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Thermo Problem

  1. Aug 28, 2008 #1
    1. The problem statement, all variables and given/known data

    Warm air is contained in a piston-cylinder assembly oriented horizontally. The air cools slowly from an initial volume of 0.003m^3 to a final volume of 0.002m^3. During the process, the spring exerts a force that varies linearly from an initial value of 900N to a final value of zero. The atmospheric pressure is 100kPa, and the area of the piston face is 0.018m^2. Friction is neglected between piston and cylinder. For the air, determine the initial and final pressures and work.

    2. Relevant equations

    Pressure x Area = Force

    Work = [tex]\int[/tex] p dV

    3. The attempt at a solution

    Initial Pressure of Air x Area = Atmospheric Pressure x Area + Spring Force
    Initial Pressure of Air x 0.018m^2 = (100x10^3Pa x 0.018m^2) + 900N
    Initial Pressure of Air = 150kPa

    Final Pressure of Air x Area = Atmospheric Pressure x Area
    Final Pressure of Air = Atmospheric Pressure
    Final Pressure of Air = 100kPa

    Work = [tex]\int[/tex] p dV

    Volume is changing from 0.003m^3 to 0.002m^3, which can be used as an interval of this integration. I am not sure about how I should use pressure to calculate the work.

    Also, I am not really sure if I got those pressures correctly.
    Thank you.
  2. jcsd
  3. Aug 28, 2008 #2


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    Physics tip - don't do the maths until you understand the problem.
    If you can solve the problem with simple maths do so.

    Pressure - at the start the pressure is atmopshere+spring, at the end it is jsut atmosphere
    Work - work is force*distance. Since the force is varying linearly you don't need to integrate just use the average/mid point force
  4. Aug 28, 2008 #3
    Did I do it wrong? I thought those calculations for pressure were quite simple...?

    Well for the work part, here's what I did.

    Since the area for both initial and final are constant, find the distance travelled by dividing volume by area. (Initial Volume/Area and Final Volume/Area)
    Once I get those, subtrace one to another to final the distance travelled.
    Then like you mentioned earlier, average force (450N) times the distance that I got to find the work done, which is 25J or 0.03kJ)
  5. Aug 28, 2008 #4


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    Sounds correct
  6. Aug 28, 2008 #5
    Thank you very much. Have a great day! :)
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