Question

What change in the boiling point of water is brought about by an increase of 0.01 bar in atmospheric pressure? Answer in Kelvin.

Answer 1

Here Atmospheric pressure P₁=1 atm, T₁=373K (standard boiling point of water)

P₂=1+0.01/1.01325 =1.0098atm, T₂=? (1atm=1.01325bar)

Using clausius claperon equation

log (P₂ /P₁)=(/2.303R)*(1/T₁-1/T2)

Here =latent heat of vapourisation of water. =40.65kJ/mol=40650J/mol

R, universal gas constant= 8.314J/mol/K

On substituting these values, we get

log (1.0098/1)=(40650/2.303*8.314)(1/373-1/T₂)

On solving we get T₂=373.3K

Change in boiling point=373.3-373=0.3K