13.10 Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 $\mathit{?}$ . Take the radius of a nitrogen molecule to be roughly 1.0 Å. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of ${\mathit{N}}_2$ = 28.0 u).

13.10 Pressure inside the cylinder containing nitrogen, P = 2.0 atm = 2 $*1.013*{10}^5$ Pa

Temperature inside the cylinder, T = 17 $?=290K$

Radius of nitrogen molecule, r = 1.0 Å = 1 $*{10}^{-10}$ m

Diameter of nitrogen molecule, d = 2 $*{10}^{-10}$ m

Molecular mass of nitrogen molecule, M = 28 u = 28 g (assume) = 28 $*{10}^{-3}$ kg

The root means square speed of nitrogen is given by the relation

R is the universal gas constant = 8.314 J/mole/K

Hence 

The mean free path $\stackrel{}{l}$ is given by

$l ?=\frac{kT}{?2?d^{2}P}$ where k = Boltzmann constant = 1.38$*{10}^{-23}$ kg-$$$m^2{s}^{-2}{K}^{-1}$ 

$\stackrel{?}{l}=\frac{1.38\mathrm{}*{10}^{-23}*290}{?2?{(2\mathrm{}*{10}^{-10})}^2*2\mathrm{}*1.013*{10}^5}$= 1.11$*{10}^{-7}$  m

Collision frequency = $\frac{V_{rms}}{\stackrel{?}{l}}$ = $\frac{508.26}{1.11\mathrm{}*{10}^{-7}}$= 4.57 ${10}^9$ /s

Collision time, T = $\frac{d}{V_{rms}}$  =$\frac{2\mathrm{}*{10}^{-10}\mathrm{}}{508.26}$S= 3.93 $*{10}^{-13}$$\frac{\mathrm{}}{}$ s

T

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13.10 Pressure inside the cylinder containing nitrogen, P = 2.0 atm = 2 $*1.013*{10}^5$ Pa

Temperature inside the cylinder, T = 17 $?=290K$

Radius of nitrogen molecule, r = 1.0 Å = 1 $*{10}^{-10}$ m

Diameter of nitrogen molecule, d = 2 $*{10}^{-10}$ m

Molecular mass of nitrogen molecule, M = 28 u = 28 g (assume) = 28 $*{10}^{-3}$ kg

The root means square speed of nitrogen is given by the relation

R is the universal gas constant = 8.314 J/mole/K

Hence 

The mean free path $\stackrel{}{l}$ is given by

$l ?=\frac{kT}{?2?d^{2}P}$ where k = Boltzmann constant = 1.38$*{10}^{-23}$ kg-$$$m^2{s}^{-2}{K}^{-1}$ 

$\stackrel{?}{l}=\frac{1.38\mathrm{}*{10}^{-23}*290}{?2?{(2\mathrm{}*{10}^{-10})}^2*2\mathrm{}*1.013*{10}^5}$= 1.11$*{10}^{-7}$  m

Collision frequency = $\frac{V_{rms}}{\stackrel{?}{l}}$ = $\frac{508.26}{1.11\mathrm{}*{10}^{-7}}$= 4.57 ${10}^9$ /s

Collision time, T = $\frac{d}{V_{rms}}$  =$\frac{2\mathrm{}*{10}^{-10}\mathrm{}}{508.26}$S= 3.93 $*{10}^{-13}$$\frac{\mathrm{}}{}$ s

Time taken between successive collision T' = $\frac{\stackrel{?}{l}}{V_{rms}}$ = $\frac{1.11\mathrm{}*{10}^{-7}}{508.26}$s = 2.18$*{10}^{-10}$
 s$\frac{\mathrm{T}\mathrm{\text{'}}}{T}$ =$\frac{2.18\mathrm{}*{10}^{-10}}{3.93\mathrm{}*{10}^{-13}}$ = 555.71

Hence, the time taken between successive collisions is 556 times the time taken for a collision.

