Question

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.

Answer 1

Correct Answer - (i) `P_(0)V_(0) (ii) (5)/(2)P_(0)V_(0), 3P_(0)V_(0) (iii) (1)/(2)P_(0)V_(0) (iv) (25)/(8) (P_(0)V_(0))/(R)`
(a)`ABCA` is a clockwise cyclic process.

`therefore` Work done by the gas
W = +Area pf triangle `ABC`
`W = (1)/(2)` (base)(height) =`(1)/(2)(2V_(0)-V_(0))(3P_(0)-P_(0)) = P_(0)V_(0)`
(b) No. of moles `n=1` and gas is monoatomic , `C_(V) = (3)/(2)R "and" C_(P) = (5)/(2)R implies (C_(V))/(R) = (3)/(2)"and" (C_(P))/(R) = (5)/(2)`
(i) Heat rejected in path `CA`
`therefore Q_(CA) = C_(P)DeltaT = C_(P)(T_(f)-T_(l) = C_(P)((P_(f)V_(f)/(R) - (P_(l)V_(l)/(R)) = (C_(P))/(R)(P_(f) - V_(f) - P_(l)V_(l))`
Substituting the values `Q_(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).`
(ii) Heat absorbed in path `AB`: `therefore Q_(AB) = C_(V)DeltaT = C_(V)(T_(f) - T_(i))`
=`C_(V)((P_(f)V_(f))/(R) - (P_(i)V_(i))/(R)) = (C_(V))/(R)(P_(f)V_(f) - P_(i)V_(i))`
=`(3)/(2)(P_(f)V_(f) - P_(i)V_(i)) = (3)/(2)(3P_(0)V_(0)-P_(0)V_(0)) = 3P_(0)V_(0)`
`therefore` Heat absorbed in the process `AB "is" 3P_(0)V_(0)`
(c) Let `Q_(BC)` be the heat absorbed in the process `BC` Total heat absorbed
`Q = Q_(CA) + Q_(AB) + Q_(BC)`
`Q = (-(5)/(2)P_(0)V_(0)) + (3P_(0)V_(0)) + Q_(BC)`
`Q = Q_(BC) + (P_(0)V_(0))/(2)`
Change in internal energy `DeltaU =0`
`Q=W " " therefore Q_(BC) + (P_(0)V_(0))/(2) = P_(0)V_(0)`
`therefore Q_(BC) = (P_(0)V_(0))/(2)`
`therefore` Heat absorbed in the process `BC` is `(P_(0)V_(0))/(2)`
(d) Maximum temperature of the gas will some where 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) therefore 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)`
`RT = -((2P_(0))/(V_(0)))V^(2) + 5P_(0)V`
`implies T = (1)/(R)[5P_(0)V - (2P_(0))/(V_(0))V^(2)] .... (i)`
For `T` to be maximum
`(dT)/(dV) = 0 implies 5P_(0) - (4P_(0))/(V_(0)) , V= 0 implies V = (5V_(0))/(4)`
i.e., at `V = (5V_(0))/(4)` (on line BC), temperature of the gas is maximum
From Equation (i) this maximum temperature will be `T_(max) = (1)/(R)[5P_(0)((5V_(0))/(4)) - (2P_(0))/(V_(0))((5V_(0))/(4))^(2)] = (25)/(8)(P_(0)V_(0))/(R)`