NEB XII Physics Questions

Statistical Mechanics

aClassical Statistical Physics Phase SpaceStatistical mechanics is a mathematical framework that applies statistical methods and probability theory to large assemblies of microscopic entities. It does not assume or postulate any natural laws, but explains the macroscopic behavior of nature from the behavior of such ensembles. In dynamical system theory, a phase space is a space in which all possible states of a system are represented, with each possible state corresponding to one unique point in the phase space. For mechanical systems, the phase space usually consists of all possible values of position and momentum variables. To locate a molecule in three dimensional, coordinate system is insufficient. It is because only three dimensional system can not give information about physical observable e.g velocity, momentum, energy etc. So three position coordinates (x,y,z) and three momentum coordinates(px, py, pz) are required to specify a molecule. Volume of phase spaceThe elimentary volume of a phase space is called cell. We consider a molecule with position coordinates dx,dy,dz,and momentum coordinates dpx, dpy,dpz. Then the volume = dx dy dz dpx dpy dpzMicrostate and MacrostateIn statistical mechanics, a microstate is a specific microscopic configuration of a thermodynamic system that the system may occupy with a certain probability in the course of its thermal fluctuations. In contrast, the macrostate of a system refers to its macroscopic properties, such as its temperature, pressure, volume and densityThe specification of a molecule in a cell is called microstate. If a molecule remains in particular region of cell, there in on change in microstate.If molecules are exchanged within a cell, there is change in microstate.If molecules are exchanged between the cells , there is change in macrostate.EnsembleEnsemble is a random collection of identitical and independent assemblies . There are three types of ensemble1. Microcanonical ensemble: An isolated system can neither exchange particles nor energy with its surroundings. The energy E, the volume and the number of particles N are constant in these systems. 2. Canonical ensemble: A closed system cannot exchange particles with its surroundings, but it can exchange energy (in form of heat or work). If the energy exchange occurs via heat but not work, the following parameters are constant: temperature T, volume V and the number of particles N3. Grand canonical ensemble: An open system exchanges particles and heat with its surroundings. The following parameters are constant temperature T, volume V and chemical potential µ.

ConstraintsThe set of conditions that must be obeyed by a system are known as constraints. For example, in distribution 3 particles in two compartments, the system must obey the constraint that total number of particles in two compartments must be three. Thermodynamical ProbabilityThermodynamical probability is defined as the number of microstate corresponding to that macrostates. For N phase points total number of permutation is N !. If n1 is the number of phase points in cell 1,n2 in the cell 2 then there are n1! permulation in cell 1, n2! permutation in cell 2 and so on. Then thermodynamic probability for such a systemΩ=N!

n1!n2!n3!FUNDAMENTAL POSTULATES OF STATISTICAL MECHANICS(1) Any gas is composed by molecules, which are in motion and behave like very small elastic spheres.(2) The size of each phase cell is same.(3) All accessible microstates corresponding to the possible macrostates are equally probable.(4) Total number of molecules is constant. Total energy of the system is constant. Let there are n1 molecules of energy 𝜀1,n2 of E2 etc. then from law of conservation of energy.E=n1𝜀1+n2𝜀2+………⇒E= ∑ini𝜀i(5) The equilibrium state of a gas corresponds to the microstate of maximum probability. BOLTZMANN CANONICAL DISTRIBUTION LAWLet n1,n2,n3,…n1 be the number of gas molecules in cell 1, cell 2, cell 3,… 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 the same. Also energy of the system remains the same. The distribution of molecules are in the following constraints.(1) Conservation of mass i.e., n1+n2+n3+…ni+…=N. or, 𝛿n1+𝛿n2+…+𝛿n1+=0 ................(i) (Since total number of molecules are constant.)(2) The conservation of energy i.e., 𝜀1n1+𝜀2n2+…𝜀n=E 𝜀1𝛿n1+𝜀2𝛿n2+…+𝜖i𝛿ni+…=0 .............(ii) (Since total energy is constant.)The probability of molecules under given restriction is,Ω=N!

n1!n2!n3!…n1!If the system is in equilibrium, the thermodynamical probability is maximum i.e., 𝛿Ω=0Since Ω is maximum logΩ is maximum ∴𝛿(logΩ)=0 ........................(iii)We have, Ω=N!

n1!n2!….ni! or, a

logΩ	=logN!-logn1!-logn2!…-logni!

