Question

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

is 47 kPa, the water will boil at
A) 0.0°C
B) 40°C
C) 80°C
D) 101.3°C

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((P_1)/(P_2))=(-\Delta H_(vap))/(R)((1)/(T_1)-(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(101.325)/(47)=(-40657)/(8.314)((1)/(373)-(1)/(T_2))\\\\0.7682=4890.1852((1)/(373)-(1)/(T_2))\\\\0.0001571=((1)/(373)-(1)/(T_2))\\\\(1)/(T_2)=0.00283\rightarrow T_2=353~K=80^oC`