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  • Can someone explain - Behaviour of Perfect Gas and Kinetic theory - NEET

At what temperature will the rms speed of
oxygen molecules become just sufficient for escaping from the Earth's atmosphere ?
(Given :
Mass of oxygen molecule (in) =. 2.76 x 10-26 kg
Boltzmann's constant kB = 1.38 x 10^{-23} J K-1)

 

  • Option 1)

    5.016 X 104 K

  • Option 2)

    8.360 X 104 K

  • Option 3)

    2.508 X 104 K

  • Option 4)

    1.254 X 104 K

 

Answers (1)

best_answer

As we have learned

Root mean square velocity -

V_{rms}= sqrt{frac{3RT}{M}}

= sqrt{frac{3P}{ ho }}
 

- wherein

R = Universal gas constant

M = molar mass

P = pressure due to gas

ho = density

 

v_{escape}=v_{rms}= \sqrt{\left ( \frac{3KT}{M} \right )}= 11.2 km/s

T= \frac{M}{3K}*(11.2*10^{3})^{2}

= \frac{2.76*10^{-26}}{3*1.38*10^{-23}}*125.44*10^{6}=8.36*10^{4} K 

 

 

 

 

 


Option 1)

5.016 X 104 K

Option 2)

8.360 X 104 K

Option 3)

2.508 X 104 K

Option 4)

1.254 X 104 K

Posted by

Avinash

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