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Physics / Sears and Zemansky's University Physics with Modern Physics 13 / Chapter 18 / Problem 40E

Sears and Zemansky's University Physics with Modern Physics | 13th Edition | ISBN: 9780321696861 | Authors: Hugh D. Young; Roger A. Freedman; A. Lewis Ford

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Answer: Smoke particles in the air typically have masses

Chapter 18, Problem 40E

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QUESTION:

Problem 40E

Smoke particles in the air typically have masses of the order of 10-10 kg. The Brownian motion (rapid, irregular movement) of these particles, resulting from collisions with air molecules, can be observed with a microscope. (a) Find the root-mean-square speed of Brownian motion for a particle with a mass of 3.00 X 10-16 kg in air at 300 K. (b) Would the root-mean-square speed be different if the particle were in hydrogen gas at the same temperature? Explain.

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QUESTION:

Problem 40E

Smoke particles in the air typically have masses of the order of 10-10 kg. The Brownian motion (rapid, irregular movement) of these particles, resulting from collisions with air molecules, can be observed with a microscope. (a) Find the root-mean-square speed of Brownian motion for a particle with a mass of 3.00 X 10-16 kg in air at 300 K. (b) Would the root-mean-square speed be different if the particle were in hydrogen gas at the same temperature? Explain.

ANSWER:

Solution 40E

Step1 :

We need to find the root mean square speed of the particle

This is given using

$1/2m({{v}_{rms})}^{2}=3/2\ K\times{}T\$

Here

$K\ -\ Boltzmann\ constant=1.3806\times{}1{0}^{-23\ }\$

$T-\ absolute\ temperature\ =\ 300\ K\$

$v-\ root\ mean\ square\ velocity\$

$m-\ mass\ of\ particle=\ 3.00\times{}1{0}^{-16}\ kg\$

We need to find $root\ mean\ square\ velocity$ hence rearrange the equation

We get

${v}_{rms}=\ \sqrt{\frac{3K\times{}T}{m\ }}$

Substituting the values

${v}_{rms}=\ \sqrt{\frac{3\times{}1.3806\times{}1{0}^{-23\ }\times{}300\ K}{3.00\times{}1{0}^{-16}\ kg}}$

${v}_{rms}=\ \sqrt{\frac{1.2425\times{}1{0}^{-20\ }}{3.00\times{}1{0}^{-16}\ kg}}$

${v}_{rms}=\ \sqrt{4.1419\times{}1{0}^{-5}\ }$

${v}_{rms}=\ 6.4357\times{}1{0}^{-3}m/s=\ 0.64\ cm/s$

Hence the root mean square velocity is obtained as $\ 0.64\ cm/s$

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