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Homework Help: Calculating q,w,H and U

  1. Oct 30, 2011 #1
    1. The problem statement, all variables and given/known data
    An ideal gas at 25degree celcius and 100kPa is allowed to expand reversibly and isothermally to 5kPa. Calculate..
    1) w per mol (J/mol)
    2) q per mol (J/mol)
    3) change in molar internal energy U (J/mol)
    4) change in molar enthalpy H (J/mol)

    2. Relevant equations

    ΔU = q + w
    = q - pextΔV
    ΔH = ΔU +p ΔV
    = ΔU + ΔngasRT

    3. The attempt at a solution

    I absolutely have no idea how to do this question. Any direction would be greatly appreciated.
  2. jcsd
  3. Oct 30, 2011 #2
    Welcome to the forum.
    A reversible isothermal process is a polytropic process where PV^n = constant. What is n for this kind of process?
  4. Oct 30, 2011 #3
    I am thinking this question should include a molar volume? (molar volume = V/n, which changes the ideal gas law as P(molar volume) = RT?)

    n is constant for this kind of process? but that sounds weird.. sorry I am lost..

    Also, since it is an isothermal process, would molar U be 0?
  5. Oct 30, 2011 #4

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    Welcome to PF, chocolatepie! :smile:

    An isothermal process is one in which T is constant.
    So molar U will not be zero, but constant.
    In other words: ΔU=0.

    With constant T=To, you get for one mole: PV=RTo.

    To calculate w you need to integrate dw=-PdV.
    Do you know how to do that?
  6. Oct 30, 2011 #5
    I assuming we need to use Wrev= nRT ln (P2/P1) to solve.
    However, we were not given n value...?

    PS. And thanks everyone for the warm welcome to PF :D
  7. Oct 30, 2011 #6

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    Yep. That will work.

    n is the number of moles.
    The question asks to calculate "per mol".
    This means you should calculate using n=1 mol.

    You're welcome! :wink:
  8. Oct 30, 2011 #7
    oh, OK!
    And for part (d) where I need to calculate for the change of molar enthalpy, should I use ΔH = qp to solve? or ΔH = ΔU + pΔV?

    I just don't know when to use each formula..
  9. Oct 30, 2011 #8

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    Actually, neither formula.
    Both have conditions attached.

    ΔH = qp only holds in an isobaric process.
    ΔH = ΔU + pΔV also only holds in an isobaric process (is this really the formula you have?)

    What else do you have for ΔH?
  10. Oct 30, 2011 #9
    The note at the bottom of the question says that the expansion does not occur at constant pressure. Thus, iti s not OK to use ΔH = ΔU + pΔV.
    I am not sure why the note says "constant pressure" when the question itself says "constant temperature" :-O

    I have one that includes heat capacity..? H=CpT (subscript p = const P)
    But I don't have the heat capacity given in the question, so I didn't use it.
  11. Oct 30, 2011 #10

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    In your relevant equations you have:
    ΔH = ΔU + ΔngasRT
    (Works only for ideal gasses.)

    What about that one?

    Btw, Cp=Cv+R for ideal gasses.
    And ΔH = ΔU + Δ(PV) is true in general.
  12. Oct 30, 2011 #11
    Aha, now I got it. Yes there was a formula on my note.
    Thank you so much!!
  13. Oct 30, 2011 #12

    I like Serena

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