Introduction
There are two types of thermodynamics one is called Classical thermodynamics and another one is called statistical thermodynamics.
In Classical thermodynamics, we study Intensive & extensive properties like energy, heat, volume, pressure, Temperature, Maximum work function, free energy function and laws of thermodynamics; whereas, statistical thermodynamics link between quantum mechanics as well as classical thermodynamics.
In statistical method elaborates on Microscopic and Macroscopic properties, which give information about how many ways a number of molecules distribute in different energy levels.
Statistical mechanics can be easily applied to the simple ideal gaseous system for monoatomic and diatomic gases.
In a Physical situation nature is described by three types of statics,
- Maxwell – Boltzman (M-B) statics
- Bose-Einstein (B-E) statics
- Fermi – Dirac (F-D) statics
Maxwell – Boltzmann (M-B) statics:
in this statics, particles are assumed to be distinguishable and any number of particles obeying M-B statics are called boltzons Or Maxellons.
Now, consider a system N distinguishable particles occupying energy levels ε0, ε1, ε2 etc.
A total number of arrangements for a number of particles placing in n0 particles in ground state energy ε0 = n1 particles in the First excited state energy level ε1 and n2 particles second excited state energy level ε2 and so on, is called thermodynamic probability (W) of the macrostate.
Here, determine W i.e. how many microstates correspond to a given macrostate W is,
\[\displaystyle \begin{array}{l}\text{W}=\frac{{\text{N}!}}{{{{\text{n}}_{1}}!~{{\text{n}}_{2}}!~{{\text{n}}_{3}}!\ldots \ldots {{\text{n}}_{\text{i}}}!}}\\\text{W}=\frac{{\text{N}!}}{{\sum{{{{\text{n}}_{\text{i}}}!}}}}\_\_\_\_\_\_\_\_(1)~\\Here,~\text{N}=\sum{{{{\text{n}}_{\text{i}}}!}}\end{array}\]
Equation (1) N = Total number of particles and summation of overall energy levels.
It means a number of energy levels in more than one way, which means more than one quantum state in the same energy.
If energy level is to be degenerate, suppose to be gi the degeneracy of the energy level is εi.
That means single-particle an energy level an is ith it is a distributing in a ways.
If two particles in have a ith energy level that means distribution is \[\displaystyle g_{i}^{{ni}}\] So, Thermodynamic probability for a system of N particles is,
\[\displaystyle \text{W}=\text{N}!\underset{i}{\mathop \sum }\,\frac{{\text{g}_{\text{i}}^{{ni}}}}{{{{\text{n}}_{\text{i}}}!}}~\text{X}~constant\_\_\_\left( 2 \right)\]
Entropy S and probability W for the given state of system the Boltzmann equation, for statistical mechanics, is,
\[\displaystyle \begin{array}{l}S~=~K~\ln ~W\_\_\_\_\_\_\_\_\left( 3 \right)\\At\text{ }equilibrium\text{ }state\text{ }probability\text{ }must\text{ }be\text{ }maximum\text{ }so,\\\text{S}~=~\text{K}~\ln ~~{{W}_{{\max }}}\_\_\_\_\_\_\_\_\_\left( 4 \right)\\For\text{ }a\text{ }closed\text{ }system\text{ }of\text{ }independent\text{ }particles\text{ }meet\text{ }\\two\text{ }requirements,\\Number\text{ }of\text{ }total\text{ }particles\text{ }is\text{ }constant\text{ }i.e.\\\text{I}\text{.}Number\text{ }of\text{ }total\text{ }particles\text{ }is\text{ }constant\text{ }i.e.\\N=\sum\limits_{i}{{{{n}_{i}}}}=\text{ constant}\_\_\_\_\_\_\left( 5 \right)\\\text{II}.Total\text{ }energy\text{ }U\text{ }of\text{ }the\text{ }system\text{ }constant\text{ }that\text{ }is\\N=\sum\limits_{i}{{{{n}_{i}}{{c}_{i}}}}=\text{ constant}\_\_\_\_\_\left( 6 \right)\\Constancy\text{ }of\text{ }number\text{ }of\text{ }total\text{ }practical\text{ }that\text{ }is,\\dN+\sum =0\_\_\_\_\_\_\_\_\_\_\_\_\_\_(7)\\Constancy\text{ }of\text{ }number\text{ }of\text{ }total\text{ }particles\text{ }that\text{ }is,\\dU=\sum{{{{\varepsilon }_{i}}=0\_\_\_\_\_\_\_\_\_\_\_(8)}}\\Where\text{ }taking\text{ }logarithms\text{ }of\text{ }both\text{ }sides\text{ }in\text{ }equation\text{ }\left( 2 \right)\text{ }we\text{ }get,\\\ln W=\ln N!+\sum\limits_{i}{{{{n}_{i}}\ln {{g}_{i}}-\sum\limits_{i}{{\ln {{n}_{i}}!+\text{ constant}}}}}\_\_\_(9)\\Now,\text{ }according\text{ }to\text{ }stirling\text{ }approximation\text{ }for\text{ }large\text{ x},\\\ln x!=x\ln x-x\_\_\_\_\_\_(10)\end{array}\]
