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.
Calculate the work done on or by the system when 4.65 mol of liquid H2O vaporizes.
Calculate the water's change in internal energy.
According to the data provided above- Volume of water vapor = nRT/P V= 1 mole * 0.0821 Latm /mole K * 373.15 K /1 atm = 30.6... View the full answer