1st law of thermodynamics (energy conservation): dE = dQ + dW_{on system}
dE = dQ - dW_{by system}

Step 1: Isothermal expansion at T_h:
W₁ = ∫ P dV = NkT_h ln(V_B/V_A) > 0
E₁ = 0
Q_h = NkT_h ln(V_B/V_A)

Step 2: Adiabatic expansion to T_c:
W₂ = ∫ P dV = a ∫ V^-γ dV = a/(1-γ) [V_C^1-γ - V_B^1-γ] > 0
= Nk/(1-γ) [T_C - T_B] (use PV^γ = a = NkT V^γ-1)
Q = 0
E₂ = - W₂ = Nk/(1-γ) [T_B - T_C]

Step 3: Isothermal compression T_c:
W₃ = ∫ P dV = NkT_c ln(V_D/V_C) < 0
E₃ = 0
Q_c = | NkT_c ln(V_D/V_C) |

Step 4: Adiabatic compression to T_h:
W₄ = ∫ P dV = ∫ V^-γ dV = b/(1-γ) [V_A^1-γ - V_D^1-γ] < 0
= Nk/(1-γ) [T_A - T_D] = Nk/(1-γ) [T_B - T_C] = - W₂
Q = 0
E₄ = - W₄ = Nk/(1-γ) [T_B - T_C] = - E₂

After completing one cycle:

W_tot = W₁ + W₃ = NkT_h ln(V_B/V_A) + NkT_c ln(V_D/V_C) = Q_h - Q_c

The efficiency is:   η

= 1 -

Qc Qh

= 1 -

NkTc ln(VC/VD) NkTh ln(VB/VA)

Now use rule for adiabats:   TV^γ-1 = constant
T_h V_B^γ-1= T_c V_C^γ-1
T_h V_A^γ-1= T_c V_D^γ-1

Dividing gives:   V_B/V_A = V_C/V_D

Final result for the efficiency:   η

= 1 -

Tc Th

Important notes:

• The Carnot cycle applies for an impractical quasi-static process
• For a car, T_c ≈ 300K, while T_h < 600K, the temperature where engine block softens.
• A Carnot car engine has an efficiency η < 0.5.
• Real engines have efficiencies η ≈ 0.2.