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Homework Help: Entropy #2

  1. Oct 19, 2005 #1
    Three moles of an ideal gas (CV,m = 7/2R) initially at 300K are taken through a series of three compression/expansion steps:

    The gas is expanded isothermally and reversibly at 300K. This process involves 10.37 kJ of heat going into the system.
    The gas is then expanded reversibly and adiabatically to a new volume and temperature. This process involves no heat.
    The gas is then compressed in a process that is neither adiabatic nor isothermal until it is back to the original state where it started. This process involves an unknown amount of heat, but we do know that for this process ?Ssurroundings = 40 J/K.
    What is dStotal for process 3?

    I tried solving this by simply saying that the sum of dS for steps 1, 2, and 3 were zero, since the process is cyclic. Therefore:

    dS1 + dS2 = -dS3

    We are told the reversible heat for step 1, so we can find it by:

    dS1 = qrev/T = 10.37 kJ / 300 K = 34.6 J/K

    The second step is adiabatic, so q = 0, and therefore dS2 = 0. This results in:

    dS3 = -dS1 = -34.6 J/K.

    My biggest concern is that I'm not using enough information provided by the question.
  2. jcsd
  3. Oct 19, 2005 #2


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    you seem to be doing fine,
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