Determine the mole fraction of dry air at the surface of a lake whose temperature is \(15^{\circ} \mathrm{C}\). Take the atmospheric pressure at lake level to be \(100 \mathrm{kPa}\).

Dry air is generally considered to be a composition of certain gases with the following approximate percentages by volume: Nitrogen (N2) - 78%, Oxygen (O2) - 21%, Argon (Ar) - 1%, and trace amounts of other gases. In this problem, we will focus on the major components (N2, O2, and Ar) and ignore the trace amounts of other gases.

Given the total atmospheric pressure at the lake surface is 100 kPa, we can calculate the partial pressures of the major constituents of dry air as follows:$$ P_{N2} = 0.78 * 100 \, \text{kPa} = 78 \, \text{kPa} \\ P_{O2} = 0.21 * 100 \, \text{kPa} = 21 \, \text{kPa} \\ P_{Ar} = 0.01 * 100 \, \text{kPa} = 1 \, \text{kPa} $$

We will use the Ideal Gas Law to convert the partial pressures of each gas to moles. The Ideal Gas Law is given by: $$ PV = nRT $$ where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin. We can rearrange the Ideal Gas Law to solve for n: $$ n = \frac{PV}{RT} $$ First, convert the temperature to Kelvin: $$ T = 15 + 273.15 = 288.15 \, \text{K} $$ Next, use the ideal gas constant R = 8.314 J/(mol·K), and assume a volume of 1 m³ since we want to find the mole fraction and not the absolute number of moles (the volume will cancel out). Calculate the moles for each gas:$$ n_{N2} = \frac{78 \, \text{kPa} \cdot 1 \, \text{m}^3}{8.314 \, \text{J/(mol·K)} \cdot 288.15 \, \text{K}} \\ n_{O2} = \frac{21 \, \text{kPa} \cdot 1 \, \text{m}^3}{8.314 \, \text{J/(mol·K)} \cdot 288.15 \, \text{K}} \\ n_{Ar} = \frac{1 \, \text{kPa} \cdot 1 \, \text{m}^3}{8.314 \, \text{J/(mol·K)} \cdot 288.15 \, \text{K}} $$

Add up the moles of each constituent gas to find the total moles of dry air:$$ n_\text{total} = n_{N2} + n_{O2} + n_{Ar} $$

Calculate the mole fraction of each constituent gas by dividing each component's moles by the total moles:$$ \text{mole fraction}_{N2} = \frac{n_{N2}}{n_\text{total}} \\ \text{mole fraction}_{O2} = \frac{n_{O2}}{n_\text{total}} \\ \text{mole fraction}_{Ar} = \frac{n_{Ar}}{n_\text{total}} $$ Since the mole fractions sum up to 1, we can say the mole fraction of dry air is equal to the sum of the mole fractions of its major components:$$ \text{mole fraction}_\text{dry air} = \text{mole fraction}_{N2} + \text{mole fraction}_{O2} + \text{mole fraction}_{Ar} $$