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Calculating Heat Energy

  1. Oct 30, 2011 #1
    Calculating Heat Energy (HELP)

    1. The problem statement, all variables and given/known data
    A mixture consisting of 1.00g oxygen gas and 1.00g neon gas is trapped within a 1.0L container at an initial temperature of 298K. Calculate the heat energy required to increase the temperature of this mixture to 398K
    (a) at a constant volume of 1.0L
    (b) under conditions where the container is allowed to expand against a constant pressure pext. The pressure pext is equal to the initial pressure of the mixture at 298K.
    * note: base your calculations on the equipartition theorem. (do not tabulate heat capacity values)

    2. Relevant equations

    C=q/ΔT
    Um(T)=3/2RT (*)

    3. The attempt at a solution
    (a) ΔU = qv when volume is constant
    When using equipartition theorem (*), what should I put for temperature?

    (b) ΔH = qp when pressure is constant.
    Should I use formula H = U + pV to calculate ΔH?

    Thanks in advance!
     
    Last edited: Oct 30, 2011
  2. jcsd
  3. Oct 30, 2011 #2

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    Hi chocolatepie! :smile:

    I'm not really sure what is intended by the equipartition theorem.
    Can you enlighten me?

    I think you're supposed to use Q = Cv ΔT for constant volume and Q = Cp ΔT for constant pressure.

    I guess you can use Um(T)=3/2RT for Neon, or rather Cv=3/2R and Cp=5/2R.
    But I suspect you should use a different formula for Oxygen since it is diatomic (Cv=5/2R and Cp=7/2R).
    And anyway, to use this, you would need to convert your masses to moles.
     
  4. Nov 1, 2011 #3
    Ok, so becuase O2 is diatomic, I am going to use Um(T) = 6/2RT = 3RT

    I don't understand why Um(T) became Q? (is Q heat? meaning q?)
     
  5. Nov 1, 2011 #4

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    Errr... no.
    Could you check your notes?

    Um(T)=Cv T.
    Cv depends on the type of molecule.
    It is Cv=f/2 R where f is the number of degrees of freedom (max 3 translational and max 3 rotational).


    Oh sorry, I always try to use the symbols from the OP, but I am used to Q myself instead of q.
    I think that in your case they mean the same thing.
    (Actually, I'm used to using a capital Q when referring to a mol, and a lowercase q when referring to a kg, but it seems you do not make that distinction.)
     
  6. Nov 1, 2011 #5
    My note says that Um(T)=(x/2)RT
    Don't we write C for heat capacity? so does it mean x/2 is a heat capacity?


    So far, I did the following:
    because V= const, ΔU = qv (v meant to remind that it was at constant V)
    ΔUm(T) = 6/2RΔT = 3RΔT = 2494.2 J for oxygen
    ΔUm(T) = 3RΔT = 1247.1J for Ne

    To calculate heat energy required, should I add these ΔUm(T) to use ΔH=ΔU +pΔV (where pΔV is gone because ΔV is 0)?
     
    Last edited: Nov 1, 2011
  7. Nov 1, 2011 #6

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    Yes, we write C for heat capacity (for 1 mol).
    More specifically, we write Cv for the heat capacity at constant volume, and Cp for the heat capacity at constant pressure.
    Furthermore the heat capacity is Cv = (x/2)R.
    (You need to include the R.)

    For an ideal gas Um(T)=Cv ΔT.
    The value of Cv depends on the type of gas.
    We distinguish 3 varieties: monatomic (like Neon), diatomic (like Oxygen), and more complex molecules that extend in 3 dimensions (like NH3).
    Each has a different Cv.
     
  8. Nov 1, 2011 #7
    Could you check my math in my previous post?
     
  9. Nov 1, 2011 #8

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    Last edited by a moderator: Apr 26, 2017
  10. Nov 1, 2011 #9
    Doesn't oxygen have 6 degrees of freedom and Ne has 3?
     
  11. Nov 1, 2011 #10
    To calculate heat energy required, should I add these ΔUm(T) to use ΔH=ΔU +pΔV (where pΔV is gone because ΔV is 0)?
     
  12. Nov 1, 2011 #11

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    No, oxygen has 5 degrees of freedom.

    And yes, Ne has 3 degrees, but you did not substitute this value correctly for x.


    Not quite.

    You have the energy of a mol, but you need the energy of a gram of each.

    Furthermore, the question asks for the heat required at constant volume.
    That is qv.
    You already had a formula for that.
    What was it?
     
  13. Nov 1, 2011 #12
    Could you explain why it is 5?
     
    Last edited: Nov 1, 2011
  14. Nov 1, 2011 #13
    At constant V, I had a formula ΔU = qv

    Does it mean I need to add ΔU for each gas since they are in a mixture?

    Also, I am using ΔU +PΔV =q for the part (b). There is no work value given.. and so is pressure
     
    Last edited: Nov 1, 2011
  15. Nov 1, 2011 #14

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    Each molecule has 3 translational degrees of freedom.
    The rotational degrees of freedom is what this is about.

    If you have a complex molecule like NH3, there are 3 rotational axes to which the molecule will resist movement so to speak.
    This gives a total of 6 degrees of freedom.

    With a diatomic molecule, the atoms are lined up on an axis.
    If you try to rotate it, there is 1 axis to which it will not "resist" movement (the axis along the atoms).
    And there are 2 axes to which it will "resist" movement.
    So a diatomic molecule has 3+2=5 degrees of freedom.

    A monatomic molecule does not "resist" any rotation, so it only has the 3 translational degrees of freedom.


    You can find it for instance here:
    http://en.wikipedia.org/wiki/Ideal_gas#Heat_capacity
    (But not so detailed I'm afraid. o:))
     
    Last edited: Nov 1, 2011
  16. Nov 1, 2011 #15

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    Yes.

    The total internal energy is the sum of the internal energy of each of the components.


    Yes, you can use that.
     
  17. Nov 1, 2011 #16
    Now I have:
    (a) qv = 3325.6J
    (b) qp = 3525.79J

    ?!
     
  18. Nov 1, 2011 #17

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    I get something different.
    How did you calculate it?

    Did you take into account that you have 1 gram of each component and not 1 mole?
     
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