91Ó°ÊÓ

The gas is Oxygen and the temperature associated with the gas is room temperature.

Assume that the room temperature is $27°\mathrm{C}\left(300\mathrm{K}\right)$.

The formula to calculate the most probable speed is given by

role="math" localid="1646968958095" $V_{\mathrm{mp}}=\sqrt{\frac{2RT}{M}}.......................\left(1\right)$

Here, $V_{\mathrm{mp}}$is the most probable speed, $R$is the universal gas constant, $T$is the absolute temperature of the gas and $M$is the molecular weight of Oxygen.

Substitute role="math" localid="1646969251456" $8.31\mathrm{J}/\mathrm{mol}·\mathrm{K}$for $R$, $300\mathrm{K}$for $T$and role="math" localid="1646969000674" $0.032\mathrm{kg}/\mathrm{mol}$for $M$into equation (1) to calculate the required most probable speed.

role="math" localid="1646969641134" $\begin{array}{rcl}{V}_{\mathrm{mp}}& =& \sqrt{\frac{2×8.31\mathrm{J}/\mathrm{mol}·\mathrm{K}×300\mathrm{K}}{0.032\mathrm{kg}/\mathrm{mol}}}\\ & ≈& 395\mathrm{m}/\mathrm{s}\end{array}$

The formula to calculate the average speed of Oxygen molecules is given by

$V_{\mathrm{avg}}=\sqrt{\frac{8RT}{\mathrm{Ï€}M}}...................\left(2\right)$

Here, $V_{\mathrm{avg}}$is the average speed.

Substitute the values of the parameters as stated in Step 2 into equation (2) to calculate the required average speed.

$\begin{array}{rcl}{V}_{\mathrm{avg}}& =& \sqrt{\frac{8×8.31\mathrm{J}/\mathrm{mol}·\mathrm{K}×300\mathrm{K}}{\mathrm{π}×0.032\mathrm{kg}/\mathrm{mol}}}\\ & ≈& 445\mathrm{m}/\mathrm{s}\end{array}$

The formula to calculate the rms speed of the Oxygen molecules is given by

$V_{\mathrm{rms}}=\sqrt{\frac{3RT}{M}}...............................\left(3\right)$

Here, $V_{\mathrm{rms}}$is the required rms speed.

Substitute the values of the parameters as stated in Step 2 into equation (3) to calculate the required average speed.

$\begin{array}{rcl}{V}_{\mathrm{rms}}& =& \sqrt{\frac{3×8.31\mathrm{J}/\mathrm{mol}·\mathrm{K}×300\mathrm{K}}{0.032\mathrm{kg}/\mathrm{mol}}}\\ & ≈& 483\mathrm{m}/\mathrm{s}\end{array}$