QJEEM23Phaa3c: JEE Main 2023 Physics - Class 11 JEEM23Phaa3c Solution: The root mean square speed of molecules of nitrogen gas at is…

To find the root mean square speed of molecules of nitrogen gas, we can use the formula:

$v_{rms} = \sqrt{\frac{3k_BT}{m}}$

where $v_{rms}$ is the root mean square speed, $k_B$ is the Boltzmann constant, $T$ is the temperature in Kelvin, and $m$ is the mass of a nitrogen molecule.

First, we need to convert the temperature from Celsius to Kelvin:

$T = 27^{\circ}\mathrm{C} + 273 = 300\,\mathrm{K}$

Now, substitute the given values of $k_B$ , $T$ , and $m$ into the formula:

$v_{rms} = \sqrt{\frac{3(1.4 \times 10^{-23}\, \mathrm{JK}^{-1})(300\,\mathrm{K})}{4.6 \times 10^{-26}\,\mathrm{kg}}}$

Simplify and calculate the root mean square speed:

$v_{rms} = 523\,\mathrm{m/s}$

The root mean square speed of molecules of nitrogen gas at $27^{\circ}\mathrm{C}$ is 523 m/s.