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  • Find out the temperature at which the rms speed of nitrogen molecules will le equal to the escape velocity from the earth's gravity. Given, mans of a nitrogen atom<img alt="=23.24 \times 10^{-2

Find out the temperature at which the rms speed of nitrogen molecules will le equal to the escape velocity from the earth's gravity. Given, mans of a nitrogen atom=23.24 \times 10^{-24} \mathrm{~g}; average radius of the earth=6390 \mathrm{~km} is =980 \mathrm{Cm} \cdot \mathrm{S}^{-2}; Boltzmann constant=1.37 \times 10^{-16} \mathrm{erg}^{\circ} \mathrm{C}^{-1}

Option: 1

1.42


Option: 2

1.42 \times 10


Option: 3

1.42 \times 10^{5} \mathrm{~K}


Option: 4

1.41 \times 10


Answers (1)

best_answer

Speed of the molecule =\sqrt{\frac{3 R T}{M}}

escape velocity=\sqrt{2 g R_{1}} where R_{1}=is radius of the earth

According to question,

\sqrt{\frac{3 R T}{M}}=\sqrt{2 g R_{1}} \quad \text { or, } T=\frac{2}{3} \frac{g^{M} R_{1}}{R}

NowR=N K and M=M N

where M = Mass of a nitrogen molecules.

=2 \times\left(23.24 \times 10^{-24}\right) g

N= number of nitrogen molecules.

So, \begin{aligned} T & =\frac{2}{3} \cdot \frac{g m N R_{1}}{N k}=\frac{2}{3} \cdot \frac{g m R_{1}}{k} \\ & =\frac{2}{3} \times \frac{980 \times\left(2 \times 23.24 \times 10^{-24}\right) \times\left(6390 \times 10^{5}\right)}{1.37 \times 10^{-16}} \\ & =1.42 \times 10^{5} \mathrm{k} \end{aligned}

 

 

 

 

Posted by

SANGALDEEP SINGH

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