Question

Assume 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 3.65 mol of liquid H2O vaporizes. 2.)Calculate the water's change in internal energy

Answer 1

1) workdonw(w) = - PDV

p = 1 atm

DV = 30.62*3.65 = 111.763 L

w = -1*111.763

   = -111.763 l.atm     (1 l.atm = 101.3 joule)

   = -111.763*101.3

   = -11.32 kj

q = heat supplied = n*DHvap

       = 3.65*40.7

       = 148.555 kj

form first law of thermodynamics

change in internal energy(DU) = q+w

     = 148.555-11.32

     = 137.235 kj