How Does Vapor Pressure Cause Water to Boil?

In summary, the conversation discusses the relationship between vapor pressure and atmospheric pressure. The participants question how increasing vapor pressure affects the boiling point of a liquid, and how atmospheric pressure affects bubbles at the bottom of a liquid. The expert explains that the pressure at the bottom of a liquid is a combination of air and liquid pressure, and that atmospheric pressure applies pressure throughout the liquid. The expert also clarifies the definition of vapor pressure and its relationship to the equilibrium between liquid and vapor at a specific temperature.
  • #1
Biker
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I have searched about this topic all over the internet and non of them seem to explain how vapor pressure is equal to atmospheric pressure.

All I need is some forces diagrams and some explanations.
How when we increase the vapor pressure it makes the liquid boil faster? Isnt the vapor pressure in all directions? Wouldn't it push molecules back?
What do they mean by the vapor pressure equaling atmospheric pressure? how does that occur?
Also, about bubbles. How does the atmospheric pressure affect the bubbles when they are at the bottom of the liquid? Like there is not connection between them..
 
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  • #2
I have read that before and it doesn't explain my point :/
Edit:^^ a reply to a deleted post.
 
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  • #3
Biker said:
How does the atmospheric pressure affect the bubbles when they are at the bottom of the liquid? Like there is not connection between them..
Suppose that you have a shallow tank of water. The pressure at the top of the tank is equal to atmospheric pressure. What is the pressure at the bottom of the tank?
 
  • #4
jbriggs444 said:
Suppose that you have a shallow tank of water. The pressure at the top of the tank is equal to atmospheric pressure. What is the pressure at the bottom of the tank?
I have read before that it is the air pressure + the liquid pressure. But I still don't get how the air pressure reaches the bottom of the liquid.. ( I guess it is just a misconception)
 
  • #5
Biker said:
I have read before that it is the air pressure + the liquid pressure. But I still don't get how the air pressure reaches the bottom of the liquid.. ( I guess it is just a misconception)
Yes, it is air pressure plus the liquid pressure. Consider for a moment what would happen if that were not so...

Suppose, for instance, that the pressure at the bottom of the liquid were from the liquid alone. For one inch depth of water, that pressure is around 1/300 of an atmosphere. This would be the downward pressure that the bottom of the water exerts on the bottom of the tank. By Newton's third law, it would also be the upward pressure that the bottom of the tank exerts on the water. So the top of the water would be under a downward pressure of one atmosphere and the bottom of the water would be under an upward pressure of 1/300 of an atmosphere. That's an unbalanced upward net force -- one would expect the water to come shooting out of that tank like a geyser. But, of course, it does not.

Now consider another example. You are holding a garden hose in your hand with your thumb over the end and the water turned off. The pressure in the hose is negligible. You turn on the tap and the pressure increases under your thumb. The water at the faucet end pushes into the hose and you feel the pressure with your thumb.

It is the same thing with the atmosphere. It pushes on the top of the water and its pressure manifests throughout the water, including the bottom.
 
  • #6
The vapor pressure increases as the liquid is heated up. At the boiling point the vapor pressure equals the atmospheric pressure.
 
  • #7
jbriggs444 said:
Yes, it is air pressure plus the liquid pressure. Consider for a moment what would happen if that were not so...

Suppose, for instance, that the pressure at the bottom of the liquid were from the liquid alone. For one inch depth of water, that pressure is around 1/300 of an atmosphere. This would be the downward pressure that the bottom of the water exerts on the bottom of the tank. By Newton's third law, it would also be the upward pressure that the bottom of the tank exerts on the water. So the top of the water would be under a downward pressure of one atmosphere and the bottom of the water would be under an upward pressure of 1/300 of an atmosphere. That's an unbalanced upward net force -- one would expect the water to come shooting out of that tank like a geyser. But, of course, it does not.

