Efficiency of Carnot Engine Question

1. Oct 30, 2009

linford86

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 [ $$PV=Nk_BT$$ ], which may be taken to be a definition of a temperature scale. Show explicitly that the the efficiency of the cycle is given by

$$\eta=1-\frac{T_2}{T_1}$$

where $$T_1>T_2$$.

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 $$\eta=\frac{W}{Q_1}$$. We also know that $$W=Q_1+Q_2$$, 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 $$T_1$$, BC is adiabatic, CD is isothermal at temperature $$T_2$$, 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:

$$dU=0=\bar{d}Q-\bar{d}W$$ and, thus, $$\bar{d}Q=\bar{d}W$$

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

$$\eta=1+\frac{\ln(\frac{V_D}{V_C}}{\ln(\frac{V_D}{V_C}}$$

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

$$\eta=1+\frac{T_2}{T_1}$$

Last edited: Oct 30, 2009
2. Oct 31, 2009

ApexOfDE

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.

3. Oct 31, 2009

linford86

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...