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Pot cooking food and being cooled, how does the pressure change?

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Homework Statement


A person cooks a meal in a 30-cm-diameter pot that is covered with a well-fitting lid and lets the food cool to the room temperature of 20°C. The total mass of the food and the pot is 8 kg. Now the person tries to open the pan by lifting the lid up. Assuming no air has leaked into the pan during cooling, determine if the lid will open or the pan will move up together with the lid.

The Attempt at a Solution



Ignore the values, my problem is with the concepts in the question. I'm going to try and explain how I think it works and then how the book solves it.
I believe that the process involved would be something along the lines of:

Sate 1: Pot at 20 C, water, food and air. Pressure roughly 1 atm.
First the water will go to its boiling temperature (more or less 99.8C) then it will start boiling and water vapor will force all air INSIDE the pot to go out. The pressure will not change.

State 2: Pot at 100 C, water (liq and vapor) and food. Pressure still 1 atm.
Then it will cool down, the vapor inside will condense and become water again, but since the air went out a vacuum will be formed inside, and the pressure will increase.

State 3: Pot at 20 C, water (liq only? liq and vapor? Can I determine that?) and food. Increased pressure.

So, my only problem was I didn't understand how to determine the new pressure inside. I don't know how to quantify the "generated vacuum" so to speak.

The book then takes the saturation pressure of water at 20 C and uses that (2.34 kPa) to calculate the rest the of the problem, it doesn't really explains why it uses this pressure, though.

So, summarizing:
  1. In State 3, is there only liquid water, only vapor water or both?
  2. How do I quantify the "vacuum"?
 

Answers and Replies

  • #2
UltrafastPED
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If you cook with a pot that can exhaust hot air during cooking - which it must if the lid is going to stay on! - then as the pot cools there will be less air inside, and the pressure will be reduced below room pressure as it cools.

This will provide a vacuum seal. It is exactly what happens to the jars in a pressure cooker; and the pressure cooker has a pressure vent on the top to bleed off the excess pressure acquired during the cooking process.
 
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If you cook with a pot that can exhaust hot air during cooking - which it must if the lid is going to stay on! - then as the pot cools there will be less air inside, and the pressure will be reduced below room pressure as it cools.

This will provide a vacuum seal. It is exactly what happens to the jars in a pressure cooker; and the pressure cooker has a pressure vent on the top to bleed off the excess pressure acquired during the cooking process.
Right. Thanks, but I understand it up to this point, a vacuum will be created (since it`s not a pressure cooker and doesn`t have a pressure vent), I just dont understand how to quantify this vacuum created.
 
  • #4
UltrafastPED
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Multiple steps:

1. PV=nRT for the original pot, room temperature ... just the air counts; we will ignore the adjustment for water

2. PV'=nRT' for the heated pot ... if the pressure was unchanged this will give you the "heated" volume.

2'. But the heated air is escaping from the pot ... so throw away whatever would not fit into the pot ... you only have the original volume left. Note that the pressure is always the original room pressure because the hot air is escaping ... equalizing the pressure! You lost a proportion of the gas ... so as n'=n*V/V' where V' is for the expanded gas volume, n'=reduced moles of gas.the loss of volume.

3. P'V=n'RT The volume of the pot is unchanged, n' is the reduced moles of gas, T is back to room temperature.

This final step gives you the "pot pressure"; this will be less than regular room pressure which you started with. It is the pressure difference, (P_room - P') which is your vacuum seal - the force required to pull of the lid is this value times the surface area of the top of the pot.
 
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  • #5
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In steps 1 and 2, you have driven all the air out of the pot, and replaced it in the gas phase with pure water vapor from the cooking fluid. There may also be liquid water in the bottom of the pan that is still boiling. The equilibrium vapor pressure of the liquid water and water vapor in the pot is 1 atm. So the vapor is at 1 atm., and the liquid water is also at 1 atm. The lid on the pot is now sealed, so that no air can get back in. You next drop the temperature of the pot and its contents from 100 C back down to 20 C. At 20 C, the equilibrium vapor pressure of water is only 2.34 kPa. So, when the system has finally equilibrated again at 20 C, the vapor pressure of the pure water vapor in the gas phase have dropped to 2.34 kPa. In order for this to happen, as you might expect, most of the water that was in the vapor phase at 100 C will have condensed out into the liquid phase. Only a small fraction of the water within the pot is now in the vapor phase. So the pressure outside the pot will be 1 atm, and the pressure inside the pot will be 2.34 kPa.
 
