**12.33.** The solubility of nonpolar gases in liquid water is typically very low. Consider nitrogen in particular. Its solubility is characterized by the equilibrium mole fraction \( x_N \) of dissolved nitrogen in the liquid phase (with water mole fraction \( x_W = 1 - x_N \)) when the system is in equilibrium with a vapor phase (with corresponding mole fractions \( y_N \) and \( y_W \)). The mole fraction \( x_N \) is typically of the order \( 10^{-5} \). The vapor pressure of water at 300 K is \( P_W^{\text{vap}} = 3.5 \) kPa, and you can assume that the impact of its temperature dependence on solubility is relatively weak. Use ideal models in what follows.(a) At 1 bar and 300 K, estimate the mole fraction of water that is present in the vapor phase, \( y_W \).(b) Show that the solubility of nitrogen is given approximately by\[x_N = C(T, P)(1 - y_W)P\]where \( C(T, P) \) is a constant that is independent of concentrations. Find an expression for \( C(T, P) \) in terms of standard and pure chemical potentials.(c) Do you expect the solubility to increase or decrease with temperature? Explain by finding the temperature dependence of \( C(T, P) \).

The objective of the question is to understand the solubility of nonpolar gases in liquid water, particularly nitrogen, and to estimate the mole fraction of water in the vapor phase at given conditions. We also need to derive an expression for the solubility of nitrogen and find the temperature dependence of the constant C(T, P).First, we need to estimate the mole fraction of water in the vapor phase, y_w. At 1 bar and 300 K, the vapor pressure of water is given as P_P = 3.5 kPa. The total pressure is 1 bar, which is equivalent to 100 kPa. Therefore, the mole fraction of water in the vapor phase can be calculated using Raoult's law, which states that the partial pressure of each component in a mixture is the product of the mole fraction in the liquid phase and the vapor pressure of the pure component. Hence, y_w = P_P / P = 3.5 kPa / 100 kPa = 0.035.Next, we need to show that the solubility of nitrogen is given by X_N = C(T,P)(1-y_w)P and find an expression for C(T, P). The solubility of nitrogen in water can be described by Henry's law, which states that the amount of dissolved gas is proportional to its partial pressure in the gas phase. Therefore, X_N = H_N * P_N, where H_N is the Henry's law constant for nitrogen and P_N is the partial pressure of nitrogen. Since the total pressure is the sum of the partial pressures of water and nitrogen, P_N = P - P_P = P(1 - y_w). Substituting this into the equation gives X_N = H_N * P(1 - y_w). Comparing this with the given equation X_N = C(T,P)(1-y_w)P, we can see that C(T, P) = H_N.Finally, we need to find the temperature dependence of C(T, P). The Henry's law constant H_N is generally temperature dependent and can be expressed as H_N = H_0 * exp(-ΔH/R*T), where H_0 is the Henry's law constant at a reference temperature, ΔH is the enthalpy change of dissolution, R is the gas constant, and T is the temperature. Differentiating this with respect to temperature gives dH_N/dT = -ΔH/R * exp(-ΔH/R*T) = -ΔH/R * H_N. Therefore, the solubility of nitrogen decreases with increasing temperature if ΔH > 0, which is typically the case for gases dissolving in water.The mole fraction of water in the vapor phase at 1 bar and 300 K is 0.035. The solubility of nitrogen in water can be described by the equation X_N = C(T,P)(1-y_w)P, where C(T, P) is the Henry's law constant for nitrogen. The solubility of nitrogen decreases with increasing temperature if the enthalpy change of dissolution is positive.