Problem — Root-mean-square speed of nitrogen

Problem

Calculate the root-mean-square speed of nitrogen molecules ($m_\text{mol} = 28\cdot 10^{-3}$ kg/mol) at $T = 300$ K.

Solution

I use the macroscopic form of the root-mean-square speed, with the molar mass and the gas constant $R = 8{,}31$ J/(mol·K): $v_\text{rms} = \sqrt{\frac{3RT}{m_\text{mol}}}$ I substitute the numerical values: $v_\text{rms} = \sqrt{\frac{3\cdot 8{,}31\cdot 300}{28\cdot 10^{-3}}}$ $v_\text{rms} = \sqrt{\frac{7479}{0{,}028}} = \sqrt{2{,}67\cdot 10^{5}}$ $\ev{v_\text{rms} \approx 517\ \text{m/s}}$

Links

Topics: Kinetic theory of gases Concepts: Root-mean-square speed · Temperature Skills: Symbolic setup