Kinetic Theory

JEE Main level, single correct, medium

40 of 168

Question 40 of 168

The root mean square speed of smoke particles of mass 5×10−17kg5\times {10}^{-17}kg in their Brownian motion in air at NTP is approximately.
[Given k=1.38×10−23JK−1k=1.38\times {10}^{-23}J{K}^{-1}]

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Answer: (C) 15mms−115mm{s}^{-1}

According to Kinetic theory,
Average kinetic energy of a gas molecule =12mvrms2=32kT=\dfrac{1}{2}m{v}_{rms}^{2}=\dfrac{3}{2}kT
Thus, vrms=3kTm{v}_{rms}=\sqrt{\dfrac{3kT}{m}}
At NTP, T=293KT=293K
Putting the values, we have
=3×1.38×10−23×2935×10−17=\sqrt{\dfrac{3\times 1.38\times {10}^{-23}\times 293}{5\times {10}^{-17}}}
≈15mms−1\approx 15mm{s}^{-1}

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