One mole of a diatomic ideal gas `(gamma = 1.4)` is taken through a cyclic process starting from point A. The process `A to B` is an adiabatic compression. `B to C` is isobaric expansion, `D to A` an adiabatic expansion and is isochoric.
The volume ratio are `V_A //V_B - 16` and `V_C //V_B - 2` and the temperature at `T_A = 300k` is
Find the temperature of the gas at the point (in K).

Answer

`T_(B) = 909K , T_(D) = 791.4K, 61.4%`

Explanation

Let `V_(B) = V_(0)`
Process `A rarr B`
`T_(A)V_(A)^(gamma-1) = T_(B)V^(gamma-1)`
`implies T_(B) = 909K`
Process `B rarr C` `(V_(B))/(T_(B)) = (V_(C))/(T_(C)) implies T_(C) = 7272K`
Process `C rarrD: T_(C)V_(C)^(gamma-1)= T_(D)V_(D)^(gamma-1) implies T_(D) = 5511.15K`
Heat flow
Process `A rarr B` `" "` `Q_(AB) = 0`
Process `D rarrA " " Q_(DA) = nCvDeltaT`
`= 1 ((5)/(2)R) (T_(A) -T_(D))`
= `-108313.753 J`
`therefore` Efficiency , `eta = ("work output ")/("Heat output") = (Q_("in") - Q_("out"))/(Q_("in"))`
=`(1- (Q_("out"))/(Q_("in"))) xx 100% = (1-(108313.753)/(185156.937)) xx 100% eta = 41.50%`