Pressure inside sealed container

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Discussion Overview

The discussion revolves around the change in pressure inside a sealed metal cylinder containing a mixture of air and water when heated from 20°C to 140°C. Participants explore the implications of temperature changes on the pressure of both the air and water phases, considering both theoretical and practical aspects of the scenario.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant suggests using the combined gas law to calculate the pressure change of the air but is uncertain about how to account for the heated water.
  • Another participant argues that at 140°C, all the water will have turned into gas, but this claim is contested by others who emphasize that the boiling point of water is affected by pressure.
  • It is proposed that a saturated steam table should be consulted to determine the pressure of saturated steam at the given temperature, which would help in calculating the volumes and masses of water and steam in the container.
  • Concerns are raised about the safety of the scenario, with warnings about the risks associated with pressure vessels and the need for understanding their operation.
  • A participant expresses interest in calculating the pressure change when the water temperature changes from 20°C to 90°C under similar conditions.
  • Another participant shares a formula for calculating pressure increase due to thermal expansion of a trapped liquid and notes the significant pressure changes even with small temperature variations.
  • There is a discussion about the complexity of the problem, with one participant indicating that they are trying to simplify the calculations and seek a more straightforward formula.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether all the water will boil at 140°C, with some arguing that the pressure prevents complete vaporization while others assert that it will. The discussion remains unresolved regarding the exact calculations and implications of the scenario.

Contextual Notes

There are limitations in the discussion regarding assumptions about the behavior of water and air under varying temperatures and pressures, as well as the dependence on definitions of boiling points and vapor pressures.

Who May Find This Useful

This discussion may be useful for individuals interested in thermodynamics, pressure vessel safety, and the behavior of gases and liquids under varying conditions, particularly in engineering and physics contexts.

Woaiyumi
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Hi all : )
I have a basic physics question:
Let's say I have a sealed metal cylinder with a volume of 1L.
It contains 1/3 air and 2/3 water.
I heat the container from 20C to 140C.
How can I determine the change in pressure or final pressure?
I guess I can calculate the pressure change of the air using the combined gas law, but I don't know how the heated water factors in.
Any comments, formulas would be much appreciated!
 
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Interesting, I will look into solving this in that direction. In the real world situation, the water in the container is not changing to gas, so the internal temperature must be lower than 100C.. I was thinking that maybe the pressure was preventing the water from boiling...
 
Fizzios said:
Notice that 140C is greater than water's boiling point, so all the water will have turned into gas.
Sorry, but that isn't true.

Chris, Fizzios; the boiling point of water depends on temperature and when the container contains that much water, it isn't all going to boil at 140C. What you need to do is find a saturated steam table and look up the pressure of saturated steam at that temperature. The steam table will also tell you the density of steam and water under those conditions. From that information you can find the volumes and masses of each by setting up an equation that sums them (and the air's volume) to the volume of the container.
 
Chris Zaic said:
Interesting, I will look into solving this in that direction. In the real world situation, the water in the container is not changing to gas, so the internal temperature must be lower than 100C.. I was thinking that maybe the pressure was preventing the water from boiling...
Real-world situation? Please stop now! What you are doing has a real chance of killing you.

What you describe is a pressure vessel that is highly pressurized and you don't understand how it works well enough to safely use such a thing.
 
Fizzios said:
Notice that 140C is greater than water's boiling point, so all the water will have turned into gas. You know the initial volume of water, so you can calculate how many moles of water there are. This is the same number of moles of water vapor (gas), and you can go on from there.
This is incorrect.

Chris: I want to clarify the initial condition slightly. The initial pressure is 1 atm, and you have 1/3 gas and 2/3 liquid water in the container at 20 C. What is the equilibrium vapor pressure of water at 20 C? What is the partial pressure of the water in the gas phase. If the total pressure is 1 atm, what is the partial pressure of the air in the gas phase? If there is 1/3 L of gas in the container, how many moles of gas are there? What is the mole fraction of water vapor in the gas, and what is the mole fraction of air in the gas? How many grams of water vapor are there in the gas? How many grams of air are there in the container? What is the total mass of water in the container?

Does the mass of water or air in the container change when you go to the new equilibrium state at 140C? What is the equilibrium vapor pressure of water at 140 C? Let x = mass of water vapor in the gas phase at 140C. In terms of x, how many grams of liquid water remain in the container? What volume does this liquid occupy? What is the new volume of gas in the container? In terms of x, how many moles of water vapor are in the container. In terms of x, using the ideal gas law, what is the partial pressure of water vapor in the gas phase? To solve for x, set this partial pressure equal to the equilibrium vapor pressure of water at 140 C. What value do you get for x? How many grams of liquid water remain in the container. Based on the volume of gas in the container and the amount of air, what is the partial pressure of the air in the head space? What is the total pressure of the gas phase? Does this exceed the equilibrium vapor pressure of water at 140 C? Does the water boil away?

Chet
 
russ_watters said:
Real-world situation? Please stop now! What you are doing has a real chance of killing you.

What you describe is a pressure vessel that is highly pressurized and you don't understand how it works well enough to safely use such a thing.
: )
Thank you for the warning. There is a safety valve. It blows when the mixture is anything but 1/3 air to 2/3 water.
All air, and the valve opens, all water, the valve opens.
I am trying to better understand what is happening inside the container, mathematically.
 
russ_watters said:
Sorry, but that isn't true.

Chris, Fizzios; the boiling point of water depends on temperature and when the container contains that much water, it isn't all going to boil at 140C. What you need to do is find a saturated steam table and look up the pressure of saturated steam at that temperature. The steam table will also tell you the density of steam and water under those conditions. From that information you can find the volumes and masses of each by setting up an equation that sums them (and the air's volume) to the volume of the container.

Thank you for this tip, I will work on that.
One more question. Suppose the water temperature is changing from 20C to 90C.
How can I solve for change in pressure, or final pressure, with all other factors the same as above?
(Suppose 1L sealed container, 1/3 air, 2/3 water)

I very much appreciate all your answers! So happy I found this forum : )
 
I found a formula for calculating "Pressure increase due to thermal expansion of a trapped liquid" at this link:
http://www.eng-tips.com/faqs.cfm?fid=1339
Made a quick excel table to play with values.
Could quickly see how even a 1 deg C change in temperature (near room temperature) causes an enormous change in pressure in a sealed container filled with water.

Still working on the 1/3 air, 2/3 water scenario...
 
Have you decided to disregard what I provided in post #5?

Chet
 
  • #10
Hi Chet, no, not at all.
Your answer posed 20 new questions, some that I didn't completely understand, so I did a little more digging to see if I could find a succinct formula to work out this kind of problem. It seems like such a common scenario to solve!
I was not able to find anything more straightforward, so I will have to slowly work through your comment, maybe build a spreadsheet around those parameters. Thank you for such a thorough response.
 
  • #11
It's not really as complicated as it seems. It's mostly a matter of using the vapor pressure vs temperature relationship and the ideal gas law. I can help wherever you get stuck.

Chet
 

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