To calculate the enthalpy change for the adsorption of nitrogen gas on the surface of activated charcoal, we can use the Clausius-Clapeyron equation, which relates the pressure, temperature, and enthalpy change of a process:ln P2/P1 = -H/R * 1/T2 - 1/T1 In this case, we have the following information:P1 = 1 atm initial pressure P2 = 0 atm final pressure, since the nitrogen gas is adsorbed onto the charcoal T1 = 298 K initial temperature T2 = 298 K final temperature, assuming no temperature change during the process R = 8.314 J/ molK universal gas constant We can plug these values into the Clausius-Clapeyron equation:ln 0/1 = -H/8.314 * 1/298 - 1/298 Since ln 0 is undefined, we cannot use the Clausius-Clapeyron equation directly in this case. However, we can make an approximation by considering the process to be isothermal constant temperature and use the isosteric heat of adsorption Qst , which is the enthalpy change for the adsorption process at constant temperature.Qst = -n * H_adsWhere n is the amount of nitrogen gas adsorbed 0.01 moles/g and H_ads is the enthalpy change for the adsorption process.We are also given the heat of combustion of the charcoal, which is -35.4 kJ/mol. Since the heat of combustion is the energy released when 1 mole of charcoal is burned, we can use this value to calculate the energy released during the adsorption process:Energy_released = -35.4 kJ/mol * 0.01 moles/gEnergy_released = -0.354 kJ/gNow, we can use the energy released and the isosteric heat of adsorption to calculate the enthalpy change for the adsorption process:Qst = -n * H_ads-0.354 kJ/g = - 0.01 moles/g * H_adsH_ads = -0.354 kJ/g / -0.01 moles/g H_ads = 35.4 kJ/molThe enthalpy change for the adsorption of nitrogen gas on the surface of activated charcoal is 35.4 kJ/mol.