Two moles of a gas `(gamma=5//3)` are initially at temperature `27^@C` and occupy a volume of 20 times. The gas is first expanded at constant pressure until the volume is doubled. Then, it is subjected to an adiabatic change until the temperature returns to its initial value.
(a) Sketch the process on a p-V diagram.
(b) What are the final volume and pressure of the gas?
(c) What is the work done by the gas?

Explanation

(a) The process is shown in figure.
During the part ab, the pressure is constant.
We have
`p_a V_a / T_a = p_b V_b / T_b`
or, T_b = V_b / V_a /T_a = 2Ta = 600 K.
During the part bc, the gas is a adiabatically returned to
The temperature T_a. The point a and the point c are on
the same isotherm. Thus, We draw an adibatic curve from b and an isotherm from a and look for the point
of intersection c. That is the final state. (fig.)
(b) From the isotherm ac,
`p_a V_a = ((p_c) (V_c))`
and from the adibatic curve bc,
`((p_b) (V_b^gamma)) (p_c)(V_b^gamma)`
or, p_a (2V_a)gamma = (p_c) (V_c^gamma)`.
Dividing (ii) by (i),
`(2gamma) (V_a ) gamma-1 = (V_c)gamma-1`
or , `(V_c )=( 2^gamma / (gamma-1) V_a) = (4sqrt(2) V_a )= 113 litres`.
From (i), `(p_c) = ((p_a) (V_a)) / (V_c) = (nRT_a) / (V_c)`
`=( 2 mol xx (8.3 J K^(-1) mol^(-1) xx (300 K)) / (113 xx 10^(-3) m^(-3))`
= 4.4 xx 10^(-4) Pa.
(c) Work done by the gas in the part ab
= `(P_a (V_b) - (V_a))`
` ((P_b) (V_b) - (p_a) (V_a))`
`nRT^(2) - nRT^(1)
= ` 2 mol xx (8.3 J K^(-1) mol ^(-1) xx (600 K - 300 K )`
`= 4980 J`.
The work done in the adiabatic part bc
`=((p_b) (V_b) -(p_c) (V_c)) / (gamma -1)
(`nR((T_2)- (T_1)))`
`4980 J / 5/ 3 -1 = 7470 J`.
The net work done by the gas
= `4980 J + 7470 J = 12450 J`.