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<p><strong>13.10&nbsp;</strong>Pressure inside the cylinder containing nitrogen, P = 2.0 atm = 2&nbsp;<span title="Click to copy mathml"><math><mo>×</mo><mn>1.013</mn><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>5</mn></mrow></mrow></msup></math></span>&nbsp;Pa</p><p>Temperature inside the cylinder, T = 17&nbsp;<span title="Click to copy mathml"><math><mi>?</mi><mo>=</mo><mn>290</mn><mi></mi><mi>K</mi></math></span></p><p>Radius of nitrogen molecule, r = 1.0 Å = 1&nbsp;<span title="Click to copy mathml"><math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>10</mn></mrow></mrow></msup></math></span>&nbsp;m</p><p>Diameter of nitrogen molecule, d = 2&nbsp;<span title="Click to copy mathml"><math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>10</mn></mrow></mrow></msup></math></span>&nbsp;m</p><p>Molecular mass of nitrogen molecule, M = 28 u = 28 g (assume) = 28&nbsp;<span title="Click to copy mathml"><math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>3</mn></mrow></mrow></msup></math></span>&nbsp;kg</p><p>The root means square speed of nitrogen is given by the relation</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1746167230php7pKJAB_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1746167230php7pKJAB.jpeg" alt="" width="88" height="53"></picture></div></div><p>R is the universal gas constant = 8.314 J/mole/K</p><p>Hence&nbsp;<span title="Click to copy mathml"><br></span></p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1746167295phpRohz5k_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1746167295phpRohz5k.jpeg" alt="" width="" height=""></picture></div></div><p>The mean free path <span title="Click to copy mathml"><math><mover accent="true"><mrow><mrow><mi>l</mi></mrow></mrow></mover></math></span>&nbsp;is given by</p><p><span title="Click to copy mathml"><math><mi>l</mi><mi>&nbsp;</mi><mo>?</mo><mo>=</mo><mi></mi><mfrac><mrow><mrow><mi>k</mi><mi>T</mi></mrow></mrow><mrow><mrow><mo>?</mo><mn>2</mn><mi>?</mi><msup><mrow><mrow><mi>d</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><mi>P</mi></mrow></mrow></mfrac></math></span>&nbsp;where k = Boltzmann constant = 1.38<span title="Click to copy mathml"><math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>23</mn></mrow></mrow></msup></math></span> kg-<math><mo></mo></math><span title="Click to copy mathml"><math><msup><mrow><mrow><mi>m</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><msup><mrow><mrow><mi>s</mi></mrow></mrow><mrow><mrow><mo>-</mo><mn>2</mn></mrow></mrow></msup><msup><mrow><mrow><mi>K</mi></mrow></mrow><mrow><mrow><mo>-</mo><mn>1</mn></mrow></mrow></msup></math></span>&nbsp;</p><p><span title="Click to copy mathml"><math><mover accent="true"><mrow><mrow><mi>l</mi></mrow></mrow><mo>?</mo></mover><mo>=</mo><mi></mi><mfrac><mrow><mrow><mn>1.38</mn><mi mathvariant="normal"></mi><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>23</mn></mrow></mrow></msup><mo>×</mo><mn>290</mn></mrow></mrow><mrow><mrow><mo>?</mo><mn>2</mn><mi>?</mi><msup><mrow><mrow><mo>(</mo><mn>2</mn><mi mathvariant="normal"></mi><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>10</mn></mrow></mrow></msup><mo>)</mo></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><mo>×</mo><mn>2</mn><mi mathvariant="normal"></mi><mo>×</mo><mn>1.013</mn><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>5</mn></mrow></mrow></msup></mrow></mrow></mfrac></math></span>= 1.11<span title="Click to copy mathml"><math><mo>×</mo><mi></mi><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>7</mn></mrow></mrow></msup></math></span> &nbsp;m</p><p>Collision frequency&nbsp;=&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><msub><mrow><mrow><mi>V</mi></mrow></mrow><mrow><mrow><mi>r</mi><mi>m</mi><mi>s</mi></mrow></mrow></msub></mrow></mrow><mrow><mrow><mover accent="true"><mrow><mrow><mi>l</mi></mrow></mrow><mo>?</mo></mover></mrow></mrow></mfrac></math></span>&nbsp;= <span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>508.26</mn></mrow></mrow><mrow><mrow><mn>1.11</mn><mi mathvariant="normal"></mi><mo>×</mo><mi></mi><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>7</mn></mrow></mrow></msup></mrow></mrow></mfrac></math></span>= 4.57 <span title="Click to copy mathml"><math><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>9</mn></mrow></mrow></msup></math></span>&nbsp;/s</p><p>Collision time, T = <span title="Click to copy mathml"><math><mfrac><mrow><mrow><mi>d</mi></mrow></mrow><mrow><mrow><msub><mrow><mrow><mi>V</mi></mrow></mrow><mrow><mrow><mi>r</mi><mi>m</mi><mi>s</mi></mrow></mrow></msub></mrow></mrow></mfrac></math></span>&nbsp; =<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>2</mn><mi mathvariant="normal"></mi><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>10</mn></mrow></mrow></msup><mi mathvariant="normal"></mi></mrow></mrow><mrow><mrow><mn>508.26</mn></mrow></mrow></mfrac></math></span>S= 3.93 <span title="Click to copy mathml"><math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>13</mn></mrow></mrow></msup></math></span><math><mfrac><mrow><mrow><mi mathvariant="normal"></mi></mrow></mrow></mfrac></math>&nbsp;s</p><p>Time taken between successive collision T’ = <span title="Click to copy mathml"><math><mfrac><mrow><mrow><mover accent="true"><mrow><mrow><mi>l</mi></mrow></mrow><mo>?</mo></mover></mrow></mrow><mrow><mrow><msub><mrow><mrow><mi>V</mi></mrow></mrow><mrow><mrow><mi>r</mi><mi>m</mi><mi>s</mi></mrow></mrow></msub></mrow></mrow></mfrac></math></span> = <span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>1.11</mn><mi mathvariant="normal"></mi><mo>×</mo><mi></mi><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>7</mn></mrow></mrow></msup></mrow></mrow><mrow><mrow><mn>508.26</mn></mrow></mrow></mfrac></math></span>s = 2.18<span title="Click to copy mathml"><math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>10</mn></mrow></mrow></msup></math></span> <span title="Click to copy mathml"><br></span>&nbsp;s<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mi mathvariant="normal">T</mi><mi mathvariant="normal">’</mi></mrow></mrow><mrow><mrow><mi>T</mi></mrow></mrow></mfrac></math></span>&nbsp;=<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>2.18</mn><mi mathvariant="normal"></mi><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>10</mn></mrow></mrow></msup></mrow></mrow><mrow><mrow><mn>3.93</mn><mi mathvariant="normal"></mi><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>13</mn></mrow></mrow></msup></mrow></mrow></mfrac></math></span> = 555.71</p><p>Hence, the time taken between successive collisions is 556 times the time taken for a collision.</p>

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