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Squeezing a piston and its effect on temperature

  1. Sep 13, 2008 #1
    1. The problem statement, all variables and given/known data
    If a piston is compressed, how is the temperature affected?

    2. Relevant equations
    U=Q + W

    3. The attempt at a solution

    My soln: I thought that U=k.e. = (3/2)nRT=(3/2) pV ?
    Since PV is constant (P1V1=P2V2), shouldnt temperature be constant?

    Answer sheet: My answer sheet says that Q=0 since piston is insulated. and U=W. Since it is compressed, work is done on the gas, increasing its U, which increases T.

    Which answer is correct???
  2. jcsd
  3. Sep 13, 2008 #2


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    Homework Helper

    Hi qazxsw11111,

    (P1V1=P2V2) is not always true; Boyle's law assumes that the temperature and amount of gas is kept constant.

    If the amount of gas is constant, then for an ideal gas you would use:

    \frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}

    which follows directly from the ideal gas law.
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