Boltzmann Canonical Distribution Law and Maxwell's Distribution of Velocities,

Boltzmann Canonical Distribution Law

Let \(n_1,n_2,...N_i\) be the number of gas molecules in cell 1, cell 2,...,cell i in the equilibrium state. The exchange of molecule may occur in the cell but the total number of molecules in the system remains constant. Also energy of the system remains constant. The distribution of molecules are in following constraints.

(1) Conservation of mass i.e., \(n_1+n_2+...n_i=N\) \begin{align*}or,\delta n_1+ \delta n_2+...+\delta n_i=0\rightarrow 1\end{align*}since total number of molecules are constant

(2)The conservation of energy i.e., \(\varepsilon_1n_1 + \varepsilon_2n_2+...\varepsilon_in_i=E\)\begin{align*}\implies \varepsilon_1\delta n_1+\varepsilon_2\delta n_2...+\varepsilon_i\delta n_i=0 \rightarrow 2\end{align*}Since total energy is constant.

The probability of molecules under given restriction is \(\Omega=\frac{N!}{n_1!n_2!n_3!...n_i!}\)

If the system is in equilibrium, the thermodynamic probability is maximum i.e., \(\delta \Omega =0\)

SInce \(\Omega\) is maximum log\(\Omega\) is maximum \begin{align*}\therefore \delta(log\Omega)=0\rightarrow 3 \end{align*}\begin{align*}We\space have\space \Omega=\frac{N!}{n_1!n_2!n_3!...n_i!}\end{align*}\begin{align*}or,\space log\Omega=logN!-logn_1!-logn_2!-...logn_i!\rightarrow 4 \end{align*}\begin{align*}=logN!-\sum_ilogn_i! \end{align*}\begin{align*}\therefore \delta(log\Omega)=\delta[logN!-\sum_ilogn_i!] \end{align*}Now applying Stirling theorem\begin{align*}logx!=xlogx-x, if\space x\space is\space large \end{align*}\begin{align*}\therefore \delta(log\Omega)=\delta[NlogN-N-\sum_i(n_ilogn_i-n_i)]\end{align*}\begin{align*}or,\space 0=0-\sum_i[logn_i\delta n_i+n_i.\frac{1}{n_i}\delta n_i-\delta n_i] \end{align*}\begin{align*}\therefore \sum_i[logn_i\delta n_i]=0 \end{align*}\begin{align*}\therefore logn_1\delta n_1+ logn_2\delta n_2+...logn_i\delta n_i=0\rightarrow 5\end{align*}

Equation 1,3 and 5 are independent. We can combine them without losing their independence multiplying by arbitary constants.

Thus by multilying (1) and (2) by \(\alpha \space and\space \beta \) respectively and adding the resulting equation (5) where \(\alpha \space and\space \beta \) are called the Lagrange's undetermined multipliers(constants).\begin{align*}\therefore \alpha(\delta n_1+ \delta n_2+...+\delta n_i) +\beta(\varepsilon_1\delta n_1+\varepsilon_2\delta n_2...+\varepsilon_i\delta n_i)+(logn_1\delta n_1+ logn_2\delta n_2+...logn_i\delta n_i)=0\end{align*}\begin{align*}\implies (\alpha+\beta\varepsilon_1+logn_i)\delta n_1+ (\alpha+\beta\varepsilon_2+logn_i)\delta n_2+(\alpha+\beta\varepsilon_i+logn_i)\delta n_i=0\end{align*}In equation 5, this is the combination of independent equation where \(\delta n_1,\delta n_2,...,\delta n_i\) cannot be zero. so,\begin{align*}\alpha+\beta\varepsilon_i+logn_i=0 \end{align*}\begin{align*}logn_i= -(\alpha+\beta\varepsilon_i)\end{align*}\begin{align*}\implies n_i=e^{ -(\alpha+\beta\varepsilon_i)} \end{align*}\begin{align*}=e^{-\alpha}.e^{-\beta \varepsilon_i} \end{align*}\begin{align*}\therefore n_i=A.e^{-\beta \varepsilon_i}\space where,\space A= e^{-\alpha}=constant\end{align*}This equation gives the number of molecules in i^th cell as a function of energy associated with each particle in that cell and is called Boltzmann's canonical distribution law.

Partition Function

We know from Boltzmann canonical distribution law, \(n_i=A.e^{-\beta \varepsilon_i}\) where n_i is the number of particle in i^th cell each having energy \(\varepsilon_i\).

So sum of all \(n_i\) must be equal to the total number N i.e., \(\sum_in_i=N=A\sum e^{-\beta \varepsilon_i}\)

The sum \(\sum e^{-\beta \varepsilon_i}\) is called the partition function or the sum of states and is represented by the letter Z i.e., \(Z=\sum e^{-\beta \varepsilon_i}\)

\begin{align*}So\space N=AZ\implies A=\frac{N}{Z}\end{align*}So the molecules in i^th cell in state of maximum thermodynamic probability is, \(n_i=\frac{N}{Z}e^{-\beta \varepsilon_i}\). Here \(\beta =\frac{1}{KT}\),K is the Boltzmann constant and T is the absolute temperature.

