Boltzmann distribution law

Boltzmann distribution law

Boltzmann distribution law is concerned with probability of distribution of the system. Mathematically, the law is expressed as,

$$\frac{N_i}{N}$$ = $$\frac{g_i e^-\frac{E_i}{KT}}{\sum_i g_i e^-\frac{E_i}{KT}}$$

Where, E_iis the number of molecule occupying the energy level E_i.

N is the total number of molecule of the system.

is the statistical weight factor

k is Boltzmann constant

T is the temperature in Kelvin scale

Derivation

The thermodynamic probability is given by,

W = $$\frac{N!}{N_0! N_1! N_2!....}$$.........(1)

Taking ln to the both sides,

$$\ln$$ w = $$\ln$$ N_i- $$\sum_i$$ $$\ln$$ N_i!.........(2)

According to Sterling's approximation,

$$\ln$$ N_i= N $$\ln$$N - N

$$\ln$$ w = N $$\ln$$N - N -$$\sum_i$$ (N_i$$\ln$$ N_i- N_i)

$$\ln$$ w = N $$\ln$$N - N -$$\sum_i$$ N_i$$\ln$$ N_i+$$\sum_i$$ N_i........(3)

This law is derived on the assumption that total number of molecules of the system remains constant and total energy of the system also remains constant.

So,$$\sum_i$$ N_i= N

Now equation (3) becomes,

$$\ln$$ w = N $$\ln$$N - N -$$\sum_i$$ N_i$$\ln$$ N_i+ N

or,$$\ln$$ w =N $$\ln$$N -$$\sum_i$$ N_i$$\ln$$ N_{i .............}(4)

Differentiating equation (4) with respect to N_i

$$\frac{d}{dN_i}$$ $$\ln$$ w =$$\frac{d}{dN_i}$$ (N $$\ln$$N) -$$\sum_i$$ ($$\frac{d}{dN_i}$$N_i$$\ln$$ N_i)

$$\frac{d}{dN_i}$$ $$\ln$$ w = 0 -$$\sum_i$$ [N_i$$\frac{d}{dN_i}$$$$\ln$$ N_i+$$\ln$$ N_i$$\frac{d}{dN_i}$$N_i]

= -$$\sum_i$$ [N_i× $$\frac{1}{N_i}$$ +$$\ln$$ N_i]

$$\frac{d}{dN_i}$$ $$\ln$$ w = -$$\sum_i$$ (1 +$$\ln$$ N_i)

-d $$\ln$$ w = -$$\sum_i$$(1 +$$\ln$$ N_i) dN_i

For the most probable distribution of particle, the derivative of$$\ln$$ w is zero,

Now, equation (5) becomes,

$$\sum_i$$(1 +$$\ln$$ N_i) dN_i = 0...............(6)

The total number of molecules of the system is assumed to be constant

$$\sum_i$$ N_i= N

or, $$\sum_i$$dN_i = 0..........(7)

Total energy of the system remains constant. So,

$$\sum_i$$ N_iE_i= constant

$$\sum_i$$E_idN_i= 0...........(8)

Here, we use Lagrange's method of multiplier (Lagrange's undetermined multiplier)

Now, Equation (7)× $$\alpha$$^'+ Equation (8)× $$\beta$$ + Equation (6)

$$\sum_i$$$$\alpha$$^' dN_i+$$\sum_i$$$$\beta$$ E_idN_i+$$\sum_i$$(1 +$$\ln$$ N_i) dN_i= 0

$$\sum_i$$ [$$\alpha$$^'+$$\beta$$ E_i + 1 +$$\ln$$ N_i]dN_i = 0........(9)

$$\sum_i$$ [$$\alpha$$+$$\beta$$ E_i +$$\ln$$ N_i]dN_i = 0........(10)

where, $$\alpha$$ =$$\alpha$$^'+ 1

From equation (10),$$\alpha$$+$$\beta$$ E_i +$$\ln$$ N_i= 0

[Each$$\sum$$ term of equation (10) should be zero for the validity of equation(10)]

$$\ln$$ N_i= -($$\alpha$$+$$\beta$$ E_i)

Taking anti $$\ln$$

N_i = e ^{-($$\alpha$$+$$\beta$$ E_i)...........(11)}

$$\beta$$ = $$\frac{1}{KT}$$

Now, equation (11) becomes,

N_i= e^{-(α + $$\frac{E_i}{KT}$$)}

N_i= e^-α. e ^{$$\frac{E_i}{KT}$$)}.........(12)

Equation (12) is called Boltzmann distribution law

We know, N =$$\sum_i$$N_i

or, N =$$\sum_i$$ e^{-($$\alpha$$+$$\beta$$ E_i)}

or, N =$$\sum_i$$ e^-$$\alpha$$. e^{-$$\beta$$ E_i}

or, e^-$$\alpha$$= $$\frac{N}{\sum_ie^ \beta E_i}$$

Putting the value of e^-αinto equation(12)

N_i= $$\frac{N}{\sum_i e^-(\beta E_i)}$$ .e^{-$$\frac{E_i}{KT}$$)}

$$\frac{N_i}{N}$$ = $$\frac{e^-\frac{E_i}{KT}}{\sum_i e^-\frac{E_i}{KT}}$$...............(13)

When there are more states corresponding to E_i, the energy level is degenreate. Then equation(13) can be written aa,

$$\frac{N_i}{N}$$ = $$\frac{g_i e^-\frac{E_i}{KT}}{\sum_i g_i e^-\frac{E_i}{KT}}$$....................(14)

References

Wikipedia. n.d. <https://en.wikipedia.org/wiki/Boltzmann_distribution>.