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Pressure Cooker-Involving Ideal Gas law and Total Pressure

  1. May 10, 2010 #1
    1. The problem statement, all variables and given/known data
    A cook puts 9.00g of water in a 2.00L pressure cooker and warms it to 500 degree celcius. What is the pressure inside the container.

    2. Relevant equations
    [tex]P_{total} = P_{1} + P_{2} +...[/tex] (eqn1)

    [tex]PV = nRT[/tex] (eqn2)

    [tex]\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{1}}[/tex] (eqn3)

    3. The attempt at a solution
    I've assumed that the initial pressure and temperature are standard conditions (101 kPa & 298k)

    Using the Total Pressure law, I know that [tex]P_{i Total} = P_{atm/air} \; and \;P_{f Total} = P_{air,\, heated\, to\, 773K} + P_{Water} [/tex]

    [tex]Total Volume = 2.00L = 0.00200m^3[/tex]

    [tex]T_{initial} = 298K \; and \; T_{final} = 773K[/tex]

    using eqn3 i found [tex]P_{air} = 261990Pa[/tex] by substituting relative values. (using P1 = 101x1^3 Pa)

    at the beginning the pressure of water can be neglected, since it's a liquid.
    However at 773K, i use PV=nRT to calculate the partial pressure of water.

    n=9.00g/18gmol^-1 = 0.5mol T = 773K V = 0.002m^3 and R = 8.314 J mol^-1 K^-1
    Partial Pressure of Water = 1606680 Pa.

    Therefore total pressure = sum of the two partial pressure = 1868670 Pa or 1869 kPa.

    I'm just wondering whether someone could check this over for me and give me any comments, since i'm a bit nervous about the huge answer :S

  2. jcsd
  3. May 11, 2010 #2
    Seems ok to me. Unless I made the same mistake you did :)
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