In a sample of chlorine gas at 300 K,the ave | vedclass

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Question

(D) The average kinetic energy $(E)$ of a gas molecule is directly proportional to the absolute temperature $(T)$: $E \propto T$.Given $E_1 = 6.21 	i

Vedclass · Accepted Answer

$12.42 	imes 10^{-21} \ J, \ 459.6 \ m/s$

Vedclass · Answer

(D) The average kinetic energy $(E)$ of a gas molecule is directly proportional to the absolute temperature $(T)$: $E \propto T$.Given $E_1 = 6.21 	imes 10^{-21} \ J$ at $T_1 = 300 \ K$. At $T_2 = 600 \ K$,the new kinetic energy $E_2$ is:$E_2 = E_1 	imes (T_2 / T_1) = 6.21 	imes 10^{-21} 	imes (600 / 300) = 12.42 	imes 10^{-21} \ J$.The root mean square speed $(
u_{rms})$ is directly proportional to the square root of the absolute temperature: $
u_{rms} \propto \sqrt{T}$.Given $(
u_{rms})_1 = 325 \ m/s$ at $T_1 = 300 \ K$. At $T_2 = 600 \ K$,the new speed $(
u_{rms})_2$ is:$(
u_{rms})_2 = (
u_{rms})_1 	imes \sqrt{T_2 / T_1} = 325 	imes \sqrt{600 / 300} = 325 	imes \sqrt{2} \approx 325 	imes 1.414 = 459.6 \ m/s$.

In a sample of chlorine gas at $300 \ K$,the average kinetic energy per molecule is $6.21 \times 10^{-21} \ J$ and the root mean square speed $\nu_{rms}$ is $325 \ m/s$. What will be the values of these quantities at $600 \ K$?

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