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Confused with the energy equation in thermodynamics

  1. Apr 25, 2016 #1
    • Homework posted in wrong forum, so no template

    I am confused with the energy conservation equation in thermodynamics.

    Is it U2 - U1 = Q - W ? This is what most books say.

    Or is it H2 - H1 = Q - W ?

    Which should I use to solve problems? For example we have a gas with U1 and P1V1 (which makes H1). After a process, we have Q transferred in or out and W transferred in or out. How can I write the equation in this situation?

  2. jcsd
  3. Apr 25, 2016 #2
    How do you understand the meaning of the quantities in the two equations?
  4. Apr 25, 2016 #3
    U is the internal energy, the energy due to the molecular motion of the gas
    H is the internal energy plus the pressure energy PV
    Q is the heat transferred in/out of the gas
    W is the work transferred in/out of the gas

    What are you trying to say? Please be clear
  5. Apr 25, 2016 #4
    That W and PV both have units of energy but are not the same. Work is not a state variable and thus depends on the process that produced it.
  6. Apr 25, 2016 #5
    So which of the two equations are correct? If both, how can that be?
  7. Apr 25, 2016 #6
    U2 - U1 = Heat added to the system - Work done by the system

    H2 - H1 = U2 - U1 + P2V2 - P1V1

    You use the one appropriate to the problem.
  8. Apr 25, 2016 #7
    So the equation H2 - H1 = Q - W is not correct?

    Also, in U2 - U1 = Q - W, you say that Q is heat ONLY added to the system and W only DONE by the system. What about heat substracted by the system and work added to the system?
  9. Apr 25, 2016 #8
    Yes W ≠ PV2 - PV1

    The first law is written as U2 - U1 = Heat added to the system - Work done by the system =Q - W where both Q and W are positive quantities. A negative Q will mean heat removed from the system and a negative W is the work done ON the system.

    You must always watch out if you are adding or removing heat to a system and whether work is done on the system versus having the system do work.
  10. Apr 25, 2016 #9
    So, there is no situation where there is no change in U, there is no heat exchanged and there is work done on or by the system? Ie. in an adiabatic expansion or compression, there is ALWAYS dU≠0?
  11. Apr 25, 2016 #10
    An adiabatic process is one that is thermally insulated so ΔQ = 0. So dU = work done by( du < 0) / on (du>0) the system.
  12. Apr 25, 2016 #11
    So you are saying that there is no possibility for work to be done adiabatically in the system and not to increase U, but rather increase pressure energy PV ?
  13. Apr 25, 2016 #12
    The first equation is always correct. The second equation is correct only for a process carried out at a constant pressure, equal to the original pressure (say by adding heat at constant pressure).

    Do you have a specific problem that we can use as an example.

    Here is a Physics Forums Insights article that should be helpful to you. Although the title refers to the second law of thermodynamics, the first section covers the first law of thermodynamics: https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/
    Last edited: Apr 25, 2016
  14. Apr 25, 2016 #13
    In the case of constant pressure H2 - H1 = Q
  15. Apr 25, 2016 #14
    Oops. Thanks. I didn't notice the W there.
  16. Apr 25, 2016 #15
    This NASA webpage says something different than what you say:
    They say that dH=Q-W (if you ignore kinetic energy). To me this is more complete, because it takes into account the change in pressure energy of the system, which can be caused by either Q or W.
  17. Apr 25, 2016 #16
    I don't have a specific problem, but if I am asked this:
    We know U1, U2, Q, W, P1, V1, V2 and they ask me P2, how do I answer?

    I do NOT want to learn equations that have limited application, ie. when P=constant or V=constant or whatever. I want to know one complete energy conservation equations that applies in EVERY situation and that I can calculate every parameter from it, if I know the rest parameters.
  18. Apr 25, 2016 #17
    This equation applies only to the open system version of the first law, in which mass is flowing into and out of the system. And H refers to the enthalpy change between the inlet and outlet streams of the flow system, not the change in enthalpy within the system. And, did you notice the subscript S on the W. That is not the total work done. It is only the "shaft work". So the NASA document does not apply to a closed system that you have been asking about.
  19. Apr 25, 2016 #18
    Mmmm now it starts to make sense.

    So there IS dH=Q-W, and it is the Steady Flow Energy Equation!

    BTW, when did I say I am asking for closed systems?

    So can you elaborate now, why in SFEE we have dH=Q-W (ignoring kinetic and potential energy) and in closed systems we have dU=Q-W ?
    And what is the Work in the SFEE that is only "part"of the work done?
  20. Apr 25, 2016 #19
    That's covered in my Physics Forums article.

    Regarding the specific question you asked, is this the actual question they asked you? What other information is given?
  21. Apr 25, 2016 #20
    As I told you I don't have a specific question. I just got confused between dU=Q-W and dH=Q-W.
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