Q. An ideal gas expands in volume from $1 \times 10^{- 3}\ m^{3}$ to $1 \times 10^{- 2}\ m^{3}$ at 300 K against a constant pressure of $1 \times 10^{5}\ Nm^{- 2}$. The work done is

A) -900 J
B) -900 kJ
C) 270 kJ
D) 900 kJ
Explanation

(a)
Work done due to change in volume against constant pressure,
$W = - p(V_{2} - V_{1})$
$= - 1 \times 10^{5}\ Nm^{- 2}\left( 1 \times 10^{- 2} - 1 \times 10^{- 3} \right)m^{3}$
$= - 900\ Nm = - 900\ J$ (1 Nm=1 J)

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.