Q. The root mean square velocity of gas molecules at 27$℃$ is $1365\ ms^{- 1}$. The gas is
A) $O_{2}$
B) $He$
C) $N_{2}$
D) $CO_{2}$
Explanation
(b) Given that, $T = 27℃ = 300\ $K
$v_{rms} = 1365\ ms^{- 1}$
We know that
$v_{rms} = \sqrt{\frac{3RT}{M}}\ \ $
$or\ \ \ \ \ \ v_{rms}^{2} = \frac{3RT}{M}$
$or\ \ \ \ \ \ \ \ \ M = \frac{3RT}{v_{rms}^{2}}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ M = \frac{3 \times 8.31 \times 300}{1365 \times 1365}$ kg
$= \ \frac{3 \times 8.31 \times 300}{1365 \times 1365} \times 1000$ g=4 g
The molecular weight of helium is 4.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.