The vapor pressure of substance x is 100. mm hg at 1080.°c. the vapor pressure of substance x increases to 600. mm hg at 1220.°c. determine the molar heat of vaporization of substance x using the derived form of the clausius-clapeyron equation given below. (include the sign of the value in your answer.) kj/mol

Explanation:

The given data is as follows.

          = 100 mm Hg or = 0.13157 atm

          = = (1080 + 273) K = 1357 K

          = = (1220 + 273) K = 1493 K

          = 600 mm Hg or = 0.7895 atm

          R = 8.314 J/K mol

According to Clasius-Clapeyron equation,

                0.77815 =

              = J/mol

                                   =

                                    = 221.9 kJ/mol

Thus, we can conclude that molar heat of vaporization of substance X is 221.9 kJ/mol.

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b. accuracy of the data matters alot more than its number of digits, organization or amount of detail. quantitative observation is more accepted than the qualitative observation.