To calculate the enthalpy change of adsorption, we can use the Clausius-Clapeyron equation, which relates the change in pressure and temperature to the enthalpy change of adsorption. The equation is:ln P2/P1 = - H_ads/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_ads is the enthalpy change of adsorption, and R is the gas constant 8.314 J/molK .In this case, the initial pressure P1 is 0.5 atm, the final pressure P2 is 1.0 atm, the initial and final temperatures are both 298 K, and we want to find H_ads.First, we need to convert the pressures from atm to Pa Pascals since the gas constant R is in J/molK. 1 atm = 101325 PaP1 = 0.5 atm * 101325 Pa/atm = 50662.5 PaP2 = 1.0 atm * 101325 Pa/atm = 101325 PaNow we can plug the values into the Clausius-Clapeyron equation:ln 101325/50662.5 = - H_ads/8.314 1/298 - 1/298 Since the temperatures are the same, the term 1/T2 - 1/T1 becomes 0, and the equation simplifies to:ln 101325/50662.5 = 0This means that the enthalpy change of adsorption, H_ads, is also 0. In this case, there is no enthalpy change when 2.5 moles of nitrogen gas are adsorbed onto 10 g of activated charcoal at 298 K.