Irodov 3.71: One Molecule in Three Thousand – Amit & Deepali Sinha Foundation

Problem Irodov 3.71

In water (permittivity $\varepsilon = 81$ , molar mass $18$ g/mol, density $1.0$ g/cm $^3$ ), a field of strength $E = 1.0$ kV/cm produces a polarization equal to what would result if one molecule out of every $N$ were turned exactly along the field and the rest stayed randomly oriented. Find $N$ . Each water molecule has the dipole moment $p = 0.62\cdot 10^{-29}$ C $\cdot$ m.

Given

$E = 1.0$ kV/cm
$p = 0.62\cdot 10^{-29}$ C $\cdot$ m
water, $\varepsilon = 81$ , $M = 18$ g/mol, density $1.0$ g/cm $^3$ (table values)

Find

N

Solution

Step 1The polarization the field produces

Water in the field E: its polarization P, the dipole moment per unit volume, drawn as the lattice of short arrows along the field. — Water in the field $\vect{E}$ : its polarization $\vect{P}$ , the dipole moment per unit volume, drawn as the lattice of short arrows along the field.

Water’s permittivity is the table value $\varepsilon = 81$ . In an isotropic dielectric the polarization is proportional to the field inside it:

P = \varepsilon_0(\varepsilon - 1)E \tag{1}

Step 2The same polarization from aligned molecules

Let $n_0$ be the number of molecules per unit volume. Randomly oriented molecules add up to no moment, so if one in $N$ points exactly along the field the moment per unit volume is:

P = \frac{n_0\,p}{N} \tag{2}

The concentration follows from water’s density $\rho = 1.0$ g/cm $^3$ and molar mass $M = 18$ g/mol, both table values, and Avogadro’s number:

n_0 = \frac{\rho N_A}{M} = \frac{1.0\cdot 10^{3}\cdot 6.02\cdot 10^{23}}{18\cdot 10^{-3}}\ \mathrm{m^{-3}} = 3.34\cdot 10^{28}\ \mathrm{m^{-3}}

Step 3Solving for

N

Equating (1) and (2):

\frac{n_0\,p}{N} = \varepsilon_0(\varepsilon - 1)E

Numerically, with $E = 1.0\cdot 10^5$ V/m:

N = \frac{3.34\cdot 10^{28}\cdot 0.62\cdot 10^{-29}}{8.85\cdot 10^{-12}\cdot 80\cdot 1.0\cdot 10^{5}} = \frac{0.207}{7.08\cdot 10^{-5}} = 2.9\cdot 10^{3}

N = \dfrac{n_0\,p}{(\varepsilon - 1)\,\varepsilon_0 E} \approx 3\cdot 10^{3}

(3)

Check

Thermal agitation sets the scale of the alignment. Freely turning dipoles at room temperature, $kT = 4.0\cdot 10^{-21}$ J, line up in a weak field by the average $\langle\cos\vartheta\rangle = pE/3kT$ :

\frac{pE}{3kT} = \frac{0.62\cdot 10^{-29}\cdot 1.0\cdot 10^{5}}{3\cdot 4.0\cdot 10^{-21}} = 5\cdot 10^{-5}

The aligned fraction from (3), $1/N = 3.4\cdot 10^{-4}$ , is of the same small order, so the result is what a thermal bias of strongly polar molecules can give; the factor of about seven between them comes from the field acting on a molecule in the liquid, which differs from the average field $E$ . Both are far below 1, as the linear law (1) requires.

Remark

Water’s large permittivity does not mean its molecules line up: at most a few in ten thousand do. It means each one carries a large permanent moment, so a tiny statistical bias toward the field already polarizes the liquid strongly. Heat the water and the bias weakens as $1/T$ , which is why the permittivity of water falls from about 88 near freezing to about 55 near boiling.