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 = 8.3J mol^(-1) K^(-1)

Find out the internal energy change for the reaction A (l) rarr A (g)

1.	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 = 8.3J mol^(-1) K^(-1)
Answer» SOLUTION :`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)`