To calculate the enthalpy of desorption for water molecules from the surface of a silica gel, we need to use the Clausius-Clapeyron equation, which relates the vapor pressure of a substance to its temperature and enthalpy of vaporization:ln P2/P1 = -Hvap/R * 1/T2 - 1/T1 Where:- P1 and P2 are the vapor pressures at temperatures T1 and T2, respectively.- Hvap is the enthalpy of vaporization.- R is the ideal gas constant 8.314 J/molK .In this case, we are given the vapor pressure of water at 25C 298.15 K , which is 3.17 kPa. We need to find the enthalpy of desorption Hdes for water molecules from the surface of the silica gel. The enthalpy of desorption is related to the enthalpy of vaporization by the following equation:Hdes = Hvap + HadsWhere Hads is the enthalpy of adsorption. Since we are not given any information about the enthalpy of adsorption, we will assume that the enthalpy of desorption is approximately equal to the enthalpy of vaporization.The enthalpy of vaporization for water is approximately 40.7 kJ/mol. We will use this value for Hdes.Now, we need to find the vapor pressure of water at the surface of the silica gel P2 . To do this, we can use the Langmuir isotherm equation, which relates the vapor pressure of a substance to its surface coverage :P = P0 * / 1 - Where P0 is the vapor pressure of the substance in the absence of the adsorbent 3.17 kPa in this case and is the surface coverage. Since we are not given any information about the surface coverage, we will assume that the vapor pressure at the surface of the silica gel is equal to the vapor pressure of water at 25C 3.17 kPa .Now, we can use the Clausius-Clapeyron equation to find the enthalpy of desorption:ln 3.17/3.17 = -Hdes/8.314 * 1/298.15 - 1/298.15 Since the vapor pressures and temperatures are the same, the left side of the equation is equal to 0:0 = -Hdes/8.314 * 1/298.15 - 1/298.15 Since the right side of the equation is equal to 0, we can conclude that the enthalpy of desorption is approximately equal to the enthalpy of vaporization:Hdes 40.7 kJ/mol