# Estimating the Properties of Water

I found a manuscript on the arXiv by Andrew Lucas about estimating macroscopic properties of materials using just a few microscopic parameters. I decided to try a version of this analysis myself. It’s based on Lucas’s work, with a few modifications. I focus exclusively on water because of its importance for biological physics, and make order-of-magnitude calculations like those Russ Hobbie and I discuss in the first section of Intermediate Physics for Medicine and Biology.

My goal is to estimate the properties of water using three numbers: the size, mass, and energy associated with water molecules. We take the size to be the center-to-center distance between molecules, which is about 3 , or 3 × 10 ^−10 m. The mass of a water molecule is 18 (the molecular weight) times the mass of a proton, or about 3 × 10^−26 kg. The energy associated with one hydrogen bond between water molecules is about 0.2 eV, or 3 × 10^−20 J. This is roughly eight times the thermal energy kT at body temperature, where k is Boltzmann’s constant (1.4 × 10^−23 J K^−1) and T is the absolute temperature (310 K). A water molecule has about four hydrogen bonds with neighboring molecules.

Estimating the density of water, ρ, is Homework Problem 4 in Chapter 1 of IPMB. Density is mass divided by volume, and volume is distance cubed

ρ = (3 × 10^−26 kg)/(3 × 10^−10 m)3 = 1100 kg m^−3 = 1.1 g cm^−3.

The accepted value is ρ = 1.0 g cm −3, so our calculation is about 10% off; not bad for an order-of-magnitude estimate.

The compressibility of water, κ, is a measure of how the volume of water decreases with increasing pressure. It has dimensions of inverse pressure. The pressure is typically thought of as force per unit area, but we can multiply numerator and denominator by distance and express it as energy per unit volume. Therefore, the compressibility is approximately distance cubed over the total energy of the four hydrogen bonds

κ = (3 × 10^−10 m)3/[4(3 × 10^−20 J)] = 0.25 × 10^−9 Pa−1 = 0.25 GPa^−1 ,

implying a bulk modulus, B (the reciprocal of the compressibility), of 4 GPa. Water has a bulk modulus of about B = 2.2 GPa, so our estimate is within a factor of two.

Once you know the density and compressibility, you can calculate the speed of sound, c, as (see Eq. 13.11 in IPMB)

c = (ρ κ)^−1/2 = 1/√[(1100 kg m^−3) (0.25 × 10^−9 Pa^−1)] = 1900 m s^−1 = 1.9 km s^−1.

The measured value of the speed of sound in water is about c = 1.5 km s^−1, which is pretty close for a back-of-the-envelope estimate.

A homework problem about vapor pressure in Chapter 3 of IPMB uses water’s latent heat of vaporization, L, which is the energy required to boil water per kilogram. We estimate it as

L = 4(3 × 10^−20 J)/(3 × 10^−26 kg) = 4.0 × 10^6 J kg^−1 = 4 MJ kg^−1.

The known value is L = 2.5 MJ kg^−1. Not great, but not bad.

The surface tension, γ, is typically expressed as force per unit length, which is equivalent to the energy per unit area. At a surface, we estimate one of the four hydrogen bonds is missing, so

γ = (3 × 10^−20 J)/(3 × 10^−10 m)2 = 0.33 J m^−2 .

The measured value is γ = 0.07 J m^−2, which is about five times less than our calculation. This is a bigger discrepancy than I’d like for an order-of-magnitude estimate, but it’s not horrible.

The coefficient of viscosity, η, has units of kg m −1s −1. We can use the mass of the water molecule in kilograms, and the distance between molecules in meters, but we don’t have a time scale. However, energy has units of kg m 2s ^−2, so we can take the square root of mass times distance squared over energy and get a unit of time, τ

τ = √[(3 × 10^−26 kg) (3 × 10^−10 m)2/4(3 × 10^−20 J)] = 0.15 × 10^−12 s = 0.15 ps.

We can think of this as a time characterizing the vibrations about equilibrium of the molecules.

The viscosity of water should therefore be on the order of

η = (3 × 10^−26 kg)/[(3 × 10^−10 m) (0.15 × 10^−12 s)] = 0.67 × 10^−3 kg m^−1 s^−1.

Water has a viscosity coefficient of about η = 1 × 10^−3kg m^−1s^−1. I admit this analysis provides little insight into the mechanism underlying viscous effects, and it doesn’t explain the enormous temperature dependence of η, but it gets the right order of magnitude.

The heat capacity is the energy needed to raise the temperature of water by one degree. Thermodynamics implies that the heat capacity is typically equal to Boltzmann’s constant times the number of degrees of freedom per molecule times the number of molecules. The number of degrees of freedom is a subtle thermodynamic concept, but we can approximate it as the number of hydrogen bonds per molecule; about four. Often heat capacity is expressed as the specific heat, C, which is the heat capacity per unit mass. In that case, the specific heat is

C = 4 (1.4 × 10^−23 J K^−1)/(3 × 10^−26 kg) = 1900 J K^−1 kg^−1.

The measured value is C = 4200 J K^−1kg^−1, which is more than a factor of two larger than our estimate. I’m not sure why our value is so low, but probably there are rotational degrees of freedom in addition to the four vibrational modes we counted.

The self- diffusion constant of water can be estimated using the Stokes-Einstein equation relating diffusion and viscosity, D = kT/(6π ηa), where a is the size of the molecule. The thermal energy kT is about one eighth of the energy of a hydrogen bond. Therefore,

D = [(3 × 10^−20 J)/8]/[(6)(3.14)(0.67 × 10^−3 kg m^−1 s^−1)(3 × 10^−10 m)] = 10^−9 m^2 s^−1.

Figure 4.11 in IPMB suggests the measured diffusion constant is about twice this estimate: D = 2 × 10^−9 m^2 s^−1.

We didn’t do too bad. Three microscopic parameters, plus the temperature, gave us estimates of density, compressibility, speed of sound, latent heat, surface tension, viscosity, specific heat, and the diffusion constant. This is almost all the properties of water discussed in Mark Denny’s wonderful book Air and Water. Fermi would be proud.