Consider a glass of water in a room at \(20^{\circ} \mathrm{C}\) and \(97 \mathrm{kPa}\). If the relative humidity in the room is 100 percent and the water and the air are in thermal and phase equilibrium, determine (a) the mole fraction of the water vapor in the air and \((b)\) the mole fraction of air in the water.

Using the Antoine equation, we can find the saturation pressure of water vapor at the given temperature (20°C). The equation is: \(P_{sat}=10^{A-\frac{B}{T+C}}\) where \(P_{sat}\) is the saturation pressure, \(T\) is the temperature in Celsius, and \(A\), \(B\), and \(C\) are constants for water. For water, the constants are \(A = 8.07131\), \(B = 1730.63\), and \(C = 233.426\). Plug the temperature value to the Antoine equation: \(P_{sat} = 10^{8.07131 - \frac{1730.63}{20 + 233.426}}\) Calculate \(P_{sat}\): \(P_{sat} = 2.339\mathrm{kPa}\)

To find the partial pressure of air, we can use Dalton's law of partial pressures: \(P_{total} = P_{H_2O} + P_{air}\) where \(P_{total}\) is the total pressure, \(P_{H_2O}\) is the partial pressure of water vapor, and \(P_{air}\) is the partial pressure of air. We are given the total pressure (\(97\mathrm{kPa}\)) and we calculated the partial pressure of water vapor in the previous step. So we can find the partial pressure of air: \(P_{air} = 97\mathrm{kPa} - 2.339\mathrm{kPa} = 94.661\mathrm{kPa}\)

Now we can find the mole fraction of water vapor in the air using the partial pressures we calculated: \(y_{H_2O} = \frac{P_{H_2O}}{P_{total}}\) Insert the known values: \(y_{H_2O} = \frac{2.339\mathrm{kPa}}{97\mathrm{kPa}}\) Calculate \(y_{H_2O}\): \(y_{H_2O} = 0.0241\) So the mole fraction of water vapor in the air is 0.0241.

Assuming that air is mostly nitrogen, we can use Henry's Law to find the mole fraction of air in the water. The equation is: \(P_{air} = K_{H} \times x_{air}\) where \(K_{H}\) is the Henry's Law constant for nitrogen in water, and \(x_{air}\) is the mole fraction of air (nitrogen) in water. The value of \(K_{H}\) for nitrogen in water at 20°C is \(76.48\frac{\mathrm{kPa}}{\mathrm{molality}}\). Rearranging the equation: \( x_{air} = \frac{P_{air}}{K_{H}}\) Insert the known values: \(x_{air} = \frac{94.661\mathrm{kPa}}{76.48\frac{\mathrm{kPa}}{\mathrm{molality}}}\) Calculate \(x_{air}\): \(x_{air} = 1.238\times10^{-3}\) So the mole fraction of air (nitrogen) in the water is \(1.238\times10^{-3}\). To summarize the results: (a) The mole fraction of water vapor in the air is 0.0241. (b) The mole fraction of air (nitrogen) in the water is \(1.238\times10^{-3}\).