# A five-liter pot is filled with 2 l of liquid water and 3 l of dry air...

## Homework Statement

1. A five-liter pot is filled with 2 l of liquid water and 3 l of dry air at the ambient pressure of 1 atm and temperature 10◦C. Subsequently, the pot is closed and heated up. What is the pressure inside the pot going to be when the temperature reaches 100◦C?

Pv=RT
Dalton's law?

## The Attempt at a Solution

So I tried to ignore the air and find the final state of water after the heating process. What I tried to do was to find out how much of the water has gone from liquid to vapor with M=PV/RT where P is the saturated pressure at T=100 degrees celsius. What i found is around 2.9 gram of vapor after the temperature reaches 100 degrees celsius. But what do I do next to get find out how the pressure has changed?

.Scott
Homework Helper
What is the partial pressure of water at 100C?
What will be the pressure of the air at 100C if it was 1 atmosphere at 10C?

Based on your calculation, you will loose 2.9cc of volume to the liquid. So the gas will have 3.029 liters instead of just 3.000 liters. But that will be a minor factor.

I might sound stupid here but will the partial pressure of water at 100C be the saturated pressure? And according to my calculations, the mixture between air+ vapor is 3.029 liters I think. Pressure of air at 100 C = RT/V where V is still 5 L and T=100 C right?

What is the partial pressure of water at 100C?
What will be the pressure of the air at 100C if it was 1 atmosphere at 10C?

Based on your calculation, you will loose 2.9cc of volume to the liquid. So the gas will have 3.029 liters instead of just 3.000 liters. But that will be a minor factor.

Oops, forgot to quote you hehe

.Scott
Homework Helper
I might sound stupid here but will the partial pressure of water at 100C be the saturated pressure? And according to my calculations, the mixture between air+ vapor is 3.029 liters I think. Pressure of air at 100 C = RT/V where V is still 5 L and T=100 C right?
Solve for air pressure independently of the water vapor pressure. Then add them.

#### Attachments

SammyS
Staff Emeritus
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Gold Member
I might sound stupid here but will the partial pressure of water at 100C be the saturated pressure? And according to my calculations, the mixture between air+ vapor is 3.029 liters I think. Pressure of air at 100 C = RT/V where V is still 5 L and T=100 C right?
5 L is not the volume to use in applying P100°C = RT/V .
Why ?

5 L is not the volume to use in applying P100°C = RT/V .
Why ?

I dont know man. But when I tried to find the pressure of air at 100 C I just got P1/T1=P2/T2... I am pretty much hard confused here View attachment 230817

Solve for air pressure independently of the water vapor pressure. Then add them.

So what I got for air pressure is the relation P1/T1=P2/T2. Do I just add this with the saturated pressure of water vapor? And what about the liquid that is left? Is it independent of how much water liquid that has vaporized? I am so confused

.Scott
Homework Helper
At any given temperature, there will be a certain partial pressure (vapor pressure) of a liquid. If the actual partial pressure is more than that, liquid will precipitate. If it is less, the will be evaporation. So in this closed system, the amount of liquid in vapor form will be a function of the temperature and the available space (3 liters).

Chestermiller
Mentor
In the final state, you will still basically have 2 l of liquid water and 3l of gas (neglecting the small amount of liquid that vaporized). If you threat the gas in the head space as an ideal gas mixture, each component can be treated separately in terms of its individual contribution to the total pressure. So the partial pressure of the water vapor will be 1 atm, and the partial pressure of the air will be given by the equation you have written (by neglecting the small change in the gas volume of about 3cc's). So what number did you come up with for the partial pressure of the air in the head space?

• SammyS
At any given temperature, there will be a certain partial pressure (vapor pressure) of a liquid. If the actual partial pressure is more than that, liquid will precipitate. If it is less, the will be evaporation. So in this closed system, the amount of liquid in vapor form will be a function of the temperature and the available space (3 liters).

But what is the total pressure then? Pressure at 100 C of air + the saturation/vapor pressure at 100 C? Can’t the amount of liquid in vapor be calculated by ideal gas eq?

In the final state, you will still basically have 2 l of liquid water and 3l of gas (neglecting the small amount of liquid that vaporized). If you threat the gas in the head space as an ideal gas mixture, each component can be treated separately in terms of its individual contribution to the total pressure. So the partial pressure of the water vapor will be 1 atm, and the partial pressure of the air will be given by the equation you have written (by neglecting the small change in the gas volume of about 3cc's). So what number did you come up with for the partial pressure of the air in the head space?

I got around 150 kPa for pressure of air at 100 C. So the total pressure inside the box is 150 + 100 kPa, aka the saturated pressure + pressure of air at 100 C?

Chestermiller
Mentor
But what is the total pressure then? Pressure at 100 C of air + the saturation/vapor pressure at 100 C? Can’t the amount of liquid in vapor be calculated by ideal gas eq?
The volume of air doesn't change, and you've already calculated the amount of water vapor in the gas as 2.9 grams (is this value correct)? The total pressure is ##P=1 +(1)(373)/283## atm. The number of moles of water vapor in the head space is $$n=\frac{pV}{RT}$$where p = 1 atm, V = 2 liters, and T = 373 K. Does this come out to 2.9 gm?

