Obtain Carnot's theorem.

Explanation

Carnot theorem is divided into two parts:
(a) No engine can have more efficiency than the reversible engine operating between the given temperature `T_1` and `T_2` of hot and cold reservoirs.
(b) The efficiency of carnot engine is independent of the nature of the working substance.
To prove the result (a), imagine a reversible carnot engine R and an irreversible engine I working between the same source and sink.

Let us couple the engines I and R in such a way so that I acts like a heat engine and R acts as a refrigerator.
Suppose I absorb heat `Q_1` from the source, deliver work W. and release the heat `Q_1-W.` to the sink.
Arrangement will made such that R returns the same heat `Q_1` to the source, taking heat `Q_2` from the sink and requiring work `W = Q_1 - Q_2` to be done on it.
`therefore` Heat given to the sink `Q_2 = Q_1 - W`, necessary work will done. Heat given to the source `Q_1` is same as from R.
Suppose `eta_R lt eta_I`, then engine R give less work than engine I.
Hence for given `Q_1 W lt W.`
With R acting like a refrigerator .
`Q_2=Q_1-W gt Q_1 - W.`
Thus , the coupled I-R system extracts heat `(Q_1-W)-(Q_1-W.)=(W. ~ W)` from the cold reservoir and delivers the same amount of work in one cycle. This is against the Kelvin Planck statement of the second law of thermodynamics. Hence `eta_I gt eta_R` is wrong, means no engine can have efficiency greater then that of the carnot engine.
A reversible engine with one particular substance cannot be more efficient than the one using another substance.
The maximum efficiency of carnot engine `eta=1-T_2/T_1` independent of the nature of the system and hence in carnot engine an ideal gas is taken as a working substance, so the maximum efficiency will obtained.
From `eta=1-T_2/T_1` and `eta=1-Q_2/Q_1`
`Q_2/Q_1=T_2/T_1 rArr Q_1/Q_2 =T_1/T_2` is independent of nature of the system .