Question

Question: The mean free path of nitrogen molecules at 0.0⁰C and 1atmis$\mathbf{0}\mathbf{.}\mathbf{80}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathit{c}\mathit{m}$. At this temperature and pressure there are$\mathbf{2}\mathbf{.}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{19}}\mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{c}\mathit{u}\mathit{l}\mathit{e}\mathit{s}\mathbf{/}\mathit{c}{\mathit{m}}^{\mathbf{3}}$. What is the molecular diameter?

Short answer

Answer

The molecular diameter of nitrogen molecule is . 0.32 nm

Step-by-step solution

Given data

• Atmospheric density

$$\begin{gathered} \frac{N}{V}=2.7\times {10}^{19}\frac{\text{molecule}}{{\text{cm}}^{\text{3}}} \\ =2.7\times {10}^{25}\frac{\text{molecule}}{{\text{m}}^{\text{3}}} \end{gathered}$$

• Temperature;

$T=0.0°\text{C}$

• Mean free path

$$\begin{gathered} \lambda =0.80\times {10}^{-5}\text{cm} \\ =0.80\times {10}^{-7}\text{m} \end{gathered}$$

• Pressure

$P=1.0\text{atm}$

Understanding the concept

The expression for the mean free path is given by,

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

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

Calculate the molecular diameter

The mean free path is given by

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

On rearranging the equation in terms of d

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

Substitute $0.80\times {10}^{-7}\text{mforλ}$and $2.7\times {10}^{25}\frac{\text{molecule}}{{\text{m}}^{\text{3}}}for\frac{\lambda }{v}$ into the above equation

$$\begin{gathered} \text{d}=\sqrt{\frac{1}{\sqrt{2}\left(3.142\right)\left(0.80\times {10}^{-7}\right)\left(2.7\times {10}^{25}\right)}} \\ \text{d}=3.22\times {10}^{-10}\text{m} \\ =0.32\text{nm} \end{gathered}$$

Therefore the molecular diameter of nitrogen molecule is 0.32 nm .