Question
A sample of 1.0 molgaseous water is condensed isothermally and reversibly to liquidwater at 100 C. The standard enthalpy of vaporization of water at100 C is 40.656 kJ/mol. Find the internal energy change for thisthermodynamic process. A sample of 1.0 molgaseous water is condensed isothermally and reversibly to liquidwater at 100 C. The standard enthalpy of vaporization of water at100 C is 40.656 kJ/mol. Find the internal energy change for thisthermodynamic process.
54.33kJ -37.55kJ -27.74kJ +38.34kJ
Explanation / Answer
1H2O (g) ...........>1H2O (l) Formula : H = U + nRT n = number of moles of gaseous products - number of moles of gaseous reactants = 0 - 1 = -1 Given , H = 40.656kJ/mol -40.656kJ =U + (-1mol)(8.314J.mol-1.K-1)(393K) -40.656kJ = U - 3.2674kJ U = -40.656kJ + 3.2674kJ =-37.40 kJ Second option is correct. Second option is correct.