Derive the expression for Carnot engine efficiency.

Explanation

Efficiency of a Carnotenginr: Efficiency is defined as the ratio of work done by the working substance in one cycle to the amount of heat extrcted from the source.
`eta=("work done" )/("heat extracted")=(W)/(Q_(H))`
From the first law of thermodynamics, `W=Q_(h)-Q_(L)`
`eta=(Q_(H)-Q_(L))/(Q_(H))=(Q_(L))/(Q_(H))`
Applying isothermal conditions, we get
`Q_(H)=muRT_(H) In((V_(2))/(V_(1)))`
`Q_(L)=muRT_(L) In((V_(3))/(V_(4)))`
`therefore (T_(L))/(T_(H))=(Q_(L) In(V_(3))/(V_(4)))/(Q_(H)In(V_(2))/(V_(1)))`
By applying adiabatic conditions we get ,
`T_(H)V_(2)^(y-1) = T_(L) V_(3)^(y-1)`
`T_(H)V_(1)^(y-1) = T_(L) V_(4)^(y-1)`
By dividing the above two equations we get,
`(V_(2)/(V_(1)))^(y-1)=(V_(3)/(V_(4)))^(y-1)`
Subsitituting equation (5) in (4), we get
`(Q_(L))/(Q_(H)) =(T_(L))/(T_(H))`
`therefore` The effiency `mu=1-(T_(L))/(T_(H))`