To calculate the enthalpy of adsorption, we first need to determine the amount of benzene adsorbed on the silica gel. The adsorption capacity is given as 0.75 g/g, which means that 0.75 grams of benzene can be adsorbed per gram of silica gel. Given the mass of adsorbent used is 0.25 g, we can calculate the mass of benzene adsorbed:Mass of benzene adsorbed = mass of adsorbent adsorption capacity Mass of benzene adsorbed = 0.25 g 0.75 g/g = 0.1875 gNow, we need to determine the moles of benzene adsorbed. The molecular weight of benzene C6H6 is 12.01 g/mol for carbon 6 + 1.01 g/mol for hydrogen 6 = 78.12 g/mol.Moles of benzene adsorbed = mass of benzene adsorbed / molecular weight of benzene Moles of benzene adsorbed = 0.1875 g / 78.12 g/mol = 0.0024 molNext, we will use the Clausius-Clapeyron equation to determine the enthalpy of adsorption. The equation is:ln P2/P1 = -H/R 1/T2 - 1/T1 Where P1 and P2 are the initial and final pressures, T1 and T2 are the initial and final temperatures, H is the enthalpy of adsorption, and R is the gas constant 8.314 J/molK .In this case, we are given the final pressure P2 as 2.0 atm. We will assume that the initial pressure P1 is 1 atm, as this is a common assumption for adsorption problems. The initial temperature T1 is 25C, which is 298.15 K. We will assume that the final temperature T2 is the same as the initial temperature since there is no mention of a temperature change.Plugging these values into the Clausius-Clapeyron equation:ln 2.0 atm / 1.0 atm = -H / 8.314 J/molK 1/298.15 K - 1/298.15 K Since the temperature does not change, the term 1/T2 - 1/T1 becomes zero, and the equation simplifies to:ln 2 = 0This means that there is no change in enthalpy during the adsorption process, so the enthalpy of adsorption H is 0 J/mol.