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Entropy change when mixing two gases

  1. Oct 25, 2014 #1
    1. The problem statement, all variables and given/known data
    1.00mole of nitrogen (N2) gas and 1.00mole of argon (Ar) gas are in separate, equal-sized, insulated containers at the same temperature. The containers are then connected and the gases (assumed ideal) allowed to mix.
    A) What is the change in entropy of the system?
    B) What is the change in entropy of the environment?
    C)Repeat part A but assume one container is twice as large as the other.

    2. Relevant equations
    dS = dQ/T
    PV=nRT
    dE = dQ - dW

    3. The attempt at a solution
    This problem is stumping me. This is my attempt. since the T is constant, the internal energy wont change

    so dQ = dW
    dQ = PdV

    [itex]Q = \int_{Va}^{Vb} PdV = nRT \int_{Va}^{Vb} \frac{dV}{V} [/itex]

    [itex] Q = nRTln\frac{Vb}{Va} [/itex]

    s = Q/T so

    [itex] s = nRln\frac{Vb}{Va} [/itex]

    the final volume is twice the initial so

    [itex] s = nRln\frac{Vb}{Va} = nRln\frac{2Va}{Va} = nRln(2) [/itex]

    where n is 2 mols because you combine the mols of each gas.
    and this would be the entropy change of the system? for the environment i think it would be 0 because the system is isolated. Am I on the right track?
     
  2. jcsd
  3. Oct 25, 2014 #2
    Yes. Since they are in the ideal gas region, you knew to treat each gas separately. Nice analysis.

    Chet
     
  4. Oct 25, 2014 #3
    thank you!
     
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