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Question

(a) Calculate the most probable speed, the mean speed, and the mean relative speed of CO2\mathrm{CO}_{2} molecules at 20C.20^{\circ} \mathrm{C}. (b) Calculate the most probable speed, the mean speed, and the mean relative speed of H2\mathrm{H}_{2} molecules at 20C.20^{\circ} \mathrm{C}.

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CO2\mathrm{CO}_{2} molecules

temperature of 20C20^{\circ} \mathrm{C} vmpv_{\mathrm{mp}} - most probable speed

vmp=(2RTMCO2)1/2v_{\mathrm{mp}}=\left(\frac{2RT}{M_{\mathrm{CO}_{2}}}\right)^{1/2}

and here MCO2M_{\mathrm{CO}_{2}} means molar mass

RR means gas constant

Molar mass for CO2\mathrm{CO}_{2} is:

MCo2=C+2(O)=12.011g/mol+(215.9994g/mol)=44.010g/mol\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:

vmp=(2(8.3145JK1mol1)(273+20)K44.010g/mol1kg1000g)1/2=333ms1\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 VmeanV_{\mathrm{mean}} whose expression is:

v mean =(8RTπMCO2)1/2v_{\text { mean }}=\left(\frac{8 R T}{\pi M_{\mathrm{CO}_{2}}}\right)^{1 / 2}

And in our case it is equal to:

vmean=(8(8.3145JK1mol1)(273+20)K(22/7)44.010g/mol1kg1000g)1/2=375ms1\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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