One mole of an ideal mono-atomic gas is taken round cyclic process `ABC` as shown in figure below. Calculate work done

Options

A. The work done by the gas.
B. The heat rejected by the gas in the path `CA`and the heat absorbed by the gas in the path `AB`.
C. The net heat absorbed by the gas in the path `BC`.
D. The maixmum temperature attained by the gas during the cycle.

Explanation

Path `AB` is isochoric `(w = 0)`, path `BC` is isothermal `(w_(2) =- ve)`, path `CA` is isobaric `(w_(3) = +ve)`.
Total work done by gas,
`w = w_(1) +w_(2) +w_(3)`
` = 0+2.303nRT log .(V_(B))/(V_(C)) +P_(0) (V_(C) - V_(A))`
`= 0 +2.303 P_(B)V_(B) log.(V_(B))/(V_(C)) +P_(0) (V_(C)-V_(A))`
`= 2.303 xx 3P_(0) xx V_(0) log.(V_(0))/(2V_(0)) +P_(0) (2V_(0)-V_(0))`
`=- 2P_(0)V_(0) +P_(0)V_(0) =- P_(0)V_(0)`
Also, `w_(2) =- 2P_(0)V_(0)` and `w_(3) = P_(0)V_(0)`
Also, For the path `AB`, i.e., isochoric
`q_(1) = n xx C_(V) xx (T_(B)-T_(A)) = 1 xx(3)/(2)R [(P_(B)V_(B)-P_(A)V_(A))/(R)]`
`= (3)/(2) [3P_(0)V_(0) - P_(0)V_(0)] =+3P_(0)V_(0)`
For the ptha `CA`, i.e., isochoric:
`q_(3) = nxx C_(P) xx (T_(A) - T_(B)) = 1xx(5)/(2)R [(P_(A)V_(A)-P_(B)V_(B))/(R)]`
`= (5)/(2) [P_(0)V_(0) - 2P_(0)V_(0)]`
`q_(3) =- (5)/(2)P_(0)V_(0)`
Also, net heat aboserbed `= 3P_(0)V_(0) - (5)/(2)P_(0)V_(0) = (P_(0)V_(0))/(2)`
`:. q_(net) = (P_(0)V_(0))/(2)`
Also, `(P_(0)V_(0))/(T_(1)) = (3P_(0)V_(0))/(T_(2)) rArr T_(2) = 3T_(1) = (3P_(0)V_(0))/(R )`