One mole of a diatomic ideal gas `(gamma=1.4)` is taken through a cyclic process starting from point A. The process `AtoB` is an adiabatic compression, `BtoC` is isobaric expansion, `CtoD` is an adiabatic expansion, and `DtoA` is isochoric. The volume ratios are `V_A//V_B=16 and V_C//V_B=2` and the temperature at A is `T_A=300K`. Calculate the temperature of the gas at the points B and D.

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%`