91影视

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;

${位}_{Ar}=9.9脳{10}^{鈭�6}\text{cm}$

Mean free path of nitrogen;

${位}_{N_2}=27.5脳{10}^{鈭�6}\text{cm}$
Temperature;

$\begin{array}{c}{T}_1=20掳\text{C}\\ =293\text{鈥块}\end{array}$

$\begin{array}{c}{T}_2=鈭�40掳\text{C}\\ =233\text{鈥块}\end{array}$

The expression for the mean free path is given by,

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

Here$位$ 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.

The mean free path is given by

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

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

Hence,

$\begin{array}{c}\text{d}鈭�\frac{1}{\sqrt{位}}\\ \frac{{\text{d}}_{Ar}}{{\text{d}}_{N_2}}=\frac{\sqrt{{位}_{N_2}}}{\sqrt{{位}_{Ar}}}\end{array}$

$\begin{array}{c}\frac{{\text{d}}_{Ar}}{{\text{d}}_{N_2}}=\frac{\sqrt{27.5脳{10}^{鈭�6}}}{\sqrt{9.9脳{10}^{鈭�6}}}\\ =1.67\\ 鈮�1.7\end{array}$

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

We know that

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

According to the gas law,

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

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

Hence,

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

$位鈭�\frac{T}{p}$

$\frac{{位}_1}{{位}_2}=\frac{T_1}{T_2}脳\frac{p_2}{p_1}$

Since,${位}_{Ar}=9.9脳{10}^{鈭�6}\text{cm}$at$\text{T}=293\text{K}$ and $\text{p}=750\text{torr}$.

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

$\begin{array}{c}\frac{{位}_1}{9.9脳{10}^{鈭�6}}=\frac{293}{293}脳\frac{750}{150}\\ {位}_1=49.5脳{10}^{鈭�6}\text{cm}\\ 鈮�5.0脳{10}^{鈭�5}\text{cm}\end{array}$

Therefore the mean free path of $A_r$at $20掳\text{C}$and $150\text{torr}$is $5.0脳{10}^{鈭�5}\text{cm}$.

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

$\begin{array}{c}\frac{{位}_1}{9.9脳{10}^{鈭�6}}=\frac{233}{293}脳\frac{750}{750}\\ {位}_1=7.87脳{10}^{鈭�6}\text{cm}\\ 鈮�7.9脳{10}^{鈭�6}\text{cm}\end{array}$

Therefore the mean free path of $A_r$at$鈭�40掳\text{C}$and $750\text{torr}$is$7.9脳{10}^{鈭�6}\text{cm}$ .