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# Indroduction to Statistical Mechanics: Building Up to the Bulk

Shaun Williams, PhD

## Atomic Structure

### Atomic Structure

• A concept key to interpreting the second law of thermodynamics is the limitation by quantum mechanics on the number of states of our system
• The number of quantum states of any real system is a finite number
• This is an important difference between our classical understanding and our modern understanding

### Quantization of Properties

• At molecular sizes, particles behave like waves
• To fulfill the mathematical rules of waves, parameters, such as energy and momentum, are only allowed certain quantized values
• For a moving particle particle, this quantization becomes measurable when the distance traveled is much greater than its de Broglie wavelength $$\lambda_{dB} = \frac{h}{mv}$$
• In this equation, called the de Broglie wavelength equation
• $$h$$ is Planck’s constant, $$6.626 \times 10^{-34}\,\mathrm{J}\cdot \mathrm{s}$$
• $$m$$ is the particles mass
• $$v$$ is the particles velocity

### Quantizer Energy

• In a classical system, energy is a continuous function
• In a quantum system, energy comes only in discrete values

### Problem with the quantum nature

• When dealing with waves, the particles/waves cannot have their position pinned down to a single location
• For any parameter, $$A\left(\vec{r}\right)$$, the average value of this parameter is given by the average value theorem $$\expect{A} = \int_{\text{all space}} \mathcal{P}_\vec{r}\left(\vec{r}\right) A\left(\vec{r}\right) d\vec{r}$$
• Where $$\mathcal{P}_\vec{r}\left(\vec{r}\right)$$ is the probability distribution

### Particle in a Three-Dimensional Box

• Consider a 3-D box with lengths of $$a$$, $$b$$, and $$c$$ along the $$x$$, $$y$$, and $$z$$ axes, respectively
• Possible energy values for a particle in the box is \begin{align} \varepsilon_{n_x,n_y,n_z} &= \frac{\pi^2 \hbar^2}{2m}\left( \frac{n_x^2}{a^2}+\frac{n_y^2}{b^2}+\frac{n_z^2}{c^2} \right) \\ &\approx \frac{\pi^2\hbar^2}{2m(abc)^\bfrac{2}{3}}\left( n_x^2+n_y^2+n_z^2 \right) = \varepsilon_0n^2 \end{align}

### States of Such a Particle

• A particle which has the lowest values of $$n_x$$, $$n_y$$ and $$n_z$$ is said to be in its ground state
• Any other state is said to an excited state
• When two or more quantum states share the same energy, we say they are degenerate
• (2,1,1), (1,2,1), and (1,1,2)
• Correspondence principle – quantum mechanics approaches classical mechanics as we increase the particle’s kinetic energy, mass, or domain

### Use of Quantum Mechanics

• Due to correspondence principle:
• Use QM when $$\lambda_{dB}$$ is comparable to the domain size
• Use classical when $$\lambda_{dB}$$ is much smaller
• Entropy allows us to indirectly measure the density of quantum states, $$W(\varepsilon)$$
• The density of states tells how many quantum states near energy $$\varepsilon$$ there are per unit energy

### Quantum States of Atoms

• The same analysis can be used on one-electron atoms yielding $$E_n = -\frac{Z^2m_ee^4}{2\left(4\pi \epsilon_0\right)^2n^2\hbar^2} \equiv -\frac{Z^2}{2n^2}E_h$$
• $$E_h$$ is the energy unit Hartree
• Hartrees are convenient units for calculations involving electrons in atoms
• Conversion: $$1\,E_h=4.360\times 10^{-18}\,\mathrm{J}$$

### Quantum State of Electrons

• The quantum state of electrons depend on three other quantum numbers
• $$l$$ – angular momentum quantum number; defines the shape
• $$m_l$$ – magnetic quantum number; defines orientation in space
• $$m_s$$ – spin angular momentum
• For each atom, we can specify the entire set of quantum numbers (electron configuration)

### Term Symbols

• In addition to electron configurations, which give $$n$$ and $$l$$ values, we can specify the term symbol which gives $$m_l$$ and $$m_s$$ values
• Ground state of nitrogen is $$1s^22s^22p^3$$ and it has a term symbol of $${}^4S$$

