The Clausius-Clapeyron equation relates the vapor pressure of a substance at different temperatures to its enthalpy of vaporization. The equation is given by:ln P2/P1 = - Hvap/R * 1/T2 - 1/T1 where P1 and P2 are the vapor pressures at temperatures T1 and T2, Hvap is the enthalpy of vaporization, and R is the ideal gas constant 8.314 J/molK .In this problem, we are given the standard boiling point T1 as 1500 K and the enthalpy of vaporization Hvap as 40.0 kJ/mol. We need to find the boiling point T2 at an atmospheric pressure of 1.0 atm 760 mmHg .At the boiling point, the vapor pressure of the substance is equal to the atmospheric pressure. Therefore, P1 is the vapor pressure of KCl at 1500 K, and P2 is the vapor pressure at the new boiling point T2 , which is equal to 1.0 atm.First, we need to convert the enthalpy of vaporization to J/mol:Hvap = 40.0 kJ/mol * 1000 J/1 kJ = 40000 J/molNext, we need to convert the pressure from atm to Pa:P2 = 1.0 atm * 101325 Pa/1 atm = 101325 PaSince we don't know the vapor pressure of KCl at 1500 K P1 , we cannot directly solve for T2. However, we can use the fact that the vapor pressure at the boiling point is equal to the atmospheric pressure. Therefore, P1 is also equal to 101325 Pa.Now we can plug the values into the Clausius-Clapeyron equation:ln 101325/101325 = - 40000 J/mol / 8.314 J/molK * 1/T2 - 1/1500 0 = - 40000 / 8.314 * 1/T2 - 1/1500 0 = 1/T2 - 1/1500 1/T2 = 1/1500T2 = 1500 KThe expected boiling point of KCl at an atmospheric pressure of 1.0 atm 760 mmHg is 1500 K. This is because the standard boiling point is already given at 1 atm, so the boiling point remains the same at the given pressure.