If temperature and volume of an ideal gas is increased to twice its values, the initial pressure P becomes

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`.