Now consider another example. You are holding a garden hose in your hand with your thumb over the end and the water turned off. The pressure in the hose is negligible. You turn on the tap and the pressure increases under your thumb. The water at the faucet end pushes into the hose and you feel the pressure with your thumb.

It is the same thing with the atmosphere. It pushes on the top of the water and its pressure manifests throughout the water, including the bottom.

Thanks, I understand it now.
But this is what I mean about vapor pressure and atmospheric pressure
http://i.imgur.com/tVfDtlb.jpg
Shouldn't when the vapor pressure increase the evaporation decrease? So why is the boiling point is when the vapor pressure is equal to the atmospheric pressure?
 
  • #8
Biker said:
Thanks, I understand it now.
But this is what I mean about vapor pressure and atmospheric pressure
http://i.imgur.com/tVfDtlb.jpg
Shouldn't when the vapor pressure increase the evaporation decrease? So why is the boiling point is when the vapor pressure is equal to the atmospheric pressure?
What is your precise definition of the term "vapor pressure?"
 
  • #9
Chestermiller said:
What is your precise definition of the term "vapor pressure?"
The pressure resulted of the vapor on the surface of the liquid when the equilibrium happen between the liquid and the vapor in a certain temperature.
 
  • #10
Biker said:
The pressure resulted of the vapor on the surface of the liquid when the equilibrium happen between the liquid and the vapor in a certain temperature.
Thanks Biker. In my judgment, this seems to place too much emphasis on what's happening at the surface at equilibrium between liquid and vapor. If you had a closed container, with liquid below and vapor in the head space (and no other substance in the container), and the system were at thermodynamic equilibrium, the pressure throughout the container (both in the liquid and the vapor) would be equal to the equilibrium vapor pressure at the system temperature (neglecting any small static gravitational pressure variation within the liquid). This is a clean-cut definition of vapor pressure.

If you had a closed container, with both air and vapor in the head space and the total pressure were 1 atm, then, at equilibrium, the pressure in the liquid would be 1 atm, the partial pressure of the vapor would be equal to the equilibrium vapor pressure at the container temperature, and the partial pressure of the air would be 1 atm minus the equilibrium vapor pressure of the liquid.

If you have an open container in a very large room, the system will typically not be at equilibrium, and the partial pressure of the vapor in the gas phase (air) will be less than the equilibrium vapor pressure. The total pressure will be 1 atm, and the pressure of the liquid in the container will be 1 atm. Vapor will be evaporating from the liquid at the surface and going into the air in the room. And the ventilation system will be circulating the air and replacing it with dehumidified air, so that the vapor does not build up in the gas phase. (So the system does not reach thermodynamic equilibrium). Now you raise the temperature of the liquid in the container, and the rate of liquid evaporation at the surface increases. But, below the liquid surface, the temperature isn't high enough for bubbles to begin to form. Now you raise the temperature of the liquid to the temperature where the equilibrium vapor pressure would be equal to the gas pressure above the liquid (1 atm). At this point you begin forming bubbles in the bulk of the liquid below the surface. There is no air in these bubbles, but only vapor. And the pressure of the vapor inside these bubble is equal to the equilibrium vapor pressure at the "boiling temperature." The reason that these bubbles are able to form below the liquid surface is that the growth of the bubbles is able to expand the total volume below the surface by the liquid pushing back the atmosphere above the surface. So the volume of liquid plus bubbles (below the surface) is larger than what the volume of the pure liquid was before boiling started. Basically, the combination of liquid plus submerged bubbles pushes back the air above. That's why the pressure of the air above determines the boiling point. Until the temperature is high enough for the equilibrium vapor pressure to equal to air pressure above, bubble can't form and expand in the bulk of the liquid below the surface.
 
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  • #11
Thanks Chest, I still don't fully grasp some minor things about the pressure but I will try to read more about it and reread your notes.

Thanks again.
 
