Classical Ideal Gas

1. Introduction

A classical ideal gas is a system of non-interacting particles that obey Maxwell–Boltzmann statistics.

The assumptions of classical ideal gas are:

Large number of particles (N ≫ 1)
No intermolecular interactions
Point particles (negligible volume)
Obeys classical mechanics

2. Hamiltonian of Ideal Gas

For N non-interacting particles:

\[ H = \sum_{i=1}^{N} \frac{p_i^2}{2m} \]

Only kinetic energy contributes.

3. Canonical Partition Function

The canonical partition function is:

\[ Z = \frac{1}{N! h^{3N}} \int e^{-\beta H} d^{3N}p \, d^{3N}q \]

Since there is no potential energy:

\[ Z = \frac{1}{N! h^{3N}} V^N \left( \int e^{-\beta \frac{p^2}{2m}} d^3p \right)^N \]

Evaluating the momentum integral:

\[ \int e^{-\beta \frac{p^2}{2m}} d^3p = (2\pi m k_B T)^{3/2} \]

Therefore,

\[ Z = \frac{1}{N!} \left( \frac{V}{\lambda^3} \right)^N \]

where thermal wavelength:

\[ \lambda = \frac{h}{\sqrt{2\pi m k_B T}} \]

4. Helmholtz Free Energy

\[ F = -k_B T \ln Z \]

Using Stirling approximation:

\[ F = -Nk_B T \left[ \ln \left( \frac{V}{N\lambda^3} \right) + 1 \right] \]

5. Equation of State

Pressure is obtained from:

\[ P = - \left( \frac{\partial F}{\partial V} \right)_{T,N} \]

Thus,

\[ P = \frac{Nk_B T}{V} \]

This gives the ideal gas law: \[ PV = Nk_B T \]

6. Internal Energy

Mean energy:

\[ U = - \frac{\partial}{\partial \beta} \ln Z \]

Result:

\[ U = \frac{3}{2} Nk_B T \]

7. Specific Heat

\[ C_V = \left( \frac{\partial U}{\partial T} \right)_V \]

\[ C_V = \frac{3}{2} Nk_B \]

For one mole:

\[ C_V = \frac{3}{2} R \]

8. Entropy (Sackur–Tetrode Equation)

\[ S = Nk_B \left[ \ln \left( \frac{V}{N\lambda^3} \right) + \frac{5}{2} \right] \]

9. Classical Validity Condition

Classical approximation is valid when:

\[ n \lambda^3 \ll 1 \]

n = number density
$\lambda$ = thermal wavelength

If $n \lambda^3 \sim 1$, quantum effects become important.

10. Important Result

The classical ideal gas provides a complete thermodynamic description of dilute gases and serves as the classical limit of quantum gases.

Classical Ideal Gas

A classical ideal gas is a theoretical gas composed of many identical point particles that do not interact with each other except through elastic collisions. It follows the laws of classical mechanics (Newtonian mechanics), and its statistical properties can be described using Maxwell-Boltzmann statistics.

Properties of a Classical Ideal Gas:

Particles are indistinguishable and non-interacting.
Each particle moves in straight lines unless it collides elastically with a wall or another particle.
The gas obeys the ideal gas law: $ PV = NkT $.
Energy is purely kinetic: $ E = \dfrac{1}{2}mv^2 $.
The probability distribution of particle speeds is given by the Maxwell-Boltzmann distribution.

Density of States:

The number of quantum states available to particles in a given energy range is:

\[ g(E) = \frac{V \cdot 2\pi(2m)^{3/2} E^{1/2}}{h^3} \]

where $ g(E) $ is the density of states, $ V $ is the volume, $ m $ is the particle mass, and $ h $ is Planck's constant.

Average Energy:

\[ \langle E \rangle = \frac{\int_0^\infty E^{3/2} e^{-\beta E} dE}{\int_0^\infty E^{1/2} e^{-\beta E} dE} = \frac{3}{2\beta} = \frac{3}{2}kT \]

This result shows that each degree of freedom contributes $ \frac{1}{2}kT $ to the average energy of the system, as predicted by the equipartition theorem.

What is Partition Function and write down its properties:

Ans:

\begin{align*} n_{i}&=g_{i} e^{\alpha} e^{-E / k T}\\ &\Rightarrow \sum n_{i}=e^{\alpha} \sum g_{i} e^{-E / k T}\\ &\Rightarrow N=e^{\alpha} \sum g_{i} e^{-E / k T}\\ &\Rightarrow \dfrac{n_{i}}{N}=\dfrac{g_{i} e^{\alpha} e^{-E / k T}}{e^{\alpha} \sum g_{i} e^{-E / k T}}\\ & \Rightarrow \dfrac{n_{i}}{N}=\dfrac{g_{i} e^{-E / k T}}{\sum g_{i} e^{-E / k T}}\\ & \Rightarrow \dfrac{n_{i}}{N}=\dfrac{g_{i} e^{-E / k T}}{Z} \end{align*}

Z is called as the partition function of the system. It represents how N particles are distributed among thier energy levels.

Features of Partition Function(Z)

It indicates the mode of distribution of energy among various energy levels.
It is a pure number, hence it is dimensionless.
It can never be zero.
Its lowest value is 1 at absolute zero temperature as molecules stay in the ground state.
It is much larger than 1 at higher temperature as fewer molecules stay in the ground state.
It is also a measure of the extent to which particles may escape from the ground state.

