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

`61.40`

Explanation

The corresponding `p - V` diagram is as shown
Given `T_(A) = 300K, n = 1, gamma - 1.4, (V_(A))/(V_(B))=16 and (V_(C ))/(V_(B)) = 2 `

Let `V_(B)=V_(0) and p_(B)=p_(0)`
Then, `V_(C )=2V_(0) and V_(A)=16V_(0)`
Efficiency of cycle
Efficiency of cycle (in percentage) is defined as
`eta=("Net work done in the cycle")/("Heat absorbed in the cycle") xx 100`
Or `eta=(W_("Total"))/(O_(+ve))xx 100`
`=(Q_(+ve)-Q_(-ve))/(Q_(+ve))xx 100 =(1-(Q_(1))/(Q_(2))) xx 100 " " ...(i)`
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)
`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 (300-791.4)J`
or `Q_(DA) = -10208.8J`
and `Q_(BC)=nc_(p) DeltaT =(1)((7)/(2)R) (T_(C )-T_(B))`
(`c_(p)=(7)/(2)R` for a diatomic gas)
`=((7)/(2))(8.31) (1818-909)J`
or `Q_(BC)=26438.3J`
Therefore, substituting `Q_(1) = 10208.8 J and Q_(2) = 26438.3 J` in Eq. (i), we get
`:.eta={1-(10208.8)/(26438.3)} xx 100`
or `eta=61.4%`