Explanation:
According to the Clausius-Clapeyron equation,
[tex]ln (\frac{P_{f}}{P_{i}}) = \frac{\Delta H_{vap}}{R}[\frac{T_{2} - T_{1}}{T_{1}T_{2}}][/tex]
The given data is as follows.
[tex]P_{f}[/tex] = 1 atm, [tex]P_{i}[/tex] = 50.0 kPa = [/tex]50.0 kPa \times \frac{9.87 \times 10^{-3}atm}{1 kPa}[/tex] = 493.5 atm
[tex]T_{1}[/tex] = 365.7 K, [tex]T_{2}[/tex] = 388.4 K
Hence, putting these values into the above equation as follows.
[tex]ln (\frac{P_{f}}{P_{i}}) = \frac{\Delta H_{vap}}{R}[\frac{T_{2} - T_{1}}{T_{1}T_{2}}][/tex]
[tex]ln (\frac{1 atm}{493.5 atm}) = \frac{\Delta H_{vap}}{8.314 J K^{-1} mol^{-1}}[\frac{388.4 K - 365.7 K}{388.4 K \times 365.7 K}][/tex]
[tex]\Delta H[/tex] = 36017.67 [tex]J mol^{-1}[/tex]
or, = 36.02 [tex]kJ mol^{-1}[/tex]
Thus, we can conclude that the enthalpy of vaporization of pyridine is 36.02 [tex]kJ mol^{-1}[/tex].
