Root mean square (rms) speed of O_2 is 500 m | vedclass

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(B) Given, root mean square $(rms)$ speed of $O_2$ is $500 \ m/s$ at a constant temperature. Root mean square speed is given by the expression $u_{rms

Vedclass · Accepted Answer

$2000 \ m/s$ and $4.0 \ kJ/mol$

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(B) Given, root mean square $(rms)$ speed of $O_2$ is $500 \ m/s$ at a constant temperature. Root mean square speed is given by the expression $u_{rms} = \sqrt{\frac{3RT}{Mw}}$. For $H_2$ gas, $u_{rms(H_2)} = \sqrt{\frac{3RT}{Mw_{H_2}}}$ where $Mw_{H_2} = 2 \ g/mol$. For $O_2$ gas, $u_{rms(O_2)} = \sqrt{\frac{3RT}{Mw_{O_2}}}$ where $Mw_{O_2} = 32 \ g/mol$. Taking the ratio: $\frac{u_{rms(H_2)}}{u_{rms(O_2)}} = \sqrt{\frac{Mw_{O_2}}{Mw_{H_2}}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$. Thus, $u_{rms(H_2)} = 4 	imes 500 \ m/s = 2000 \ m/s$. Average kinetic energy per mole is given by $KE_{avg} = \frac{3}{2} RT$. First, find $T$ from $u_{rms(O_2)} = \sqrt{\frac{3RT}{Mw_{O_2}}}$. $500 = \sqrt{\frac{3 	imes 8.33 	imes T}{0.032 \ kg/mol}}$. $250000 = \frac{24.99 	imes T}{0.032} \implies T = \frac{250000 	imes 0.032}{24.99} \approx 320 \ K$. $KE_{avg} = \frac{3}{2} 	imes 8.33 	imes 320 \approx 4000 \ J/mol = 4 \ kJ/mol$.

Root mean square $(rms)$ speed of $O_2$ is $500 \ m/s$ at a constant temperature. Calculate the $rms$ speed and the average kinetic energy of $H_2$ at the same temperature. (Consider, $R=8.33 \ J \ K^{-1} \ mol^{-1}$)

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