Question

In an interstellar gas cloud at$\mathbf{50}\mathbf{.0}\mathbf{\text{\hspace{0.17em}K}}$,the pressure is$\mathbf{1}\mathbf{.00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{}\mathbf{\text{Pa}}$. Assuming that the molecular diameters of the gases in the cloud are all$\mathbf{20}\mathbf{.0}\mathbf{\text{\hspace{0.17em}nm}}$, what is their mean free path?

Short answer

The gas cloud has a mean free path equal to$38.8\text{\hspace{0.17em}m}$

Step-by-step solution

The given data

1. The temperature of the interstellar gas cloud,$T=50.0\text{\hspace{0.17em}K}$$T=50.0\text{\hspace{0.17em}K}$
2. The pressure on the gas,$P=1.00\times {10}^{-8}\text{Pa}$
3. The molecular diameter of a gas in the cloud,$d=20\text{\hspace{0.17em}nmor}20\times {10}^{-9}\text{m}$

Understanding the concept of mean free path

The mean free path is defined as its average distance covered by the moving particle before undergoing any substantial change in direction or energy due to one or more collisions with the present particles in the system. As the density of the gas increases, the molecules come closer to each other, thus increasing the number of collisions taking place, and decreasing the mean free path. As per the ideal gas equation, the pressure is inversely proportional to the volume of the gas, thus less the volume; the more will be the pressure that implies a decrease in the mean free path.

The mean free path is given as, $\lambda =\frac{1}{\sqrt{2}\pi d^2\left(\frac{N}{V}\right)}$ (i)

where,is the number of molecules,is the diameter of the gas,is the volume of the gas.

The ideal gas equation, $PV=nRT$ (ii)

where, P is the pressure applied on the gas, V is the volume of the gas, is the number of moles of the gas,R is the gas constant, T is the temperature of the gas.

Calculation of the mean free path of the gas

Using the given data in equation (ii), the volume of the interstellar gas cloud can be given as follows:

$\begin{array}{c}V=\frac{nRT}{P}\\ =\frac{1\times \left(8.31{\text{m}}^3\text{.Pa/K.mol}\right)\times 50\text{K}}{1.00\times {10}^{-8}\text{Pa}}\\ =4.16\times {10}^{10}{\text{\hspace{0.17em}m}}^3\end{array}$

Number of molecules per unit volume will be can be given as:

$\begin{array}{c}\frac{N}{V}=\frac{n{N}_A}{V}(\because \text{Numberofmolucules=No.ofmoles}\times \text{Avogardo'snumber})\\ =\frac{1\times 6.02\times {10}^{23}\text{\hspace{0.17em}molecules}}{4.16\times {10}^{10}{\text{\hspace{0.17em}m}}^3}\\ =1.45\times {10}^{13}{\text{molecules/m}}^3\end{array}$

Now, substituting the given data and the above value in equation (i), the mean free path of the gas can be given as follows:

$\begin{array}{c}\lambda =\frac{1}{\sqrt{2}\pi \times {(20.0\times {10}^{-9}\text{m})}^2\times 1.45\times {10}^{13}{\text{molecules/m}}^3}\\ =38.8\text{\hspace{0.17em}m}\end{array}$

The mean free path of the gas cloud is $38.8\text{\hspace{0.17em}m}$.