Dry air whose molar analysis is \(78.1\) percent \(\mathrm{N}_{2}\), \(20.9\) percent \(\mathrm{O}_{2}\), and 1 percent Ar flows over a water body until it is saturated. If the pressure and temperature of air remain constant at \(1 \mathrm{~atm}\) and \(25^{\circ} \mathrm{C}\) during the process, determine (a) the molar analysis of the saturated air and \((b)\) the density of air before and after the process. What do you conclude from your results?

First, write down the given molar analysis of dry air: - 78.1% N₂ - 20.9% O₂ - 1% Ar Let's assume that we have a total of 100 moles of dry air. Then, the moles of each component will be as follows: - \(\mathrm{N}_{2}\): \(78.1\) moles - \(\mathrm{O}_{2}\): \(20.9\) moles - Ar: \(1\) mole Total moles of dry air: \(78.1 + 20.9 + 1 = 100\) moles

Now, the air flows over a water body until it is saturated. Under given conditions, refer to saturated vapor pressure tables to find the pressure of the water vapor, \(P_{\mathrm{H}_2\mathrm{O}} = 0.0313 \ \mathrm{atm}\). With the ideal gas law, we can calculate the added moles of water vapor using the equation: $$n_{\mathrm{H}_2\mathrm{O}} = \frac{P_{\mathrm{H}_2\mathrm{O}} \times V_{\mathrm{mixture}}}{RT},$$ where \(P_{\mathrm{H}_2\mathrm{O}} = 0.0313 \ \mathrm{atm}\), \(V_{\mathrm{mixture}}\) is the volume of the mixture, R is the ideal gas constant \(0.0821 \ \mathrm{\frac{atm \cdot L}{mol \cdot K}}\), and \(T = 25 + 273.15 = 298.15 \ \mathrm{K}\). Since we are considering the air under constant pressure and temperature, we can simplify the equation by dividing it by the total moles of dry air: $$\frac{n_{\mathrm{H}_2\mathrm{O}}}{n_{\mathrm{air}}} = \frac{P_{\mathrm{H}_2\mathrm{O}}}{P_{\mathrm{air}}}$$ From the given data, we know that the total pressure of the air, \(P_{\mathrm{air}} = 1 \ \mathrm{atm}\). Thus, we get: $$n_{\mathrm{H}_2\mathrm{O}} = 100 \times \frac{0.0313}{1} = 3.13 \ \mathrm{moles}$$ Now, to find the total moles of the saturated air: Total moles of saturated air = Total moles of dry air + Moles of water vapor = \(100 + 3.13 = 103.13 \ \mathrm{moles}\)

Now that we have the total moles of saturated air, we can find the mole percent for each component. Divide the moles of each component by the total moles of the saturated air and multiply by 100. - N₂: \(\frac{78.1}{103.13} \times 100 \approx 75.7 \%\) - O₂: \(\frac{20.9}{103.13} \times 100 \approx 20.3 \%\) - Ar: \(\frac{1}{103.13} \times 100 \approx 1 \%\) - H₂O: \(\frac{3.13}{103.13} \times 100 \approx 3 \%\) So, the molar analysis of the saturated air is approximately: - 75.7% N₂ - 20.3% O₂ - 1% Ar - 3% H₂O

To find the density of the air before and after the process, use the general equation for the density of an ideal gas: $$\rho = \frac{n \cdot M}{V} \times \frac{R \cdot T}{P}$$ where \(n\) is the number of moles, \(M\) is the molar mass, \(R\) is the ideal gas constant, \(T\) is the temperature in Kelvin, and \(P\) is the pressure. First, we need to find the molar mass of the mixture. For that, we can find the weighted average of the molar mass of each air component multiplied by its mole fraction. The molar masses are: \(28 \ \mathrm{g/mol}\) for \(\mathrm{N}_{2}\), \(32 \ \mathrm{g/mol}\) for \(\mathrm{O}_{2}\), and \(39.95 \ \mathrm{g/mol}\) for Ar. For dry air: Molar mass of dry air (M) = \(0.781 \times 28 + 0.209 \times 32 + 0.01 \times 39.95 \approx 28.8 \ \mathrm{g/mol}\) For saturated air: Molar mass of saturated air (M') = \(\frac{75.7}{100} \times 28 + \frac{20.3}{100} \times 32 + \frac{1}{100} \times 39.95 + \frac{3}{100} \times 18 \approx 29.1 \ \mathrm{g/mol}\) Now we can calculate the density of the air before and after the process. Before (dry air): $$\rho = \frac{28.8}{29.1} \times \frac{0.0821 \cdot 298.15}{1} \approx 1.18 \ \mathrm{kg/m^3}$$ After (saturated air): $$\rho' = \frac{29.1}{29.1} \times \frac{0.0821 \times 298.15}{1} \approx 1.19 \ \mathrm{kg/m^3}$$

From our calculations, we can conclude the following: 1. The molar analysis of the saturated air is around 75.7% N₂, 20.3% O₂, 1% Ar, and 3% H₂O (by mole). This means that air becomes less concentrated in nitrogen and oxygen, and holds more water vapor when it is saturated. 2. The density of air increases slightly after the process, from 1.18 kg/m³ to 1.19 kg/m³. This suggests that saturated air is slightly more dense than dry air, due to the added water vapor.