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Thermo AP

  1. Jan 5, 2007 #1
    1. The problem statement, all variables and given/known data
    Ive got a question regarding a recent AP free response, http://www.collegeboard.com/prod_downloads/ap/students/physics/b_physics_b_frq_03.pdf
    Problem 5 is about a simple refrigerator, work is applied and heat is removed keeping temperature constant.

    My question is on part (e). I think heat being removed from the gas would cause entropy to decrease. However, the answer key states that because the gas returns to the same state, entropy is the same. Who is right and if it is constant, why?

    2. Relevant equations
    change in entropy = Q/T

    3. The attempt at a solution
    Less heat, less entropy
  2. jcsd
  3. Jan 5, 2007 #2

    Andrew Mason

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    This is one of the conceptual problems that occurs because we talk about heat and not heat flow. We now think of heat as energy but in thermodynamics we use terminology from the mid 1800s as if it were a fluid flowing through a substance. In thermodynamics, Q is heat flow.

    In one cycle, heat is added to the gas (AB) then work is done on the gas (BC) and then heat is removed from the gas (CA). The internal energy of the gas is the same when it returns to A. The energy that is removed as heat from C to A consists of the heat added from A to B + the work done in compressing from B to C. Qh = Qc + W, where Qh is the heat delivered to the hot reservoir. Since Qh > Qc, there is heat removed (better to say: a greater flow of heat out of the gas than the heat flow into the gas). However this does not mean that there is energy removed. The energy at A is the same.

    Since entropy is a state variable, the entropy of the gas is the same from A back to A. The entropy of the system (consisting of the gas + the hot and cold reservoirs) continually increases, however.

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