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Gas at a constant pressure: using Cv.dT correctly

  1. Jun 30, 2017 #1
    Hello, my questions is not so much homework, but a request for a definition. When we use Cv*dT to solve for dU (internal energy) in a constant pressure example, what is the order of the temperatures entered into dT?
    Is one to assume its final temp minus initial temp? I ask because it leads to sign convention and I want to make sure I understand it properly.

  2. jcsd
  3. Jun 30, 2017 #2


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    Yes, dT represents final temperature minus initial temperature for an infinitesimal change in temperature.

    Positive dT corresponds to a temperature increase, which also corresponds to an increase in internal energy of the gas. It doesn't matter whether the pressure is constant or not.
  4. Jun 30, 2017 #3
    Thank you, that helped me a lot!

    Could you point me in the right direction for the last part of this problem? I tried to simply sum up the dQ's, dU's and dW's, but noticed that there was overlap where in some cases dU = dW... I then thought out the process and how the gas starts at a certain temperature and pressure, the gas loses heat while expanding and regains said heat while contracting. I thought that they might cancel each other out. The gas returns to its original starting point, temperature and pressure - no net W done.

    Thanks in advance.

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  5. Jul 1, 2017 #4
    In step 1, ##\Delta U## is negative, since the temperature decreased. If you correct this, then you will find that ##\Delta U## for the cycle is equal to zero. You should also then find that W for the cycle is equal to Q for the cycle.
  6. Jul 1, 2017 #5
    I was close!
    I missed the fact that the calculation was final temperature minus initial temperature for dT which yielded the negative value for dU and dW for process one.

    Process 1: Adiabatic

    dQ = 0 No heat added in an Adiabatic expansion, the gas expands using internal energy.

    dU = dW = Cv×dT = 12.471 JK-1 × (389K-588K) --> -2,482 J

    But now that I look at the other parts, my process for isobaric would need to yield a positive value to have the whole system balance out with 993J Q and W
    My understanding of the equation was that final volume minus initial volume in this case will yield a negative value.....
    Where have I gone wrong?

    Process 2: Isobaric

    W = P(V2-V1)--> 3.59x10^4Pa (0.0484m3- 0.0899m3) --> -1,490 J work done on the gas - final volume is less.

    dU = Cv×dT à 12.471 JK-1 × (209.4K-389K) = -2,239 J

    dQ = -2,239 J + -1,490 J = -3,729 J
  7. Jul 1, 2017 #6

    No. $$\Delta U=Q-W$$so, if Q = 0, $$W=-\Delta U=+2482 J$$
    Process 2 calc is correct.
  8. Jul 1, 2017 #7
    Ok, I see the manipulation, simply restating the Q=U+W equation, solve for U instead. What would lead me to do that though? The lecture notes showed us many derivations and the one that came to mind when solving this problem was dU = dW = Cv×dT (for adiabatic process). Is this equation wrong? Forgive my questioning, but I am trying to understand the why. I don't want to simply plod through the mechanics of the math.

    (Looking at it from a strict definition standpoint it makes sense - for an adiabatic process - internal energy changes as much as work done, but since the work done come from the gas itself, the value changes!)
  9. Jul 1, 2017 #8
    There are two sign conventions used in the literature for the first law: $$\Delta U=Q-W$$where W is the work done by the system on the surroundings, and $$\Delta U=Q+W$$whereW is the work done by the surroundings on the system? Which sign convention are you using?
  10. Jul 1, 2017 #9
    Can we use dW = -dU?

    Process 1: Adiabatic

    dQ = 0 No heat added in an Adiabatic expansion, the gas expands using internal energy.

    dW = -Cv×dT = 12.471 JK-1 × (389K-588K) à +2,482 J

    dU = -dW = -2,482 J
  11. Jul 1, 2017 #10
    Only if Q = 0. Otherwise $$\Delta U=Q-W$$and$$W=Q-\Delta U$$

    (Please don't use d's when you mean ##\Delta##'s or, in the case of path-dependent quantities like heat and work, capitalized symbols Q and W).
  12. Jul 1, 2017 #11
    Got it! only for adiabatic. Sorry, I was using what the professor taught us. He did mention not to call it 'd' but delta. Maybe he had a hard time with his software ;-)
    Thank you for helping me understand this better. I was having fits with it!
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