Standard enthalpy of vapourisation D vapH Q for water at 100° C is 40.66 kJmol–1. The internal energy of vapourisation of water at 100°C (in kJmol–1) is ?

Dear student!
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 10-3 KJ/Kmol  x 373K) = 40.66 - 3.10 KJ/mol = 37.55 KJ/mol

  • 299
What are you looking for?