Q. When $1\ g$ of water at $0℃$ and $1 \times 10^{5}\ N/m^{2}$ pressure is converted into ice of volume $1.091\ cm^{3}$, the external work done will be
A) $0.0091\ joule$
B) $0.0182\ joule$
C) $- 0.0091\ joule$
D) $- 0.0182\ joule$
Explanation
(a) It is an isothermal process. Hence work done
$= P(V_{2} - V_{1})$
$= 1 \times 10^{5} \times (1.091 - 1) \times 10^{- 6} = 0.0091\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.