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The value closest to the thermal velocity of a Helium atom at room temperature (300 K) in ms^{-1} is : [ k_{B} =1.4 × 10^{-23}J/K; m_{He} = 7 ×10^{-27} kg ]

  • Option 1)

    1.3 ×10^{^{4}}

  • Option 2)

    1.3 ×10^{^{3}}

  • Option 3)

    1.3 ×10^{5}

  • Option 4)

    1.3 ×10^{^{2}}

 

Answers (2)

best_answer

As we learnt

Root mean square velocity -

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

= \sqrt{\frac{3P}{\rho }}
 

- wherein

R = Universal gas constant

M = molar mass

P = pressure due to gas

\rho = density

 

 v = \sqrt{\frac{3kT}{m}}=\sqrt{\frac{3*0.2*10^{-23}*300}{4*10^{-27}}} 

v = \sqrt{1.8*10^{6}}=1.3*10^{3}m/s

 


Option 1)

1.3 ×10^{^{4}}

Option 2)

1.3 ×10^{^{3}}

Option 3)

1.3 ×10^{5}

Option 4)

1.3 ×10^{^{2}}

Posted by

Aadil

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