linford86
Oct30-09, 10:29 PM
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}
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}