### Electron Spin

• Spin is really an intrinsic property, like mass and charge
• Particles called Fermions all have spins that are half-integers
• Electrons, protons, neutrons
• Particles called Bosons all have integer spins
• $$\chem{{}^4He}$$ nucleus

### The Pauli Exclusion Principle

• No two individual Fermions can have the same set of quantum numbers
• Unlike Fermions, according to the Pauli exclusion principle, multiple Bosons can have the same set of quantum numbers
• This gives rise to very interesting effects (chapter 4)

### Degrees of Freedom

• Atoms in molecules attract and repel.
• Energy added to a molecule goes into one of 4 forms of motion
• Electronic – fast motion
• Vibrational
• Rotational
• Translational – slow motion
• Note: increasing mass of the moving particle

### Translational Energy

• Due to its large mass and slow speed, translational energy can normally be treated classically $$K=\frac{mv^2}{2}$$
• We don’t need to worry too much about the other forms yet
• When energy is added to a system, it distributes itself into some or all of these degrees of freedom

### Electronic State Transitions (UV-vis)

• Quantum states of molecules is not so easy
• Exciting from the electronic ground state to excited state gives us information
• This is done with electromagnetic (EM) radiation
• A photon has energy given by Planck’s law $$E_{photon} = h\nu = \frac{hc}{\lambda}$$
• $$c$$ – speed of light: $$c=2.9979\times 10^8\,\mathrm{m}\,\mathrm{s}^{-1}$$

### Vibrational State Transitions (IR)

• After electronic, the remaining contributions come from the motion of the nuclei
• For $$N_{atom}$$ in the molecule, there are a total of $$3N_{atom}$$ degrees of freedom
• Number of vibrational degrees of freedom:
• For non-linear molecules: $$3N_{atom}-6$$
• For linear molecules: $$3N_{atom}-5$$

### Vibrational Modes

• For each vibrational coordinate we can define a vibrational constant, $$\omega_e$$
• Depends on mass of atoms and rigidity of bond
• Harmonic approximation defined the constant $$\omega_e(J)=\hbar \sqrt{\frac{k}{\mu}}$$
• The reduced mass is $$\mu=\frac{m_Am_B}{m_A+m_B}$$

### Vibrational Energies

• The quantum states for vibrational motion have vibrational energies approximately equal to $$E_v=\left(v+\frac{1}{2}\right)\omega_e$$
• $$v$$ can be any integer 0 or greater
• Vibrational energies correspond to the infrared region of the EM spectrum

### Rotational States Transions (microwave)

• Non-linear molecules have three rotational coordinates
• Linear molecules have two rotational coordinates
• For a simple linear molecule the rotational energies are given by $$E_J=BJ(J+1)$$
• $$B$$ is the rotational constant
• $$J$$ can be any integer 0 or greater

### Rotational State Degeneracies

• Rotational states have degeneracies corresponding to different orientations of rotational motion
• For linear molecules, $$g_{rot}=2J+1$$
• We will eventually need to know the degeneracy to determine how a molecule distributes energy

## Bulk Properties

### Some Definitions

• System (or sample) – the thing we want to study
• Surroundings – everything outside the system
• Boundary – what separates the system from the surroundings
• Universe – everything: the system, the surroundings, and the boundary between them

### Balloon Example

• If we discuss the properties of the gas in a balloon
• The gas in the balloon is the system
• The atmosphere around the balloon is the surroundings
• The elastic material of the walls of the balloon is the boundary
• The boundary defines how the system interacts with the surroundings

### Bulk

• In bench-top chemistry, we deal with so many particles that removal of 30 or addition of 10000 will not change the properties
• A cluster of 8 sodium atoms has an ionization energy 20% higher than the bulk metal value
• This drops to 10% if we use 9 sodium atoms
• Bulk limit is much smaller than a visible sample
• $$\sim 10^{10}\,\mathrm{molecules} < 10^{-13}\,\mathrm{moles}$$

