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Reversible adiabatic expansion of ideal gas, entropy change?

  1. Jan 31, 2010 #1
    1. The problem statement, all variables and given/known data
    One mole of an ideal gas at 0 celcius is subjected to changes below, calculate the change in entropy of the gas:

    i) Gas is expanded reversibly and isothermally to twice its initial volume. DONE - 5.76 J/K
    ii)A similar expansion to i. is performed reversibly and adiabatically [hint, for an ideeal gas undergoing reversible adiabatic expansion TV^(gamma - 1) = constant]

    2. Relevant equations

    I know delta Q = -nRT Ln (Vf/Vi) for a reversible isothermal expansion

    delta S = delta Q / absolute T

    dE = dW + dQ

    3. The attempt at a solution

    for a isolated system i know the entropy change would be zero, but as it doesnt state this i dont feel that i can just state "assuming system is isolated entropy change is zero"
    so im stuck how to go about this..

  2. jcsd
  3. Jan 31, 2010 #2

    Andrew Mason

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    The entropy change of the gas is not zero.

    Work it out (Hint: you don't actually have to do any integration).

    [tex]\Delta S = \int_1^2 dQ/T + \int_2^3 dQ/T[/tex]

  4. Jan 31, 2010 #3
    i know
    \Delta S = \int_1^2 dQ/T + \int_2^3 dQ/T
    but dQ is zero for an adiabatic process isnt it?

  5. Nov 15, 2010 #4


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    Sorry to bring this up again, but doing a similar question...

    Surely here dQ = 0 since the gas is expanded adiabatically?
  6. Nov 16, 2010 #5


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    heloooo? *bump*
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