Question

Find out the internal energy change for the reaction `A (l) rarr A (g)` at 373 K . Heat of vaporisation is 40.66 kJ `//` mol and `R = 8.3 J mol^(-1) K^(-1)`

Answer 1

`A (l) rarr A(g), Delta n_(g) = n_(p) - n_(r) = 1 -0 = 1`
`Delta H = Delta U + Delta n_(g) RT`
or `Delta U = Delta H - Delta n_(g) RT = 40660 J - 1 mol xx 8.314 J K^(-1) mol^(-1) xx 373 K `
`= 40660J - 3101 J = 37559 J mol^(-1) = 37.56 k J mol^(-1)`