### Difficulty with QM

• QM accurately predicts the properties of individual molecules
• QM will work as the system becomes bigger
• It is no longer practical to analyze a mole of particles with QM, simply to complicated
• With so many molecules, we typically don’t care how a single one of them behaves

### Macroscopic Properties

• We seek macroscopic properties of the system
• Two group of macroscopic properties:
1. Extensive – obtained by summing together contributions from all the molecules in the system
2. Intensive – obtained by averaging the contributions from all the molecules in the system

### Parameters

• Extensive Parameters
• $$N$$, the total number of molecules in the system or the total number of moles, $$n$$
• $$V$$, the total space occupied
• $$E$$, the sum of the translational, rotational, vibrational, and electron energies
• Intensive Parameters
• $$P$$, the pressure, which is the average force per unit area exerted by the molecules on their surroundings
• $$\rho$$, the number density, which is the average number of molecules per unit volume
• Related to mass density $$\rho_m=\rho\frac{\mathcal{M}}{\mathcal{N}_A}$$

### Statistical Mechanics

• Statistical mechanics used probability theory to describe the macroscopic groups of molecules
• Based on common characteristics of individual molecules
• Statistical mechanics extends the results of QM to the behavior of bulk quantities of substances
• The things we want to study are perfect statistical samples

### Common Assumptions in Statistical Mechanics

1. Chemically identical molecules share the same physics
2. Macroscopic variables are continuous variables
3. Measured properties reflect the ensemble average
• If we fix some values (e.g. volume and moles)
• Ensemble - Set of all quantum states, same values
• Microstate – a unique quantum state
• $$\Omega$$ – number of microstates in the ensemble

### Ergodic Hypothesis

• If we fix $$E$$, $$N$$, and $$V$$
• Let the system find its own pressure
• The result will be the average of $$P$$ over all microstates
• The ensemble average

### 12-Member Ensemble

• Here are 12 distinct microstates
• All have the same properties
• Let the particle exert a pressure of p0 on the walls next to it
• Half the top walls have two particles next to it and half have one
• $$\expect{p_{top}} = 1.5p_0$$

### The Microcanonical Ensemble

• Approaching a problem, we must decide which parameters are allowed to vary
• More flexible is more realistic but more difficult to solve
• If we fix all the extensive variables, $$E$$, $$V$$, and $$N$$ we have our first ensemble – the microcanonical ensemble

### Properties of Microcanonical Ensemble

• If we fix all extensive variables then intensive variables that are ratios of extensive variables must also be fixed
• Density, $$\rho=\frac{N}{V}$$
• What can change?
• Microscopic properties that we’re not bothering to measure
• Some intensive properties (such as pressure) that are not ratios of extensive variables

### Advantages of Microcanonical Ensemble

• Energy and mass are rigorously conserved
• There system must be completely isolated from the rest of the universe
• Constant volume means that no mechanical force pushes on the surroundings
• Isolation greatly simplifies the system

## Entropy

### Boltzmann Entropy

• The Boltzmann entropy is given by $$S\equiv k_\mathrm{B}\ln \Omega$$
• Boltzmann constant is $$k_\mathrm{B}=1.381\times 10^{-23}\,\mathrm{J}\,\mathrm{K}^{-1}$$
• This provides a different definition of the relation between heat and temperature
• This will be our rigorous definition of entropy since it works under and circumstances
• The entropy counts the total number of distinct microstates of the system

### $$\Omega$$ versus $$g$$

• For a microcanonical ensemble
• $$\Omega$$ is the total number of microstates that have the same energy
• Under these conditions $$\Omega$$ is the same as $$g$$
• $$\Omega$$ is the total number of microstates in an ensemble
• $$g$$ is the degeneracy of quantum states in some particular individual particle

### Size of $$\Omega$$

• $$\Omega$$ is enormous
• However, $$\Omega$$ is always finite
• The number of arrangements of system is countable
• In a classical system, $$\Omega$$ would be infinite due to the continuous nature of classical mechanics
• Quantization makes $$\Omega$$ finite

