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Thermo physics

  1. Mar 29, 2008 #1
    Last edited: Mar 29, 2008
  2. jcsd
  3. Mar 30, 2008 #2


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    On your TS-diagram, you might want to reverse the arrow so it runs from C to D...

    Don't forget to include the universal gas constant, R, or your units won't come out right (or are units defined in some way so that R = 1?). [EDIT: As I go back and look at the problem, this is a fussy point. It would matter if you had to state the values for W or Q anywhere, but for the efficiency, it drops out at the end...]

    Your work result looks all right. You could exploit properties of logarithms to simplify it to

    NR · (T2 + T1) · ln(V2/V1).

    Heat input occurs for processes A->B (raise temperature) and B->C (expand gas isothermally). So the term for A->B should be (3/2)NR · (T2 - T1), shouldn't it, since T2 > T1 ? The term for B->C will be NR · T2 · ln(V2/V1) .

    In the efficiency quotient (which should read 'net work'), you have N moles in every term, so you can divide it out. (The efficiency is independent of the scale of the process. Oh yeah, the R will cancel out also.) Except for the small corrections I suggested, this result looks OK.
    Last edited: Mar 30, 2008
  4. Mar 30, 2008 #3
    thanks for the comments. It seems I am not familiar with this R constant. I am using Thermal Physics by Charles Kittel and they did a similar calculation without using the R term. This calculation can be seen of pg. 238, or:


    Why is this?

    Also, I more or less guessed in some sense for the PV and TS graphs. I'm not sure why certain parts of the graph is curved. Is it safe to assume that a straight line in the PV graph correlates to a curved line on the TS graph? So for example, the Carnot cycle, the TS graph is a box, and this, the PV graph is made up of curved lines.
    Last edited: Mar 30, 2008
  5. Mar 30, 2008 #4


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    Interesting. I own a copy of the second edition, but I didn't use the book for a course. I'll have to look through it to see why he omits R (I suspect he does something like make a choice of units for which R = 1). But it's the 'R' in PV = nRT, so if you are giving values for W, Q, etc., in any set of standard units (SI, cgs metric, British Engineering), R needs to be there.

    For the Carnot cycle on a T-S diagram, the processes are isothermal, thus vertical lines [constant T], or reversible adiabatic, where Q = 0 and so dS = 0 [thus constant S]. (Note: there are also irreversible adiabatic processes, in which dS is not zero.)

    For the cycle you're asked to work on, the isovolumic (also called isochoric) processes involve no work, so

    dU = dQ + dW = dQ + 0 and dU = (3/2) · nR · dT , for the monatomic gas.

    So the change in entropy is found by integrating

    dS = dQ / T = (3/2) · nR · ( dT / T ) ,

    in which case delta-S is going to involve a logarithmic function of T. So your concave downward curves are correct.
  6. Mar 30, 2008 #5
    thanks for the explanation. I'm also wondering why for heat in, Kittel doesnt use [tex]dQ=\tau d\sigma[/tex]

    I guess i'm not sure why I assumed that the heat taken up and the work done are equal (hence using dQ=dW)
    Last edited: Mar 30, 2008
  7. Mar 30, 2008 #6
    So the "heat in" would would only be from A->B since Q=0 for B->C in the Carnot cycle.

    But for the question at hand,weI have to consider A->B->C because S always increases from A->B->C so for total heat in, we would consider A->B->C instead of just A->B in the Carnot cycle. Is this the right thought process?
  8. Mar 31, 2008 #7


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    This is true for the isothermal processes where dU = 0 (I presume you're using the form of the First Law, dU = dQ - dW , W being the work done by the system).
  9. Mar 31, 2008 #8


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    Yes, because the cycle is isothermal expansion, adiabatic expansion, isothermal compression, adiabatic compression. Heat would have to be introduced during the expansions (and removed during the compressions), but the reversible adiabatic expansion involves no heat input. So that's one less term in the denominator of the efficiency quotient, thus the efficiency is higher.

    Yes. The isovolumic process requires a temperature increase to raise the gas pressure, so we have a heating step that doesn't occur in the Carnot cycle.

    I'll mention that a reversible adiabatic process is also called "isentropic". (There are, as I'd said earlier, irreversible adiabatic processes, such as a violent explosion or sudden compression, where there is no time (realistically, very little time) for heat to be exchanged with the environment, but the shock occurring in the gas clearly causes a considerable entropy change...)
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