One mole of an ideal monoatomatic gas is taken round the cylic process `ABCA` as shown Then

Options

A. the work doen 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.

Answer

A, B, C, D

Explanation

`ABCA` is a clockwise cyclic process.
`:.` Work done by the gas
`W =+` Area fo triangle `ABC`
`= (1)/(2) (base)(height)`
`= (1)/(2) (2V_(0)-V_(0)) (3P_(0)-P_(0))`
`W = P_(0)V_(0)`

(b) Number of moles `n = 1` and gas monoatomic, therefore
`C_(V) = (3//2) R` and `C_(P) =(5//2) R`
`rArr (C_(V))/(R) = (3)/(2)` and `(C_(P))/(R) = (5)/(2)`
Heat rejected in path `CA:` (process is isobaric)
`:. dQ_(CA) = C_(p)dT = C_(P) (T_(A)-T_(C))`
`= C_(P) ((P_(A)V_(A))/(R)-(P_(C)V_(C))/(R)) - (C_(P))/(R) (P_(A)V_(A)-P_(C)V_(C))`
Subsituting the values
`dQ_(CA) = (5)/(2) (P_(0)V_(0) - 2P_(0)V_(0)) =- (5)/(2)P_(0)V_(0)`
Therefore, heat rejected in the process `CA` is `(5)/(2)P_(0)V_(0)`
Heat absorbed in path `AB:` (process is isochoric)
`:. dQ_(AB) = C_(V)dT`.
`= C_(V) (T_(B) -T_(A))`
`= C_(V) ((P_(B)V_(B))/(R)-(P_(A)V_(A))/(R)) = (C_(V))/(R) (P_(B)V_(B)- P_(A)V_(A)) = (3)/(2) (P_(B)V_(B)-P_(A)V_(A))`
`= (3)/(2) (3P_(0)V_(0)-P_(0)V_(0))`
`dQ_(AB) = 3P_(0)V_(0)`
`:.` Heat absorbed in the process `Ab` is `3P_(0)V_(0)`
(c) Let `dQ_(BC)` be the heat absorbed in the process `BC`,
Total heat absorbed, `dQ = dQ_(CA) +dQ_(AB) +dQ_(BC)`
`dQ = (-(5)/(2)P_(0)V_(0)) +(3P_(0)V_(0)) +dQ_(BC)`
`dQ = dQ_(BC) +(P_(0)V_(0))/(2)`
Change in internal energy, `dU = 0`
`:. dQ = dW`
`:. dQ_(BC) +(P_(0)V_(0))/(2) = P_(0)V_(0)`
`:. dQ_(BC) =(P_(0)V_(0))/(2)`
`:.` Heat absorbed in the process `BC` is `(P_(0)V_(0))/(2)`
`:. BC = (P_(0)V_(0))/(2)`
(d) Maximum temperature of the gas will be somewhere between `B` and `C`. Line `BC` is a straight line. Therefore `P-V` equation for the process `BC` can be written as:
`P = - mV +c`
`(y = mx +c)`
Here `m = (2P_(0))/(V_(0))` and `C = 5P_(0)`
`P =- ((2P_(0))/(V_(0))) V +5P_(0)`
Multiplying the equation by `V`.
`PV =- ((2P_(0))/(V_(0))).V^(2) +5P_(0)V`
`(PV =RT for n = 1)`
`R =- ((2P_(0))/(V_(0))) V^(2) +5P_(0)V`
or `T = (1)/(R) [5P_(0)V-(2P_(0))/(V_(0))V^(2)] ...(1)`
For `T` to be maximum, `(dT)/(dV) = 0`
`5P_(0) -(4P_(0))/(V_(0)).V = 0 V = (5V_(0))/(4)`
i.e at `V = (5V_(0))/(4)`, (on line `BC)` temperature of the gas is maximum. From equation (1), this maximum temperature will be
`T_(max) = (1)/(R) [5P_(0)((5V_(0))/(4)) -((2P_(0))/(V_(0))) ((5V_(0))/(4))^(2)]`
`rArr T_(max) = (25)/(8) (P_(0)V_(0))/(R)`