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5.5: Partition Functions can be Decomposed

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    204463
  • 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.

    Contributors and Attributions

    • www.chem.iitb.ac.in/~bltembe/pdfs/ch_3.pdf