At a temperature of 300 K,the kinetic energy | vedclass

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Question

(D) The average kinetic energy of a gas molecule is given by $E = \frac{3}{2} k_B T$,which implies $E \propto T$.Given $T_1 = 300 \ K$ and $T_2 = 600

Vedclass · Accepted Answer

$12.42 	imes 10^{-21} \ J, \ 684 \ m/s$

Vedclass · Answer

(D) The average kinetic energy of a gas molecule is given by $E = \frac{3}{2} k_B T$,which implies $E \propto T$.Given $T_1 = 300 \ K$ and $T_2 = 600 \ K$,the new kinetic energy $E_2 = E_1 	imes (T_2 / T_1) = 6.21 	imes 10^{-21} 	imes (600 / 300) = 12.42 	imes 10^{-21} \ J$.The $rms$ speed is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$,which implies $v_{rms} \propto \sqrt{T}$.Given $v_{rms,1} = 484 \ m/s$,the new $rms$ speed $v_{rms,2} = v_{rms,1} 	imes \sqrt{T_2 / T_1} = 484 	imes \sqrt{600 / 300} = 484 	imes \sqrt{2} \approx 484 	imes 1.414 = 684.37 \ m/s \approx 684 \ m/s$.Thus,the correct option is $D$.

At a temperature of $300 \ K$,the kinetic energy of $O_2$ is $6.21 \times 10^{-21} \ J$ and its $rms$ speed is $484 \ m/s$. What will be the kinetic energy and $rms$ speed at $600 \ K$?

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