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Kinetic theory of gases

  1. May 4, 2006 #1
    Sir,
    Please help me with this problem.
    # A fixed mass of gas at constant pressure occupies a volume V. The gas undergoes a rise in temperature so that the root mean squared velocity(c) of the molecule is doubled. What is the new volume?
    I solved it in the following way:-
    Let V1 be the initial volume & V2 be the final volume. Assuming the pressure to be constant,
    c = (3PV1/M)^1/2 --------- (1)
    2c = (3PV2/M)^1/2 --------- (2)
    Dividing equation (1) by (2) we get,
    ½ = (V1/V2)^1/2
    Squaring on both sides we get,
    ¼ = (V1/V2)
    V2 = 4V1
    But the answer given in my book is V2 = V1/[(2)^1/2] read as V1 divided by root 2.Here the symbol ^ represents power.
     
  2. jcsd
  3. May 4, 2006 #2

    Doc Al

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    Staff: Mentor

    Your answer seems OK to me. Since rms speed is proportional to the square root of the temperature, the temperature (and volume) must increase by a factor of four.

    The book's answer makes no sense, since it implies that the volume decreases.
     
  4. May 4, 2006 #3

    Andrew Mason

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    Science Advisor
    Homework Helper

    Where did you get your equations (1) and (2)?

    How is temperature related to the vrms of the gas? Temperature is a measure of the internal (kinetic) energy of the gas molecules. So how is the kinetic energy of the molecules related to speed?

    If vrms of the gas doubles, what is the increase in temperature (internal kinetic energy)?

    Use PV=nRT with P constant to determine how Volume changes with temperature.

    AM
     
  5. May 5, 2006 #4
    Thank you Sir for your help.
     
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