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Isobaric Compression

  1. Dec 12, 2011 #1
    Howdy all, I have a final coming up, and I'm having a very difficult time grasping a few concepts about the ideal gas laws, specifically a problem with isobaric compression.

    1. The problem statement, all variables and given/known data

    Alright, the intro to the problem is:

    A quantity of ideal gas is slowly compressed to 1/3 of its original volume. In this compression, the work done on the gas has a magnitude of 800J. For the gas, Cp= 7R/2

    It then breaks up into several problems based on a situation
    a.) If the problem is isothermal, what is the heat flow (Q) for the gas? Does the heat flow into or out of the gas?

    b.) If the problem is isobaric, what is the change in internal energy of the gas? Does the internal energy increase or decrease?


    2. Relevant equations
    P1V1=P2V2
    T1V1=T2V2

    W=p(dV) (for an isobaric problem only?)

    Cp= Cv + R

    dU= Q - W

    dU(isothermic) = 0

    W = nRTln(V2/V1)

    pV= nRT


    3. The attempt at a solution

    a.) Ok, because isothermic reactions have a dU of 0, this one was pretty simple

    dU = Q - W
    (0) = (Q) - (-800J)

    Q = -800J, I know that heat flows out of the gas, but I don't know why?
    -----------------------------------------------------------
    b.) This one is the one I'm so confused on, and I've gone through every equation and it feels like I need 1 more variable.

    Ok, so for isobaric, P remains constant, so dP = 0

    Here's what I've worked through.

    Q = dU + W
    -----------------
    Q = n(Cp)(dT)
    Q = n(7R/2)(dT)
    Q = n(29.1)(dT)
    ------------------

    T1V1 = T2V2
    T1V1 = T2(1/3 V1)
    ((T1V1)/T2) = 1/3 V1 or ((T1V1)/(1/3V1) = T2
    -------------------------
    p(dV) = W = 800J
    p(V2-V1) = W = 800J
    p(1/3V1 - V1) = W = 800J
    -------------------------------

    From the answer I have, I know that dU is supposed to be -2000 J.....problem is, I don't know how to get it. It seems like I need just 1 more variable to figure out the entire thing!!

    Any advice?

    Thanks,

    Garrett
     
  2. jcsd
  3. Dec 13, 2011 #2

    ehild

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    In the equation dU=Q-W, means the heat added to the gas, and W is the work done by the gas. If the added heat is negative does it mean heat flowing in or out?

    Check the last equation, it is not correct.

    You can find useful the relation between internal energy and temperature,

    U=nCvT.

    ehild
     
  4. Dec 13, 2011 #3
    Ok, I suppose the biggest thing I'm wondering (most of those calculations were desperate attempts to find something =P) is about the "n". Because the number of moles of gas stays the same, can we assume that there is one mole of gas? Or maybe not that there is one mole, but that n remains one because there is no change? It just seems like I need one more variable to complete everything =/
     
  5. Dec 13, 2011 #4

    ehild

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    You do not need the number of moles. The change of internal energy ΔU=nCvΔT is related to the work done. The number of moles do not change. Solve the problem symbolically. If you are incapable to calculate with symbols, use 1 mole of gas.

    ehild
     
    Last edited: Dec 13, 2011
  6. Dec 13, 2011 #5
    Alright, maybe I'm slow, maybe it's late so I'm not thinking real well, but I am still just as stuck. Trying to solve it symbolically sent me down a long rabbit run of equations.....I'll try and clean them up for you, but there's a lot to fix, I'm sure.

    Ok,

    dU = Q - W = Q - (-800)
    also,
    dU = nCv(dT)

    So,

    nCv(dT) = Q - W = Q - (-800)

    Q = nC(dT)
    W = nRTln(V2/V1)

    Therefore,

    nCv(dT) = nC(dT) - nRTln(V2/V1)

    Since we know that W = -800 J, we know that nRTln(V2-V1) = -800 J

    In the beginning we were told that V2 = 1/3V1,
    So: n(8.314)Tln(1/3) = -800
    nT = (-800)/(8.314*-1.1)
    nT = 87.47

    Moving back to this equation:
    nCv(dT) = nC(dT) - nRTln(V2/V1)

    We now know that the Work portion, nRTln(V2/V1) is equal to (87.47)nT

    nCv(dT) = nC(dT) - (87.47)nT; Divide out an N

    Cv(dT) = C(dT) - (87.47)T; Subtract and move over, then divide out the (dT)

    (dT)(Cv-C) = (87.47)T

    About here is where I completely run out of ideas.....This was a stretch, so I wouldn't be surprised if it's completely wrong =/

    From there, the only thing I can see doing is replacing Cv with (5R/2), which is 20.785, so you'd have

    (dT)(20.785 - C) = (87.47)(T) = 800J

    I feel like this is all wrong, to be honest, but I just can't figure out where to go otherwise. It just seems like I'm beating around the problem in a bunch of big circles!
     
  7. Dec 13, 2011 #6

    ehild

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    Combine these equations above in bold, you get


    nCv(dT)=nCp(dT)+800


    You have written in your first post already that

    Cp= Cv + R.

    Cp was given, Cp=(7/2)R

    Can you isolate dT (it is better to write it ΔT)?. Do not calculate anything, n and R will cancel...

    The change of internal energy is ΔU=CvnΔT.

    Substitute the expression of ΔT into the equation for ΔU above.

    I did not want to hurt you, but this problem can be solved symbolically and the solution is short and elegant. You just need to collect and manipulate the relevant equations and know data. No need and no possibility to find P, V, and the initial and final temperatures. You can go round and round... there are no data. But you can determine the change of internal energy from the work done and from Cp.

    Your derivation was wrong as you assumed that this process was also isotherm. It was not not.

    ehild
     
    Last edited: Dec 13, 2011
  8. Dec 13, 2011 #7
    Alright, that's fantastic. To be honest, I see what the real problem was. In the equation sheet I was given for the exam, it has Q = nC(dT), not Q = nCv(dT). If I had known it was Cv instead of some other C that I didn't really know, this would have been MUCH simpler! Thanks for all the help, it's very much appreciated!


    (Oh, and I know it's easier to write it with the symbol, but I just now looked and saw the menu on the right when you post =P)
     
  9. Dec 13, 2011 #8

    ehild

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    Q=nCΔT is correct, but that C depends on the process. If it is isobaric, you use Cp. For a constant volume process, it is Cv. You can imagine any kind of process, and they will have different specific heat capacities.

    As for ΔT, I meant that dT and ΔT are not the same. We use dt for an infinitesimal change of T, in calculus. ΔT=Tfinal -Tinitial, change of temperature in a process.

    The symbols on the right are very handy. You can also click on the Ʃ above, the menu provides you to write in TeX.

    And it is my signature, you can copy-paste symbols also from there. You can copy the whole thing into your signature, and then it is available any time.


    ehild
     
    Last edited: Dec 13, 2011
  10. Dec 13, 2011 #9
    Appreciate the help a lot, bud! I'll probably be on here tomorrow with another question, so keep an eye open ;)
     
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