At what temperature in K will the r.m.s. spee | vedclass

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(D) The escape velocity from the Earth is $v_{e} = 11.2 \, km/s = 11.2 	imes 10^{3} \, m/s$.The $r.m.s.$ speed of a gas molecule is given by $v_{rms}

Vedclass · Accepted Answer

$10063$

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(D) The escape velocity from the Earth is $v_{e} = 11.2 \, km/s = 11.2 	imes 10^{3} \, m/s$.The $r.m.s.$ speed of a gas molecule is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.Equating $v_{rms}$ to the escape velocity $v_{e}$:$v_{e} = \sqrt{\frac{3RT}{M}}$Squaring both sides:$v_{e}^{2} = \frac{3RT}{M}$Solving for $T$:$T = \frac{M v_{e}^{2}}{3R}$Given for hydrogen $(H_{2})$,the molar mass $M = 2 	imes 10^{-3} \, kg/mol$ and the universal gas constant $R = 8.31 \, J/(mol \cdot K)$.Substituting the values:$T = \frac{(2 	imes 10^{-3}) 	imes (11.2 	imes 10^{3})^{2}}{3 	imes 8.31}$$T = \frac{2 	imes 10^{-3} 	imes 125.44 	imes 10^{6}}{24.93}$$T = \frac{250.88 	imes 10^{3}}{24.93} \approx 10063 \, K$.

At what temperature in $K$ will the $r.m.s.$ speed of a hydrogen molecule be equal to the escape velocity from the Earth?

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