An ideal gas is taken from the state A (P, V) to the state B `(P//2, 2 V)` along a st. line path as shown in Fig. Select the correct statement from the following:

Options

A. work done by the gas in going from A to B exceed the work done in going from A to D under isothermal conditions,
B. in the T - V diagram, part AB would become a parabola
C. in the T - V diagram, part AB would become a hyperbola
D. in going from A to D the Temp. Of gas first increases to a max value and then decreases

Answer

A, B, D

Explanation

Isothermal curve from A to B will be parabolic with lesser area under the curve than the area under st. Line AB. Therefore, work done by the gas in going straight from A to B is more.
`:.` Choice (a) is correct.
If `P_(0), V_(0)` be the intercepts of curve on P and V axes, then its eqn. is obtained from = `y =mx + c`
i.e., `P =(P_0)/(V_0) V+P_(0) or (RT)/(V) = (P_(0)V)/(V_0) +P_(0)`
or, `T = (P_0)/(V_(0)R) V^(2)+(P_(0)V)/(R)`
Which is the eqn. of a parabola. hence T-V curve is parabolic. therefore Choice (b) is correct. Also `(P//2)(2V) = PV` = const. i.e., process is isothermal.