\[\displaystyle \begin{array}{l}\ln W=\ln N!+\sum\limits_{i}{{{{n}_{i}}\ln {{g}_{i}}-\sum\limits_{i}{{{{n}_{i}}\ln {{n}_{i}}+\sum\limits_{i}{{{{n}_{i}}+\text{ constant}}}}}}}\\=(N\ln N-N)+\sum\limits_{i}{{{{n}_{i}}}}\ln {{g}_{i}}-\sum\limits_{i}{{{{n}_{i}}\ln {{n}_{i}}+N+\text{constant}}}\\=N\ln N+\sum\limits_{i}{{{{n}_{i}}}}\ln {{g}_{i}}-\sum\limits_{i}{{{{n}_{i}}}}\ln {{n}_{i}}+\text{constant }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ (11)}\\\text{Differniating in above equation is N and }{{\text{g}}_{i}}\text{ are constant, }\\\text{it is,}\\d\ln W=\sum\limits_{i}{{\ln {{g}_{i}}d{{n}_{i}}-}}\sum\limits_{i}{{\ln {{n}_{i}}d{{n}_{i}}}}-\sum{{{{n}_{i}}d\ln {{n}_{i}}\_\_\_(12)}}\\Now,\\\sum\limits_{i}{{{{n}_{i}}d\ln {{n}_{i}}=\sum\limits_{i}{{{{n}_{i}}}}\frac{{d{{n}_{i}}}}{{{{n}_{i}}}}}}\\=\sum\limits_{i}{{d{{n}_{i}}=0}}\_\_\_\_\_(13)\\\text{At equilibrium,}\\d\ln W=\sum{{\ln {{g}_{i}}d{{n}_{i}}-\sum{{\ln {{n}_{i}}d{{n}_{i}}=0}}}}\_\_\_\_(14)\\\text{Above equation }\left( {\text{14}} \right)\text{ change the lnW it result }\\\text{number of particles in each energy level is }\\\text{different}\text{.}\\\text{For open system, }{{\text{n}}_{i}}\text{ particles free with any restriction }\\\text{not dependent to one another}\text{. So, in equation (14) each }\\\text{coefficient of d}{{\text{n}}_{i}}=0\\\text{For the solution used lagrange }\!\!’\!\!\text{ s undetermined multipliers}\\\text{and again write eqation(14),}\\\sum\limits_{i}{{\ln }}\frac{{{{g}_{i}}}}{{{{n}_{i}}}}d{{n}_{i}}=0\_\_\_\_\_\_\_(15)\\\text{Now , Multiplying equation }\left( \text{7} \right)\text{ }\!\!\And\!\!\text{ }\left( \text{8} \right)\text{ for arbitrary }\\\text{constant }\alpha \text{ and }\beta \text{ weget,}\\\text{dN=}\sum{{d{{n}_{i}}\times \alpha }}=0\\dU=\sum{{{{\varepsilon }_{i}}\times \beta =0}}\\\text{And this equation subtracting from equation }\left( {\text{16}} \right)\text{ we get,}\\\sum\limits_{i}{{\left[ {\ln \frac{{{{g}_{i}}}}{{{{n}_{i}}}}-\alpha -\beta {{\varepsilon }_{i}}} \right]d{{n}_{i}}=0}}\_\_\_(16)\\\text{Here, }\alpha \text{ and }\beta \text{ is lagrange }\!\!’\!\!\text{ s undetermined multipliers}\\\text{Select the values of }\!\!~\!\!\text{ }\alpha \text{ and }\beta \text{ }\!\!~\!\!\text{ that one of the terms in the }\sum{{}}\text{is}\\\text{Zero, and the value of }\!\!~\!\!\text{ d}{{\text{n}}_{i}}\text{ is irrelevant}\text{.}\\\text{All other d}{{\text{n}}_{i}}\text{ terms that not dependent to one another, since }\!\!~\!\!\text{ d}{{\text{n}}_{i}}\\\text{can be obtained from this equation }\left( \text{7} \right)\text{ }\!\!~\!\!\text{ in }\!\!~\!\!\text{ d}{{\text{n}}_{i}}\\\text{terms,Coefficients of }\!\!~\!\!\text{ d}{{\text{n}}_{i}}\text{ in equation }\left( {\text{16}} \right)\text{ equal to zero}\text{.}\\\ln \left( {\frac{{{{g}_{i}}}}{{{{n}_{i}}}}} \right)-\alpha -\beta {{\varepsilon }_{i}}=0\\\ln \frac{{{{g}_{i}}}}{{{{n}_{i}}}}=\alpha +\beta {{\varepsilon }_{i}}\\or\\\ln {{n}_{i}}=\ln {{g}_{i}}-\alpha -\beta {{\varepsilon }_{i}}\\or\\{{n}_{i}}={{g}_{i}}{{e}^{{-\alpha -\beta {{\varepsilon }_{i}}}}}\_\_\_\_\_(17)\\\begin{array}{*{20}{l}} \begin{array}{l}\text{Equation }\left( {\text{17}} \right)\text{ is a Boltzman distribution law, and gives most }\\\text{probable distribution in a macrosate}\text{.}\end{array} \\ ~ \end{array}\end{array}\]
Bose-Einstein Statistics
Here, consider a one system for N indistinguishable particles is like ni particles are in a energy level that genenerey gi .This particles distributed in a state.
