17.8: Partition Functions can be Decomposed

Express the partition function of a collection of N molecules $$Q$$ in terms of the molecular partition function $$q$$. Assuming the N molecules to be independent, the total energy $$E_{tot}$$ of molecules is a sum of individual molecular energies

$E_{tot} = \sum_i E_i$

and all possible

$Q = \sum _{\text{all possible energies}} e^{-E/k_BT} = \sum _i e^{-E_i/k_BT} \sum _j e^{-E_j/k_BT} \sum _k e^{-E_k/k_BT} ... \sum _i e^{-E_i/k_BT}$

$Q = q \times q \times q \times ... q^N$

Here $$\epsilon_i^{(1)}$$, $$\epsilon_i^{(2)}$$, $$\epsilon_i^{N}$$ are energies of individual molecules and a sum of all energies can only come from summing over all $$\epsilon_i$$. Gibbs postulated that

$Q = \dfrac{q^N}{N!}$

where the $$N!$$ in the denominator is due to the indistinguishability of the tiny molecules (or other quantum particles in a collection).

Maxwell Boltzmann Distribution

The Maxwell Boltzmann distribution of molecular speeds can be derived from the Boltzmann distribution. If we represent a molecular velocity by $$\vec{v}$$, it has three independent components in 3-D space ($$v_x$$, $$v_y$$, and $$v_z$$ in the three directions $$x$$, $$y$$ and $$z$$. Let us consider only monatomic gas of mass $$m$$. The probability $$F(x,y,z)$$ that given molecule will have velocity components lying between $$x$$ and $$x + dx$$, $$y$$ and $$y + dy$$ and $$z$$ and $$z + dz$$ can be written as

$F(x, y, z) \,dz\, dy\, dz = f (x) f (y) f(z)\, dx \,dy\, dz \label{MB1}$

$$F$$ is written as a product of three functions $$f$$ because $$x$$, $$y$$, and $$z$$ are independent and since nature is isotropic and does not distinguish between $$x$$, $$y$$ and $$z$$ (unless directional fields like gravitational or electromagnetic are present), the form of $$f$$ is the same in the three directions. Again, since there is no distinction between positive and negative $$x$$, $$f$$ depends on $$| x |$$ or $$x^2$$. We can rephrase Equation $$\ref{MB1}$$ as $F( ) = f ( ) f ( ) f( )$ The only function that satisfies the above equation is an exponential function since

$e^{x^2 + y^2 + x^2} = e^{x^2} + e^{y^2 }+ e^{x^2}$

and so we conclude that $$f ( )$$ may be written as

$f( ) = C e^{\pm v_x^2} → C e^{- v_x^2}$

We take only the negative exponent ($$C$$ and $$b$$ are positive) because a positive exponent implies that very large velocities have very high probabilities which is highly unlikely. To evaluate $$C$$, We invoke the physical argument that the velocity has to lie somewhere between $$-\infty$$ to $$+\infty$$ and that the total probability is 1 i.e.

$\int _{ -\infty}^{+\infty} f(v_x^2) \,d v_x = C \int _{ -\infty}^{+\infty} e^{- v_x^2} \,d v_x = 1$

The above integral is a standard Gaussian integral for the following form

$\int _{ -\infty}^{+\infty} e^{-ax^2} dx = \dfrac{\sqrt{\pi}}{2}$

But since we want the right side to be unity, therefor

$C \sqrt{ \dfrac{π}{b}} = 1$

or

$C = \sqrt{\dfrac{b}{π}}$

and f ( x ) = (b / π) 1/2 2 - b v x e From a probability distribution such as f ( x), average quantities can be determined. The averages of x and x 2 are given by

< x > = ∞ − ∞ ∫ (b / π) 1/2 x 2 - b v x e d x = 0 (3.31)

< x 2 > = ∞ − ∞ ∫ x 2 (b / π) 1/2 2 - b v x e d x =1/2b (3.32)

The averages have been denoted by < > . We have also used another standard integral, 2 2 a x 0 x e ∞ − ∫ dx = 1/4 (π/a 3 ) 1/2 (3.33) The integral for < x > is zero because the value of the integrand for positive x is equal and opposite to its value at – x. Thus the area on the left of x = 0 is equal and opposite in sign to the area on the right. This is a special case of a general result that the integral of 69 the product of an even function and an odd function of x is zero over a symmetric interval around zero. To evaluate b, we take the help of the kinetic theory of gases. Do look up the details. The pressure of a gas is given in terms of the mean square velocity (speed) as

p = 1/3 (N/ V) m

Where N/ V = (number of molecules of the gas / volume) = the density of the gas. But N = nN A where n = number of moles and N A , the Avogadro number; and since pV = nRT, we have

pV = 1/3 Nm < v 2 > = 1/3 n N A m < v 2 > = nRT (3.34)

< v 2 > = 3 RT / mN A = 3 k B T / m (3.35)

Where k B = R / N A is the Boltzmann constant, 1.3806 X 10 -23 J / K. Since = 3 k B T / m, we have = k B T/ m and substituting, we get b = m / 2 k B T. Equations for f and F now become f(v x ) = (m / 2 k B T ) 1/2 - m v / 2 k T B 2 x e (3.36) F(v x , v y , v z ) = (m / 2 k B T) 3 / 2 B 2 2 2 m (v + v + v )/2k T x y z e − (3.37) This is the Maxwell - Boltzmann distribution of molecular speeds.

F( v x , v y , v z ) d v x dv y dv z gives the probability of finding an arbitrary molecule with a velocity ( v x , v y , v z ) in the volume element d v x dv y dv z . A more appealing interpretation of the same is that it is the fraction (of the total molecules) of molecules having velocities ( v x , v y , v z ). Analogous to the radial probability distribution in coordinate space, we can now estimate the probability of finding a particle in a spherical shell of volume 2 4 v dv π . This probability in such a spherical shell of radius v and thickness dv is given by 2 4 v dv π (m / 2 k B T) 3 / 2 B 2 2 2 m (v + v + v )/2k T x y z e − (3.