Respuesta :
Explanation:
(a) The given data is as follows.
[tex]P_{1}[/tex] = 92.0 torr, [tex]T_{1} = 23^{o}C[/tex] = (23 + 273)K = 296 K
[tex]P_{2}[/tex] = 378.0 torr, [tex]T_{2} = 45^{o}C[/tex] = (45 + 273)K = 318 K
According to Clasius-Clapeyron equation,
[tex]ln(\frac{P_{2}}{P_{1}}) = \frac{\Delta H_{vap}}{R}[\frac{1}{T_{2}} - \frac{1}{T_{1}}[/tex]
Putting the given values into the above formula as follows.
[tex]ln(\frac{P_{2}}{P_{1}}) = \frac{\Delta H_{vap}}{R}[\frac{1}{T_{2}} - \frac{1}{T_{1}}[/tex]
[tex]ln(\frac{378}{92}) = \frac{\Delta H_{vap}}{8.314 J/mol K}[\frac{1}{318} - \frac{1}{296}[/tex]
1.413 = [tex]\frac{\Delta H_{vap}}{8.314 J/mol K}[\frac{1}{318} - \frac{1}{296}[/tex]
[tex]\Delta H_{vap}[/tex] = 2926063.008 J/mol
or, = 2926.063 kJ/mol
Hence, heat of vaporization for this liquid is 2926.063 kJ/mol .
(b) It is known that normal atmospheric pressure is 760 torr. Hence, we will calculate the normal boiling point as follows.
Hence, putting this value into Clasius-Clapeyron equation as follows.
[tex]ln (\frac{760 torr}{378 torr}) = -\frac{2926063.008 J/mol }{8.314 J/mol K} [\frac{1}{T} - \frac{1}{318}][/tex]
0.698 = 351944.07 [\frac{(318 - T)}{318T}][/tex]
[tex]1.98 \times 10^{-6} = \frac{(318 - T)}{318T}[/tex]
6.306 \times 10^{-6} \times T = (318 - T)
T = 317.998 K
Therefore, normal boiling point of this liquid is 317.998 K.
The heat of vaporization of the liquid is 39.2 kJ and the normal boiling point is 330 K or 57 ∘C.
Using the Clausius - Clapeyron equation;
ln(P2/P1) = -ΔH∘vap/R (1/T2 - 1/T1)
P2 = 378.0 Torr
P1 = 92.0 Torr
T2 = 318 K
T1 = 296 K
R = 8.314 J/K
ΔH∘vap = ?
ln(378.0 /92.0) = -ΔH∘vap/8.314 (1/318 - 1/296)
1.413 = 0.0003ΔH∘vap/8.314
ΔH∘vap = 1.413 × 8.314/0.0003
ΔH∘vap = 39159 J or 39.2 kJ
The boiling point occurs when P2 = 760 torr (atmospheric pressure)
ln(760/378) = 39159/8.314 (1/T2 - 1/378)
2.01 = 4710 (1/T2 - 1/378)
2.01 / 4710 = (1/T2 - 1/378)
0.00043 = 1/T2 - 0.0026
1/T2 = 0.00043 + 0.0026
T2 = 330 K or 57 ∘C
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