Estimate the average distance between air molecules at $0.0^{\circ} \mathrm{C}$ and 1.00 atm.

First, we need to convert 0.0°C to Kelvin and 1.00 atm to Pascal. To convert from Celsius to Kelvin: \(K = ^{\circ}C + 273.15\) So the temperature in Kelvin is: \(0.0 + 273.15 = 273.15\) K. To convert from atm to Pascal: \(Pa = atm × 101325\) So the pressure in Pascal is: \(1.00 × 101325 = 101325\) Pa.

The ideal gas law states: \(PV = nRT\) Where: - P is pressure (in Pa) - V is volume (in m³) - n is the amount of substance (in moles) - R is the ideal gas constant (8.314 J/(mol·K)) - T is the temperature (in K) We want to find the density (\(\rho\)) of air, which is the mass (m) per unit volume (V): \(\rho = \frac{m}{V}\). To do this, we need the molar mass (M) of air and the amount of substance (n) of air. The molar mass of air is approximately \(M = 29 g/mol ≈ 0.029 kg/mol\). The mass (m) of air can be determined by multiplying the molar mass (M) by the amount of substance (n): \(m = nM\). We can rewrite the density equation as: \(\rho = \frac{nM}{V}\). Now let's substitute this equation into the ideal gas law, and solve for the density (\(\rho\)): \(PV = nRT \longrightarrow V = \frac{nRT}{P} \longrightarrow \rho = \frac{nM}{\frac{nRT}{P}} \longrightarrow \rho = \frac{MP}{RT}\) Using the values we found for temperature, pressure, and molar mass, we can calculate the density of air: \(\rho = \frac{(0.029 kg/mol)(101325 Pa)}{(8.314 J/(mol·K))(273.15 K)} ≈ 1.29 kg/m³\)

Now that we know the density of air, we can estimate the average distance between air molecules. A simple model for estimating the average distance assumes that the air molecules are equally spaced in a perfect cubic arrangement. Hence, we can calculate the volume of such a cube, with each molecule at one corner, by dividing the total volume by the number of molecules: \(volume\;of\;cube = \frac{1}{(density)(molecules\;per\;kg)}\) Where \(molecules\;per\;kg\) can be calculated as follows: \( molecules\;per\;kg = \frac{N_A}{molar\;mass}\), where \(N_A = 6.022 × 10^{23} particles/mol\) (Avogadro's number) \(molecules\;per\;kg = \frac{6.022 × 10^{23} mol^{-1}}{0.029 kg/mol} ≈ 2.08 × 10^{25} molecules/kg\) Now, calculating the volume of the cubic cell: \(volume\;of\;cube = \frac{1}{(1.29 kg/m³)(2.08 × 10^{25} molecules/kg)} ≈ 3.01 × 10^{-26} m³\) Finally, to find the average distance between air molecules, we take the cubic root of the volume: \(average\;distance = \sqrt[3]{volume\;of\;cube} = \sqrt[3]{3.01 × 10^{-26} m³} ≈ 3.1 × 10^{-9} m\) So, the estimated average distance between air molecules at \(0.0^{\circ}C\) and 1.00 atm is approximately \(3.1 × 10^{-9}\) meters.