Maxwell's Distribution Law of Velocities

Let us consider an ideal gas in a vessel of volume V. If the gas is in state of equilibrium then according to Boltzmann cannonical distribution law, the number of molecules in a cell of energies \(\varepsilon_i\) will be \(n_i=A.e^{-\beta \varepsilon_i}\). The number of molecules having energy\(\varepsilon_i\) and having position co-ordinates betwee x and x+dx, y and y+dy, z and z+dz and velocity components \(V_x\space and\space V_x+dV_x,V_y\space and\space V_y+dV_y and\space V_z\space and\space V_z+dV_z\) is given by

\begin{align*}n_idxdydzdV_xdV_ydV_z=A.e^{-\beta \varepsilon_i}dxdydzdV_xdV_ydV_z\end{align*}But \(\varepsilon_i\)=Energy of particle =\(\frac12mV^2=\frac12(dV_x^2+dV_y^2+dV_z^2)\)\begin{align*}which\space yeilds\space n_idxdydzdV_xdV_ydV_z=A.e^{-\beta \frac12(dV_x^2+dV_y^2+dV_z^2)}dxdydzdV_xdV_ydV_z\rightarrow 1\end{align*}Integrating 1 overall available volume and all ranges of velocities i.e.,\begin{align*}N=\int\int\int\int\int\int A.e^{-\beta \frac12(dV_x^2+dV_y^2+dV_z^2)}dxdydzdV_xdV_ydV_z\end{align*}\begin{align*}=AV\int_{-\infty}^{\infty}e^{\frac{-m\beta V_x^2}{2}}dV_x\int_{-\infty}^{\infty}e^{\frac{-m\beta V_y^2}{2}}dV_y\int_{-\infty}^{\infty}e^{\frac{-m\beta V_z^2}{2}}dV_z\end{align*}\begin{align*}=AV(\frac{2\pi}{m\beta})^{1/2}(\frac{2\pi}{m\beta})^{1/2}(\frac{2\pi}{m\beta})^{1/2}\end{align*}\begin{align*}=AV(\frac{2\pi}{m\beta})^{3/2}\end{align*}\begin{align*}\therefore A=\frac{N}{V}(\frac{m}{2\pi KT})^{3/2}\rightarrow 2\end{align*}Therefore, with the help of equation 2, equation 1 becomes,\begin{align*}n_idxdydzdV_xdV_ydV_z=\frac{N}{V}(\frac{m}{2\pi KT})^{3/2}e^{-m(dV_x^2+dV_y^2+dV_z^2)/2KT}dxdydzdV_xdV_ydV_z\rightarrow 3\end{align*}Therefore, the number of molecules having velocity coordinates in the range \(V_x\space to\space V_x+dV_x,V_y\space to\space V_y+dV_y and\space V_z\space to\space V_z+dV_z\) irrespective to the position co-ordinates can be found by integrating equation 3 with respect to position co-ordinates.\begin{align*}\therefore n_idV_xdV_ydV_z=\frac{N}{V}(\frac{m}{2\pi KT})^{3/2}e^{-m(dV_x^2+dV_y^2+dV_z^2)/2KT}dV_xdV_ydV_zV\end{align*}\begin{align*}=N(\frac{m}{2\pi KT})^{3/2}e^{-m(dV_x^2+dV_y^2+dV_z^2)/2KT}dV_xdV_ydV_z\rightarrow 4\end{align*}The number of molecules having velocity components in the range \(V_x\space to\space V_x+dV_x\) irrespective of \(V_x,V_y,V_zx,y,z\)is obtained by integrating equation 4 with respect to V_x and V_y i.e.,\begin{align*}n_idV_x=N(\frac{m}{2\pi KT})^{3/2}\int\int e^{-m(dV_x^2+dV_y^2+dV_z^2)/2KT}dV_xdV_ydV_z\end{align*}\begin{align*}=N\left [ (\frac{m}{2\pi KT})^{3/2}e^{\frac{-mV_x^2}{2KT}}dV_x+\int_{-\infty}^{\infty} e^{\frac{-mV_y^2}{2KT}}dV_y+\int_{-\infty}^{\infty} e^{\frac{-mV_z^2}{2KT}}dV_z\right ]\end{align*}\begin{align*}=N (\frac{m}{2\pi KT})^{3/2}e^{\frac{-mV_x^2}{2KT}}dV_x\left ( \sqrt{\frac{2\pi KT}{m}}\sqrt{\frac{2\pi KT}{m}}\right )\end{align*}\begin{align*}\therefore n_idV_x=N (\frac{m}{2\pi KT})^{1/2}e^{\frac{-mV_x^2}{2KT}}dV_x\rightarrow 5\end{align*}Therefore the probability that a molecule will have x component of velocity in the range \(V_x\space to\space V_x+dV_x\) is given by,\begin{align*}P(V_x)dV_x=\frac{n_i}{N}dV_x= (\frac{m}{2\pi KT})^{1/2}e^{\frac{-mV_x^2}{2KT}}dV_x\rightarrow 6\end{align*}Equation 5 and 6 represent Maxwell's distribution of velocities.

