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

Since triangle `O A V_(0)` and OC V are similar therefore
`(2P_(0))/(V) = (P_(0))/(V_(0)) implies V = 2 v_(0)`
b. 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 ) = 2 T_(B) = 4 T_(0)`
Since process is cyclic therefore
`Delta Q = Delta w`= area under the cycle ` =(1)/(2) P_(0) V_(0)`.
d. Since `Delta u` and `Delta W` both are negative in process CA
`:. Delta Q` is negative in process ca and heat is rejected in process CA
`Delta Q_(CA) = Delta W_(CA) + Delta U_(CA)`
`= - (1)/(20) [P_(0) + 2 P_(0)] V_(0) - (5)/(2) nr (T_(c ) - T_(a))`
`= -(1)/(2) [P_(0) + 2 P_(0)] V_(0) - (5)/(2) nR ((4 P_(0) V _(0))/(nR) - (P_(0) V_(0))/(nR))`
`= - 9 P_(0) V_(0) =` Heat injected.
e. `eta = ` efficiency of the cycle
`= ("work done by the gas")/("heat injected") = eta = (P_(0) V_(0)//2)/(Q_("injected") xx 100`
`Delta Q_("inj") = Delta W_(AB) + Delta U_(BC)`
` = [(5)/(2) nr (2T_(0) - T_(0))] + [(5)/(2) nRT (2 T_(0)) + 2P_(0) (2V_(0) - V_(0))]`
`(19)/(2) P_(0) V_(0)`