The water will boil at C) 80°C
Further explanation
Given
Vapour pressure of water = 47 kPa
Required
Boiling point of water
Solution
We can use the Clausius-Clapeyron equation :
[tex]\tt ln(\dfrac{P_1}{P_2})=\dfrac{-\Delta H_{vap}}{R}(\dfrac{1}{T_1}-\dfrac{1}{T_2})[/tex]
Vapour pressure of water at boiling point 100°C=101.325 kPa
ΔH vap for water at 100°C=40657 J/mol
R = 8.314 J/mol K
T₁=boiling point of water at 101.325 kPa = 100+273=373 K
Input given values :
[tex]\tt ln\dfrac{101.325}{47}=\dfrac{-40657}{8.314}(\dfrac{1}{373}-\dfrac{1}{T_2})\\\\0.7682=4890.1852(\dfrac{1}{373}-\dfrac{1}{T_2})\\\\0.0001571=(\dfrac{1}{373}-\dfrac{1}{T_2})\\\\\dfrac{1}{T_2}=0.00283\rightarrow T_2=353~K=80^oC[/tex]
