Respuesta :
Explanation:
The given data is as follows.
[tex]P_{1}[/tex] = 2.34 kPa = 2.34 \times 1000 Pa = 2340 Pa = 0.0231 atm
[tex]T_{1}[/tex] = (20 + 273) K = 293 K
[tex]\Delta H_{vaporization}[/tex] = 2537.4 kJ/kg = 2537400 J/kg
[tex]P_{2}[/tex] = ?, [tex]T_{2}[/tex] = (40 + 273) K = 313 K
According to Clausius-Clapeyron equation,
[tex]P_{313 K}[/tex] = [tex]P_{293} Exp[\frac{-\Delta H_{v}}{R} (\frac{1}{313} - \frac{1}{293})][/tex]
= [tex]0.0213 Exp [\frac{-2537400 J/kg}{8.314 J/mol} (\frac{1}{313} - \frac{1}{293})K][/tex]
= [tex]1.86 \times 10^{27}[/tex] atm
or, = 18846.45 kPa
Thus, we can conclude that the vapor pressure at [tex]40^{o}C[/tex] is 18846.45 kPa.
