1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper

    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.

  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.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook