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Calculating q,w,H and U [Practical Setting]

  1. Oct 30, 2011 #1
    1. The problem statement, all variables and given/known data
    Background info: Evaporation of NH3(l) at its boiling point of -33.4degrees celcius and at a pressure of 100kPa requires heat absorption of 23.3x10^3 J mol-1.

    Question: NH3(g) is condensed at its boiling point at a pressure that is infinitesimally greater than 100kPa. Calculate..
    a) w per mole (J mol-1)
    b) q per mole (J mol-1)
    c) change in molar internal energy U (J mol -1)
    d) change in molar enthalpy H (J mol-1)

    Thanks for your help.

    2. Relevant equations

    ΔU = q + w
    ΔU = q - pextΔV
    ΔH = ΔU +p ΔV
    ΔH = ΔU + ΔngasRT
    PV=nRT
    Um(T)=3/2RT

    3. The attempt at a solution
    P is also constant? because it's a condensation process?

    (a) irreversible work?
    (b) I would know when I figure out what w is.. (reverse)
    (c) zero because T stayed the same
    (d) isn't this already given in the question? (23.3 x 103)?
     
  2. jcsd
  3. Oct 30, 2011 #2

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    Hi chocolatepie! :smile:

    You won't be able to use these formulas, since they only apply to ideal gasses.
    A chemical making a phase transition is definitely not an ideal gas.


    Yes, P is constant.


    No. A phase transition is reversible.

    Actually I'm not sure what's is intended here.
    We do have w=PΔV, where ΔV would be minus the molar volume at -33.4 °C, since in liquid form the volume is negligible and in gas form, the volume of 1 mol is the molar volume.


    For an isobaric process ΔH=q.
    (You can also deduce this from your relevant equations. :wink:)


    No. That's only for an ideal gas.

    You do have ΔU = ΔH - PΔV (for an isobaric process).
    But for a phase transition the difference between ΔU and ΔH should be negligible.


    Yes, I believe so, although you will need a minus sign.
     
    Last edited: Oct 30, 2011
  4. Nov 1, 2011 #3

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    I asked someone else and apparently you can use the ideal gas laws, but only if NH3 is in gaseous form.
    From the ideal gas laws you can calculate ΔV (which is the molar volume in this case), with PV=RT (for 1 mol).


    Btw, your equation Um(T)=3/2RT is not right for NH3.
    It should be Um(T)=(6/2)RT according to the equipartition theorem.
    But you don't need that for this problem.
     
  5. Nov 1, 2011 #4
    Hi there again! :)

    I was just wondering why H has a negative sign? It is weird that the answer is right in the question.

    I thought I have to use a formula or something.
     
  6. Nov 1, 2011 #5
    So here is my attempt:

    (a) W= PΔV (ΔV derived from ideal gas law). Use 100kPa for P.
    (b) change the sign of the answer from (a) and the value is exactly the same
    (c) 0
    (d) -23.3 x 10^3 J mol-1 (I was curious why this is the answer though.)

    A BIG thanks to you!
     
  7. Nov 1, 2011 #6

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    Well, these questions seem to be intended to make you aware of what the physical quantities mean.
    And anyway, some of the formulas are trivial too. The challenge is to find the proper formula.
    As such some questions will be trivial, as long as you know what you're doing.

    It takes energy to evaporate liquid NH3, but the problem asks for condensation, so energy is released, meaning q is negative.


    Yes (but note that the resulting W must be negative).
    So what is the answer?


    No. ΔH=q.


    No. ΔU=q+w


    Yep! :smile:
     
  8. Nov 1, 2011 #7


    So I got:
    (a) w= -1993 J/mol
    (b) I thought ΔU should be zero because it's isothermal and the reaction is reversible?
    Thus q= 1993J/mol?
     
  9. Nov 1, 2011 #8

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    Yep! :wink:


    Errr... no.

    ΔU is not zero.
    In an isothermal reaction of an ideal gas, ΔU would be zero.
    It does not depend on being reversible or not.
    A phase transition is not ideal gas behaviour.
    To behave as an ideal gas, it needs to be gaseous with a low density.

    Note that in my first response I marked 3 of your formulas as being specific for ideal gases.


    Furthermore, ΔH=q and you already "calculated" ΔH.
     
  10. Nov 1, 2011 #9
    Okay, that makes sense.

    Now I have the following:
    (a) -1993J/mol
    (b) -23.3 x 10^3 J/mol
    (c) -25293J/mol
    (d) -23.3 x 10^3 J/mol

    yes? :)
     
  11. Nov 1, 2011 #10

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    Yep! :cool:
     
  12. Nov 1, 2011 #11
    Awesome!! thank you so much! Now I only need to deal with the other question.. :)
     
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