Example
Let us analyze an ideal gas undergoing a Carnot cycle between two temperatures TH and TL.
- 1 to 2, isothermal expansion, DU12 = 0
QH = Q12 = W12 = ?PdV = mRTHln(V2/V1)
- 2 to 3, adiabatic expansion, Q23 = 0
(TL/TH) = (V2/V3)k-1 ? (1)
- 3 to 4, isothermal compression, DU34 = 0
QL = Q34 = W34 = - mRTLln(V4/V3)
- 4 to 1, adiabatic compression, Q41 = 0
(TL/TH) = (V1/V4)k-1 ? (2)
From (1) & (2), (V2/V3) = (V1/V4) and (V2/V1) = (V3/V4)
?th = 1-(QL/QH )= 1-(TL/TH) since ln(V2/V1) = ln(V4/V3)
It has been proven that ?th = 1-(QL/QH )= 1-(TL/TH) for all Carnot engines since the Carnot efficiency is independent of the working substance.