Question

Class 11 Physics
Question 49 of 50
1.5 pc
The root mean square speed of a nitrogen
molecule at 300 Kis
2672m/s
2.672m/s​

Answer 1

Answer:

$\boxed{\mathfrak{ v_{rms} = 517 \: m {s}^{ - 1} }}$

Explanation:

Temperature (T) = 300 K

Gas Constant (R) = 8.314 J/mol.K

Molar mass of $\rm N_2$ (M) = $\rm 28 \times 10^{-3}$ kg/mol

Root mean square speed:

$\boxed{ \bold{v_{rms} = \sqrt{\dfrac{3RT}{M}}}}$

By substituting value in the formula we get:

$\rm \implies v_{rms} = \sqrt{ \dfrac{3 \times 8.314 \times 300}{28 \times {10}^{ - 3} } } \\ \\ \rm \implies v_{rms} = \sqrt{26.72 \times {10}^{ 4} } \\ \\ \rm \implies v_{rms} = 5.17 \times 10^2 \\ \\ \rm \implies v_{rms} = 517 \: m {s}^{ - 1}$