Question

find the latent heat of vapour at normal atmospheric pressure while given that entropy of water is 0.30 and of vapour is 1.75 at normal boiling point​

Answer 1

Answer:

plz subtract the 0.03 from 1.75 you will get the answer

Explanation:

may God bless you

Answer 2

ΔH vap = ΔU vap +pΔV

3.57J/kg

Explanation:

The term latent heat of vaporization can be defined as the amount of heat needed for the conversion of 1kg of liquid at its boiling point to gas at the same temperature.  It differs among different liquids.

Particles of water vapor at 100°C (373K) have more energy than liquid water at the same temperature. this due to the absorption of extra energy in the form of latent heat of vaporization.

= 22.5X10⁵ X (1.75-0.30)

= 3.57J/Kg.