Two moles of an ideal monoatomic gas is taken through a cycle ABCA as shown in the P-T diagram. During the process AB , pressure and temperature of the gas very such that `PT=Constant`. It `T_1=300K`, calculate

(a) the work done on the gas in the process AB and
(b) the heat absorbed or released by the gas in each of the processes. Give answer in terms of the gas constant R.

Answer

(i) ` 1200R` (ii) `Q_(AB) = -2100R, Q_(BC) = 1500R, Q_(CA) = 1200Rln2`

Explanation

(i) Number of moles
`n=2 , T_(1) = 300K`
During the process `A rarr B`
`PT` = constant or `P^(2)V` = constant = K(say)
`thereforeP = (sqrtK)/(sqrtV)`
`therefore W_(A rarr B) = underset(V_(A))overset(V_(B))intP.dV = underset(V_(A))overset(V_(B))intsqrt(K)/(sqrtV)dV
=`2sqrtK[sqrtV_(B)-sqrtV_(A)]=2[sqrtKV_(B) - sqrtKV_(A)]`
=`2[sqrt((P_(B)^(2)V_(B))V_(B))-sqrt((P_(A)^(2)V_(A))V_(A))] (K = P^(2)V)`
=`2(P_(B)V_(B) -P_(A)V_(A)] = 2[nRT_(B) -nRT_(A)]`
=`2nR [T_(1) - 2T_(1)] = (2)(2)(R)[300-600]`
=`-1200R`.
`therefore` Work done on the gas in the process `AB "is" 1200R`.
(ii) Heat absorbed //released in different processes.
Since, the gas is monoatomic.
Therefore , `C_(V) =(3)/(2)R "and" C_(P)= (5)/(2)R "and" gamma=(5)/(3)`
Process A-B : ` DeltaU = nC_(V)DeltaT = (2)((3)/(2)R) (T_(B) - T_(A)) = (2)((3)/(2)R)`
`(300-600) = -900R`
`therefore Q_(A rarrB) = W_(A rarrB) +DeltaU = (-1200R) - (900R)`
`Q_(A rarrB) =-2100R` (Released)
Process B-C:
Process is isobaric
`therefore Q_(B rarrC) = nC_(P)DeltaT`
=`(2)((5)/(2)R)(2T_(1) - T_(1))`
=`(5R) (600-300)`
`Q_(B rarrC) = 1500R` (absorbed)
Process C-A :
Process is isothermal
`DeltaU = 0 "and" Q_(C rarrA) = W_(CrarrA) = nRT_(C)ln((P_(C))/(P_(A)))`
=`nR(2T_(1))ln((2P_(1))/(P_(1))) = (2)(R)(600)ln(2)`
`Q_(C rarrA) = 831.6R` (absorbed)