State and explain Carnot's theorem.

(N/A) Carnot's theorem consists of two main parts:
$(a)$ No heat engine operating between two given heat reservoirs at temperatures $T_{1}$ and $T_{2}$ can be more efficient than a reversible Carnot engine operating between the same two reservoirs.
$(b)$ All reversible engines operating between the same two heat reservoirs have the same efficiency,regardless of the working substance used.
To prove part $(a)$,consider an irreversible engine $I$ and a reversible Carnot engine $R$ operating between a source at temperature $T_{1}$ and a sink at temperature $T_{2}$.
Assume $I$ absorbs heat $Q_{1}$ from the source,performs work $W'$,and rejects heat $Q_{1} - W'$ to the sink. Let $R$ be coupled to $I$ such that $R$ acts as a refrigerator,taking heat $Q_{2}$ from the sink and requiring work $W$ to return heat $Q_{1}$ to the source.
If we assume $\eta_{I} > \eta_{R}$,then for the same heat $Q_{1}$ absorbed from the source,the work done by $I$ is $W' > W$. The net work output of the combined system is $W' - W$,and the net heat extracted from the cold reservoir is $(Q_{1} - W) - (Q_{1} - W') = W' - W$. This implies the system converts heat directly into work without any other effect,which violates the Kelvin-Planck statement of the second law of thermodynamics. Thus,$\eta_{I} \leq \eta_{R}$.
For part $(b)$,if we have two reversible engines $R_{1}$ and $R_{2}$,we can run one in reverse to form a combined system. If one were more efficient,it would violate the second law. Therefore,all reversible engines must have the same efficiency,given by $\eta = 1 - \frac{T_{2}}{T_{1}}$,which is independent of the working substance.