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Multiple steps:

1. PV=nRT for the original pot, room temperature ... just the air counts; we will ignore the adjustment for water

2. PV'=nRT' for the heated pot ... if the pressure was unchanged this will give you the "heated" volume.

2'. But the heated air is escaping from the pot ... so throw away whatever would not fit into the pot ... you only have the original volume left. Note that the pressure is always the original room pressure because the hot air is escaping ... equalizing the pressure! You lost a proportion of the gas ... so as n'=n*V/V' where V' is for the expanded gas volume, n'=reduced moles of gas.the loss of volume.

3. P'V=n'RT The volume of the pot is unchanged, n' is the reduced moles of gas, T is back to room temperature.

This final step gives you the "pot pressure"; this will be less than regular room pressure which you started with. It is the pressure difference, (P_room - P') which is your vacuum seal - the force required to pull of the lid is this value times the surface area of the top of the pot.
Thanks, that's a clever way of doing it.

In steps 1 and 2, you have driven all the air out of the pot, and replaced it in the gas phase with pure water vapor from the cooking fluid. There may also be liquid water in the bottom of the pan that is still boiling. The equilibrium vapor pressure of the liquid water and water vapor in the pot is 1 atm. So the vapor is at 1 atm., and the liquid water is also at 1 atm. The lid on the pot is now sealed, so that no air can get back in. You next drop the temperature of the pot and its contents from 100 C back down to 20 C. At 20 C, the equilibrium vapor pressure of water is only 2.34 kPa. So, when the system has finally equilibrated again at 20 C, the vapor pressure of the pure water vapor in the gas phase have dropped to 2.34 kPa. In order for this to happen, as you might expect, most of the water that was in the vapor phase at 100 C will have condensed out into the liquid phase. Only a small fraction of the water within the pot is now in the vapor phase. So the pressure outside the pot will be 1 atm, and the pressure inside the pot will be 2.34 kPa.
Hmm, that's exactly what I need, but I am highly confused, I don't understand why the pressure drops to 2.34 kPa if it also condenses and I don't think I understand equillibrium vapor pressure very well.

1) Let's say the water went to 95C (compressed liquid, not saturated) and I pumped the air out somehow and then dropped the temperature to 20C, would the same be true? The pressure would be 2.34 kPa?

2) Let's say that I simply start at 20C and never increase the temperature, if I simply pump all the air now, the resulting pressure will be 2.34 kPa?
 
  • #7
UltrafastPED
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Chestermiller has proposed a simpler solution, but it requires knowledge of how the water behaves, including its vapor pressure at STP.

But I specifically ignored the water - see if the two methods come close!
 
  • #8
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Hmm, that's exactly what I need, but I am highly confused, I don't understand why the pressure drops to 2.34 kPa if it also condenses and I don't think I understand equillibrium vapor pressure very well.
You need to brush up on equilibrium vapor pressure.
1) Let's say the water went to 95C (compressed liquid, not saturated) and I pumped the air out somehow and then dropped the temperature to 20C, would the same be true? The pressure would be 2.34 kPa?
Yes. The liquid in the pot at 20 C and the vapor in the head space would, in the end, be at equilibrium with one another, and both would be at a pressure of 2.34 kPa. That's what equilibrium vapor pressure is all about.
2) Let's say that I simply start at 20C and never increase the temperature, if I simply pump all the air now, the resulting pressure will be 2.34 kPa?
Yes.
 
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  • #9
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But I specifically ignored the water - see if the two methods come close!
I would be surprised if the values are close to each other. Vapor pressure is roughly exponential in temperature, the ideal gas law is linear.

I think you have to assume that the boiling water pushed out all air inside. Then you just have 100% water vapor - and you can calculate its pressure based on the known temperature.
 

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