An ideal gas is taken from the state A (pressure p, volume V) to the state B (pressure p/2, volume 2V) along a straight line path in the p-V diagram. Select the correct statement from the following

Options

A. The work done by the gas in the process `A` to `B` exceeds the work done that would be done by it if the system were taken from `A` to `B` along an isotherm.
B. In the `T-V` diagram, the path `AB` becomes a part of a parabola.
C. In the `P-T` diagram, the path `AB` becomes a part of hyperboal.
D. In going from `A` to `B`, the temperature `T` of the gas first increase to a maximum value and then decreases

Answer

A, B, D

Explanation

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.