............(iv) =logN!- ∑ilogni!∴𝛿(logΩ)=𝛿[logN!- ∑ilogni!]Now applying Striling theorem logx!=xlogx-x, if x is large∴𝛿(logΩ)=𝛿(NlogN-N- ∑i(nilogni-ni)∣| or, 0=0- ∑i[logni𝛿ni+ni⋅1

ni𝛿ni-𝛿ni]∴ ∑i[logn1𝛿ni]=0∴logn1𝛿n1+logn2𝛿n2+…+log2𝛿4𝛿1=0.....................(v)Equation (i), (ii) and (v) are independent. We can combine them without losing their independence multiplying by arbitrary constants. Thus multiplying (i) by 𝛼 and (ii) by 𝛽 and adding the resulting oquation (v) where 𝛼 and 𝛽 are called Lagrange's undetermined multipliers (constants).∴𝛼(𝛿n1+𝛿n2+…𝛿ni)+𝛽(𝜀1𝛿n1+𝜀2𝛿n2+…+𝜀3𝛿n3)+(logn2𝛿n1+logn2𝛿n2+…+logni𝛿ni) = 0⇒(𝛼+𝛽𝜀1+logni)𝛿n1+(𝛼+𝛽𝜀2+logn2)𝛿n2+(𝛼+𝛽𝜀i+logni)𝛿ni=0In equation (v), this is the combination of independent equation where 𝛿n1, 𝛿n2 ... . 𝛿ni, cannot be zero so.a

𝛼+𝛽𝜀i+logni	=0
⇒-(𝛼+𝛽𝜀i)	=logni
⇒ni	=e-(𝛼+𝛽𝜀i)=e-𝛼e-𝛽𝜀i
ni	= A e-𝛽𝜀i is where A=e-𝛼= constant

[ni= A e-𝛽𝜀i]This equation gives the number of molecules in ith cell as a function of cnergy associated with cach particle in that cell and is called Bolizmann's canonical distribution law. PARTITION FUNCTIONWe know from Boltzmann canonical distribution law, ni=Ae-𝛽𝜀i where ni is number of particles in ith cell each having energy 𝜀i.So the sum of all ni most be equal to the total number N i.e. ∑ni= N=A𝛴 e-𝛽𝜀i The sum 𝛴 e-𝛽𝜀i, is called partition function or sum of states and is represented by the letter Z. i.e. Z= 𝛴 e-𝛽𝜀iso aN=AZ⇒A=N

ZSo number of molecules in ith cell in state of maximum thermodynamic probability is,

ni=N

Ze-𝛽𝜀i

, Here 𝛽=1

KT , K is Boltzmann constant and T is absolute temperature. MAXWELL'S DISTRIBUTION LAW OF VELOCITIES Consider an ideal gas in a vessel of volume V. If the gas is equilibrium then according to Maxwell's Boltzmann canonical distributio law, the number of molecules in a cell of energies 𝜀i will be, ni=Ae-𝛽𝜀iThe number of molecules having energy 𝜀i and having position coordinates between x and x+dx,y+dy,z and z+dz and velocity components vx and vx+dvx,vy and vy+dvy and vz and vz+dvz is given by,nidxdyd zdvxdvydv2=Ae𝛽𝜖idxdydzdvxdvydvz. a

But 𝜀i= Energy of particle

=1 2mv2

=1 2m(vx2+vy2+vz2 )

so, nidxdydzdvxdvydvz=Ae-𝛽(1 2m(v2x+v2y+v2z))dxdydzdvxdvydvx ..............(i)

Integrating (i) overall available volume and all ranges of velocitiesN=∭ AVe-m𝛽

2(v2x+v2y+v2z)dvxdvydvz||=AV∞∫-∞e-m𝛽v2x

2dvx|∞∫-∞e-m𝛽v2y

2dvy∞∫-∞e-m𝛽v2z

2dvz =AV(2𝜋

m𝛽)1

2(2𝜋

m𝛽)1

2(2𝜋

m𝛽)1

2 [∵∞∫-∞e-𝛼x2dx=𝜋

𝛼]N=AV(2𝜋

m𝛽)3

2 ∴A=N

V(m𝛽

2𝜋)3/2=N

V(m

2𝜋KT)3/2.......................(ii)With the help of (ii), equation (i) becomes.nidxdydzdv2dvydvz=N

V(m

2𝜋KT)3/2e-m𝛽

2(vx2+v2y+xz2)dxdydzdvxdvydvx................(iii) ∴ The number of molecules having velocity coordinates in the range vx to vx+dvx,vy to vy+dvy,vz to vz+dvz irrespective of the position coordinates can be found by integrating equation (iii) with respect to position coordinates.∴nidvxdvydvz=N

V(m

2𝜋KT)3/2e-m𝛽

2(v2x+v2y+v2z)dvxdvydvz∫∫∫dxdydzor, nidvxdvydvz=N

V(m

2𝜋KT)3/2e-m𝛽

2(v2x+v2y+v2z)dvxdvydvzV=N(m

2𝜋KT)3/2e-m𝛽

2(v2x+v2y+v2z)dvxdvydvz...............................(iv)The number of molecules having velocity components in the range vx to vx+dvx irrespective of vy,vz,x,y,z is obtained by integrating equation (iv) with respect to vy and vz i.e.nidvx=N(m

