Question

The vaporization of 1 mole of liquid water (the system) at 100.9 degrees C, 1.00 atm, is endothermic.

H2O(l) + 40.7kJ ---> H2O (g)

Assume at exactly 100.0 degrees C and 1.00 atm total pressure, 1.00 mole of liquid water and 1.00 mole of water vapor occupy 18.80 mL and 30.62 L, respectively.

1. Calculate the work done on or by the system when 3.65 mol of liquid H2O vaporizes. Answer in J

2. Calculate the water's change in internal energy. Answer in kJ

Answer 1

1) Work done , W = -P x V

Where P is the external pressure = 1 atm

Here, initial volume = 18.8 mL= 0.0188 L

Final volume = 30.62 L

Therefore V = 3.65 x (30.62 - 0.018)

= 111.72 L

Therefore work done , W = -P x V

= -1 atm x 111.72 L = - 111.72 L atm

=(-111.72 L atm) x (101.3 J/L atm) = -11317 J =-11.31KJ

W = -11.31 KJ

2) Change in internal energy,

E = H_reaction +W

We have H_reaction = (40.7 KJ/mol) x (3.65 mol) = 148.55 KJ

Therefore, E =148.55 KJ - 11.31KJ =137.24 KJ

E= 137.24 KJ