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n moles of a diatomic gas has undergoes ...

`n` moles of a diatomic gas has undergoes a cyclic process `ABCA` as shown in figure. Temperature at `A` is `T_(0)`. Find
(i) Volume at `C`?
(ii) maximum temperature?
(iii) Total heat given to gas?
(iv) Is heat rejected by the gas, if yes how much heat is rejected?
(v) Find out the efficiency

Text Solution

Verified by Experts

(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)%`
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