The relationship between enthalpy and internal energy change is

Explanation

We know, Path `CA:` Isothermal compression
Path `AB`: Isobaric expansion
Path `BC`: Isochroic change
Let `V_(i)` and `V_(f)` be initial volume and final volume at repective points.
For temperature `T_(1) ("for" C): PV - nRT_(1)`
`2 xx 10 = 1 xx 0.0821 xx T_(1)`
`T_(1) = 243.60K`
For temperature `T_(2)` (For `C` and `B): (P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))`
`(2xx10)/(T_(1)) = (20 xx 10)/(T_(2))`
`(T_(2))/(T_(1)) = 10`
`T_(2) = 243.60 xx 10 = 2436.0K`
Path `CA: w = + 2.303 nRT_(1) "log"(V_(i))/(V_(f))`
`= 2.303 xx 1xx2xx 243.6"log" (10)/(1)`
`=+1122.02 cal`
`DeltaE = -0` for isotermal compression, also `q = W`
Path `AB, w =-P (V_(f) - V_(1))`
`=- 20 xx (10 - 1) =- 180 L-atm`
`= (-180 xx 2)/(0.0821) =- 4384.9 cal`
Pth `BC: w =- P(V_(f) - V_(1)) = 0`
(`:' V_(f) - V_(1) = 0)` since volume is constant
For monoatomic gas, heat chnage at constant volume `=qv = DeltaU`
Thus, for path `BC, q_(v) = C_(v) xx n xx DeltaT = DeltaU`
`:.q_(v) = (3)/(2)R xx1 (2436 - 243.6)`
`= (3)/(2) xx2xx1xx (2192.4) = 65772.2 cal`
Since process involves cooling `:.q_(v) = DeltaU =- 6577.2 cal`
Also in path `AB`, the initial enegry in state `A` and state `C` is same. Thus,, during path `AB`, an increase in internal enegry equivalent of change in internal enegry during path.
`BC` should take place. Thus, `DeltaU` for path `AB = + 6577.2 cal`
Now `q` for path `AB = DeltaU - w_(AB) = 6577.2 + 6577.2 cal`
`= 10962.1 cal`
Cycle: `DeltaU = 0`,
`q =- w =- [w_(path CA) +w_(pathAB) +w_(pathBC)]`
`=- [+1122.02 +(-4384.9+0]`
`:. q =- w = + 3262.88 cal`