### Two Non-Interacting Subsystems

• The degeneracy is the product of degeneracies
• The entropy is the sum of the entropies \begin{align} \Omega_{\chem{A+B}} &= \Omega_\chem{A}\Omega_\chem{B} \\ \ln \Omega_{\chem{A+B}} &= \ln \left(\Omega_\chem{A}\Omega_\chem{B}\right) = \ln \Omega_\chem{A}+\ln \Omega_\chem{B} \\ S_\chem{A+B} &= S_\chem{A}+S_\chem{B} \end{align}

### Gibbs Definition of Entropy

• Gibbs developed a general equation for estimating the entropy in any ensemble using probability rather than $$\Omega$$
• Derivation:
• The total number of ways of arranging the labels on $$N$$ molecules is $$N!$$
• There are a total of $$N_i!$$ ways of rearranging the labels

### The Ensemble Size

• These definitions mean that we get $$\Omega=\frac{N!}{N_1!N_2!N_3!\dots N_k!} = \frac{N!}{\prod_{i=1}^kN_i!}$$
• Example: 6 particles with 4 possible states with 3 in state 1, 2 in state 2, 1 in state 3 and 0 in state 4 $$\Omega = \frac{6!}{(3!)(2!)(1!)(0!)}=60$$
• Using this definition of the size of the ensemble we find \begin{align} S &= k_\mathrm{B} \ln \Omega = k_\mathrm{B} \ln \left( \frac{N!}{\prod_{i=1}^kN_i!} \right) \\ &= k_\mathrm{B} \left[ \ln N! - \ln \prod_{i=1}^k N_i! \right] \end{align}

### Using Stirling's Approximation

$$S = k_\mathrm{B} \left[ \ln N! - \ln \prod_{i=1}^k N_i! \right]$$

• Now we can use Stirling’s approximation: $$\ln N! \approx N\ln N$$ to yield \begin{align} S &= k_\mathrm{B} \left[ N\ln N - N - \left( \sum_{i=1}^k N_i \ln N_i - \sum_{i=1}^k N_i \right) \right] \\ &= k_\mathrm{B} \left[ \sum_{i=1}^k N_i \ln N - N - \sum_{i=1}^k N_i \ln N_i +N \right] \\ &= k_\mathrm{B} \sum_{i=1}^k \left[ N_i \ln N - N_i \ln N_i \right] \\ &= k_\mathrm{B} \sum_{i=1}^k N_i \left[ \ln N - \ln N_i \right] \end{align}

### Rewriting Entropy in Terms of Probability

$$S = k_\mathrm{B} \sum_{i=1}^k N_i \left[ \ln N - \ln N_i \right]$$

• We can write this in terms of probabilities $$\mathcal{P}(i)=\frac{N_i}{N}$$ \begin{align} S &= Nk_\mathrm{B} \sum_{i=1}^k \frac{N_i}{N} \ln \frac{N}{N_i} \\ &= Nk_\mathrm{B} \sum_{i=1}^k \mathcal{P}(i) \ln \frac{1}{\mathcal{P}(i)} \\ &= -Nk_\mathrm{B} \sum_{i=1}^k \mathcal{P}(i) \ln \mathcal{P}(i) \end{align}

## Temperature and the Partition Function

### The Ideal Gas

• A collection of particles
• Don’t interact with each other at all
• Elastically bounce off walls
• We want to use statistical mechanics to find the pressure exerted by an ideal gas on the walls of the container as a function of the macroscopic variables

### System and Reservoir

• The box has:
• Fixed volume, $$V$$
• Fixed number of molecules, $$N$$
• Energy will no longer be held constant
• Reservoir strongly interacts with system

### The Reservoir

• The reservoir has:
• $$E_r \gg E$$
• $$E+E_r = E_T \gg E$$
• $$\Omega_r\left(E_r\right) \gg \Omega\left(E\right)$$
• The ensemble size of the universe is $$\Omega_T\left(E_T\right) = \sum_E \Omega\left(E\right) \Omega_r \left(E_r\right)$$

