Question

At$\mathbf{20}\mathbf{°}\mathit{C}$and $\mathbf{750}\mathbf{}\mathit{t}\mathit{o}\mathit{r}\mathit{r}$pressure, the mean free paths for argon gas (Ar) and nitrogen gas (${\mathit{N}}_{\mathbf{2}}$) ARE${\mathit{\lambda }}_{\mathbf{A}\mathbf{r}}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{}\mathit{c}\mathit{m}$and ${\mathit{\lambda }}_{\mathbf{N}\mathbf{2}}\mathbf{=}\mathbf{27}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathit{c}\mathit{m}$.

a) Find the ratio of the diameter of ${\mathit{A}}_{\mathbf{r}}$an atom to that of an ${\mathit{N}}_{\mathbf{2}}$ molecule.

b) What is the mean free path of argon at $\mathbf{20}\mathbf{°}\mathit{C}$and$\mathbf{150}\mathbf{}\mathit{t}\mathit{o}\mathit{r}\mathit{r}$?

c) What is the mean free path of argon at$\mathbf{-}\mathbf{40}\mathbf{°}\mathit{C}$and$\mathbf{750}\mathbf{}\mathit{t}\mathit{o}\mathit{r}\mathit{r}$?

Short answer

a) The ratio of the diameter of an $A_r$atom to that of an $N_2$molecule is $1.7$.

b) Mean free path of $A_r$at$20°\text{C}$ and $150\text{torr}$is $5.0\times {10}^{-5}\text{cm}$.

c) Mean free path of$A_r$ at $-40°\text{C}$and$750\text{torr}$ is$7.9\times {10}^{-6}\text{cm}$ .

Step-by-step solution

Given data

Pressure;

$\begin{array}{c}{p}_1=150\text{torr}\\ =19998.3\text{Pa}\end{array}$

$\begin{array}{c}{p}_2=750\text{torr}\\ =99991.8\text{Pa}\end{array}$

Mean free path of argon;

${\lambda }_{Ar}=9.9\times {10}^{-6}\text{cm}$

Mean free path of nitrogen;

${\lambda }_{N_2}=27.5\times {10}^{-6}\text{cm}$
Temperature;

$\begin{array}{c}{T}_1=20°\text{C}\\ =293\text{\hspace{0.17em}K}\end{array}$

$\begin{array}{c}{T}_2=-40°\text{C}\\ =233\text{\hspace{0.17em}K}\end{array}$

Understanding the concept

The expression for the mean free path is given by,

$\lambda =\frac{1}{\sqrt{2}\pi d^2\left(\frac{N}{V}\right)}$

Here$\lambda$ is the mean free path,$\frac{N}{V}$ is the molecule density in$molecule/c{m}^3$ .

The expression for the ideal gas equation is given by,

$pV=nRT$

Here $p$is the pressure, $V$is the volume, $n$is the number of moles, $R$is the universal gas constant,$T$ is the temperature.

The relation between universal gas constant and Boltzmann constant is given by,

$k=\frac{R}{N_A}$

Here$k$is the Boltzmann constant.

(a) Calculate the ratio of the diameter of an Ar atom to that of an N2 molecule 

The mean free path is given by

$\lambda =\frac{1}{\sqrt{2}\pi d^2\left(\frac{N}{V}\right)}$

At constant temperature and pressure,$V$is constant.

Hence,

$\begin{array}{c}\text{d}\propto \frac{1}{\sqrt{\lambda }}\\ \frac{{\text{d}}_{Ar}}{{\text{d}}_{N_2}}=\frac{\sqrt{{\lambda }_{N_2}}}{\sqrt{{\lambda }_{Ar}}}\end{array}$

$\begin{array}{c}\frac{{\text{d}}_{Ar}}{{\text{d}}_{N_2}}=\frac{\sqrt{27.5\times {10}^{-6}}}{\sqrt{9.9\times {10}^{-6}}}\\ =1.67\\ \approx 1.7\end{array}$

Therefore the ratio of the diameter of an $A_r$atom to that of an $N_2$molecule is$1.7$ .

(b) Calculate the mean free path of argon at  20°C  and  150 torr.

We know that

$\lambda =\frac{1}{\sqrt{2}\pi d^2\left(\frac{N}{V}\right)}$

According to the gas law,

$\text{pV}=\text{NkT}$

$\frac{N}{V}=\frac{p}{kT}$

Hence,

$\lambda =\frac{1}{\sqrt{2}\pi d^2\left(\frac{p}{kT}\right)}$

$\lambda \propto \frac{T}{p}$

$\frac{{\lambda }_1}{{\lambda }_2}=\frac{T_1}{T_2}\times \frac{p_2}{p_1}$

Since,${\lambda }_{Ar}=9.9\times {10}^{-6}\text{cm}$at$\text{T}=293\text{K}$ and $\text{p}=750\text{torr}$.

Mean free path of$A_r$at$20°\text{C}$and$150\text{torr}$ .

$\begin{array}{c}\frac{{\lambda }_1}{9.9\times {10}^{-6}}=\frac{293}{293}\times \frac{750}{150}\\ {\lambda }_1=49.5\times {10}^{-6}\text{cm}\\ \approx 5.0\times {10}^{-5}\text{cm}\end{array}$

Therefore the mean free path of $A_r$at $20°\text{C}$and $150\text{torr}$is $5.0\times {10}^{-5}\text{cm}$.

(c) Calculate the mean free path of argon at -40°C and750 torr

Similarly, mean free path of$A_r$ at$233\text{K}$and$750\text{torr}$ .

$\begin{array}{c}\frac{{\lambda }_1}{9.9\times {10}^{-6}}=\frac{233}{293}\times \frac{750}{750}\\ {\lambda }_1=7.87\times {10}^{-6}\text{cm}\\ \approx 7.9\times {10}^{-6}\text{cm}\end{array}$

Therefore the mean free path of $A_r$at$-40°\text{C}$and $750\text{torr}$is$7.9\times {10}^{-6}\text{cm}$ .