Question 269: Thermodynamics & KTG

Question

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

Accepted Answer

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

Answer

91 m/s

Answer

1260 m/s

Answer

27.4 m/s

Answer

523 m/s

Thermodynamics & KTG

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

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