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

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The work done by the gas in the process $A$ to $B$ exceeds the work that would be done by it if the system were taken from $A$ to $B$ along the isotherm.

In the $T - V$ diagram, the path $A B$ becomes a part of a parabola

In the $P - T$ diagram, the path $A B$ becomes a part of a hyperbola

In going from $A$ to $B$ , the temperature $T$ of the gas first increases to a maximum value and then decreases

answer is A, B, D.

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Detailed Solution

The area under isotherm is smaller than the area under the straight-line process.

Equation of the given process is , p = $\frac{3 P}{2} - \frac{P}{2 V} v$ ………..(1) and equation of state is , pv =nRT' ………….(2)

Combining (1) and (2) we get

nRT' = $\frac{3 P}{2} v - \frac{P}{2 V} v^{2} a n d \frac{P n R}{2 V} T' = \frac{3 P}{2} p - p^{2} . T h e s e t w o e q u a t i o n s r e p r e s e n t p a r a b o l a .$

Also from the diagram shown below , it is clear that deviation of the given process from the isothermal process is maximum at the point of the process AB .So temperature of the gas first increases then decreases.