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Efficiency of Carnot Engine Question

  1. Oct 30, 2009 #1
    1. The problem statement, all variables and given/known data

    This is from "Equilibrium Statistical Physics" by Plischke and Bergerson, problem 1.1:

    "Consider a Carnot engine working between reservoirs at temperatures T1 and T2. The working substance is an ideal gas obeying the equation of state [ [tex]PV=Nk_BT[/tex] ], which may be taken to be a definition of a temperature scale. Show explicitly that the the efficiency of the cycle is given by


    where [tex]T_1>T_2[/tex].

    2. Relevant equations

    The first law of thermodynamics, the ideal gas law, and a few other things. I'll introduce these in my partial solution; I hope that's not inappropriate.

    3. The attempt at a solution

    Well, I began by noting that the efficiency is defined as [tex]\eta=\frac{W}{Q_1}[/tex]. We also know that [tex]W=Q_1+Q_2[/tex], using the convention that heat leaving the system is negative. For the sake of reference, I will define that the Carnot engine moves through points A, B, C, and D, and that AB is isothermal at temperature [tex]T_1[/tex], BC is adiabatic, CD is isothermal at temperature [tex]T_2[/tex], and DA is adiabatic. For the process from A to B, since it is isothermal, the internal energy does not change. By the first law of thermodynamics, we have that:

    [tex]dU=0=\bar{d}Q-\bar{d}W[/tex] and, thus, [tex]\bar{d}Q=\bar{d}W[/tex]

    Next, we have that [tex]\int_A^B \! dQ=\int_A^B P \, dV=Nk_BT\ln(\frac{V_B}{V_A}[/tex]. Likewise, [tex]Q_2=Nk_BT\ln(\frac{V_D}{V_C})[/tex]. Simplifying the equation for [tex]\eta[/tex], we have:


    I understand that, for an adiabatic process, [tex]TV^\gamma=(constant)[/tex] and that [tex]\gamma=\frac{C_D}{C_V}[/tex], but I have no idea how to use this information to further reduce [tex]\eta[/tex] to the desired result --

    Last edited: Oct 30, 2009
  2. jcsd
  3. Oct 31, 2009 #2
    From adiabatic process (BC and DA), we have:

    T_{2} V_{b} ^\gamma ^- ^1 = T_{1} V_{c} ^\gamma ^- ^1

    T_{2} V_{a} ^\gamma ^- ^1 = T_{1} V_{d} ^\gamma ^- ^1

    From 2 eq, we shall get:

    \frac{V_{b}}{V_{a}} = \frac{V_{c}}{V_{d}}

    Plug it to efficiency to get a result.
  4. Oct 31, 2009 #3
    Ha, that's so obvious!! Thanks!

    How do I mark this thread as "Solved"? I tried looking under "Thread Tools" but I didn't see an option for doing that...
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