On mole of a monoatomic ideal gas is taken through the cycle shown in figure .
`A rarr B` : Adiabatic expansion `" "` `BrarrC` : Cooling at constant volume
`CrarrD`: Adiabatic compression `" "` `D rarr A` : Heating at constant volume
The pressure and temperature at `A , B`, etc, are denoted by `P_(A) , T_(A) , P_(B) , T_(B)` etc., respectively . Given that `T_(A) = 1000K, P_(B) = ((2)/(3))P_(A)"and"P_(C) = ((1)/(3))P_(A)`. Calculate the following quantities: (i) The work done by the gas in the processs `A rarr B` (ii) The heat lost by the gas in the process `B rarr C` (iii) The temperature `T_(D). ("Given" : ((2)/(3))^(2//5) = 0.85`)

Answer

(i)`1869.75 J (ii) -52.97.6 J (iii) 500K`

Explanation

Given that
`T_(A)=1000K n=1`
`P_(B)=(2/3)P_(A),lambda=5/3`
`P_(C)=(1/3)P_(A),(2/3)^(2//5)=0.85`
For Process `AtoB`
`P_(A)^(1-y)T_(A)^(y)=P_(B)^(1-y)T_(B)^(y)`
`T_(B)=T_(A)(P_(A)/P_(B))^((1-lambda)/lambda)=1000xx(3/2)^(-2//5)`
`=1000xx0.85=850 K`
For process `B to C `
`P_(B)/T_(B)=P_(C)/P_(c)RightarrowT_(c)=T_(B)(P_(C)/P_(B))=850xx1/2=425K`
(i) `W_(AtoB)=(nR(T_(A)-T_(B)))/(lambda-1)`
`=(1xx8.134 xx (1000-850))/((5)/(3) - 1) = 1870.2J`
(ii) `DeltaQ_(BrarrC) = DeltaU_(BrarrC) + DeltaW_(BrarrC) = nC_(V)DeltaT + 0`
= `(3)/(2)R(T_(C)-T_(B)) = 1 xx (3)/(2) xx 8.314(425-850)`
=`-5300.175J`
(iii) For process `A rarr B P_(A)V_(A)^(gamma) = P_(B)V_(B)^(gamma) implies ((V_(B))/(V_(A))^(gamma) = (P_(A))/(P_(B)) = (3)/(2)`
For process ` C rarr D P_(C)V_(C)^(gamma) = P_(D)V_(D)^(gamma)`
`implies ((V_(C))/(V_(D)))^(gamma) = ((V_(B))/(V_(A)))^(gamma) = (3)/(2) =(P_(D))/(P_(C)) implies P_(D) = (3)/(2)P_(C)`
At end points A and D `implies (P_(A))/(T_(A)) = (P_(D))/(T_(D)) implies (3P_(C))/(1000) = ((3P_C)/(2))/(T_(D)) implies T_(D) = 500K`