At what temperature would the root-mean-square speed (thermal speed)of oxygen molecules be 13.0 m/s? Assume that oxygen approximates an ideal gas.The mass of one O₂molecule is 5.312 × 10^-26 kg.The Boltzmann constant is 1.38 × 10^-23 J/K. A)0.217 K B)1800 K C)5410 K D)0.0666 K

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Quizplus · Accepted Answer

The root-mean-square speed (v_rms) of a gas molecule is given by the formula $v_{rms} = \sqrt{\frac{3kT}{m}}$, where $k$ is the Boltzmann constant, $T$ is the temperature in Kelvin, and $m$ is the mass of a molecule. Rearranging for $T$, we get $T = \frac{m(v_{rms})^2}{3k}$. Substituting the given values, $T = \frac{(5.312 \times 10^{-26} \, \text{kg}) \times (13.0 \, \text{m/s})^2}{3 \times (1.38 \times 10^{-23} \, \text{J/K})}$, which calculates to approximately 0.217 K.

At What Temperature Would the Root-Mean-Square Speed (Thermal Speed)of Oxygen