Problem — Temperature from v_rms

Problem

For the molecules of a gas with $M = 32$ g/mol a root-mean-square speed of $480$ m/s is measured. What is the temperature of the gas? Identify the likely gas.

Solution

I start from the relation $v_\text{rms} = \sqrt{\dfrac{3RT}{M}}$ and invert it for $T$: $T = \frac{M\,v_\text{rms}^{2}}{3R}$ I convert the molar mass: $M = 32\ \text{g/mol} = 0{,}032\ \text{kg/mol}$. $T = \frac{0{,}032 \cdot 480^{2}}{3 \cdot 8{,}31}$ $\ev{T \approx 296\ \text{K}}$ The molar mass $M = 32$ g/mol and a temperature close to room temperature indicate that the gas is likely oxygen $\text{O}_2$.

Links

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