A monatomic gas undergoes a cycle consisting of two isothermals and two isobarics. The minimum and maximum temperatures of gas during the cycle are `T_(1)= 400 K` and `T_(2)=800 K`, respectively, and the ratio of maximum to minimum volume is 4.

The efficiency of the cycle is

Options

A. `(3 1n 2)/( 5 +4 1n 2) xx 100 %`
B. `(2 1n 2)/( 5 +4 1n 2) xx 100 %`
C. `(3 1n 3)/( 5 +4 1n 2) xx 100 %`
D. `(2 1n )/( 3 +4 1n 2) xx 100 %`

Answer

B

Explanation

Process `CB` is done at minimum temperature, while process `DA` is done at maximum temperature.
Process
`A rarr b(` isobaric `)`
`(V_(A))/(V_(B))=(T_(A))/(T_(B))=(800)/(400)=2`
hence` V_(B)=2V_(0)`
`B rarr C(` isothermal `)`
`C rarr D (` isobaric `)`
`(V_(D))/(V_(C))=(T_(D))/(T_(C))implies V_(D)=2V_(0)`
`D rarr(` isothermal `)`
`W_(AB)=P(V_(B)-V_(A))=nRDT`
`=nR(-400)=-400nR`
`W_(BC)=nRT1n(V_(C))/(V_(B))=nR4001n(V_(0))/(2V_(0))`
`=-400nR 1n2`
`W_(CD)=P(V_(D)-V_(C))=nR(400)=400nR`
`W_(DA)=nRT1n(V_(A))/(V_(D))`
`=nRxx800 xx 1 n (2V_(0))/(V_(0))=800nR1n2`
`Delta W=400nR1n2`
`DeltaQ=` heat is extracted
`Q_(CD)=nC_(p) DeltaT=n(5)/(4)R400=1000nR`
`=800 n R 1 n 2= 1000 n R+800 nR 1 n 2`
Efficiency `=(DeltaW)/( Delta Q)`
`eta =(2 1n 2)/(5 +4 1n 2 ) xx 100%`