Respuesta :
Answer:
P = 2.65 E-3 mm Hg
Explanation:
liquid-vapor equilibrium:
⇒ ec. Clausius-Clapeyron:
- Ln (P2/P1) = - ΔHv/R [ (1/T2) - (1/T1) ]
∴ R = 8.314 E-3 KJ/K.mol
∴ ΔHv = 59.11 KJ/mol
∴ T2 = 25°C ≅ 298 K
∴ T1 = 356.7°C = 629.7 K
normal boiling point:
∴ P = 1 atm = P1 = 101.325 KPa
vapor pressure (P2):
⇒ Ln P2 - Ln (101.325) = - (59.11 KJ/mol)/(8.314 E-3 KJ/K.mol)*[ (1/298) - (1/629.7) ]
⇒ Ln P2 - 4.618 = - (7109.694 K)*[ 1.7676 E-3 K-1 ]
⇒ Ln P2 = 4.618 - 12.567
⇒ P2 = e∧(-7.9494)
⇒ P2 = (3.5313 E-4 KPa)×(7.50062mm Hg/KPa) = 2.65 E-3 mm Hg
The vapor pressure is caused by the vaporized component of a liquid on the walls of the closed container.
A Clausis Clayperon equation's formula is as follows:
[tex]\ln \frac{P1}{P2}=-\Delta \frac{H}{R} - (\frac{1}{T1} - \frac{1}{T2})[/tex]
where
[tex]P1 = 760 \ mm Hg \\\\ \Delta H vap = 59.11 \ \frac{kJ}{mol}\\\\ R= 8.314 \frac{J}{Kmol}\\\\T1= 356.7 ^{\circ} C = 629.85\ K \\\\T2 = 25 ^{\circ} C = 298.15\ K\\\\[/tex]
putting the value into the formula:
[tex]\ln \frac{760}{P2}=-\frac{59110}{8.314}-(\frac{1}{629.8} - \frac{1}{298.15})\= 12.557\\\\\to \frac{760}{P2 }= 284076.828\\\\\to P2 = 2.675 \times 10^{-3} \ mmHg[/tex]
Therefore, the final answer is "[tex]2.675 \times 10^{-3} \ mmHg[/tex]" .
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