Question

The collisional cross section of \(\mathrm{N}_{2}\) is \(0.43 \mathrm{nm}^{2}\). What is the diffusion coefficient of \(\mathrm{N}_{2}\) at a pressure of 1 atm and a temperature of \(298 \mathrm{K} ?\)

Short answer

The diffusion coefficient of \(\mathrm{N}_{2}\) at a pressure of 1 atm and a temperature of \(298 \mathrm{K}\) is approximately \(1.72 \times 10^{-5} \mathrm{m}^{2}/\mathrm{s}\).

Step-by-step solution

Convert collisional cross section to \(\mathrm{m}^{2}\)

To convert the cross section from \(\mathrm{nm}^{2}\) to \(\mathrm{m}^{2}\), multiply by \(10^{-18}\): \(\sigma = 0.43 \mathrm{nm}^{2} \times 10^{-18} \mathrm{m}^{2/nm}^{2} = 4.3 \times 10^{-21} \mathrm{m}^{2}\)

Calculate the concentration of the gas using the ideal gas law

We will use the ideal gas law equation to find the concentration of the gas: \(n = \frac{P}{kT}\) where: \(n\) is the concentration (in particles per cubic meter) \(P\) is the pressure in Pascals \(k\) is the Boltzmann constant (\(1.38 \times 10^{-23} \mathrm{J / K}\)) \(T\) is the temperature in Kelvin Note that we need to convert the pressure from atm to Pascal, multiply by \(101325 \mathrm{Pa/atm}\): \(P = 1 \mathrm{atm} \times 101325 \mathrm{Pa/atm} = 101325 \mathrm{Pa}\) Now, substitute the values: \(n = \frac{101325 \mathrm{Pa}}{(1.38 \times 10^{-23} \mathrm{J / K})(298\mathrm{K})} \approx 2.46 \times 10^{25} \mathrm{m}^{-3}\)

Calculate the mean molecular speed of \(\mathrm{N}_{2}\)

We will use the equation for the mean molecular speed: \(\bar{c} = \sqrt{\frac{8kT}{\pi m}}\) where: \(m\) is the mass of an \(\mathrm{N}_{2}\) molecule The molar mass of \(\mathrm{N}_{2}\) is \(28 \mathrm{g/mol}\). We need to convert this to mass in kg: \(m = \frac{28 \mathrm{g/mol}}{1000 \mathrm{g/kg} \times N_A} = \frac{28}{1000 \times 6.02 \times 10^{23}} \mathrm{kg} \approx 4.65 \times 10^{-26} \mathrm{kg}\) Now, calculate the mean molecular speed: \(\bar{c} = \sqrt{\frac{8(1.38 \times 10^{-23} \mathrm{J / K})(298 \mathrm{K})}{\pi(4.65 \times 10^{-26} \mathrm{kg})}} \approx 511 \mathrm{m/s}\)

Calculate the diffusion coefficient using the Chapman-Enskog equation

We will use the Chapman-Enskog formula for the diffusion coefficient: \(D = \frac{3}{16}\frac{\bar{c}\sigma}{n}\) Plug in the values: \(D = \frac{3}{16}\frac{(511\mathrm{m/s})(4.3 \times 10^{-21} \mathrm{m}^{2})}{2.46 \times 10^{25} \mathrm{m}^{-3}} \approx 1.72 \times 10^{-5} \mathrm{m}^{2}/\mathrm{s}\) The diffusion coefficient of \(\mathrm{N}_{2}\) at a pressure of 1 atm and a temperature of \(298 \mathrm{K}\) is approximately \(1.72 \times 10^{-5} \mathrm{m}^{2}/\mathrm{s}\).

Key concepts

Collisional Cross Section

The collisional cross section is a measure of how likely gas molecules are to collide with each other. It's a crucial factor when calculating properties like diffusion. In this exercise, the collisional cross section of nitrogen (\(\mathrm{N}_2\)) is given as \(0.43 \, \mathrm{nm}^2\). To use it in calculations, we convert it to square meters. This involves multiplying by \(10^{-18}\), resulting in \(4.3 \times 10^{-21} \, \mathrm{m}^2\).

This value represents the effective area one molecule presents as a target to others. It helps understand how molecules spread through a medium. Despite being small in scale, collisional cross section plays a significant role in gas behavior and properties.

Ideal Gas Law

The Ideal Gas Law is a fundamental principle for understanding gas behavior: \(PV = nRT\). In this context, we use the equation in a slightly different form to find the concentration of the gas: \(n = \frac{P}{kT}\), where \(k\) is the Boltzmann constant.

With the pressure and temperature provided, we first convert pressure from atmospheres to Pascals. Then, by substituting the values into the equation, we calculate concentration as approximately \(2.46 \, \times \, 10^{25} \, \mathrm{m}^{-3}\). This helps determine how densely the molecules are packed, which is important for understanding diffusion and other properties.

Mean Molecular Speed

Mean molecular speed offers insight into how fast molecules move in a gas. It's calculated using the formula: \(\bar{c} = \sqrt{\frac{8kT}{\pi m}}\). You need the mass of a single molecule, which is derived from its molar mass. For \(\mathrm{N}_2\), this conversion gives \(4.65 \times 10^{-26} \, \text{kg}\).

By substituting known values, the mean molecular speed is approximately \(511 \, \mathrm{m/s}\). This speed offers insights into both the energy and collision frequency of gas particles, contributing to our understanding of how diffusion occurs in gases.

Chapman-Enskog Equation

The Chapman-Enskog equation is an important tool for calculating diffusion coefficients of gases. The formula is \(D = \frac{3}{16}\frac{\bar{c}\sigma}{n}\). It combines mean molecular speed, collisional cross section, and concentration to give the diffusion coefficient \(D\).

By substituting all required values, the diffusion coefficient for \(\mathrm{N}_2\) at 1 atm and 298 K is found to be approximately \(1.72 \times 10^{-5} \, \mathrm{m}^{2}/\mathrm{s}\). This coefficient is vital for predicting how gases mix and spread in various environments, providing insight into molecular movement and reaction rates.