Calculate the work done on or by the system on 5.25 mol of liquid H2O vaporizes in J. Calculate the water s change in internal energy in kJ.?

Question

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

H2​O (l) + 40.7 kJ ---> H2​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.

Answer 1

Use calculation formula: ΔΕ = q + w to calculate the internal energy

H₂​O (l) + 40.7 kJ ---> H₂​O (g)

q = 40.7 kJ (endothermic)

w = - P ΔV (gas expends)

w = - 1.0 atm x (30.62 - 0.0188) L =30.60 atm-L

1.0 atm-L = 101.3 J therefore: -30.60 atm-L x 101.3 J / 1.0 atm-L = -3099.78 J = -3.10 kJ

ΔΕ = 40.7 kJ + ( -3.10 kJ) = 37.6 kJ