`1mol` of a mono-atomic gas is subjected to following cyclic process:

a. Calculate `T_(1)` and `T_(2)`.

Explanation

a. At A: (Using ideal gas equation)
`pV = nRT`
`20 xx 1 = 1xx 0.082 xx T_(1)`
`T_(1) = 243.6K`
At B: `PV = nRT`
`20 xx 10 = 1 xx 0.0820 xx T_(2)`
`T_(2) = 2436.05K`
b. Path `AB:` Isobaric process `(DeltaU = 0,q = W)`
`W = P DeltaV = 20 xx 9 = 180L atm = (108 xx 101.3)/(4.185)cal`
`= 4356.9cal`
(work in compression is positive)
Path `BC:` Isochoric process
`W = 0`
`q_(v) = DeltaU = nC_(v)DeltaT = 1xx(3)/(2) R xx (2436 - 243.6)`
`=(3)/(2) xx2xx2192.4 = 6577.2cal`
It is cooling process: `q_(v) =- 6577.2cal`
Path `CA:` It is an isothermal compression `DeltaU =0`
`q = w = 2.303nRT log .(V_(2))/(V_(1))`
`= 2.303 xx 1xx 2xx log .(10)/(1) = 1122.02cal`