2𝜋KT)3/2∫∫e-m𝛽

2(v2x+v2y+v2z)dvxdvydvx=N(m

2𝜋KT)3/2e-m𝛽v2x

2dvx∞∫-∞e-m𝛽v2y

2dvy∞∫-∞e-m𝛽v2z

2dvz=N(m

2𝜋KT)3/2e-m𝛽v2x

2dvx(2𝜋

m𝛽⋅2𝜋

m𝛽)=N(m

2𝜋KT)3/2e-mv2x

2KTdvx(2𝜋KT

m⋅2𝜋KT

m) [∵𝛽=1

KT] = N(m

2𝜋KT)3/2e-mv2x

2KTdvx(m

2𝜋KT)-1

2(m

2𝜋KT)-1

2 a∴[nidvx=N(m

2𝜋KT)1/2e-mv2x

2KTdvx] ni

Ndvx=(m

2𝜋KT)1/2e-mv2x

2KTdvx ..............(v)velocity in the probability that a molecule will have x component of velocity in the range vx to vx + dvx is given by,aP(vx)dvx=ni

Ndvx[P(vx)dvx=(m

2𝜋KT)1/2e-mv2x

2KTdvx]..................................(vi)Equation (v) and (vi) represent Maxwell's distribution of velocities. DEGREE OF FREEDOM AND LAW OF EQUIPARTITION OF ENERGY The degree of freedom of a dynamical system may be defined as the total number of independent coordinates required to specify completely its position and configuration. A molecule ina gas can move along any of the three coardinate ares. It has three degrees of freedom. A monatomic gas molecule has three degree of freedom. A diatomic gas molecule has three degree of translation and two degree of freedom of rotation, in total five. A rigid rigid body has six degrees of freedom (thee rotational and three translational). According to kinetic theory of gas, the mean K. E of molecule of temperature ' T is given by,1

2mc2=3

2KT ....................(i)Here, K is Boltzmann constant and c is root mean square speed. But c2=u2+v2+w2As x,y,z are all equivalent, the mean square velocities along the three axes are equal, i.e., u2=v2=w2.a

∴1 2mu2	=1 2mv2=1 2mw2
∴1 2mc2	=3[1 2mu2]=3[1 2mv2]=3[1 2mw2]=3 2KT
∴1 2mu2	=1 2KT
1 2mv2	=1 2KT
1 2mw2	=1 2KT

Thus the average energy associated with each degree of freedom (whether translatory or rotatory) =1

2KT. MAXWELL'S BOLTZMANN STATISTICSFollowing are the conditions of Maxwell Boltzmann statistics:1. Any number of particles (n=0,1,2,3…) can be accommodated in a quantum state.2. The particles are considered to be distinguishable.3. The sum of particles in each quantum state is the total number of particles (Conservation of mass)4. The sum of the energy of each particle in the quantum state is the total energy (i.e, conservation of energy). Let us consider a system of N particles n1,n2,…ni consisting energy 𝜀1,𝜀2,….𝜀i,…. The number of particles can exchange the state so that the particles in each state be the same. In a collection of particles n1, one of them can be accommodated in g1 Ways, second of them in g2 ways. Thus n1 particle can be accommodated in g1n, ways.If gi the probability of location, a particle in a certiain energy state 𝜀i then the probability of locating two particles in the same state is gi ⨯ gi=g2i. For ni particles, the probability is gnii.∴ The number of eigen state of the N particles is, a

G	=N!gni1g2n2..........gini n1!n2!n3!..........ni!
	=N! ∏igini ni! .................(i)

∴ The probability of given state is,Ω=N! ∏ignii

ni!× constant. ...........(ii)Taking log on (ii),logΩ=logN!+ ∑i{niloggi-logni!}+ const. Using Stirling approximationa

logΩ	=NlogN-N+ ∑i(niloggi-nilogni+ni)+constant

Now,a

𝛿logΩ

=𝛿(NlogN-N)+ ∑i𝛿(niloggi-nilogni+ni)

=constant - ∑i𝛿(niloggi-nilogni+ni)= ∑i( log gi𝛿ni -ni .1

ni𝛿ni -logni 𝛿ni + 𝛿ni)= ∑i( log gi𝛿ni -𝛿ni -logni 𝛿ni + 𝛿ni)=- ∑i( -log gi𝛿ni +logni 𝛿ni )= - ∑i( logni

gi )𝛿ni For Maxima, 𝛿logΩ = 0 or ∑i[logni

gi]𝛿ni=0 ...............(iii) Now aswe know , 𝛴𝛿ni=0 .............(iv) and 𝛴𝜖i𝛿i=0........................(v)Multiply (iv) by 𝛼 and (v) by 𝛽 and adding resulting with (iii),a

∑i[logni gi]𝛿ni+ ∑i𝛼𝛿ni+ ∑i𝛽𝜀i𝛿ni=0

or, ∑i[logni gi+𝛼+𝛽𝜀i]𝛿ni=0

For each independent 𝛿ni logni

gi+𝛼+𝛽𝜀i=0(...)a

∴logni gi	=-(𝛼+𝛽𝜀i)
⇒ ni gi	=e-(𝛼+𝛽𝜖i)
∴ ni	=gie-(𝛼+𝛽𝜀i)=gi e𝛼+𝛽𝜀i
[ni=gi e𝛼+𝛽𝜀i]

This gives the number of particles in ith cell following the Maxwe Boltzmann Statistics. This is Maxwell Boltzmann distribution law.