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Physics LibreTexts

7.8: Examples

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7.8.1: Osmotic Pressure

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Figure 7.8.1: Illustrating the set-up for calculating osmotic pressure

An example of the use of the idea of the partition function in a very simple way is provided by the osmotic pressure. Here one considers a vessel partitioned into two regions, say, I and II, with a solvent (labeled A) on one side and a solution of the solvent plus a solute (labeled B) on the other side. The separation is via a semipermeable membrane which allows the solvent molecules to pass through either way, but does not allow the solute molecules to pass through. Thus the solute molecules stay in region II, as the Fig. 7.8.1. When such a situation is set up, the solvent molecules pass back and forth and eventually achieve equilibrium with the average number of solvent molecules on each side not changing any further. What is observed is that the pressure in the solution pII is higher than the pressure pI in the solvent in region I. Once equilibrium is achieved, there is no further change of volume or temperature either, so we can write the equilibrium condition for the solvent as

μ(I)A=μ(II)A

Correspondingly, we have z(I)A=z(II)A, for the fugacities. The partition function has the form

Z=NA,NBzNAAzNBBQNA,NB=NAzNAAQNA,0+NAzNAAzBQNA,1+···=ZA(1+1ZANAzNAAzBQNA,1+···)

Here ZA is the partition function for just the solvent. For simplicity, let us take the volumes of the two regions to be the same. Then we may write

logZA=pIVkT

even though this occurs in the formula for the full partition function in region II, since zA is the same for regions I and II. Going back to Z, we expand log in powers of zB, keeping only the lowest order term, which is adequate for dilute solutions. Thus

logZlogZA+zB(1ZANAzNAAQNA,1)

The derivative of the partition function with respect to µB is related to NB, as in Equation 7.4.16, so that

(1ZANAzNAAQNA,1)=NB

Further, logZ is given by pIIVkT. Using these results, Equation ??? gives

pII=pI+NBVkT=pI+nBkT

where nB is the number density of the solute. The pressure difference pIIpI is called the osmotic pressure.

7.8.2: Equilibrium of a Chemical Reaction

Here we consider a general chemical reaction of the form

νAA+νBBνCC+νDD

If the substances A, B, C, D can be approximated as ideal gases, the partition function is given by

logZ=zAQ1A+zBQ1B+zCQ1C+zDQ1D

For the individual chemical potentials, we can use the general formula in 7.5.10 but with the correction due to the βmc2 factor as in Equation 7.7.5, since we have different species of particles here. Thus

βμ=logn+βmc2log(mkT2πħ2)32

The condition of equilibrium of the reaction is given as νAμA+νBμBνCμCνDμD. Using Equation ???, this becomes

log(nνAAnνBBnνCCnνDD)=βc2(νAμA+νBμBνCμCνDμD)+νAfA+νBfBνCfCνDfDf=log(mkT2πħ2)32

The total pressure of the mixture of the substances is given from pV=kTlogZ as

p=(nA+nB+nC+nD)kT

So if we define the concentrations,

xi=nijnj,i=A,B,C,D

then we can rewrite Equation ??? as

xνAAxνBBxνCCxνDDpνA+νBνCνD=exp(βϵ+νAϕA+νBϕBνCϕCνDϕD)Kϵ=(νAmA+νBmBνCmCνDmD)c2ϕ=f+logkT

With our interpretation of the masses as rest energy, we see that ϵ is the heat of reaction, i.e., the total energy released by the reaction. ϵ is positive for an exothermic reaction and negative for an endothermic reaction. K, in Equation ???, is known as the reaction constant and is a function only of the temperature (and the masses of the molecules involved, but these are fixed once a reaction is chosen). The condition in Equation ??? on the concentrations of the reactants is called the law of mass action.

7.8.3 Ionization Equilibrium

Another interesting example is provided by ionization equilibrium, which is of interest in plasmas and in astrophysical contexts. Consider the ionization reaction of an atom X

XX++e

There is a certain amount of energy ϵI needed to ionize the atom X. Treating the particles involved as different species with possible internal states, we can use Equation 7.7.7 to write

logZ=zXgXQ1X+zegeQ1e+ZX+gX+Q1X+

By differentiation with respect to µ, we find, for each species,

z=ngeβmc2(2πħ2mkT)32

The condition for equilibrium µXµX+µe=0 is the same as (zXzX+ze)=1. Using Equation ???, this becomes

1=nXgXgX+nX+geneeβ(mXmX++me)c2(mX+memX)32(kT2πħ2)32

The mass of the atom X is almost equal to the mass of X+ and the electron; the difference is the binding energy of the electron in X. This is the ionization energy ϵI, (mXmX+me)c2=ϵI. Using this, Equation ??? can be rewritten as

(nX+nenX)=(gX+gegX)eβϵI(mX+memX)32(kT2πħ2)32

This is known as Saha’s equation for ionization equilibrium. It relates the number density of the ionized atom to that of the neutral atom. (While the mass difference is important for the exponent, it is generally a good approximation to take mXmX+ in the factor (mX+mX)32. So it is often omitted.) The degeneracy for the electron states, namely ge, is due to the spin degrees of freedom, so ge=2. The degeneracies gX and gX+ will depend on the atom and the energy levels involved.

The number densities can be related to the pressure by the equation of state; they are also important in determining the intensities of spectral lines. Thus by observation of spectral lines from the photospheres of stars, one can estimate the pressures involved.


This page titled 7.8: Examples is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by V. Parameswaran Nair via source content that was edited to the style and standards of the LibreTexts platform.

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