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) Since triangle `OAV_(0)` and `OCA` are similar therefore `" "` `(2P)/(V)=(P_(0))/(V_(0)) implies V = 2V_(0)`
Since process `AB` is isochoric hence `" "` `(P_(A))/(T_(A)) = (P_(B))/(T_(B)) implies T_(B) = 2T_(0)`
Since process `BC` is isobaric therefore `" "` `(T_(B))/(V_(B)) = (T_(C))/(V_(C)) implies T_(C) = 2T_(B) = 4T_(0)`
(iii) Since process is cyclic therefore `Q=W` = area under the cycle = `(1)/(2)P_(0)V_(0)`
(iv) Since `U` and `W` both are negative in process `CA`
`therefore Q` is negative in process `CA` and heat is rejected in process `CA`
`Q_(CA) = W_(CA) + U_(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))`
=`9P_(0)V_(0)` = Heat injected .
(v) `eta = "efficieny of the cycle" = ("work done by the gas ")/("heat injected ") = (P_(0)V_(0)//2)/(Q_("injected")) xx100`
where `Q_("inj") = Q_(AB) + Q_(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)`.
Therefore `eta = (100)/(19)%`