The root mean squar speed nitrogen |MotionStraighJEEMAIN2023

JEE MAIN 2023

11-4-2023

Question

The root mean square speed of molecules of nitrogen gas at $27^{\circ} \mathrm{C}$ is approximately: (Given mass of a nitrogen molecule $=4.6 \times 10^{-26} \mathrm{~kg}$ and take Boltzmann constant $\mathrm{kB}=1.4 \times 10^{-23} \mathrm{JK}^{-1}$ )

Select the correct option:

$523 \mathrm{~m} / \mathrm{s}$

$1260 \mathrm{~m} / \mathrm{s}$

$91 \mathrm{~m} / \mathrm{s}$

$27.4 \mathrm{~m} / \mathrm{s}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

$V_{r m s}=\sqrt{\frac{3 k_B T}{m}}=\sqrt{\frac{3 \times 1.4 \times 10^{-23} \times 300}{4.6 \times 10^{-26}}}=523 \mathrm{~m} / \mathrm{s}$

Question Tags

JEE Main

Chemistry