Question

Consider a glass of water in a room at \(27^{\circ} \mathrm{C}\) 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.

Short answer

Determine the mole fraction of water vapor in the air and the mole fraction of air in the water when a glass of water in a room at 27 °C temperature and 97 kPa is in thermal and phase equilibrium with 100% relative humidity. The mole fraction of the water vapor in the air is approximately 0.0365, and the mole fraction of air in the water is approximately 2.18 × 10^-5.

Step-by-step solution

Find the saturation pressure of the water at the given temperature

To find the saturation pressure of the water at a given temperature, we use the Antoine equation: \(\log_{10}(P_\text{sat}) = A - \frac{B}{T+C}\), where \(P_\text{sat}\) is the saturation pressure in \(\text{kPa}\), \(T\) is the temperature in \(\text{Celsius}\), and \(A\), \(B\), \(C\) are constants for the substance. For water, \(A = 8.07131\), \(B = 1730.63\), and \(C = 233.426\). Now, let's plug in the provided temperature (\(27^\circ \mathrm{C} = T\)) and find the saturation pressure of water: \(\log_{10}(P_\text{sat}) = 8.07131 - \frac{1730.63}{27+233.426}\) \(P_\text{sat} = 10^{(8.07131 - \frac{1730.63}{27+233.426})}\) \(P_\text{sat} ≈ 3.54\,\mathrm{kPa}\)

Calculate the mole fraction of water vapor in air

The relative humidity is given as 100%, which means the partial pressure of water vapor in the air is equal to the saturation pressure: \(P_\text{water} = P_\text{sat} ≈ 3.54\,\mathrm{kPa}\) The total pressure of the air is given as \(97\,\mathrm{kPa}\). Hence, the partial pressure of dry air can be calculated as: \(P_\text{air} = P_\text{total} - P_\text{water} ≈ 93.46\,\mathrm{kPa}\) Now, we can calculate the mole fraction of water vapor, \(y_\text{water}\), in the air using the partial pressure: \(y_\text{water} = \frac{P_\text{water}}{P_\text{total}} ≈ \frac{3.54\,\mathrm{kPa}}{97\,\mathrm{kPa}}\) \(y_\text{water} ≈ 0.0365\) (a) The mole fraction of the water vapor in the air is approximately \(0.0365\).

Calculate the mole fraction of air in the water

To find the mole fraction of air dissolved in the water, we can use Henry's Law, which states that the amount of a gas dissolved in a liquid is directly proportional to the partial pressure of the gas above the liquid: \(c_\text{air} = k_\text{Henry}\cdot P_\text{air}\), where \(c_\text{air}\) is the concentration of dissolved air, and \(k_\text{Henry}\) is the Henry's Law constant for the gas in the liquid. At room temperature, the value of \(k_\text{Henry}\) for air in water is roughly \(1.295\times10^{-5}\,\mathrm{mol/L\,kPa}\). Now, let's calculate the concentration of dissolved air in the water: \(c_\text{air} ≈ (1.295\times10^{-5}\,\mathrm{mol/L\,kPa})\cdot(93.46\,\mathrm{kPa}) ≈ 0.00121\,\mathrm{mol/L}\) To find the mole fraction of air in the water, we also need the concentration of water in the liquid state, which can be calculated from the molar density of water, \(\rho_\text{water} ≈ 55.5\,\mathrm{mol/L}\), as: \(y_\text{air} = \frac{c_\text{air}}{c_\text{water} + c_\text{air}}\) \(y_\text{air} ≈ \frac{0.00121\,\mathrm{mol/L}}{0.00121\,\mathrm{mol/L} + 55.5\,\mathrm{mol/L}}\) \(y_\text{air} ≈ 2.18\times10^{-5}\) (b) The mole fraction of air in the water is approximately \(2.18\times10^{-5}\). So, our final answers are: (a) The mole fraction of the water vapor in the air is approximately \(0.0365\). (b) The mole fraction of air in the water is approximately \(2.18\times10^{-5}\).