Suppose energy level have gi – 1 partitions that are enough to distributed energy level have a intervals.
Here, a number of possible distributions of ni particles gi in state conclude of partitions and particles.
Now, number of total permutations ni particles and (gi – 1) partitions that is ( ni + gi – 1)! ; whereas particles and partitions are identical. It means, Alternations of two partitions do not switch an arrangement; as well as alternations of two particles does not switch an arrangement.
It is compulsory divide ( ni+ gi – 1)! number of permutations of the (gi – 1) partition, viz ( gi – 1)! and particles number of permutation viz ni! to get the possible number of arrangements of the ni particles in the level of energy so,
\[\displaystyle \text{The number of arrangements = }\frac{{\left( {{{n}_{i}}+{{g}_{i}}-1} \right)!}}{{{{n}_{i}}({{g}_{i}}-1)!}}\_\_\_(1)\]
According to Maxwell Boltzman statics, suppose number of total particles is constant also a system of total energy constant it means,
\[\displaystyle \begin{array}{l}\sum\limits_{i}{{{{n}_{i}}=\text{constant}}}\\\sum{{{{n}_{i}}{{\varepsilon }_{i}}=\text{constant}}}\end{array}\]
Here, W has thermodynamics probability of the system have N particles is given by,
\[\displaystyle W=\sum\limits_{i}{{\left[ {\ln ({{n}_{i}}+{{g}_{i}}-1)!-\ln {{n}_{i}}!-\ln ({{g}_{i}}-1)!} \right]}}+\text{consrant }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ (3)}\]
But, ni and gi is very large numbers, and according to stirling’s approximation
\[\displaystyle \begin{array}{l}\ln x!=x\ln x-x,\text{ to get}\\\ln W=\sum\limits_{i}{{[({{n}_{i}}+{{g}_{i}})\ln ({{n}_{i}}+{{g}_{i}})-{{n}_{i}}\ln {{n}_{i}}-{{g}_{i}}in{{g}_{i}}]+\text{constant }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ (4)}}}\end{array}\]
Here, we have set ni + gi – 1 = ni + gi as well as gi – 1 = gi , Because ni is very large, it behave like continuous variable.
Differentiations of above equation (4) with respect to ni and differential equal to zero we get most probable thermodynamic state of system,
\[\displaystyle \begin{array}{l}\delta \ln W=[\ln {{n}_{i}}-\ln ({{n}_{i}}+{{g}_{i}})]\delta {{n}_{i}}=0\\\delta \ln W=\sum\limits_{i}{{\left[ {\ln \frac{{{{n}_{i}}}}{{({{n}_{i}}+{{g}_{i}})}}} \right]\delta {{n}_{i}}=0}}\_\_\_\_(5)\\\text{accroding M-B sataics e}{{\text{q}}^{n}}\\\delta N=\sum\limits_{i}{{\delta {{n}_{i}}=0}}\_\_\_\_\_(6)\\\delta U=\sum\limits_{i}{{{{\varepsilon }_{i}}d{{n}_{i}}=0}}\_\_\_\_(7)\end{array}\]
In equation 5, 6 & 7 using method of Lagrange’s undetermined multipliers,
\[\displaystyle \sum\limits_{i}{{\left[ {\ln \frac{{{{n}_{i}}}}{{({{n}_{i}}+{{g}_{i}})}}+\alpha +\beta {{\varepsilon }_{i}}} \right]}}\delta {{n}_{i}}=0\_\_\_(8)\]
The variations δni are not dependent to one another so,
\[\displaystyle \ln \frac{{{{n}_{i}}}}{{({{n}_{i}}+{{g}_{i}})}}+\alpha +\beta {{\varepsilon }_{i}}=0\_\_\_\_(9)\]
Where,
\[\displaystyle \begin{array}{l}\ln \left[ {\frac{{{{g}_{i}}}}{{{{n}_{i}}}}+1} \right]=\alpha +\beta {{\varepsilon }_{i}}\\or\\\frac{{{{g}_{i}}}}{{{{n}_{i}}}}+1={{e}^{{\alpha +\beta {{\varepsilon }_{i}}}}}\_\_\_\_(10)\end{array}\]
∴\[\displaystyle {{n}_{i}}=\frac{{{{g}_{i}}}}{{\left[ {exp\left( {\alpha +\beta {{\varepsilon }_{i}}} \right)-1} \right]}}\_\_(11)\]$
For, equation (11) is the most probable distributation of particles N having number of ways energy levels according to Bose –Einstein statics.
