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

Question

H2​O(l)+40.7kJH2​O(g)

Assume that at exactly 100.0°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.

Calculate the work done on or by the system when 3.25 mol of liquid H₂O vaporizes.

Answer 1

w = -P∆V where w = work; P = pressure and ∆V = change in volume.

∆V = change in volume = 30.62 L - 18.80 L = 11.82 L

w = -(1.00 atm)(11.82 L) = - 11.82 L-atm. To change this to a more familiar unit of work, i.e. joules/kjoules...

w = -11.82 L-atm x 101.325 J/L-atm = -1200.7 J = -1.20 kJ (this is for 1 mole H2O(l) --> 1 mole H2O(g)

For 3.25 mol H2O(l), ... -1.20kJ/mol x 3.25 mol = -3.90 kJ of work done by the system

This would be the answer if the system was initially at 100.0ºC and 1 atm. But since the initial condition was at 100.0ºC and 1.00 atm, we need to add/subtract the work provided/used during this transition which is given in the equation

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

For 3.25 mol H2O, it would be 3.25 mol x 40.7 kJ/mol = 132.3 kJ

This has a positive value as heat was added to the system

w = -3.90 kJ + 132.2 kJ = 128.3 kJ (positive - heat done on the system).