Chestermiller
Mentor
I got around 150 kPa for pressure of air at 100 C. So the total pressure inside the box is 150 + 100 kPa, aka the saturated pressure + pressure of air at 100 C?
The 150 kPa is high. I get about 130 kPa.

The 150 kPa is high. I get about 130 kPa.

Yeah my bad, it was around 133 kPa for pressure of air at 100 C. When it comes to the amount of water vapor, M=PV/RT right? Where P equals saturated pressure and V = 2L then? But the most important question: Is the pressure going to be saturated pressure (1 atm ) at 100 C + 133 kPa which is the pressure of air at 100 C?

Chestermiller
Mentor
Yeah my bad, it was around 133 kPa for pressure of air at 100 C. When it comes to the amount of water vapor, M=PV/RT right? Where P equals saturated pressure and V = 2L then? But the most important question: Is the pressure going to be saturated pressure (1 atm ) at 100 C + 133 kPa which is the pressure of air at 100 C?
Yes. That is the total pressure. It is also the pressure that will be imposed on the liquid (and, of course, it will be the pressure in the liquid).

.Scott
Homework Helper
Yeah my bad, it was around 133 kPa for pressure of air at 100 C. When it comes to the amount of water vapor, M=PV/RT right? Where P equals saturated pressure and V = 2L then? But the most important question: Is the pressure going to be saturated pressure (1 atm ) at 100 C + 133 kPa which is the pressure of air at 100 C?
Yes. They add together to get the total pressure.
The partial pressure of water will be 1 atmosphere. The air will be at (273+100)/(273+10) = 1.32 atmospheres. Total is 2.32 atmospheres (1760 mmHg, 235KPa).

Yes. That is the total pressure. It is also the pressure that will be imposed on the liquid (and, of course, it will be the pressure in the liquid).

Wow finally, thank you so much! Last questions: Do I just clarify my answer that because at 100 degrees celsius by Pv-diagram we see that the state of water at this temperature is saturated liquid/vapor and this only occurs at the saturation pressure, therefore the partial pressure of vapor equals 1 atm. Since the pressure of air is 133kPa at 100 C, according to Dalton's law the total pressure is the sum of those two? Is that good enough as an explanation? And the partial pressure of vapor is always equal to saturated pressure independent of how much vapor there is? Thank you so much in advance

• Chestermiller
Yes. They add together to get the total pressure.
The partial pressure of water will be 1 atmosphere. The air will be at (273+100)/(273+10) = 1.32 atmospheres. Total is 2.32 atmospheres (1760 mmHg, 235KPa).

Thank you so much man. May I ask why the partial pressure of vapor (or water cause it is the same) is independent of how much liquid that has evaporated to vapor (like in this problem there are only a few grams of vapor when heating to 100 C). Cause a few grams of vapor do not occupy that much volume so the total pressure is over 2 atm? Sorry if this question sounds so silly cause I haven't found anything related to this in my professor's lecture notes.

haruspex
Homework Helper
Gold Member
Cause a few grams of vapor do not occupy that much volume
Vapor will occupy whatever volume is available. Did you mean something else?

Chestermiller
Mentor
Since the pressure of air is 133kPa at 100 C, according to Dalton's law the total pressure is the sum of those two? Is that good enough as an explanation?
Yes.
And the partial pressure of vapor is always equal to saturated pressure independent of how much vapor there is? Thank you so much in advance
Yes, if the system has come to equilibrium.

Chestermiller
Mentor
Thank you so much man. May I ask why the partial pressure of vapor (or water cause it is the same) is independent of how much liquid that has evaporated to vapor (like in this problem there are only a few grams of vapor when heating to 100 C). Cause a few grams of vapor do not occupy that much volume so the total pressure is over 2 atm?
Sure, if that's what the ideal gas law predicts for the water vapor. I haven't checked that 2.9 grams of water vapor is the amount at 1 atm partial pressure, 100 C, and 3 liters volume.

.Scott
Homework Helper
May I ask why the partial pressure of vapor (or water cause it is the same) is independent of how much liquid that has evaporated to vapor (like in this problem there are only a few grams of vapor when heating to 100 C). Cause a few grams of vapor do not occupy that much volume so the total pressure is over 2 atm? Sorry if this question sounds so silly cause I haven't found anything related to this in my professor's lecture notes.
It's not that they are independent, it's that the equilibrium reaches a balance based on the vapor pressure. If the vapor pressure is too high, then you will get more condensation than evaporation - so it will come down. If the vapor pressure is too low, there will be more evaporation than condensation - so it will rise. So what you end up with is the vapor pressure for water at that temperature. Once you have that value, you can figure how much water has evaporated given the volume.

The amount of pressure exerted by a few grams of water vapor will depend on the temperature and volume.

haruspex
Homework Helper
Gold Member
It's not that they are independent, it's that the equilibrium reaches a balance based on the vapor pressure. If the vapor pressure is too high, then you will get more condensation than evaporation - so it will come down. If the vapor pressure is too low, there will be more evaporation than condensation - so it will rise. So what you end up with is the vapor pressure for water at that temperature. Once you have that value, you can figure how much water has evaporated given the volume.

The amount of pressure exerted by a few grams of water vapor will depend on the temperature and volume.
Not sure, but I felt @iamnotsmart was concerned with the increased volume above the liquid resulting from the evaporation.

.Scott