Fermi – Dirac Statics
Consider the particles are distributed in sates (ni < gi ) Here, gi that is degeneracy of the energy level. Suppose particles are distinguishable that may be first particle placed in one of the state and second particles placed in remaining gi – 1 state and so on. Expression for number of arrangements is
gi!(gi-ni)!
However, particles are indistinguishable and above equation divided number of possible permutation ni of particles that is ni!.
Arrangement of a number of ni particles in ith energy level is given by,
\[\displaystyle \frac{{{{g}_{i}}!}}{{\left( {{{n}_{i}}!\left( {{{g}_{i}}-{{n}_{i}}} \right)!} \right)}}\]
Thermodynamic probability W for the system of N particles, we get,
\[\displaystyle W=\sum\limits_{i}{{\frac{{{{g}_{i}}!}}{{{{n}_{i}}!\left( {{{g}_{i}}-{{n}_{i}}} \right)!}}\times \text{constant }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ (1)}}}\]
Taking logarithm both sides in above equation (1) obtained,
\[\displaystyle \ln W=\sum\limits_{i}{{\left[ {\ln {{g}_{i}}!-\ln {{n}_{i}}!-\ln \left( {{{g}_{i}}-{{n}_{i}}} \right)!} \right]}}+\text{constant }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ (2)}\]
Suppose ni , gi and gi – ni was very large, and apply stirling’s approximation we get
as,
\[\displaystyle \begin{array}{l}N=\sum\limits_{i}{{{{n}_{i}}=\text{contant and}}}\\U=\sum\limits_{i}{{{{n}_{i}}{{\varepsilon }_{i}}}}=\text{constant,}\end{array}\]
and
\[\displaystyle \delta N=\sum\limits_{i}{{\delta {{n}_{i}}}}\text{ and }\delta U=\sum\limits_{i}{{d{{n}_{i}}}}=0\_\_\_(4)\]
For applying Lagrange’s method of undetermined multipliers, we obtain
\[\displaystyle \sum\limits_{i}{{\left[ {\frac{{\ln {{n}_{i}}}}{{\left( {{{g}_{i}}-{{n}_{i}}} \right)+\alpha +\beta {{\varepsilon }_{i}}}}} \right]\delta }}{{n}_{i}}=0\_\_\_\_(5)\]
As the variations δni are not dependent to one another so,
\[\displaystyle \frac{{\ln {{n}_{i}}}}{{\left( {{{g}_{i}}-{{n}_{i}}} \right)+\alpha +\beta {{\varepsilon }_{i}}}}=0\]
or
\[\displaystyle \ln \left( {\frac{{{{g}_{i}}}}{{{{n}_{i}}}}} \right)-1=\alpha +\beta {{\varepsilon }_{i}}\]
or
\[\displaystyle \left( {\frac{{{{g}_{i}}}}{{{{n}_{i}}}}} \right)-1={{e}^{{\alpha +\beta {{\varepsilon }_{i}}}}}\_\_\_\_\_(6)\]
∴\[\displaystyle {{n}_{i}}=\frac{{{{g}_{i}}}}{{\left[ {e\times p\left( {\alpha +\beta {{\varepsilon }_{i}}} \right)+1} \right]}}\_\_\_\_(7)\]
Above equation (7) is most probable distribution having N particles of different energy levels according to Fermi – Dirac statistics.
Citation & Reference:
|
(Sheadiya, 2021) Sheladiya, B. s. (2021, December 29). STATISTICAL THERMODYNAMICS. Retrieved from www.purechemistry.org: https://www.purechemistry.org/statistical-thermodynamics/ |
