Chapter 13, Problem 13.58P

Interpretation Introduction

Interpretation:

Pressure of ${\text{H}}_2$ molecule when the mean path length is $1\text{ μm}$, $1\text{ mm}$ and $1\text{ m}$ at $\text{20 }°\text{C}$ has to be calculated.

Concept Introduction:

Gaseous molecules frequently collide and follow a zig-zag pathway. The mean path length thus tranversed by gaseous particles is called the mean free path. Mathematical expression of mean free path is as follows:

  $l=\frac{RT}{\left(\pi \sqrt{2}\right)d^2{N}_{\text{A}}P}$

Here,

$d$ represents diameter of the molecule.

$N_{\text{A}}$ represents Avogadro’s number.

$P$ represents pressure.

$R$ represents gas constant.

$T$ represents temperature.

$l$ represents mean free path.

This expression can be rearranged to evaluate pressure as follows:

  $P=\frac{\rho RT}{M}$

Answer to Problem 13.58P

Pressure of ${\text{H}}_2$ molecule when the mean path length is $1\text{ μm}$, $1\text{ mm}$ and $1\text{ m}$ is $\text{12498}{\text{.4 J/m}}^3$, $\text{12}{\text{.4984 J/m}}^3$ and $0.01249845{\text{ J/m}}^3$ respectively.

Explanation of Solution

The formula to convert degree Celsius to kelvin is as follows:

  $T\left(\text{K}\right)=T\left(°\text{C}\right)+273\text{ K}$        (1)

Substitute $\text{20 }°\text{C}$ for $T\left(°\text{C}\right)$ in equation (1).

  $\begin{array}{c}T\left(\text{K}\right)=\text{20 }°\text{C}+273\text{ K}\\ =293\text{ K}\end{array}$

The formula to convert $\text{μm}$ to $\text{m}$ is as follows:

  $1\text{ μm}={10}^{-6}\text{ m}$

The formula to convert $\text{mm}$ to $\text{m}$ is as follows:

  $1\text{ mm}={10}^{-3}\text{ m}$

Refer to table 13.4 for diameter of molecules.

The molecular diameter of ${\text{H}}_2$ is $\text{270 pm}$.

The formula to convert $\text{pm}$ to $\text{m}$ is as follows:

  $1\text{ pm}={10}^{-12}\text{ m}$

Thus, $\text{270 pm}$ is converted to $\text{m}$ as follows:

  $\begin{array}{c}\text{Diameter}=\left(\text{270 pm}\right)\left(\frac{{10}^{-12}\text{ m}}{1\text{ pm}}\right)\\ =270\times {10}^{-12}\text{ m}\end{array}$

The formula to calculate pressure is as follows:

  $P=\frac{RT}{\left(\pi \sqrt{2}\right)d^2{N}_{\text{A}}l}$        (2)

Substitute $\text{8}\text{.3145 J/mol}\cdot \text{K}$ for $R$, $\text{293 K}$ for $T$ , $270\times {10}^{-12}\text{ m}$ for $d$, $6.022\times {10}^{23}{\text{ mol}}^{-1}$ for $N_{\text{A}}$ and ${10}^{-6}\text{ m}$ for $l$ in equation (2) evaluate pressure.

  $\begin{array}{c}P=\frac{\left(\text{8}\text{.3145 J/mol}\cdot \text{K}\right)\left(\text{293 K}\right)}{\left(4.43996\right){\left(270\times {10}^{-12}\text{ m}\right)}^2\left(6.022\times {10}^{23}{\text{ mol}}^{-1}\right)\left({10}^{-6}\right)}\\ =0.01249845\times {10}^6\\ =\text{12498}{\text{.4 J/m}}^3\end{array}$

Substitute $\text{8}\text{.3145 J/mol}\cdot \text{K}$ for $R$ , $\text{293 K}$ for $T$ , $270\times {10}^{-12}\text{ m}$ for $d$, $6.022\times {10}^{23}{\text{ mol}}^{-1}$ for $N_{\text{A}}$ and ${10}^{-3}\text{ m}$ for $l$ in equation (2) evaluate pressure.

  $\begin{array}{c}P=\frac{\left(\text{8}\text{.3145 J/mol}\cdot \text{K}\right)\left(\text{293 K}\right)}{\left(4.43996\right){\left(270\times {10}^{-12}\text{ m}\right)}^2\left(6.022\times {10}^{23}{\text{ mol}}^{-1}\right)\left({10}^{-3}\right)}\\ =0.01249845\times {10}^3\\ =\text{12}{\text{.4984 J/m}}^3\end{array}$

Substitute $\text{8}\text{.3145 J/mol}\cdot \text{K}$ for $R$ , $\text{293 K}$ for $T$ , $270\times {10}^{-12}\text{ m}$ for $d$, $6.022\times {10}^{23}{\text{ mol}}^{-1}$ for $N_{\text{A}}$ and $1\text{ m}$ for $l$ in equation (2) evaluate pressure.

  $\begin{array}{c}P=\frac{\left(\text{8}\text{.3145 J/mol}\cdot \text{K}\right)\left(\text{293 K}\right)}{\left(4.43996\right){\left(270\times {10}^{-12}\text{ m}\right)}^2\left(6.022\times {10}^{23}{\text{ mol}}^{-1}\right)\left(1\right)}\\ =0.01249845\times 1\\ =0.01249845{\text{ J/m}}^3\end{array}$