Chapter 8, Problem 119QRT

(a)

Interpretation Introduction

Interpretation:

Mean free path of Argon gas has to be calculated.

Concept Introduction:

Mean free path: It is the average distance travelled by a particle between two collisions. It can be calculated using the given formula.

  $\lambda \text{ = }\frac{1}{\sqrt{2}\pi N{\sigma }^2}$

Here, $\sigma$ is the diameter of the atom and N is the number of atoms (or molecules) per cubic centimetre.

Ideal gas Equation:

Any gas is described by using four terms namely pressure, volume, temperature and the amount of gas. Thus combining three laws namely Boyle’s, Charles’s Law and Avogadro’s Hypothesis the following equation could be obtained. It is referred as ideal gas equation.

  $\text{PV = nRT}$

Here,

  n is the moles of gas

    P is the Pressure

  V is the Volume

    T is the Temperature

    R is the gas constant

Answer to Problem 119QRT

Mean free path of Argon gas is $2.53\times {10}^{-5}\text{cm}\text{.}$

Explanation of Solution

Diameter of the atom is determined as follows,

  $\begin{array}{c}1{\text{ pm = 10}}^{-10}\text{ cm}\\ \\ \sigma =2r\text{ = 2}\times \text{91 pm }\\ \\ \text{= 182 pm}\times \frac{{\text{10}}^{-10}\text{ cm}}{1\text{ pm}}\\ \\ =\text{ 182}\times {\text{10}}^{-10}\text{ cm}\end{array}$

Volume occupied by $1\text{ mol}$ of an atom at STP is $22.4\text{}\text{ L}\text{. }$

Therefore, number of molecules of Argon gas is calculated as given below,

  $\begin{array}{c}\text{No}\text{. of molecules}=\frac{1\text{L}}{1000{\text{ cm}}^3}\left(\frac{1\text{mol}}{22.4\text{ L}}\right)\left(\frac{6.022\times {10}^{23}\text{Ar molecules }}{1\text{mol}}\right)\\ \\ =\text{ 2}\text{.69}\times {\text{10}}^{19}\text{ molecules}\end{array}$

Mean free path of Argon gas is calculated as follows,

  $\begin{array}{c}\lambda =\frac{1}{\sqrt{2}\pi \text{N}{\sigma }^2}\\ \\ =\frac{1}{\sqrt{2}\left(3.14\right)\left(2.69\times {10}^{19}{\text{cm}}^{-3}\right){\left(\text{182}\times {\text{10}}^{-10}\text{ cm}\right)}^2}\\ \\ =2.53\times {10}^{-5}\text{cm}\end{array}$

(b)

Interpretation Introduction

Interpretation:

Mean free path is how many times higher than the diameter of Argon atom has to be calculated.

Answer to Problem 119QRT

Mean free path is 1390 times higher than the diameter of Argon atom.

Explanation of Solution

Diameter of the atom is determined as follows,

  $\begin{array}{c}1{\text{ pm = 10}}^{-10}\text{ cm}\\ \\ \sigma =2r\text{ = 2}\times \text{91 pm }\\ \\ \text{= 182 pm}\times \frac{{\text{10}}^{-10}\text{ cm}}{1\text{ pm}}\\ \\ =\text{ 182}\times {\text{10}}^{-10}\text{ cm}\end{array}$

Mean free path of Argon gas is $2.53\times {10}^{-5}\text{cm}\text{.}$

Mean free path can be compared with the diameter as shown below,

  $\frac{2.53\times {10}^{-5}\text{cm}}{\text{182}\times {\text{10}}^{-10}\text{ cm}}=\text{ 1390}$

Therefore, mean free path is 1390 times higher than the diameter of Argon atom.

(c)

Interpretation Introduction

Interpretation:

Pressure required to change the mean free path of Argon has to be calculated.

Concept Introduction:

Mean free path: It is the average distance travelled by a particle between two collisions. It can be calculated using the given formula.

  $\lambda \text{ = }\frac{1}{\sqrt{2}\pi N{\sigma }^2}$

Here, $\sigma$ is the diameter of the atom and N is the number of atoms (or molecules) per cubic centimetre.

Ideal gas Equation:

Any gas is described by using four terms namely pressure, volume, temperature and the amount of gas. Thus combining three laws namely Boyle’s, Charles’s Law and Avogadro’s Hypothesis the following equation could be obtained. It is referred as ideal gas equation.

  $\text{PV = nRT}$

Here,

  n is the moles of gas

    P is the Pressure

  V is the Volume

    T is the Temperature

    R is the gas constant

Answer to Problem 119QRT

Pressure required to change the mean free path of Argon is $1.0\times {10}^{-4}\text{atm}\text{.}$

Explanation of Solution

Diameter of the atom is determined as follows,

  $\begin{array}{c}1{\text{ pm = 10}}^{-10}\text{ cm}\\ \\ \sigma =2r\text{ = 2}\times \text{91 pm }\\ \\ \text{= 182 pm}\times \frac{{\text{10}}^{-10}\text{ cm}}{1\text{ pm}}\\ \\ =\text{ 182}\times {\text{10}}^{-10}\text{ cm}\end{array}$

Number of molecules of Argon gas present in $1\text{ cm}$ is calculated as follows,

  $\begin{array}{c}\lambda =\frac{1}{\sqrt{2}\pi \text{N}{\sigma }^2}\\ \\ 1\text{}\text{ cm}=\frac{1}{\sqrt{2}\left(3.14\right)\times N\times {\left(\text{182}\times {\text{10}}^{-10}\text{ cm}\right)}^2}\\ \\ N=\frac{1}{\sqrt{2}\left(3.14\right){\left(\text{182}\times {\text{10}}^{-10}\text{ cm}\right)}^2\left(1\text{ cm}\right)}\\ \\ =\text{ 6}\text{.8}\times {\text{10}}^{14}{\text{ molecules per cm}}^3\end{array}$

Ratio of moles to liters is calculated to obtain the pressure.

  $\begin{array}{c}\frac{\text{n}}{\text{V}}=\left(\frac{6.8\times {10}^{14}\text{ Ar molecules }}{1{\text{ cm}}^3}\right)\times \left(\frac{1\text{ mol}}{6.022\times {10}^{23}\text{ Ar molecules }}\right)\times \left(\frac{1000{\text{ cm}}^3}{1\text{L}}\right)\\ \\ =4.5\times {10}^{-6}\text{mol}/\text{L}\end{array}$

Pressure required to change the mean free path of Argon is calculated using Ideal Gas Law as follows,

  $\begin{array}{c}T\text{ = 0}°\text{C = 273 K}\\ \\ PV\text{ = nRT}\\ \\ \text{P}=\left(\frac{\text{n}}{\text{V}}\right)\text{RT}\\ \\ =\left(4.5\times {10}^{-6}\text{mol/L}\right)\left(0.0821\text{ L}\text{.atm}{\text{.mol}}^{-1}.K^{-1}\right)\left(273.15\text{K}\right)\\ \\ =1.0\times {10}^{-4}\text{atm}\end{array}$