One mole of a diatomic ideal gas `(gamma=1.4)` is taken through a cyclic process starting from point A. The process `AtoB` is an adiabatic compression, `BtoC` is isobaric expansion, `CtoD` is an adiabatic expansion, and `DtoA` is isochoric. The volume ratios are `V_A//V_B=16 and V_C//V_B=2` and the temperature at A is `T_A=300K`. Calculate the temperature of the gas at the points B and D and find the efficiency of the cycle.

Answer

`T_(B) = 600 xx 2^(3//5)K`,
`T_(D) = 1200 xx 2^(-3//5) K, eta = 61.37%`

Explanation

The corresponding `P- V` diagram is as shown
Given : `T_(A) = 300 K, n = 1 , gamma = 1.4, V_(A)//V_(B) = 16`
`{:(,and,V_(C)//V_(B)=2),(LetV_(B)=V_(0),and,P_(B)=P_(0)),("Then"V_(C)=2V_(0),and,V_(A)=16V_(0)),("Temperatureat" B B,,):}`
Process `A -B` is adiabatic. Hence
`T_(A)V_(A)^(gamma-1) = T_(B)V_(B)^(gamma-1)`
or `T_(B) = T_(A) ((V_(A))/(V_(B)))^(gamma-1)`
`= (300) (16)^(1.4-1)`
`=600 xx 2^(3//5) K`,

Temperature at `D`
`B rarr C` is an isobaric process `(P =` constant)
`:. T prop V`
`V_(C) = 2V_(B)`
`:. T_(C) = 2T_(B) = (2) 600 xx 2^(3//5) K`
`T_(C) = 1200 xx 2^(3//5)K`
Now the process `C-D` is adiabatic.
Therefore
`:. T_(D) = T_(C) ((V_(C))/(V_(D)))^(gamma-1) = 1200 xx 2^(3//5) ((2)/(16))^(1.4-1)`
`T_(D) = 1200 xx 2^(3//5) K`,
Efficiency of cycle-
Efficiency of cycle (in percentage) is defined as
`eta = ("Net work doen in the cycle")/("Heta absorbed in the cycle") xx 100`
`eta = (W_(Total))/(Q_(+ve)) xx 100`
`= (Q_(+ve)=-Q_(-ve))/(Q_(+ve)) xx 100 =(1-(Q_(1))/(Q_(2))) xx 100 ...(1)`
where `Q_(1) =` Negative heat in the cycle (heat released)
and `Q_(2) =` Positive heat in the cycle (heat absorbed) In the cycle-
`Q_(AB) = Q_(CD) = 0` (Adiabatic process)
In process `DA: DeltaQ = DeltaU + DeltaW (DeltaW = 0)`
`Q_(DA) = nC_(v)DeltaT = (1) ((5)/(2)R) (T_(A)-T_(D)) (C_(v) =(5)/(2)R` for a diatomic gas)
`= (5)/(2) xx 8.31 (300 - 791.7)J`
`[T_(D) = 1200 xx 2^(3//5) K = 791.K]`
or `Q_(DA) = - 10215.06 J` (Release)
and `Q_(BC) = n C_(P) DeltaT = (1) ((7)/(2)R) (T_(C)-T_(B)) (C_(P) = (7)/(2)R` for a diatomic gas)
`= ((7)/(2)) (8.31) (181.8 - 909.4)J`
`[T_(D) = 1200 xx 2^(3//5) K ~~ 1818.8 K, T_(B) = 600 xx 2^(3//5) ~~ 909.4 K]`
or `Q_(BC) = 26449.89 J`
Therefore, substituting `Q_(1) = 10215.06 J` and `Q_(2) = 26449.89J` in diatomic (1) we get-
`:. eta = {1-(10215.06)/(26449.89)} xx 100`
or `eta = 61.37%`