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Thermodynamics: cylindrical figures contains a gas

  1. Jan 11, 2012 #1
    1. The problem statement, all variables and given/known data

    A cylindrical figures contains a gas and we have a piston that is capable of changing the gases volume.
    Initial state: V0=1l P0=10^5p T=300°K

    We pressurize on the piston that has a mass m=300g and a surface S=20cm^2.
    The new temperature T1=540°K

    2. Relevant equations

    Calculate the change in internal energy.

    3. The attempt at a solution

    Well I know that the change in internal energy is: ΔU=W+Q or ΔU=-P1(V1-V0)

    and that p=F/S. but since my teacher didn't explain the lesson well enough I have no clue on how to answer it. Any help would be very well appreciated.
    Last edited by a moderator: Jan 11, 2012
  2. jcsd
  3. Jan 11, 2012 #2

    Simon Bridge

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  4. Jan 11, 2012 #3
    Re: Thermodynamics

    i believe i would have to use this one right?


    I would get: ΔU=p1(V0-V2)+Q right?
  5. Jan 11, 2012 #4

    Simon Bridge

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    OK, I'll let you off the hook:
    Internal energy for an ideal gas depends only on it's temperature.
    [tex]\Delta U = \frac{3}{2}nR\Delta T[/tex]

    You know the change in temperature - it's given to you.
    You need to find nR.
  6. Jan 11, 2012 #5
    alright i'll give it a try right now.
  7. Jan 11, 2012 #6
    I got n=.0401 by using the equation of ideal gases and i got a change of internal energy as 55J correct?
    Last edited: Jan 11, 2012
  8. Jan 12, 2012 #7
    Is this correct?
  9. Jan 12, 2012 #8

    Simon Bridge

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    I've not done the math so I cannot tell you.
    But it's numbers-in-numbers-out, hard to go wrong.
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