Question

Take the volume change into account and calculate ΔH and ΔU for exactly 1 g of ice melting into 1 g of water at standard pressure. The density of ice at 0˚C is 0.9168 g/mL; the density of water at 0˚C is 0.99984 g/mL.

Answer 1

Solution :-

Enthalpy of fusion of ice is 6.01 kJ/mol

So lets calculate the heat needed to melt 1 g ice

1 g ice * 6.01 kJ / 18.0148 g per mol = 0.3336 kJ

0.3336 kJ* 1000 J / 1 kJ =333.6 J

So delta H = 333.6 J

Now lets calculate the delta U

Delta U =delta H + w

W= - p*delta V

Volume of water = 1 g /0.99984 g per ml = 1.00016 ml

Volume of the ice = 1 g/ 0.9168 g per ml = 1.09075 ml

So the difference in the volume delta V =1.09075 ml – 1.00016 ml

                                                                       = 0.09059 ml

0.09059 ml * 1 L / 1000 ml = 9.059*10^-5 L

W= - 1 atm * 9.059*10^-5 L

= -9.059*10^-5 L atm

-9.059*10^-5 L atm * 101.325 J / 1 L atm = -0.00918 J

So the delta U = delta H + w

                          = 333.6 J + (-0.00918 J)

                         = 333.59 J