### $$\mathcal{P}(i)$$ and Number of Microstates

• Choose a single microstate, $$i$$, with energy $$E_i$$
• If there are 9 distinct microstates with energy the reservoir must be changing
• Thus we can determine the system number from the reservoir number ($$E_r=E_T-E$$)
• The probability of the system being in state $$i$$ is $$\mathcal{P}(i)=\frac{\Omega_r\left(E_r\right)}{\Omega_T\left(E_T\right)}$$

### A Problem

• The problem is we don’t want to have to measure the reservoir instead of the system
• First we will rewrite this in terms of entropy (Taylor series expansion) $$S_r \left( E_r \right) = S_r \left( E_T-E_i \right) = S_r \left( E_T \right) - \left. \left( \frac{\partial S_r}{\partial E_r} \right)_{V,N} \right|_{E_T} E_i + \cdots$$
• For a fixed total energy we know that $$dE_T = d\left(E_T-E\right)=-dE$$
• We also know that $$\Omega_T\left(E_T\right)=\Omega_r\left(E_r\right)\Omega\left(E\right)$$

### Continuing to Solve the Problem

• We also know that \begin{align} dS_r &= k_\mathrm{B} d \ln \Omega_r \left(E_r\right) = k_\mathrm{B} d\left[ \ln \left( \frac{\Omega_T\left(E_T\right)}{\Omega\left(E\right)} \right) \right] \\ &= k_\mathrm{B} d \left[ \ln \Omega_T \left(E_T\right) - \ln \Omega\left(E\right) \right] = -k_\mathrm{B} d\ln \Omega\left(E\right) \\ &= -dS \end{align}
• Combining these equations we find that $$\left. \left( \frac{\partial S_r\left(E_r\right)}{\partial E_r} \right) \right|_{E_T} = \left. \left( \frac{\partial S(E)}{\partial E} \right) \right|_{E_T}$$

### Rewriting the Entropy Equation

• Using these previous equation we find $$S_r\left(E_r\right) = S_r\left(E_T-E_i\right) = S_r\left(E_T\right) - \left. \left( \frac{\partial S(E)}{\partial E} \right) \right|_{E_T} E_i$$
• This introduces temperature $$T \equiv \left( \frac{\partial E}{\partial S} \right)_{V,N}$$
• Using temperature in the entropy equation $$S_r\left(E_r\right) = S_r\left(E_T-E_i\right)=S_r\left(E_T\right)-\frac{E_i}{T}$$

### Solving the Issues

• We can now go back to $$\Omega$$ $$k_\mathrm{B} \ln \left[ \Omega_r\left(E_r\right) \right] = k_\mathrm{B} \ln \left[ \Omega_r\left(E_T\right) \right] -\frac{E_i}{T}$$
• Solving for the number of reservoir states we get $$\Omega_r\left(E_r\right)=\Omega_r\left(E_T\right)e^{-\bfrac{E_i}{k_\mathrm{B}T}}$$

### A New Ensemble

• In typical experiments, the energy is not fixed
• The system and the surroundings/reservoir can exchange energy
• If the reservoir is big enough, its properties are constant, so we fix $$T$$
• So a canonical ensemble has $$T$$, $$V$$, and $$N$$ fixed and $$E$$ as a variable

### More Equations, Now Within the Canonical Ensemble

• Combining some previous equations $$\mathcal{P}(i)=\frac{\Omega_r\left(E_T\right)}{\Omega_T\left(E_T\right)} e^{-\bfrac{E_i}{k_\mathrm{B}T}}$$
• The $$\Omega$$ are constants
• We can eliminate them by requiring that the probability be normalized $$\sum_{i=1}^\infty \mathrm{P}(i) = \frac{\Omega_r\left(E_T\right)}{\Omega_T\left(E_T\right)} \sum_{i=1}^\infty e^{-\bfrac{E_i}{k_\mathrm{B}T}} = 1$$

### Solving the Equation

$$\sum_{i=1}^\infty \mathrm{P}(i) = \frac{\Omega_r\left(E_T\right)}{\Omega_T\left(E_T\right)} \sum_{i=1}^\infty e^{-\bfrac{E_i}{k_\mathrm{B}T}} = 1$$

