An ideal gas taken from state `A(P,V)` to the state `B(P//2,2V)` along a straight path in PV diagram. Select the correct option(s).

Options

A. In PT diagram path AB becomes a part of hyperbola.
B. Work done by the gas In the process A to B is more than the work done if the system were taken from A to B along an isotherm.
C. In `TV` diagram,t he path `AB` becomes a part of parabola.
D. in going from A to B the temperature T of the gas first increases to a maximum value and then decreases.

Answer

A, B, D

Explanation

Area under PV curve is work done.
Area under AB straight line is greater than area under isotherm therefore `W_("process")gtW_("isotherm")`
let `P=alphaV+beta` (from equation of straight line AB)
`PV-alphaV^(2)+betaV=nRT`
`impliesT=(1)/(nR)(-alphaV^(2)+betaV)` equation of parabola
`(d^(2)T)/(dV^(2))=-2alpha=-ve`
`TpropPV`
`(PV_(A)=(P.V)BimpliesT_(A)=T_(B)`
since `T_(A)=T_(B)` increases b/w A and B reaches a maximum and again decreases to reach the same temperature.