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

• iamnotsmart
In summary: But what is the total pressure then? Pressure at 100 C of air + the saturation/vapor pressure at 100 C?The total pressure is 6.8 L at 100 degrees Celsius.
iamnotsmart

## 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?

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?

.Scott said:
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

iamnotsmart said:
I might sound stupid here but will the partial pressure of water at 100C be the saturated pressure?

iamnotsmart said:
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.

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iamnotsmart said:
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 ?

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

I don't 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

.Scott said:
View attachment 230817Solve 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

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).

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
.Scott said:
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?

Chestermiller said:
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?

iamnotsmart said:
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?

iamnotsmart said:
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.

Chestermiller said:
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?

iamnotsmart said:
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).

iamnotsmart said:
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).

Chestermiller said:
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
.Scott said:
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.

iamnotsmart said:
Cause a few grams of vapor do not occupy that much volume
Vapor will occupy whatever volume is available. Did you mean something else?

iamnotsmart said:
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.

iamnotsmart said:
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.

iamnotsmart said:
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.

.Scott said:
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.

I don't think that the author of the problem intended for the student to consider secondary effects - or at least not to attempt to refine his answer in response to them.

But, if secondary effects are to be considered, here are some of them:
1) At 10C, the water would have held some air - most of which would be driven out on heating it to 100C.
2) The volume of the water would decrease because of evaporation.
3) The volume of the water would increase (by about 4%) because of the change in temperature.
4) Neither water nor air act as perfect gases.

## 1. How much liquid water is in the pot?

2 liters. The pot is filled with 2 l of liquid water and 3 l of dry air, so the amount of liquid water is equal to the amount of space left in the pot after the dry air is added.

## 2. How much dry air is in the pot?

3 liters. The pot is filled with 2 l of liquid water and 3 l of dry air, so the amount of dry air is equal to the amount added to the pot.

## 3. What is the total volume of the pot?

5 liters. The pot is a five-liter pot and it is completely filled with a combination of 2 l of liquid water and 3 l of dry air.

## 4. What is the ratio of liquid water to dry air in the pot?

The ratio is 2:3. This means that for every 2 liters of liquid water, there are 3 liters of dry air in the pot.

## 5. How is the liquid water and dry air distributed in the pot?

The liquid water will settle at the bottom of the pot, while the dry air will occupy the space above it. This is due to the difference in density between the two substances. However, over time, some of the dry air may dissolve into the liquid water, making the distribution less distinct.

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