• Solving for our fraction $$\frac{\Omega_r\left(E_T\right)}{\Omega_T\left(E_T\right)} = \frac{1}{\sum_{i=1}^\infty e^{-\bfrac{E_i}{k_\mathrm{B}T}}} = \frac{1}{Q(T)}$$
• We have made a definition of the partition function

### Canonical Partition Function and Distribution

• So, the canonical partition function is $$Q(T)=\sum_{i=1}^\infty e^{-\bfrac{E_i}{k_\mathrm{B}T}}$$
• The canonical distribution can be written as $$\mathcal{P}(i) = \frac{e^{-\bfrac{E_i}{k_\mathrm{B}T}}}{Q(T)}$$

### Meaning

$$\mathcal{P}(i) = \frac{e^{-\bfrac{E_i}{k_\mathrm{B}T}}}{Q(T)}$$

• We see the importance of $$T$$
• The likelihood of getting the system into a particular state $$i$$ drops exponentially with respect to the ratio of the microstate energy $$E_i$$ to the thermal energy $$k_\mathrm{B}T$$
• The temperature establishes how likely the system is to be found at any particular energy.

### Finding Another Likelihood

• Let’s find the likelihood of finding the system t any particular energy E $$\mathcal{P}(E)=\Omega(E)\mathcal{P}(i) = \frac{\Omega(E) e^{-\bfrac{E_i}{k_\mathrm{B}T}}}{Q(T)}$$
• In all these equation the beauty of the canonical ensemble is that is allows us to ignore how $$\Omega$$ depends on $$V$$ and $$N$$

### Assertion

• The probability extends to the probability of a particle within our system having some particular energy $$\varepsilon$$
• Say we have a particular quantum state $$i$$ and degeneracy $$g$$ \begin{align} \mathcal{P}(\varepsilon) &= \frac{g(E) e^{-\bfrac{\varepsilon}{k_\mathrm{B}T}}}{q(T)} \\ q(T) &= \sum_{\varepsilon=0}^\infty g(E) e^{-\bfrac{\varepsilon}{k_\mathrm{B}T}} \end{align}

### Check of Zero Energy

• If we measure all energy relative to $$E_0$$ then \begin{align} \mathcal{P}(E) &= \frac{ge^{-\bfrac{\left(E-E_0\right)}{k_\mathrm{B}T}}}{\sum_E ge^{-\bfrac{\left(E-E_0\right)}{k_\mathrm{B}T}}} \\ &= \left( \frac{ge^{-\bfrac{E}{k_\mathrm{B}T}}}{\sum_E ge^{-\bfrac{E}{k_\mathrm{B}T}}} \right) \left( \frac{e^{-\bfrac{E_0}{k_\mathrm{B}T}}}{e^{-\bfrac{E_0}{k_\mathrm{B}T}}} \right) = \frac{ge^{-\bfrac{E}{k_\mathrm{B}T}}}{\sum_E ge^{-\bfrac{E}{k_\mathrm{B}T}}} \end{align}

### Example 2.1

At 298 K, calculate the ratio of the number of $$\chem{NH_3}$$ molecules in the excited state to the number in the ground state, where the excited state is (a) the $$0^-$$ state of the inversion, which lies $$0.79\,\mathrm{cm}^{-1}$$ above the $$0^+$$ ground state; and (b) the $$1^+$$ state, which lies $$932.43\,\mathrm{cm}^{-1}$$ above the $$0^+$$. (The wavenumber unit, $$\mathrm{cm}^{-1}$$, is conventionally used by spectroscopist as an energy unit, based on the relation between the transition energy in the experiment and the reciprocal wavelength of the photon that induces the transition: $$E_{photon}=\frac{hc}{\lambda}$$. Because the energy is inversely proportional to the wavelength, energy is given in units of $$\frac{1}{\text{distance}}$$.)

### Example 2.2

At a temperature of 1000 K, how many vibrational states of $$\chem{H_2}$$ are populated by at least 1% of the molecules, given the vibrational constant $$\omega_e=4395\,\mathrm{cm}^{-1}$$ and the vibrational energy (relative to the ground state)of approximately $$E_{vib}=\omega_e\nu$$?

