Question

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

H₂O(l)+40.7kj -------> H₂O(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.

Part 1

Calculate the work done on or by the system when 1.25 mol of liquid H2O vaporizes.

Part 2

Calculate the water's change in internal energy

Answer 1

Part (1) : work done, W--ΡχΔν Here P is the extemal pressure 1 atm Initial volume = 18.8 mL= 0.0188 L Final volume = 30.62 L Therefore, AV-1.25 x (30.62-0.018) 38.26 L Now Work done . W--P x Δ V ー-1 atm x 38.26 L-ー38.26 L atm (-38.26 L at) x (101.3 J/L atm) --3874.74J3.876 KJ -- 3.876KJ Pait (2) :- Clhange in internal energy We have ΔΗ,eaction-(40.7 KJ/mol) x ( 1.25 mol)-50.875 KJ Therefore, ΔΕ-50.875 KJ-3.875 KJ 47.0 KJ