Respuesta :
Explanation:
The given data is as follows.
[tex]P_{1}[/tex] = 100 mm Hg or [tex]\frac{100}{760}atm[/tex] = 0.13157 atm
[tex]T_{1}[/tex] = [tex]1080 ^{o}C[/tex] = (1080 + 273) K = 1357 K
[tex]T_{2}[/tex] = [tex]1220 ^{o}C[/tex] = (1220 + 273) K = 1493 K
[tex]P_{2}[/tex] = 600 mm Hg or [tex]\frac{600}{760}atm[/tex] = 0.7895 atm
R = 8.314 J/K mol
According to Clasius-Clapeyron equation,
[tex]log(\frac{P_{2}}{P_{1}}) = \frac{\Delta H_{vap}}{2.303R}[\frac{1}{T_{1}} - \frac{1}{T_{2}}[/tex]
[tex]log(\frac{0.7895}{0.13157}) = \frac{\Delta H_{vap}}{2.303 \times 8.314 J/mol K}[\frac{1}{1357 K} - \frac{1}{1493 K}][/tex]
[tex]log (6) = \frac{\Delta H_{vap}}{19.147}[\frac{(1493 - 1357) K}{1493 K \times 1357 K}][/tex]
0.77815 = [tex]\frac{\Delta H_{vap}}{19.147J/K mol} \times 6.713 \times 10^{-5} K[/tex]
[tex]\Delta H_{vap}[/tex] = [tex]2.219 \times 10^{5}[/tex] J/mol
= [tex]2.219 \times 10^{5}J/mol \times 10^{-3}\frac{kJ}{1 J}[/tex]
= 221.9 kJ/mol
Thus, we can conclude that molar heat of vaporization of substance X is 221.9 kJ/mol.