  • #12
"It is the same thing with the atmosphere. It pushes on the top of the water and its pressure manifests throughout the water, including the bottom." jbriggs444 said that.

"The atmosphere creates the pressure only at the upper surface of the liquid" - You

Okay maybe you will understand my point of view with this. So imagine we have a closed cylinder and a liquid inside it. The atmospheric pressure inside it is 1 atmosphere. So as you guys said the vapor pressure must equal the atmospheric pressure so the total pressure is 2 right?

Now You guys said
"If you had a closed container, with both air and vapor in the head space and the total pressure were 1 atm, then, at equilibrium, the pressure in the liquid would be 1 atm,"

In this case I have 2 atm above the liquid and the same should apply to the liquid 2 atmosphere.
Now you said bubbles form if the pressure insides equal to its surrounding. The pressure inside it should be 2 atmosphere so it is different from the vapor pressure created on the top of the liquid?

Second things is, In cooking with pressure.The water evaporates at the beginning increasing the vapor above it which results in the high pressure. So at the first the boiling point would be 100 then more water will change to vapor as a result increasing the total pressure. My point is the gasses of the atmosphere have 1 atmosphere pressure then the total changed until the vapor pressure equals the atmosphere (100 c) so as I said above more water will change to vapor as a result increasing the pressure above. So it will just start countering it is own vapor? That is why the boiling point is 120 not 100 ?
 
  • #13
Doing a closed (constant volume) cylinder is too complicated for you to start with. I'm going to try to specify a sequence of problems with increasing complexity for you to analyze. These focusing problems should help you. Besides, if you can't solve the simplest ones, you won't be able to solve the more complicated ones.

Problem 1:
You have a cylinder with a frictionless piston (axis of cylinder is vertical). The cross sectional area of the cylinder (and piston) is 0.01 m^2. The piston weighs 100 N. There is 1 liter of water (no air) in the cylinder at 20 C. The cylinder and piston are enclosed in a vacuum chamber, so that the outside pressure is 0 bars. The system is in thermodynamic equilibrium. What is the pressure exerted by the piston on the water? What is the pressure throughout the water within the cylinder? What is the equilibrium vapor pressure of water at 20 C? Is the water in the chamber (a) 100% liquid, (b) 100% vapor or (c) part liquid and part vapor (head space)?
 
  • #14
Biker,

According to what you have told me privately, you have already covered "Newtons laws and its applications" in your course. From what you have learned about Newtons laws and its applications, if you draw a free body diagram of the piston, what are the two forces acting on the piston? What is the force exerted by the water on the piston if the piston is in equilibrium? What is the force per unit area (i.e., the pressure) exerted by the water on the piston? From Newton's 3rd law, what is the force per unit area (pressure) exerted by the piston on the water?

Chet
 
  • #15
Chestermiller said:
Doing a closed (constant volume) cylinder is too complicated for you to start with. I'm going to try to specify a sequence of problems with increasing complexity for you to analyze. These focusing problems should help you. Besides, if you can't solve the simplest ones, you won't be able to solve the more complicated ones.

Problem 1:
You have a cylinder with a frictionless piston (axis of cylinder is vertical). The cross sectional area of the cylinder (and piston) is 0.01 m^2. The piston weighs 100 N. There is 1 liter of water (no air) in the cylinder at 20 C. The cylinder and piston are enclosed in a vacuum chamber, so that the outside pressure is 0 bars. The system is in thermodynamic equilibrium. What is the pressure exerted by the piston on the water? What is the pressure throughout the water within the cylinder? What is the equilibrium vapor pressure of water at 20 C? Is the water in the chamber (a) 100% liquid, (b) 100% vapor or (c) part liquid and part vapor (head space)?

