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.

Calculate the water's change in internal energy.

Answer 1

H2​O(l) + 40.7kJ ===> H2​O(g)

Not sure if the 100.9ºC in the question was a typo and should be 100.0ºC. Answer below is based on the assumption of the typo. If it is meant to be 100.9º, then q should be re-calculated using C_steam and ∆T = 0.9 and also using ∆Hvap for phase change.

w = -P∆V

w= work = ?

P = pressure = 1.00 atm

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

w = -(1.00 atm)(11.82 L) = - 11.82 L-atm

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 (negative sign)

Change in internal energy ∆U = q + w

q = 40.7 kJ/mol x 3.25 mol = 132 kJ

w = - 3.90 kJ

∆U = 132 kJ - 3.90 kJ = 128 kJ