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

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)`

Explanation

(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)`