Q. An ideal gas expands isothermally from a volume $V_{1}$ to $V_{2}$ and then compressed to original volume $V_{1}$ adiabatically. Initial pressure is $p_{1}$ and final pressure is $p_{3}$. Total work done is $W$. Then
A) $p_{3} > p_{1};W > 0$
B) $p_{3} < p_{1};W < 0$
C) $p_{3} > p_{1};W < 0$
D) $p_{3} = p_{1};W = 0$
Explanation
(c) As slope of adiabatic process at a given state is more than the slope of isothermal process, therefore, in figure $AB$ is an isotherm and $BC$ is an adiabat
[image:12_image52.emf]
In going from $A$ to $B$, volume is increasing
$\therefore\ W_{AB} =$positive
In going from $B$ to $C$ volume is decreasing
$\therefore W_{BC} =$negative
As work done is area under $p - V$ graph, therefore,
$\left| W_{BC} \right| > |W_{AB}|$
$\therefore\ W = W_{AB} + W_{BC} = Negative\ ie\ W < 0$
From the graph, it is clear that $p_{3} > p_{1}$. Choice (c) is correct
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.