Thermodynamics - A Real Pressure Cooker

  1. Ok I am getting frustrated by these two questions in my physics assignment that seem simple but I just can't get the correct answer for.

    1. If an otherwise empty pressure cooker is filled with air of room temperature and then placed on a hot stove, what would be the magnitude of the net force on the lid when the air inside the cooker had been heated to 120ºC? Assume that the temperature of the air outside the pressure cooker is 20ºC (room temperature) and that the area of the pressure cooker lid is A. Take atmospheric pressure to be P_a.
    Treat the air, both inside and outside the pressure cooker, as an ideal gas obeying pV=NRT.


    So I found an equation in my textbook F=pA
    I am looking for force and I know the area so then I just have to calculate the pressure.
    since v, N, and R constant:
    (Pi/Ti) = (Pf/Tf)
    Pf = (Pi*Tf)/(Ti) or Pf = Pi*(Tf)/(Ti) and if I know the that Ti = 20ºC(293K) and Tf 120ºC(393K)
    then Pf = Pi*(393K/293K) where pi = P_a according to the question then my answer should be:
    F=-(P_a*(393K/293K))*A
    F=-(P_a*1.34)*A
    However when I enter that as my answer it says that I am off by a multiplication factor.

    2. Five grams of nitrogen gas at an initial pressure of 3.0 atm and at 20ºC undergo an isobaric expansion until the volume has tripled. What is the gas volume after the expansion?

    So I started off by calculating N (number of mols)
    5.0g *(1mol/28.0g) = 0.179mol

    Then I used Pv=NRT to solve for the initial volume
    v=1.43*10^-3cubic meters

    Then P, N, and R constant:
    (vi/Ti) = (vf/Tf)
    vf=(vi*Tf)/Ti
    vf=((1.43*10^-3)*879)/293
    vf=4.29*10^-3cubic meters

    However when I enter that as my answer it says that I need to check my signs.

    Any help to as where I might have gone wrong would be highly appreciative.
     
  2. jcsd
  3. Andrew Mason

    Andrew Mason 6,880
    Science Advisor
    Homework Helper

    You have to take into account the air pressure in the room that is pressing down on the lid. The net force is the F you found less P_aA. So F_net = .34 P_aA

    You don't really need to involve temperature here, although it is correct. You just need to multiply vi by 3. You have done it correctly but there may be a slight rounding error. I get 4.30e-3 using 1 atm = 101325 Pa.

    AM
     
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