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Pressure inside sealed container

  1. Oct 13, 2014 #1
    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!
     
  2. jcsd
  3. Oct 13, 2014 #2
    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...
     
  4. Oct 13, 2014 #3

    russ_watters

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    Staff: Mentor

    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.
     
  5. Oct 13, 2014 #4

    russ_watters

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    Staff: Mentor

    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.
     
  6. Oct 13, 2014 #5
    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
     
  7. Oct 14, 2014 #6
    : )
    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.
     
  8. Oct 14, 2014 #7
    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 : )
     
  9. Oct 14, 2014 #8
    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...
     
  10. Oct 14, 2014 #9
    Have you decided to disregard what I provided in post #5?

    Chet
     
  11. Oct 14, 2014 #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.
     
  12. Oct 14, 2014 #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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