A monoatomic ideal gas of two moles is taken through a cyclic process starting from `A` as shown in figure. The volume ratios are `(V_(B))/(V_(A))=2` and `(V_(D))/(V_(A))=4`. If the temperature `T_(A)` at `A` is `27^(@)C`, calculate:

(a) The temperature of the gas at point `B`.
(b) Heat absorbed or released by the gas in each process.
(c) The total work done by the gas during complete cycle.

Answer

(i) `600K` (ii) `1500R, 831.6K, -900R, -831.6 R` (iii) `600R`

Explanation

Given:
No. of moles , `n=2`
`C_(V) = (3)/(2)R` & `C_(P) = (5)/(2)R`
(Monoatomic)
`T_(A) = 27^(@)C = 300K`
Let `V_(A) =V_(0)`
then `V_(B)=2V_(0) "and" V_(D) = V_(C) = 4V_(0)`
(i) Process `A rarrB ` :
`V alpha T implies (T_(B))/(T_(A)) = (V_(B))/(V_(A))`
`therefore T_(B) = T_(A)((V_(B))/(V_(A))) = (300)(2) = 600K`
`therefore T_(B) = 600K`
(ii) Process `A rarrB` :
`V alpha T implies P` = constant
`therefore Q_(AB) =nC_(P)dT = nC_(P)(T_(B)-T_(A))`
=`(2)((5)/(2)R)(600-300)`
`therefore Q_(AB) = 1500R` (absorbed)
Process `B rarrC` :
`T` = constant `" "` `therefore dU = 0`
`therefore Q_(BC) = W_(BC) = nRT_(B)ln((V_(C))/(V_(B))) = (2)(R)(600)ln((4V_(0))/(2V_(0)))`
=`(1200R)ln(2) = (1200R) (0.693)`
`implies Q_(BC) approx 831.6R`(absorbed)
Process `C rarrD`
`V` = constant
`therefore Q_(CD) = nC_(V)dT = nC_(V)(T_(D) -T_(C))`
=`n((3)/(2)R)(T_(A) - T_(B))(T_(D) = T_(A) "and" T_(C) = T_(B))`
=`(2)((3)/(2)R)(300-600)`
`implies Q_(CD) = -900R` (released)
Process `D rarrA`
`T` = constant `implies dU = 0`
`therefore Q_(DA) = W_(DA) = nRT_(D)ln((V_(A))/(V_(D)))`
=`(2)(R)(300) ln ((V_(0))/(4V_(0))) = 600Rln((1)/(4))`
`Q_(DA) approx -831.6R` (Released)
(iii) In the complete cycle: `dU =0`
Therefore , from conservation of energy
`W_(net) = Q_(AB) + Q_(BC) + Q_(CD) + Q_(DA)`
`W_(net) = 1500R + 831.6R -900R - 831.6R`
`implies W_(net) = W_(total) = 600R`