The temperature at which the root mean square | vedclass

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Question

(A) The root mean square velocity of gas molecules is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$,where $R = k_B N_A$ is the universal gas constant,$T$

Vedclass · Accepted Answer

$800 \, K$

Vedclass · Answer

(A) The root mean square velocity of gas molecules is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$,where $R = k_B N_A$ is the universal gas constant,$T$ is the temperature,and $M$ is the molar mass of hydrogen $(2 	imes 10^{-3} \, kg/mol)$.The escape velocity from the Earth is given by $v_e = \sqrt{2gR_e}$.Equating the two: $\sqrt{\frac{3RT}{M}} = \sqrt{2gR_e}$.Squaring both sides: $\frac{3RT}{M} = 2gR_e$.Solving for $T$: $T = \frac{2gR_e M}{3R}$.Given $R = k_B N_A \approx 8.314 \, J/(mol \cdot K)$.Substituting the values: $T = \frac{2 	imes 10 	imes 6.4 	imes 10^6 	imes 2 	imes 10^{-3}}{3 	imes 8.314} = \frac{25600}{24.942} \approx 1026 \, K$.Among the given options,the value closest to $1026 \, K$ is $800 \, K$ (Option $A$).

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