Just giving it a shot, From what I understood from your recent comments.
So pressure is = F/A Where A is the area
So the pressure exerted by the piston on the water is 100/0.1 gives me 100 N/m^2
as you mentioned before the pressure manifests through the water, Rationally thinking because if I push the molecules at the top they will push the others down with the same pressure ( Just an assumption ). So it is 100 N/m^2 before any water evaporates
Now for the equilibrium vapor pressure, I found it through a table that it should be equal 2.3388 kPa So it is 2338.8 Pa
And Pa is a unit that represents N/m^2
Hmm so I believe that pressure is exerted in all directions so the vapor pressure should counter the piston and it will. So it will push it upwards ( I don't know for how much but the vapor molecules should increase the pressure exerted down by them and the piston so it will eventually balance out?)

So there will probably be vapor at the head space and liquid at the bottom not all the liquid should evaporate. I hope that is right...
 
  • #16
Biker said:
Just giving it a shot, From what I understood from your recent comments.
So pressure is = F/A Where A is the area
So the pressure exerted by the piston on the water is 100/0.1 gives me 100 N/m^2
as you mentioned before the pressure manifests through the water, Rationally thinking because if I push the molecules at the top they will push the others down with the same pressure ( Just an assumption ). So it is 100 N/m^2 before any water evaporates
Nicely done conceptually. But, to work this problem correctly, you got to get the math right. I said that the area of the piston is 0.01 m^2, not 0.1 m^2. And, in addition, 100/0.01 gives 10000 N/m^2.

Now, how does this affect your conclusions?

Chet
 
  • #17
Chestermiller said:
Nicely done conceptually. But, to work this problem correctly, you got to get the math right. I said that the area of the piston is 0.01 m^2, not 0.1 m^2. And, in addition, 100/0.01 gives 10000 N/m^2.

Now, how does this affect your conclusions?

Chet
Oh hhhhhh, Sorry didnt notice it was 0.01. When I thought about it I thought it is 0.1 Anyway.

The answer that will change is that it will be only 100% liquid
 
  • #18
Biker said:
Oh hhhhhh, Sorry didnt notice it was 0.01. When I thought about it I thought it is 0.1 Anyway.

The answer that will change is that it will be only 100% liquid
OK. How much does the 1 liter of water weigh? So what is the combined weight of the water plus the piston? If you draw a free body diagram on the combination of water and piston, showing the forces acting on the combination, what is the upward force that the base of the cylinder has to exert on the water immediately above to satisfy the overall force balance? What is the pressure of the water at the bottom of the cylinder?
 
  • #19
Yep, and I missed that part too... Sorry
1 L is equal to 1 kg
So about 10 N on the bottom. Both of them (Piston and water) are acting down. So a net force of 110 N is pushing the the bottom of the cylinder. The pressure at the bottom of the cylinder is 110/0.01 = 11000 N/m^2 and there is an opposite force that the bottom of the cylinder exerts to balance out forces.
 
  • #20
Biker said:
Yep, and I missed that part too... Sorry
1 L is equal to 1 kg
So about 10 N on the bottom. Both of them (Piston and water) are acting down. So a net force of 110 N is pushing the the bottom of the cylinder. The pressure at the bottom of the cylinder is 110/0.01 = 11000 N/m^2 and there is an opposite force that the bottom of the cylinder exerts to balance out forces.
Nicely done. OK. Suppose we heat the contents of the cylinder to a given temperature and allow the contents to reach thermodynamic equilibrium. To what temperature would we have to heat the contents in order for the piston to begin to "levitate." What would be happening at this point?

Chet
 
  • #21
Chestermiller said:
Nicely done. OK. Suppose we heat the contents of the cylinder to a given temperature and allow the contents to reach thermodynamic equilibrium. To what temperature would we have to heat the contents in order for the piston to begin to "levitate." What would be happening at this point?

Chet
Okay, I believe that it should be the temperature where the vapor pressure equals to the surrounding pressure. So from the table it should be between 50-45
But that made me think a little bit about few things:
1)Would the force exerted by the bottom of the cylinder lowers overtime? as the vapor pressure counters now the pressure resulted of both liquid and the piston..
2) Does the molecules at the surface feel a bit less pressure as they are not as deep as at the bottom?
 
