For given mass of an ideal gas, consider the following diagrams
What is variation of volume in (a), pressure in (b) and temperature in (c).

Explanation

(a) Slope of incline line
`tantheta=(P)/(T)=(nR)/(V)`
In `1rarr2`, `theta` is decreasing or `tantheta` is decreasing
`(P)/(T)=(nR)/(V)` is decreasing hence volume is increasing
(b) Slope of inclined line
`tantheta=(V)/(T)=(nR)/(P)`
In `1rarr2`, `theta` is increasing or `tantheta` is increasing
`(V)/(T)=(nR)/(P)` is increasing hence pressure is decreasing.
(c) The slope of straight line is negative and intercept is positive.
Positive.
The equation of line `1rarr2` will be
`y=mx+CimpliesP=-aV+b` .(i)
`PV=nRTimpliesP=(nRT)/(V)` in (i)
`(nRT)/(V)-aV+b`
`T=-(a)/(nR)V^2+(b)/(nR)V`
The T v/s V diagram will be parabola, open downward
During expansion, temperature increases, reaching maximum value and then decreases.
T will be maximum.
When `(dT)/(dV)=0implies-(2aV)/(nR)+(b)/(nR)=0impliesV=(b)/(2a)`
`T_(max)=-(a)/(nR).(b^2)/(4a^2)+(b)/(nR).(b)/(2a)=(b^2)/(4nRa)`