Q. At what temperature is the kinetic energy of a gas molecule double that of its value of $27℃$
A) $54℃$
B) $300\ K$
C) $327℃$
D) $108℃$
Explanation
(c) $\frac{E_{1}}{E_{2}} = \frac{T_{1}}{T_{2}} \Rightarrow \frac{E}{2E} = \frac{(273 + 27)}{T_{2}} \Rightarrow T_{2} = 600K = 327℃$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.