  • #22
Biker said:
Okay, I believe that it should be the temperature where the vapor pressure equals to the surrounding pressure. So from the table it should be between 50-45

This is correct. But I need you to interpolate in the table more accurately. Do you use the 10000 N/M^2, or do you use the 11000 N/M^2 to get the initial levitation temperature?
But that made me think a little bit about few things:
1)Would the force exerted by the bottom of the cylinder lowers overtime? as the vapor pressure counters now the pressure resulted of both liquid and the piston..
Does the weight of the water change when part of it becomes vapor and part of it remains liquid, or does its total weight remain the same? How would the answer to this question figure in answering your question?
2) Does the molecules at the surface feel a bit less pressure as they are not as deep as at the bottom?
You already answered this question. The pressure at the bottom never changes. At the initial levitation point, what is the pressure at the top?

Chet
 
  • #23
Chestermiller said:
This is correct. But I need you to interpolate in the table more accurately. Do you use the 10000 N/M^2, or do you use the 11000 N/M^2 to get the initial levitation temperature?

Does the weight of the water change when part of it becomes vapor and part of it remains liquid, or does its total weight remain the same? How would the answer to this question figure in answering your question?

You already answered this question. The pressure at the bottom never changes. At the initial levitation point, what is the pressure at the top?

Chet
First question: 10000 N/m^2 about 46-47 c.

Second question: Mass doesn't vanish so still the weight of the liquid molecules in the gas phase still acts. Total mass and weight is constant.

You have really cleared out a lot of things up to now
Third question: Hmm Vapor pressure will exert pressure in all directions (That makes the piston go up) and also makes the pressure on the top of the liquid increase.
 
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  • #24
Biker said:
First question: 10000 N/m^2 about 46-47 c.
Correct.
Second question: Mass doesn't vanish so still the weight of the liquid molecules in the gas phase still acts. Total mass and weight is constant.
Yes. So pressure on the bottom is always the same.
You have really cleared out a lot of things up to now
Third question: Hmm Vapor pressure will exert pressure in all directions (That makes the piston go up) and also makes the pressure on the top of the liquid increase.
Really? Is the piston in force equilibrium or isn't it? If the forces on the piston are balanced, how can the force on the bottom of the piston increase? How can it be anything but 10000 N/m^2?

Next questions:
If we raise the temperature enough for half the liquid water to evaporate, what will be the liquid pressure at the base of the cylinder? What will be the pressure on the upper surface of the liquid? What will be the gas pressure at the piston face? What will be the temperature of the contents of the cylinder when this happens?
 
  • #25
Chestermiller said:
Correct.
Really? Is the piston in force equilibrium or isn't it? If the forces on the piston are balanced, how can the force on the bottom of the piston increase? How can it be anything but 10000 N/m^2?

Next questions:
If we raise the temperature enough for half the liquid water to evaporate, what will be the liquid pressure at the base of the cylinder? What will be the pressure on the upper surface of the liquid? What will be the gas pressure at the piston face? What will be the temperature of the contents of the cylinder when this happens?
Hmm, So vapor pressure is acting upward? 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?)

Solutions to the question:
1) 11000 N/m^2
2) Piston pressure is 100/0.01 which is 10000 N/M^2 + 0.5*10/0.01 = 10500 N/M^2 Or Pa
3) I need some answers for the questions above to continue
 
  • #26
Biker said:
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
Chestermiller said:
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
Biker said:
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
 
  • #29
Biker said:
First question: 10000 N/m^2 about 46-47 c.

I get 45.7 C for the initial levitation temperature at 10000 Pa.
 
  • #30
Chestermiller said:
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 don't 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
Biker said:
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 don't 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
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
Chestermiller said:
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
Biker said:
ً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
Chestermiller said:
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.
 

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