Derive the expression for Carnot engine efficiency.

Explanation

Efficiency of a Carnot engine: Efficiency is defined as the ratio of work done by the working substance in one cycle to the amount of heat extracted from the source.
`eta=("Work done")/("Heat extracted")=(W)/(Q_(H))" ...(1)"`
From the first law of thermodynamics, `W=Q_(H)-Q_(L)`
`eta=(Q_(H)-Q_(L))/(Q_(H))=1-(Q_(L))/(Q_(H))" ...(2)"`
Applying isothermal conditions, we get,
`Q_(H)=muRT_(H) ln ((V_(2))/(V_(1)))`
`Q_(L)=muRT_(L)ln((V_(3))/(V_(4)))" ...(3)"`
Here we omit the negative sign. Since we are interested in only the amount of heat `(Q_(1))` ejected into the sink, we have
`(Q_(L))/(Q_(H))=(T_(L)ln ((V_(3))/(V_(4))))/(T_(H)ln((V_(2))/(V_(1))))" ...(4)"`
By applying adiabatic conditions, we get
`T_(H)V_(2)^(gamma-1)=T_(L)V_(3)^(gamma-1)`
`T_(H)V_(1)^(gamma-1)=T_(L)V_(4)^(gamma-1)`
By dividing the above two equations, we get
`((V_(2))/(V_(1)))^(gamma-1)=((V_(3))/(V_(4)))^(gamma-1)`
Which implies that `(V_(2))/(V_(1))=(V_(3))/(V_(4))" ..(5)"`
Substituting equation (5) in (4), we get
`(Q_(L))/(Q_(H))=(T_(L))/(T_(H))" ...(6)"`
`therefore " The efficiency "eta =1-(T_(L))/(T_(H))" ...(7)"`
Note : `T_(L) and T_(H)` should be expressed in Kelvin scale.
Important results:
1. n is always less than 1 because `T_(L)` is less than `T_(H)`. This implies the efficiency cannot be `100%`.
2. The efficiency of the Carnot.s engine is independent of the working substance. It depends only on the temperatures of the source and the sink. The greater the difference between the two temperatures, higher the efficiency.
3. When `T_(H) = T_(L)`, the efficiency `eta= 0`. No engine can work having source and sink at the same temperature.