Respuesta :
Answer:
liquid's heat of vaporization = 38.4 kJ/mol
Explanation:
given data
vapor pressure P1 = 6.91 mmHg
at temperature = 0 °C = 273.15 K
boiling temperature = 105 °C
solution
for vapor pressure and temperature we get here
P2 = 760.0 mmHg
T2 = 68.73°C = 378.15 K
we use here the Clausius-Clapeyron Equation that is
ln [tex]\frac{P1}{P2}[/tex] = [tex]\frac{\Delta H}{R} (\frac{1}{T2} -\frac{1}{T1} )[/tex] .................1
put here value
In [tex]\frac{6.91}{760} = \frac{x}{8.31447} (\frac{1}{378.15} -\frac{1}{273.15} )[/tex]
solve it we get
x = 38445 J/mol
liquid's heat of vaporization = 38.4 kJ/mol
