Question

5. Surface energy (2 pts). Let’s use the concept of surface tension as surface energy per unit area to see if we can estimate, at least to the correct order of magnitude, the surface tension of water.

a) Water has a molar mass of 18 g/mol and a density of 1000 kg/m3 (or 1 g/cm3). Based on this data, estimate the number of water molecules per unit surface area of water.

b) The coordination number of water (i.e., the average number of “neighbors” each water molecule has) in the liquid state is 4. Neighboring water molecules attract each other via hydrogen bonds, each of which has a binding energy of roughly 10–20 J (although this number depends relatively strongly on temperature). Use this information to estimate the surface tension of water. How does your estimate compare to the observed figure ( γ water = 0.072 N/m ) (hints: Keep in mind that we can think of surface tension as surface energy per unit area, and consider the energy needed to bring a molecule from the bulk to the surface)?

Answer 1

a) We know that the density of water is 1g/cm³

1 mole of water will have 6.023 x 10²³ molecules weight of which will be 18 g.

1 cm³ will have 6.023 x 10²³/18 = 3.34 x10²² molecules or surface coverage of 1cm² will have 3.34 x10²² molecules

b) surface tension as surface energy per unit area

Surface energy = joules/area = joules/meter²

Surface energy is the amount of work done per unit area extended.

If the area of the surface is to be extended, one has to bring more molecules from the bulk of the liquid to its surface. This requires expenditure of some energy because work has to be done in bringing the molecules from the bulk against the inward attractive forces.

When we attempt to bring a molecule of water from the bulk to the surface 4 attractive interaction are replace by 3 so the loss of energy is assuming 10J /hydrogen bond 30/40 = 0.75 Joules or 0.75 N/m This number is one order higher than the actual surface tension of water.