Standard enthalpy of vaporisation ∆_vap H^° for water at 100°C is 40 .,66kj /mol. Internal energy of vaporisation of water at 100°C is

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Solution

We know that, for gaseous reactant and products, we have a relation between standard enthalpy of vaporization( ΔvapH0) and standard internal energy( ΔU0) as :

ΔvapH0 = ΔU0 + Δng RT

where,Δng = n2- n1 i.e. difference between no. of moles of reactant and product.

Here, for vaporization of water ,

H2O (l) -> H2O (g)

So, change in no. of moles ,Δng = 1-0 =1

Thus, ΔvapH0 = ΔU0 + RT

ΔU0 = ΔvapH0 - RT = 44.66 - ( 8.314 x 10-3KJ/Kmol x 373K) = 40.66 - 3.10 KJ/mol = 37.55 KJ/mol

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