The root mean square speed of molecules of nitrogen gas at 27°C is approximately : (Given mass of a nitrogen molecule = \(4.6 × 10^{–26}\) kg and take Boltzmann constant KB = \(1.4 × 10^{–23}\) JK–1 )

Question

Find net kinetic energy (maximum possible) associated with 20 diatomic molecules (Here k_B is Boltzmann constant and T is absolute temperature of diatomic gas).

Answer 1

The root mean square (rms) speed of gas molecules is given by the formula:

v_{\text{rms}} = \sqrt{\frac{3k_{B}T}{m}}

where v_{\text{rms}} is the root mean square speed, k_{B} is the Boltzmann constant, T is the absolute temperature in Kelvin, and m is the mass of one molecule of the gas.

Given:

Mass of a nitrogen molecule, m = 4.6 \times 10^{-26} kg
Boltzmann constant, k_{B} = 1.4 \times 10^{-23} JK^–1
Temperature, T = 27^{\circ}C = 27 + 273 = 300 \text{ K}

Substituting these values into the formula:

v_{\text{rms}} = \sqrt{\frac{3 \times 1.4 \times 10^{-23} \times 300}{4.6 \times 10^{-26}}}

Calculating inside the square root:

v_{\text{rms}} = \sqrt{\frac{1.26 \times 10^{-20}}{4.6 \times 10^{-26}}}

v_{\text{rms}} = \sqrt{2.7391 \times 10^{5}}

After calculating the square root:

v_{\text{rms}} \approx 523 \text{ m/s}

Thus, the root mean square speed of nitrogen molecules at 27°C is approximately 523 m/s.

This matches the given correct answer option.

Answer 2