One mole of an ideal monatomic gas is taken round the cyclic process ABCA as shown in figure. Calculate

(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 maximum temperature attained by the gas during the cycle.

Explanation

a. The work done by the gas is equal to the area under the closed curve. Thus, work done in cycle is
`W = (1)/(2) (2 V_(0) - V_(0)) (3 P_(0) - P_(0))`
`= (1)/(2) V_(0) xx 2 P_(0) = P_(0) V_(0)`
b. Heat rejected in path CA is given as
`Q_(CA) = nC_(P) Delta T = nC_(P) (T_(C ) - T_(A))`
`= 1 xx (5//2) R [(P_(0) 2 V_(0))/(R ) - (P_(0) V_(0))/(R )] = (5)/(2) P_(0) V_(0)`
(as n = 1 mole)
Heat absorbed in path AC is
`Q_(AC) = nC_(P) (T_(B) - T_(A))`
`= 1 xx (3)/(2) R xx [(3 P_(0) V_(0))/(R ) - (P_(0) V_(0))/(R )]= 3 P_(0) V_(0)`
For cycle ABC, we have
heat supplied = work done by the gas
or `- ((5)/(2)) P_(0) V_(0) + 3 P_(0) V_(0) + Q_(BC) = P_(0) V_(0)`
Heat supplied to path BC is given by
`Q_(BC) = P_(0) V_(0) + ((5)/(2)) P_(0) V_(0) - 3 P_(0) V_(0) = P_(0) V_(0) //2`
d. We know that `PV//T` = constant. So, when PV is maximum, T is also maximum. PV is maximum for part BC. Hence, temperature will be maximum between B and C.
Let equation of BC be
`P = kV + k'`
satisfying both the point B and C
Fro point C,
`3 P_(0) = V_(0) + k'`
For point C,
`P_(0) = k (2 V_(0)) + k'`
Solving these equations, we get
`k = - 2 (P_(0)//V_(0))` and `k' = 5 P_(0)`
So the equation for time BC is
`P = - 2 ((P_(0))/(V_(0))) xx V + 5 P_(0)`
or `((RT)/(V)) = - ((2 P_(0) V)/(V_(0))) + 5 P_(0)`
or `T = (P_(0))/(R ) [5 V - 2 ((v_(2))/(V_(0)))] = 0`
For maximum, `dT//dV = 0`
So `((dT)/(dV)) = ((P_(0))/(R )) [5 - ((4 V)/(V_(0)))] = 0`
Hence, `5 - ((4 V)/(V_(0))) = 0` or `5 V_(0) - 4 V = 0`
or `V = ((5)/(4)) V_(0)`
Substituting the value fo V from Eq. (ii) in Eq. (i) we get
`Tan_(max) = (P_(0))/(R) [5 xx ((5)/(4) V_(0)) - 2 ((5 V_(0))/(4))^(2) (1)/(V_(0))] = (P_(0))/(R ) [(25 V_(0))/(4) - (25 V_(0))/(8)]`
`= (P_(0))/(R ) xx (25 V_(0))/(8) = (25 P_(0) V_(0))/(8 R)`