Q. When the temperature of a gas is raised from $27℃$ to $90℃,$ the percentage increase in the $r.m.s.$ velocity of the molecules will be
A) 10%
B) 15%
C) 20%
D) 17.5%
Explanation
(a) $v_{rms} = \sqrt{\frac{3RT}{M}} \Rightarrow \frac{v_{2}}{v_{1}} = \sqrt{\frac{T_{2}}{T_{1}}} = \sqrt{\frac{(273 + 90)}{(273 + 27)}} = 1.1$
% increase $= \left( \frac{v_{2}}{v_{1}} - 1 \right) \times 100 = 0.1 \times 100 = 10\%$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.