An ideal gas is taken form the state `A(P,V)` to the state `B((P)/(2),2V)` along a straight line path in the `P-V` diagram.
Select the correct options
(i) The work done by the gas in the process A to B exceeds the work that would be done by it if the system were taken from A to B along the isotherm
(ii) In the `T-V` diagram, the path AB becomes part of a hyperbola
(iii) In the `P-T` diagram, the path AB becomes part of a hyperbola
(iv) In going from A to B, the temperature T of the first increases to maximum value and then decreases

Options

A. (i),(iii)
B. (ii),(iii)
C. (iii),(iv)
D. (i),(ii),(iv)

Answer

D

Explanation

(i) The area below the process along `ArarrB` exeefs the isotherm between A and B.
(ii) The equation of straight line AB is
`P=-(a)/(nR)V^2+(b)/(nR)V`
T v/s V graph is parabola
(ii) is O.K.
(iii) `P=-aV+b`
`P=-a(nRT)/(P)+b`
`P^2=-anRT+bP`
`T=(P^2)/(anR)+bP`
T v/s P graph is parabola (iii) is wrong.
(iv) `T=-(a)/(nR)V^2+(b)/(nR)V`
for T to be maximum
`(dT)/(dV)=-(a)/(nR).2V+(b)/(nR)=0`
`V=(b)/(2a)`
`T_(max)=-(a)/(nR)((b)/(2a))^2+(b)/(nR).(b)/(2a)`
`=(b^2)/(4nRa)`
(iv) is O.K.