Question
A molecule of molecular mass 102 g/mol is placed in anelectrically heated container. When the pressure was 650 torr, the liquid boiled at78ºC. After the boiling point had been reached, it was found that a current of .232A from a 12Vsupply passed for 650s vaporized the sample. Calculate the molar enthalpy andinternal energy of vaporization. A molecule of molecular mass 102 g/mol is placed in anelectrically heated container. When the pressure was 650 torr, the liquid boiled at78ºC. After the boiling point had been reached, it was found that a current of .232A from a 12Vsupply passed for 650s vaporized the sample. Calculate the molar enthalpy andinternal energy of vaporization.
Explanation / Answer
Because the vaporization occurs at constant pressure, thechange in enthalpy is equal to the heat supplied by the heater.Therefore, the startegy is to calculate the heat supplied (from q=IVt), expresses that as an enthalpy change, and then convert theresult ot a molar enthalpy change by division by the amountof moles of compound. To convert from enthalpy change to internalenergy change, we assume that the vapor is a perfect gas. The enthalpy change is H = qp = (0.323A)(12V)(650s) =210.028J Hm = H / n = 210.028J / (numbermoles of compound) number of moles of compound = mass of one molecule / molarmass of compound = 16.94*10^-23 g / 102g/mol = 0.167*10^-23 mol Hm = 210.028J / 0.167*10^-23mol -------------- Um = Hm - RT =(210.028J / 0.167*10^-23mol) - (8.314J/K.mol)(78+273)K