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Pressure in cooker

  1. Dec 6, 2011 #1
    1. A pressure cooker with a volume of 6 L is filled with 27 grams of water and heated to 390.2 K. What is the pressure (atm) in the cooker



    2. 3) PV = nRT



    3. P(6)=27*0.082*390.2
    P=144
     
  2. jcsd
  3. Dec 6, 2011 #2


    At these conditions water is not an ideal gas. It is a saturated mixture of liquid and vapor. You need to look up the saturation pressure that goes with that temperature, and also confirm that the specific volume falls between vf and vg.

    BBB
     
  4. Dec 6, 2011 #3
    I think that the problem is probably asking for a much simpler approach. Since 390.2 K=117.2 degrees Celsius, the problem likely assumes that all the water is evaporated in the cooker. You have the equation PV=nRT, which can be rewritten as P=(nRT)/V. For n, you need to convert 27 g water to moles of water.
     
  5. Dec 6, 2011 #4
    No, that is not correct. Water at these conditions is a mixture of saturated liquid and vapor and the ideal gas law does not apply. You are trying to reason from the boiling point at atmospheric pressure, not understanding that the pressure inside the fixed-volume cooker is considerably higher than atmospheric and has a much higher boiling point (for the given specific volume, somewhere around 170-175 deg C).

    Two thermodynamic parameters are given: the temperature T=390.2 K and the specific volume = 0.006/0.027 m3/kg = 0.222 m3/kg. This is enough to determine the state completely. In particular, it is not hard to compute that the quality is 0.22745, i.e. the mixture is 22.75% liquid and 77.25% vapor.

    So to solve the problem correctly, the OP needs to find the saturation pressure and verify that the quality is between 0% and 100%.

    BBB
     
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