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Change in internal thermal energy.

  1. Oct 5, 2009 #1
    Q: 20 J of heat energy is added to an ideal gas while -40J of work is done BY the same gas. The total change in internal thermal energy of the system is what?



    Equations I used: [tex]\Delta[/tex]Eth = Heat + Work
    [tex]\Delta[/tex]Eth = Heat - Ws

    This was a problem on a recent physics test I took and I said the answer was 60J. However, my instructor insists that the correct answer is -20J. He says that -40J of work is energy leaving the gas irregardless of whether the gas is doing the work or the environment is doing the work. I don't see how this is possible. My reasoning is that (-40J) of work is done by the gas = 40J of work done ON the gas. Thus, [tex]\Delta[/tex]Eth= 20 J + 40J.

    Can anyone enlighten me on why this is?

    Thanks a lot!
     
  2. jcsd
  3. Oct 5, 2009 #2

    Mapes

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    I'm on your side. Essentially, the question contains a double negative (a negative amount of exiting energy) which your instructor is interpreting as a negative amount and you are interpreting as a positive amount.

    Or look at it this way: most people would take the opposite statement, "40J of work is done by a gas," to mean that 40J of energy left the system. Your instructor is claiming, then, that replacing "40J" with "-40J" in the above sentence makes no difference, which is illogical.
     
  4. Oct 5, 2009 #3

    Andrew Mason

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    You are right. The convention is to write the first law as:

    dU = dQ - dW; where dQ is the heat flow INTO the gas and dW is the work done by the gas.

    If you are using that convention, a negative work term means that positive work is being done ON the gas so it increases internal energy.

    As Mapes noted, if the question had stated that +40J of work is done by the gas, then the change in internal energy would be dQ - dW = 20 - 40 = -20J. So it cannot be the same result if -40J is done by the gas.

    Your teacher may be confusing Q with work. Work is not something that "flows" into or out of the gas. That is the case with Q (at least the model on which thermodynamics is based) but not W. Work is only done ON or BY the gas.

    AM
     
  5. Oct 5, 2009 #4
    Yeah, these are the same things I told my professor, but he stood firm on his answer. I guess I'll let it slide for now and if my grade is affected at the end of the semester I'll look back into it.

    Thanks
     
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