## Related questions with answers

(a) Calculate the most probable speed, the mean speed, and the mean relative speed of $\mathrm{CO}_{2}$ molecules at $20^{\circ} \mathrm{C}.$ (b) Calculate the most probable speed, the mean speed, and the mean relative speed of $\mathrm{H}_{2}$ molecules at $20^{\circ} \mathrm{C}.$

Solution

Verified$\mathrm{CO}_{2}$ molecules

temperature of $20^{\circ} \mathrm{C}$ $v_{\mathrm{mp}}$ - most probable speed

$v_{\mathrm{mp}}=\left(\frac{2RT}{M_{\mathrm{CO}_{2}}}\right)^{1/2}$

and here $M_{\mathrm{CO}_{2}}$ means molar mass

$R$ means gas constant

Molar mass for $\mathrm{CO}_{2}$ is:

$\begin{align*} M_{\mathrm{Co}_{2}}&=\mathrm{C}+2 \cdot(\mathrm{O})\\ &=12.011 \mathrm{g} / \mathrm{mol}+(2 \cdot 15.9994 \mathrm{g} / \mathrm{mol})\\ &=44.010 \mathrm{g} / \mathrm{mol}\\ \end{align*}$

Most probable speed is:

$\begin{align*} v_{\mathrm{mp}}&=\left(\frac{2 \cdot\left(8.3145 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\right) \cdot(273+20) \mathrm{K}}{44.010 \mathrm{g} / \mathrm{mol} \cdot \frac{1 \mathrm{kg}}{1000 \mathrm{g}}}\right)^{1 / 2}\\ &=333 \mathrm{ms}^{-1}\\ \end{align*}$

Next step is the mean speed $V_{\mathrm{mean}}$ whose expression is:

$v_{\text { mean }}=\left(\frac{8 R T}{\pi M_{\mathrm{CO}_{2}}}\right)^{1 / 2}$

And in our case it is equal to:

$\begin{align*} v_{\mathrm{mean}}&=\left(\frac{8 \cdot\left(8.3145 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\right) \cdot(273+20) \mathrm{K}}{(22 / 7) \cdot 44.010 \mathrm{g} / \mathrm{mol} \cdot \frac{1 \mathrm{kg}}{1000 \mathrm{g}}}\right)^{1 / 2}\\ &=375 \mathrm{ms}^{-1}\\ \end{align*}$

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