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Homework Help: Thermodynamics Question

  1. Feb 14, 2016 #1
    This is a question on my review package, but I still cannot understand the solution and do not know why my solution is wrong.

    1. The problem statement, all variables and given/known data
    The PV diagram is attached.
    Problem: During change AB, 300 J of thermal energy is supplied to the gas. During change BC, 250 J of thermal energy is transferred. The area ABC on the PV diagram represents 120 J of energy.

    Calculate the thermal energy transfer during the stage CA.

    I am using equation Q = U+W

    W>0 when gas expands and DOES work and W<0 when work is DONE ON gas

    2. The attempt at a solution

    I did this question in two ways, arriving at the same answer, but neither is correct.

    Way 1:
    Since CA is an isochoric change, W=0 and so Q=change in U
    To find change in U, first consider change AB:
    Q= deltaUAB + W, Q=300J, W=120J (from graph)
    deltaUAB = 180J

    Then consider change BC:
    Q= deltaUBC + W, Q=250J, W= -(120+120) = -240 J (area under curve)
    deltaUBC = 490J

    deltaUCA = deltaUAB + deltaUBC = 670 J = Q

    Way 2:
    Total energy put in = 300J + 250 J = 550J
    120 J is done ON the gas, Wnet = - 120J
    Q= deltaU + W
    deltaU = Q - W = 550 - (-120) = 670 J

    However, the answer is 550 - 120 = 430J without much explanation. I am confused...

    Attached Files:

  2. jcsd
  3. Feb 14, 2016 #2


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    Welcome to PF!

    I agree with your answer except for the sign. Shouldn't the answer be QCA = -670 J?

    The graph doesn't seem to agree with the statement that the area enclosed by the cycle represents 120 J. Each large block of the grid represents 20 J. The number of blocks enclosed in the cycle is about 12, which corresponds to an energy of about 240 J.
  4. Feb 14, 2016 #3


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    I want to make sure I'm following your second way.
    This is the total heat added for just the two steps AB and BC, not the total heat for the cycle. Right?
    Wnet here is the work done BY the gas for steps AB and BC together and it is also the work done by the gas for the complete cycle. Is that right?
    Here's where I'm a little uncertain. I think you're calculating the total change in internal energy for steps AB and BC together. If so, you still need to state how to use this to get QCA.
    Last edited: Feb 14, 2016
  5. Feb 14, 2016 #4
    Change CA is indeed a loss of thermal energy. If we regard loss as negative, then yes, the answer should be negative.

    For the second way maybe I didn't explain as thoroughly.

    I think that it should also be the total heat input because no heat is added during change CA.

    I think that 120 J is the work done ON the gas because the graph shows that during BC the volume is decreases so the gas is compressed. This is also why I used -120J instead of postive.

    I should have explained this more. As shown in the first way, CA is isochoric, and W=0. Therefore Q=deltaUCA

    Now deltaU should be numerically the same as Q lost during change CA.

    I hope this clears up the confusion.

    But the overall question is: I am pretty confident that my derivation is reasonable but it does not match the answer key, which should be pretty reliable because it is from past phyiscs exam. I attached a photo of the actual answer.

    Attached Files:

  6. Feb 15, 2016 #5


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    But you are trying to find the head added during change CA
    OK, good.
    QCA=deltaUCA, where deltaUCA = -(deltaUAB+deltaUBC).
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