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Understanding H=U+PV

  1. Aug 12, 2011 #1
    it was stated on a cambridge notes that, H is a state function.
    though H depends on U, which is a state function, however P and V also affect the value of H,
    how can H be a state function?
  2. jcsd
  3. Aug 12, 2011 #2


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    P and V are also state functions, aren't they?
  4. Aug 12, 2011 #3
    yes they are, but then why isn't work done a state function?
  5. Aug 12, 2011 #4

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    I think the introductory paragraph from wikipedia states it pretty well:
  6. Aug 12, 2011 #5
    The product PV is a state function because it doesn't depend on the path.
  7. Aug 12, 2011 #6
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    does it mean that work done is a path function, because for the same value of work done, the might be a lot of combination of pressure and volume values?

    then why are you saying that PV(the work done) is a state function?
  8. Aug 12, 2011 #7

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    It means that you can execute a process doing work, while starting and ending at the same state.

    Consider for instance a combustion engine.
    It makes the following so called Otto cycle:

    During the cycle a number of transitions are made, beginning and ending at the same p-V point, which is the same state.
    However, since the path through which work is done, has an enclosed surface, the work done is not zero, but an amount that depends on the path.
  9. Aug 12, 2011 #8
    alright, then H=U+PV.
    we know that U is state function while PV is a path function. then how can the sum making H a state function?
  10. Aug 12, 2011 #9

    PV is itself a state function, not a path one. No matter how a system gets to a particular Pressure and Volume, the value PV will always be the same; it is a function only of the state variables P and V.

    In fact, the equation PV = constant is a state equation known a Boyle's Law.
  11. Aug 13, 2011 #10

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    Wow! PV is not a function, and actually it should be pV (pressure is usually denoted using the lower case letter p).
    It is simply p times V, pressure times volume, which are both state variables.

    And yes, pV = constant is Boyle's law, but that law is only a special case of the ideal gas law, and assumes temperature is constant.
    In other words, in general pV is not constant.

    Furthermore pV is not the work done.
    Last edited: Aug 13, 2011
  12. Aug 13, 2011 #11
    pV = RT is a state equation known as boyle's law.hence, pV must be a state function and so (U+pV) is a state function on account of being the sum of 2 state functions or variables.
  13. Aug 13, 2011 #12
    refer hess,s theorem on constant heat summation. enthalpy or H can be calculated using this law for any chemical equation depicting that H does not depend on how many intermediate equations were considered to obtain the result.

    2C +H2>C2H2 ; H=a(say)
    C2H4 + H2>C2H4 ; H=B(say)

    2C+2H2>C2H4 ; H=a+b.
    even if it is independent of the prev. 2 steps"
    simply, 2H+2H2>C2H4 ; H=a+b
    so the eqn. as well as the enthalpy,H is independent of the path taken.It is a state function.
  14. Aug 13, 2011 #13

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    Again, you cannot use Boyle's law in general, but since you do not use it in your argument, the conclusion is correct. ;)
  15. Aug 13, 2011 #14
    I suggest you take that up with my Thermodynamics textbook, and wikipedia for that matter. They both state that PV is a state function. The textbook was Carter's “Classical and Statistical Thermodynamics” 2nd edition.

    And as far as p versus P, the textbook I used had used P for pressure. However, I have seen various other textbooks use one or the other.
  16. Aug 13, 2011 #15

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    All right. I concede! :smile:
    I see indeed that P is also used for pressure and that PV is referred to as a state function.
    Note that PV is still simply P times V, which are both state variables.
  17. Aug 13, 2011 #16
    As a mathematician, ILS may have been thinking of composition of functions when discussing PV.

    However there is no reason why the composite function (PV) could not be considered as a (state) function in its own right. For instance the function 'work' is also known as Fd.

    go well
  18. Aug 13, 2011 #17
    Mathematically, you could think a state function as resultant of a exact differential, i.e., the integral form of your function is dependent only on the final and initial states.
  19. Aug 13, 2011 #18


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    Work done is not PV, it is PdV
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