Correct Answer - A::B::D
Figure, shows the straight line path along with the corresponding isothermal path. Since the work done by the gas is equal to area under the curve (such as shown in the figure by the shaded portion for the isothermal path ), it is obvious that the gas does more work along the straight line path as compared with that for the isothermal path.
As the volume is increased from `V` to `2V`, the difference of pressure between the straight line path and isothermal path intially increases and then decreases after attaining a maximum value. The same trend is observed in the case of temperature `(P prop T, :. V` is constant )
Now, the slope of straight line path is
`m=(P-P//2)/(V-2V)=(P)/(2V)`
or ` P-2 Vm`
Putting this in the ideal gas equation,
`PV=nRT`
or `[ -2Vm]V=nRT`
`V^(2)=-(nR)/( 2 m ) T`
`V^(2)= k T`
Which is the equation of a parabola.
Similarly, eliminating `V` from ideal gas equation, we get
`P[-(P)/(2m)]=nRT`
or `P^(2)=` (constant) `T`
Which is again an equation of a parabola.