Answer to Question #99584 in General Chemistry for Beverlie

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.
1. Calculate the work done on or by the system when 2.25 mol of liquid H2O vaporizes. in J
2. Calculate the water's change in internal energy. in KJ

Answer 1

1.      work done, $w =pdv$ where pressure p is 101325pa and change in volume dv is (0.03062*2.25) m3

w =101325*0.068895

w =6980.79 J

2.     since dv=o, w=o hence;

 change in internal energy, $dU=Q$ where Q is heat of the system.

 Q=40.7KJ

 Hence dU=40.7KJ