If the kinetic energy of O_2 gas is 4.0 kJ | vedclass

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(C) The kinetic energy $(KE)$ of an ideal gas is given by $KE = \frac{3}{2} RT$. The root mean square speed $(v_{rms})$ is given by $v_{rms} = \sqrt{\

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$5.0 	imes 10^4$

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(C) The kinetic energy $(KE)$ of an ideal gas is given by $KE = \frac{3}{2} RT$. The root mean square speed $(v_{rms})$ is given by $v_{rms} = \sqrt{\frac{3 RT}{M}}$. Substituting $\frac{3}{2} RT = KE$,we get $v_{rms} = \sqrt{\frac{2 KE}{M}}$. Given $KE = 4.0 \ kJ \ mol^{-1} = 4000 \ J \ mol^{-1}$ and molar mass $M$ of $O_2 = 32 	imes 10^{-3} \ kg \ mol^{-1}$. $v_{rms} = \sqrt{\frac{2 	imes 4000 \ J \ mol^{-1}}{32 	imes 10^{-3} \ kg \ mol^{-1}}} = \sqrt{\frac{8000}{32 	imes 10^{-3}}} = \sqrt{250000} = 500 \ m \ s^{-1}$. Converting to $cm \ s^{-1}$: $500 \ m \ s^{-1} 	imes 100 \ cm \ m^{-1} = 5.0 	imes 10^4 \ cm \ s^{-1}$.

If the kinetic energy of $O_2$ gas is $4.0 \ kJ \ mol^{-1}$,its $RMS$ speed in $cm \ s^{-1}$ is

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