n moles of a diatomic gas has undergone a cyclic process ABC as shown in figure. Temperature at A is `T_(0)`. Find
a. Volume at C ?
b. Maximum temperature ?
c. Total heat given to gas ?
d. Is heat rejected by the gas, if yes how much heat is rejected ?
e. Find out the efficiency

Explanation

(i) For process `AC, P alpha V`
`(2P_(0))/(V_(c)) = (P_(0))/(V_(0)) rArr V_(c) = 2V_(0)`
(ii) Since process `AB` is isochoric hence
`(P_(A))/(T_(A)) = (P_(B))/(T_(B)) rArr T_(B) = 2T_(0)`
since process `BC` is isobaric therefore `(T_(B))/(V_(B)) = (T_(C))/(V_(C))`
`rArr T_(C) = 2T_(B) = 4T_(0)`
(iii) Since process is cyclic therefore
`DeltaQ = W =` area under the cycle `= (1)/(2) P_(0)V_(0)`.
(iv) Since `DeltaU` and `DeltaW` both are negative in process `CA`
`:. DeltaQ` is negative in process `CA` and heat is rejected in process `CA`
`DeltaQ_(CA) = W_(CA) +DeltaU_(CA)`
`= -(1)/(2) [P_(0) +2P_(0)] V_(0) - (5)/(2) nR (T_(C)-T_(A))`
`= - (1)/(2) [P_(0) +2P_(0)] V_(0) - (5)/(2) nR ((4P_(0)V_(0))/(nR)-(P_(0)V_(0))/(nR))`
`=- 9 P_(0)V_(0)` (Heat rejected)
(v) `eta =` efficiency of the cycle `= ("work done by the gas")/("heat injected") rArr eta = (P_(0)V_(0)//2)/(Q_("injected")) xx 100`
`DeltaQ_(inj) = DeltaQ_(AB) +DeltaQ_(BC)`
`= [(5)/(2)nR (2T_(0)-T_(0))] +[(5)/(2)nR(2T_(0))+2P_(0)(2V_(0)-V_(0))] = (19)/(2)P_(0)V_(0)`.
`eta = (100)/(19) %`