One mole of a monoatomic ideal gas is taken through the cycle shown in Fig:
`AtoB`: adiabatic expansion
`BtoC`: cooling at constant volume
`CtoD`: adiabatic compression
`DtoA`: 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 process `AtoB`
(ii) The heat lost by the gas in the process `BtoC`.
(iii) The temperature `T_D`. [Given :`(2//3)^(2//5)=0.85`]

Answer

A, B

Explanation

Given `T_A=1000K`
`P_B=2/3P_A`
`P_C=1/3P_A`

(i) `W_(AB) ( adiabatic expansion)`
`W_(AB)=(nR[T_A-T_B])/(gamma-1)`
Here, `n=1,R=8.32Jmol^-1k^-1,T_A=1000K`
`gamma=5/3 (For monoatomic gas)`
To find `T_B,` we use
`T_A^(gamma)P_A^(gamma-1)=T_B^(gamma)P_B^(gamma-1) rArr ((P_A)/(P_B))^(gamma-1)=((T_A)/(T_B))^(gamma)....(i)`
`rArrT_B=T_A[((P_A)/(P_B))^((gamma-1)/gamma)=1000[3/2]^((1-5//3)/(5//3))=850K`
`:. W_(AB)=(1xx8.31[1000-850])/(5//3-1)=1870J`
(ii) Heat Lost `BtoC`
`Q=nC_vDeltaT=nC_v(T_B-T_C)`
Here, `n=1,C_v=3/2R(For monoatomic gas),`
`T_B=850K`
To find `T_C,` we use `(P_B)/(T_B)=(P_C)/(T_C)` (volume constant)
`(P_C)/(P_B)=(T_C)/(T_B)......(ii)`
`rArrT_C=(P_C)/(P_B)xx(T_B)=1/2xx850=425K[:' (P_C)/(P_A)=(1/3P_A)/(2/3P_A)=1/2]`
`:. Q=1xx3/2xx8.31[425-850]=-5298J`
(ii) Temperature `T_D` : C to D is adiabatic compression
`((P_C)/(P_D))^(gamma-1)=((T_C)/(T_D))^(gamma)......(iii)`
D to A is isochoric process `(P_D)/(T_D)=(P_A)/(T_A)`
`rArr(P_A)/(P_D)=(T_A)/(T_D) ......(iv)`
Multiplying (i) and (iii)
`((P_CP_A)/(P_DP_B)^(gamma-1)=((T_C)/(T_D)xx(T_A)/(T_B))^(gamma)......(v)`
Multiplying (ii) and (iv)
((P_AP_C)/(P_BP_D)=((T_CT_A)/(T_BT_D)).....(vi)`
From (v) and (vi)
`((T_CT_A)/(T_BT_D))^(gamma-1)=((T_CT_A)/(T_BT_D))^(gamma) rArr (T_AT_C)/(T_BT_D)=1`
`rArrT_D=(T_AT_C)/(T_B)=(1000xx425)/850=500K`