### Example 2.3

You discover a molecular system having the energy levels and degeneracies $$\varepsilon = c\left(n-1\right)^6;\;g=n;\;\text{for }n=1,2,3,\dots\text{ and }k_\mathrm{B}T=400c$$ Evaluate the partition function and calculate $$\mathcal{P}(\varepsilon)$$ for each of the four lowest energy levels.

## The Ideal Gas Law

### Moving A Wall

• We momentarily free one wall
• It reduces its forces therefore its pressure on the gas, $$P_{min}$$
• The gas pushes the wall out infinitesimally, $$ds$$
• The energy spent moving the wall is $$dE=-F\,dS = -P_{min} A\,dS = -P_{min} \,dV$$

### The Pressure Equation

• If we hold $$N$$ and $$S$$ constant then $$P_{min} = -\left( \frac{\partial E}{\partial V}\right)_{S,N}$$
• Mostly we will deal with the case where $$P\cong P_{min}$$ $$\text{if}\,P=P_{min}:P=-\left(\frac{\partial E}{\partial V}\right)_{S,N}$$

### Cyclic Rule for Partial Derivatives

• The cyclic rule for partial derivatives is $$\left( \frac{\partial X}{\partial Y}\right)_Z = -\left( \frac{\partial X}{\partial Z}\right)_Y\left(\frac{\partial Z}{\partial Y}\right)_X$$
• This allows us to write $$P=-\left(\frac{\partial E}{\partial V}\right)_{S,N} = \left(\frac{\partial E}{\partial S}\right)_{V,N} \left(\frac{\partial S}{\partial V}\right)_{E,N}$$
• Another equation for temperature $$T=\left(\frac{\partial E}{\partial S}\right)_{V,N}$$

### Determining Entropy Change with Volume

• Let the translational states of the gas correspond to all the possible ways the gas particles can be arranged inside our box, regardless of what energy the gas has
• The box has a volume $$V$$
• Break up the volume into units $$V_0$$, each able to hold a single particle
• We have $$M$$ number of units so $$V=MV_0$$

### More On Our Gas

• The particles are indistinguishable
• This reduces the number of distinct states
• If we had two particles the number of distinct states would be $$\bfrac{M\left(M-1\right)}{2}$$
• Extending this to $$N$$ particles $$\frac{M(M-1)(M-2)\dots (M-N)}{N!}$$
• For a gas $$M \gg N$$

### Even More on Our Gas

• Introduce a constant $$A$$ to absorb any other parameter’s effects, such as energy $$\Omega = \lim_{M\gg N} A\frac{1}{N!} M(M-1)(M_2)\dots (M_N)=A\frac{1}{N!}M^N$$
• Because $$V=MV_0$$ $$\Omega\frac{1}{N!}\left(\frac{V}{V_0}\right)^N = A\frac{1}{V_0^NN!} V^N$$

### The Entropy

• Ignoring everything but the volume dependence \begin{align} S &= k_\mathrm{B} \ln \Omega = k_\mathrm{B} \ln \left[ \text{constant}\cdot V^N \right] \\ &= k_\mathrm{B} \left[ \ln \text{constant} + N\ln V\right] \end{align}
• This gives $$\left(\frac{\partial S}{\partial V}\right)_{E,N} = k_\mathrm{B}N\left(\frac{\partial \ln V}{\partial V}\right)_{E,N} = k_\mathrm{B} \frac{N}{V}$$
• So $$P=\left(\frac{\partial E}{\partial S}\right)_{V,N}\left(\frac{\partial S}{\partial V}\right)_{E,N} = \frac{k_\mathrm{B}TN}{V}$$

### So What?

• Remember that $$k_\mathrm{B}\mathcal{N}_\mathrm{A}=R$$
• So \begin{align} P &= \frac{k_\mathrm{B}TN}{V} = \frac{\frac{R}{\mathcal{N}_\mathrm{A}}TN}{V} = \frac{RT\frac{N}{\mathcal{N}_\mathrm{A}}}{V} = \frac{RTn}{V} \\ PV &= nRT \end{align}
• We managed to derive the classical ideal gas law equation from statistical mechanics using a canonical ensemble!

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