posted by mel on .
One mole of H2O(g) at 1.00 atm and 100.°C occupies a volume of 30.6 L. When one mole of H2O(g) is condensed to one mole of H2O(l) at 1.00 atm and 100.°C, 40.66 kJ of heat is released. If the density of H2O(l) at this temperature and pressure is 0.996 g/cm3, calculate E for the condensation of one mole of water at 1.00 atm and 100.°C.
Delta E = q + w
q = -40.66 kJ.
w = -p(delta V) = -p(V2-V1)
p = 1.00 atm.
V1 = 30.6 L
V2 = calculated from density. 1 mol H2O has mass of 18.015 g, density is 0.996, V2 = ??
Don't forget to change w from L*atm units to J by multiplying w*101.3 J/L*atm. Also remember w is in Joules and q is in kJ so make them the same unit.
Post your work if you get stuck.
i don't get how to convert the density to V2...
density=mass/volume so volume=mass/density
how do you convert cm^3 to L so you can subtract V1 from V2
100 Cm^3=1 L
1000 cm^3 = 1 L since 1 cm^3 = 1 mL and since there are 1000 mL in one L.