Question

If the pressure on the surface of water in the liquid stato

Answer 1

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 :

$\tt ln(\dfrac{P_1}{P_2})=\dfrac{-\Delta H_{vap}}{R}(\dfrac{1}{T_1}-\dfrac{1}{T_2})$

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 :

$\tt ln\dfrac{101.325}{47}=\dfrac{-40657}{8.314}(\dfrac{1}{373}-\dfrac{1}{T_2})\\\\0.7682=4890.1852(\dfrac{1}{373}-\dfrac{1}{T_2})\\\\0.0001571=(\dfrac{1}{373}-\dfrac{1}{T_2})\\\\\dfrac{1}{T_2}=0.00283\rightarrow T_2=353~K=80^oC$