At T(K),the u_{rms} of CO_2 is 412 ms^{-1}. | vedclass

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(A) The root mean square velocity $(u_{rms})$ is given by the formula $u_{rms} = \sqrt{\frac{3RT}{M}}$.Given $u_{rms} = 412 \ ms^{-1}$ and molar mass

Vedclass · Accepted Answer

$3.7343$

Vedclass · Answer

(A) The root mean square velocity $(u_{rms})$ is given by the formula $u_{rms} = \sqrt{\frac{3RT}{M}}$.Given $u_{rms} = 412 \ ms^{-1}$ and molar mass $M = 44 	imes 10^{-3} \ kg \ mol^{-1}$.Squaring both sides: $u_{rms}^2 = \frac{3RT}{M} \implies \frac{3}{2}RT = \frac{1}{2} M u_{rms}^2$.The kinetic energy $(KE)$ of $1 \ mol$ of an ideal gas is given by $KE = \frac{3}{2}RT = \frac{1}{2} M u_{rms}^2$.Substituting the values: $KE = \frac{1}{2} 	imes (44 	imes 10^{-3} \ kg \ mol^{-1}) 	imes (412 \ ms^{-1})^2$.$KE = 0.5 	imes 0.044 	imes 169744 \ J \ mol^{-1} = 3734.368 \ J \ mol^{-1}$.Converting to $kJ \ mol^{-1}$: $KE = \frac{3734.368}{1000} \approx 3.7344 \ kJ \ mol^{-1}$.Thus,the correct option is $A$.

At $T(K)$,the $u_{rms}$ of $CO_2$ is $412 \ ms^{-1}$. What is its kinetic energy (in $kJ \ mol^{-1}$) at the same temperature? $(CO_2 = 44 \ u)$.

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