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Homework Help: Thermodynamics: 3-chambered system

  1. Dec 8, 2013 #1
    1. The problem statement, all variables and given/known data
    In a 3-chambered system, separated by 2 pistons, all of the walls are adiabatic, except for the wall on the outside of chamber C (the wall marked }).

    [A | B | C}

    The pistons are frictionless, the gas fills all 3 chambers and is ideal, and [itex]\bar{}[/itex] = R. Total volume = 12 m3

    Chamber A:
    300K --> 261.856 K
    4 bar --> 3.047 bar
    3 m3 --> 3.437 m3

    Chamber B:
    500 K --> 436.364 K
    4 bar --> 3.047 bar
    6 m3 --> 6.875 m3

    Chamber C:
    700 K --> 300 K
    4 bar --> 3.047 bar
    3 m3 --> 1.688 m3

    Calculate q, w, ΔU, ΔH, and ΔS for each chamber.

    3. The attempt at a solution

    So part (a) wanted us to calculate all of the final conditions after equilibrium, given only that Tf in chamber C is 300 K. Somewhat tricky, but I'm fairly sure the values there are correct.

    I was also able to find all of the values needed for chambers A and B:

    Chamber A:
    q = 0 J
    w = -152.575 kJ
    ΔU = -152.575 kJ
    ΔH = -305.150 kJ
    ΔS = 0 J

    Chamber B:
    q = 0 J
    w = -305.452 kJ
    ΔU = -305.452 kJ
    ΔH = -610.903 kJ
    ΔS = 0 J

    So far, for chamber C, I found that:
    ΔU = -685.710 kJ
    ΔH = -1371.421 kJ

    But I can't seem to find out what w, q, and ΔS would be. A classmate of mine contends that because we don't know the nature of chamber A (adiabatic, isothermal, isobaric, etc...), but because it is all reversible, we can find ΔS through first an isothermal, reversible expansion, followed by isochoric heating, then we can find q and then w. I thought perhaps that wA would simply be -(wB + wC), because that would work for a 2-chambered piston, but I'm not sure if it would work for a 3-chambered one. Any input would be helpful!
  2. jcsd
  3. Dec 8, 2013 #2
    You're right. Because no work is done by the 3-chamber system (since its volume is constant), wC=-(wA+wB). You also know the change in internal energy for the 3-chamber system, so, from the first law, you know the amount of heat q. This heat all applies to chamber C. The change in specific entropy for an ideal gas is determined by ΔS = CvΔln(T)+RΔln(V).
  4. Dec 8, 2013 #3

    So, it wouldn't be correct to use q = nRT ln (Vf/Vi) + n [itex]\int_{700}^{300} \bar{C_{v}}dT[/itex]? That's what my friend thinks is correct; essentially it's the same way we found ΔS, but we aren't dividing by T, because ΔS = q/T.

    Wouldn't there be expansion work against volume in chamber B and A that would diminish the amount of work done on C? That's the main argument I can think of against just using the negative of works A and B.

    So, essentially, I've come down to 2 methods of calculating q:

    q = -wC = - [-(wA + wB)]


    q = nRT ln (Vf/Vi) + n [itex]\int_{700}^{300} \bar{C_{v}}dT[/itex]

    And I'm not sure which is correct.
    Last edited: Dec 8, 2013
  5. Dec 8, 2013 #4
    No. This is not correct. The first term is wrong because T is not constant.
    There is work against volume in chambers B and A. But the total amount of work done by the gas in the closed constant-volume 3-compartment container is zero. Writing wC = -(wA + wB) is how you take this into account.
    Neither method is correct. You know that, for the overall container, w = 0, and you have calculated ΔU for each of the three containers. So you know the total change in internal energy of the combined 3 chambers: ΔUTotal=ΔUA+ΔUB+ΔUC
    So, from the first law applied to the overall combination of 3 chambers: q = ΔUTotal. This is not only the heat added to the combination of the three chambers (actually, it's negative, so heat is removed), but also the heat added to chamber C, since chambers A and B are adiabatic.

  6. Dec 8, 2013 #5
    Ah, sorry, for the first one I meant q = ΔU - wC = ΔU - [- (wA + wB)].

    For the second, his rationale is that you can take the process stepwise, and calculate the q for first the isothermal, reversible expansion (q = nRT ln (Vf/Vi), where T = 300 K), and then calculate q for the isochoric heating (q = n ∫ Cv dT), and then add them together. I do know this works for ΔS, because it's a state function, but adding the q's together like this doesn't work, correct? And so the only method viable is to find it from ΔU - wC, which we find as the negative of works A and B?
  7. Dec 8, 2013 #6
    Yes. That's correct. q is path-dependent, as is w. Only q-w is not path dependent for a closed system.
    There is another viable method. That's the method I described in my previous post of getting the total ΔU for the combined system. You should verify that both these methods give the same answer for q.
  8. Dec 9, 2013 #7
    Actually, the two methods don't give me the same answer. If I use wC = -(wA + wB), I get wC = 458.027 kJ, and q = ΔU - w = -227.683 kJ.
    If I take ΔU of the entire system to be qC, then I get -1143.737 kJ. I'm not sure, but why would w = 0 for the entire system?

    EDIT: I just re-calculated it, they are indeed equal. That's nice to see. Thanks for the help!
    Last edited: Dec 9, 2013
  9. Dec 10, 2013 #8
    Hello Kyle.

    How are you doing on this exam?
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