Q. At what temperature the kinetic energy of gas molecule is half of the value at 27$℃$?
A) 13.5$℃$
B) 150$℃$
C) 75 K
D) $- 123℃$
Explanation
(d) Kinetic energy of a gas molecule
$\ E = \frac{3}{2}kT$
where $k$ is Boltzmann’s constant.
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ E \propto T$
or $\frac{E_{1}}{E_{2}} = \frac{T_{1}}{T_{2}}\ \ \ \ \ \ or\ \ \ \ \ \ \frac{E}{(\frac{E}{2})} = \frac{300}{T_{2}}$
or $T_{2} = 150$ K
$T_{2} = 150 - 273 = - 123℃$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.