Question

The vaporization of 1 mole of liquid water (the system) at 100.9°C, 1.00 atm, is endothermic. $$ 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 4.65 mol of liquid H2O vaporizes.

Answer 1

The process is irrversible.

Then , work done= - P_ex ( V_steam - V_water)

Volume of 4.65 mol steam = 4.65*30.62 L.

Volume of 4.65 mol water = 4.65*18.80*10^-3 L .

External pressure (P_ex) = 1.00 atm.

Then , W = - 1.00*4.65(30.62 - 18.80*10^-3)

Or, W = - 1.00*4.65*30.6012 L-atm.

Or, W = - 142.295 L-atm.

Or, w = - 142.295*101.3 J = - 14414 J = - 14.414 KJ.

Therefore as sign of W is negative then work done by the system when 4.65 moles liquid H₂O vaporises = 14.414 KJ.