One mole of a monoatomic ideal gas is taken through the cycle shown in figure.
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`ArarrB` Adiabatic expansion
`BrarrC` Cooling at constant volume
`CrarrD` Adiabatic compression.
`DrarrA` 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, `T_A=1000K`, `p_B=(2/3)p_A` and `p_C=(1/3)p_A`. Calculate
(a) the work done by the gas in the process `ArarrB`
(b) the heat lost by the gas in the process `BrarrC`
Given, `(2/3)^0.4=0.85` and `R=8.31J//mol-K`

Explanation

a. As for adiabatic change `PV^(gamma)=` constant
i.e.`P((muRT)/P)^(gamma)=` constant [as `PV=muRT`]
i.e. `(T^(gamma))/(P^(gamma-1))=` constant so `((T_(B))/(T_(A)))^(gamma)=((P_(B))/(P_(A)))^(gamma-1)` with `gamma=5/3`
i.e. `T_(B)=T_(A)(2/3) ^(1-1/T)=1000(2/3)^(2//5)=850K`
So `W_(AB)=(muR[T_(B)-T_(A)])/([1-gamma])=(1xx8.31[1000-850])/([(5//3)-1])`
i.e. `W_(AB)=(3//2)xx8.31xx150=1869.75J`
b. for `B to C` V= constant so `DeltaW=0`
So from first law of thermodynamics
`DeltaQ=DeltaU+DeltAW=muC_(v)DeltaT+0`
or `DeltaQ=1xx(3/2R)(T_(c)-850)` as `C_(v)=3/2R`
Now along path BC, V= constant `P prop T`
i.e. `(P_(C))/(P_(B))=(T_(C))/(T_(B)), T_(C)=((1//3)P_(A))/((2//3)P_(A))xxT_(B)=(T_(B))?2=850/2=425K`
So `DeltaQ=1xx3/2xx8.31(425-850)=-5297.625J`
[Negative heat means, heat is lost by the system]
c. As A and D are on the same isochoric
`(P_(D))/(P_(A))=(T_(D))/(T_(a)) i.e. P_(D)=P_(A)(T_(D))/(T_(A))`
or `(T_(D))^(1//2)=T_(C)[(P_(A))/(P_(C)T_(A))]^(1-1/gamma)` i.e. `T_(D)^(3/5)=((T_(B))/2)[(P_(A))/((1//3)P_(A)1000)]^(2/5)`
i.e. `T_(D)^(3/5)=[1/2(2/3)^(2/5)xx1000 ][3/1000]^(2/5)` i.e. `T_(D)=500K`