperature" relation must be satisfied is at the interface between the vapor and liquid. The vapor at the piston face is "superheated," in that its pressure is less than the equilibrium vapor pressure at the system temperature.
In a mixture of ideal gases, each gas exerts a partial pressure as if the other gases were not even present. Certainly, at the interface with the liquid water, the partial pressure of the water vapor is equal to the equilibrium vapor pressure, 2338.8 Pa. Since we are neglecting gravitational effects, the partial pressure of the water vapor throughout the head space is 2338.8 Pa. So, at the piston face, it will also be equal to2338.8 Pa. Since the total pressure is 100 kPa, and the partial pressure of the water vapor in the head space is 2.3388 kPa, the partial pressure of the air in the head space must be 97.66 kPa.
6)Hmm think it should be 100 KPa
This is correct. So, even through the partial pressure of the water vapor is equal to the equilibrium vapor pressure 2.3388 kPa, the pressure in the liquid water is 100 kPa.
7)
10^5 * 0.01 = n 8.3 * (273+25)
n = 0.404 moles
Correct.


Next questions:
1. Given the partial pressures of water vapor and air in the head space, what are the mole fractions of water vapor and air?
2. How many moles of water are in the head space? How many grams of water are in the head space?
3. How many moles of air are in the head space? How many grams of air are in the head space?
4. How many grams of water total are in the container, assuming that the density of the liquid is 1000 gm/liter?
 

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