Degree of Freedom and Law of Equipartition of Energy

The degree of freedom can be defined as the total number of independent coordinates required to specify completely its position and configuration. The degree of freedom can also be defined as the number of independent ways in which a molecule can have energy. The degree of freedom is given by the formula, f=3A-R, where A=number of atoms in molecules and R=number of independent relation among atom.

(a)For monatomic gasses, A=1, R=0

\(\therefore f=3\) i.e. only translational motion.

(b)For diatomic gasses, A=2, R=1

\(\therefore f=5\) i.e. 3 degrees of freedom of translational motion and 2 degrees of freedom of rotational motion

(c) For triatomic (non-linear) gas molecule, A=3, R=3

\(\therefore f=6\)

According to kinetic theory of gas, the mean K.E> of molecule of a temperature T is given by

\(\frac12mc^2=\frac32KT\)

Here K is the Boltzmann constant and c is the root mean square speed.

But \(c^2=u^2+v^2+w^2\)

As x,y,z are all equivalent, the mean square velocities along three axes are equal i.e.,\(u^2=v^2=w^2\)

\begin{align*}\therefore \frac12mu^2=\frac12mv^2=\frac12mw^2\end{align*}\begin{align*}\therefore \frac12mc^2=3[\frac12mu^2]=3[\frac12mv^2]=3[\frac12mw^2]=\frac32KT\end{align*}\begin{align*}\therefore \frac12mu^2=\frac12mv^2=\frac12mw^2=\frac12KT\end{align*} Thus the average kinetic energy associated with each degree of freedom (whether translatory or rotatory)\(=\frac12KT\).

Maxwell-Boltzmann Statistics

Conditions of Maxwell Boltzmann statistics

(1) Any number of particles (n=0,1,2,...) can be accomodated in quantum state,

(2)The particles are considered to be distinuishable,

(3)The sum of particles in each quantum state is the total number of particles(COnservation of mass)

(4) The sum of energy of each particle in the quantum state is the total energy (Conservation of energy).

Consider a system of N particles \(n_1, n_2,...,n_i\) consisting energy \(\varepsilon_1,\varepsilon_2...\varepsilon_i\). The number of particles can exchange the state so that the particles in each state be the same.

In a collection od particles \(n_1\), one of them can be accomodated in \(g_1\) ways, second of them in\(g_2\) ways. Thus \(n_1\) can be accomodated in \(g_n\) ways.

If \(g_i\) is the probability of finding a particle in certain energy state \(\varepsilon_1\), then the probability of finding two particles in the same state is \(g_i\times g_i=g_i^2\). For \(n_i\) particles, the probability is \(g_i^{n_i}\).

Therefore the number of eigenstate(measured state of some object possessing quantifiable characteristics such as position, momentum, etc.) of N particles is,

\begin{align*}G=\frac{N!g_1^{n_1}g_2^{n_2}g_i^{n_i}}{n_1!n_2!n_i!}=N!\prod_i\frac{g_i^{n_i}}{n_i!}\rightarrow 1\end{align*}\(\therefore \) The probability of given state is,\begin{align*}\Omega=N!\prod_i\frac{g_i^{n_i}}{n_1!n_2!...n_i!}\times constant\rightarrow 2\end{align*}Taking log on 2,\begin{align*}log\Omega=logN!+\sum_i(n_ilogg_i-logn_i!)+constant\end{align*}Using Stirling approximation,\begin{align*}log\Omega=NlogN-N+\sum_i (n_ilogg_i-logn_i+n_i)+constant\end{align*}\begin{align*}=constant-\sum_i[n_ilogn_i-n_ilogg_i!]\rightarrow 3\end{align*}\begin{align*}\therefore\delta log\Omega=-\sum_i[n_i\frac{1}{n_i}\delta n_i+logn_i\delta n_i-logg_i\delta n_i]\end{align*}\begin{align*}For\space maxima,\space \delta log\Omega=0\space i.e. \sum_i[n_i\frac{1}{n_i}\delta n_i+logn_i\delta n_i-logg_i\delta n_i]=0\rightarrow 4\end{align*}\begin{align*}Now,\sum \delta n_i=0\rightarrow 5\end{align*}\begin{align*}and\space \sum\varepsilon_i\delta_i=0\rightarrow 6\end{align*}Multiplying 5 by \(\alpha\) and 6 by \(\beta\) and adding resultant with 4,\begin{align*}\sum_i[log\frac{n_i}{g_i}+\alpha +\beta\varepsilon_i]\delta n_i=0\end{align*}For each independent \(\delta n_i\),\begin{align*}log\frac{n_i}{g_i}+\alpha+\beta\varepsilon_i=0\end{align*}\begin{align*}\therefore log\frac{n_i}{g_i}=-(\alpha+\beta\varepsilon_i)\end{align*}\begin{align*}\implies \frac{n_i}{g_i}=e^{-(\alpha+\beta\varepsilon_i)}\end{align*}\begin{align*}\therefore n_i=g_ie^{-(\alpha+\beta\varepsilon_i)}\end{align*}This gives the number of particles in i^th cell following the Maxwell Boltzmann statistics. This is the Maxwell BOltzmann law.