Q. A gas at pressure $6 \times 10^{5}\ Nm^{- 2}$ and volume $1\ m^{3}$ and its pressure falls to $410^{5}\ Nm^{- 2}$, When its volume is $3m^{3}$. Given that the indicator diagram is a straight line, work done by the system is
A) $6 \times 10^{5}\ J$
B) $3 \times 10^{5}\ J$
C) $4 \times 10\ J$
D) $10 \times 10^{5}\ J$
Explanation
(a) [image:12_image55.emf]
Work done by the system
= area under $p - V$ diagram
= area of rectangle $BCDE$+ area of $\mathrm{\Delta}ABC$
$= 4 \times 10^{5} \times 2 + \frac{2 \times 10^{5} \times 2}{2}$
$W = 10 \times 10^{5}$ J
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.