Partition Function for Canonical Ensembles with no degeneracy
Partition Function for Canonical Ensembles with degeneracy
Partition Function for Grand Canonical Ensembles with

Thermodynamic quantities in terms of Partition Function

Relation between Mean energy $\langle E\rangle$ and Z :

$$ \begin{aligned} \langle E\rangle&=\frac{E}{N} \\ & =\dfrac{\sum n_{i} E_{i}}{\sum n_{i}}\\ &=\dfrac{\sum(N / Z) g_{i} e^{-E / k T} E_{i}}{\sum n_{i}} \\ &= \dfrac{\sum(N / Z) g_{i} e^{-\beta E} E_{i}}{N}\\ &=\sum \dfrac{1}{Z} g_{i} e^{-\beta E} E_{i} \\ & =\dfrac{1}{Z}\left(\frac{-\partial Z}{\partial \beta}\right) \\ \Rightarrow\langle E\rangle & =-\dfrac{\partial}{\partial \beta}(\ln Z) \end{aligned} $$

Entropy in terms of Z :

\[ \begin{align*} S &= k \ln W = k \ln(Z_N e^{E/kT}) = k \ln Z_N + \frac{E}{T} \\ S &= k \ln\left(n! \prod_{i=1}^{\infty} \frac{g_i^{n_i}}{n_i!}\right) \\ &= k\left[n \ln n - n + \sum\left(n_i \ln g_i - n_i \ln n_i + n_i\right)\right] \\ &= k\left[n \ln n + \sum\left(n_i \ln g_i - n_i \ln n_i\right)\right] \\ &= k\left[n \ln n + \sum\left(n_i \ln \frac{g_i}{n_i}\right)\right] \\ &= k\left[n \ln n + \sum\left(n_i \ln \left(\frac{Z}{n} e^{\beta E_i}\right)\right)\right] \\ &= k\left[n \ln n + \sum\left(n_i \ln Z - n_i \ln n + \beta n_i E_i\right)\right] \\ &= k\left[n \ln n + \sum n_i \ln Z - \sum n_i \ln n + \beta \sum n_i E_i\right] \\ &= k[n \ln n + n \ln Z - n \ln n + \beta E] \\ &= kn \ln Z + k \cdot \frac{1}{kT} \cdot \frac{3}{2} n k T \\ &= kn \ln Z + k \cdot \frac{1}{kT} E \\ &= kn \ln Z + \frac{E}{T} \\ &= k \ln Z + \frac{3}{2} nk \end{align*} \]

HelmHoltz Free Energy F

$$ \begin{aligned} F & =E-T S \\ & =E-T\left(k \ln Z+\frac{E}{T}\right)\\ F&=-k T \ln Z \end{aligned} $$

Gibbs Free Energy G

$$ \begin{aligned} & G=H-T S\\ &=E+PV-T S \\ & =E+R T-T(n k \ln Z+E / T) \\ & =E+R T-n k T \ln Z-E \\ & =R T-n k \ln Z \end{aligned} $$

Total Energy of the system E :

$$ \begin{aligned} E=N\langle E\rangle \\ & =N\left(\frac{1}{Z}\left(-\frac{\partial Z}{\partial \beta}\right)\right) \\ = & \frac{N}{Z} \frac{\partial Z}{\partial T} \frac{\partial T}{\partial \beta}\\ &=\frac{N}{Z} \frac{\partial Z}{\partial T} \frac{\partial}{\partial \beta}\left(\frac{1}{k \beta}\right) \\ &= \frac{N}{Z} \frac{\partial Z}{\partial T}\left(-\frac{1}{k \beta^{2}}\right) \\ &= -\frac{N}{Z} \frac{\partial Z}{\partial T} \frac{k^{2} T^{2}}{k}\\ & =-N k T^{2} \frac{1}{Z} \frac{\partial Z}{\partial T} \\ & =-N k T^{2} \frac{\partial}{\partial T}(\ln Z) \end{aligned} $$

Enthalpy H :

$$ \begin{aligned} H & =U+P V\\ &=E+R T\\ &=N\langle E\rangle+R T \\ & =N K T^{2} \dfrac{\partial}{\partial Z}(\ln Z)+R T \end{aligned} $$

Pressure of the gas $\mathbf{P}$ :

$$ \begin{aligned} & \because F=U-T S\\ &=E-T S \\ & \Rightarrow d F=d E-S d T-T d S \\ &\Rightarrow d F=d E-S d T-d Q \\ & \Rightarrow d F=d E-S d T-d Q \\ &\Rightarrow d F=d U-S d T-d Q \\ \quad \because d Q &=d U+d W=d E+d W \\ & \Rightarrow d F=-d W-S d T \\ &\Rightarrow d F=-P d V-S d T \\ & \Rightarrow-\left(\frac{d F}{d V}\right)_{T}=P \\ & \Rightarrow P=-\left(\frac{d}{d V}(-k T \ln Z)\right)_{T} \\ & \Rightarrow P=k T\left(\frac{d}{d V} \ln Z\right)_{T} \end{aligned} $$

Specific heat in terms of Z

$$ \begin{aligned} & C_{V}=\left(\frac{\partial E}{\partial T}\right)_{V}\\ &=\frac{\partial}{\partial T}\left(N k T^{2} \frac{\partial}{\partial T} \ln Z\right) \\ & =N k\left(2 T \frac{\partial}{\partial T} \ln Z+T^{2} \frac{\partial^{2}}{\partial T^{2}} \ln Z\right) \end{aligned} $$

Prove that $\beta=\frac{1}{k T}$

$$\begin{aligned} \because d Q &=d U+d W\\ &=d U+P d V\\ \Rightarrow T d S&=d U+P d V\\ \Rightarrow\left(\frac{d S}{d U}\right)_{V}&=\frac{1}{T}.\\ but d(\ln W)&=0 \quad \& \quad d(\beta U)=0\\ \Rightarrow d(\ln W)&=\beta d U\\ \Rightarrow \beta &=\frac{d(\ln W)}{d U}\\ &=\frac{1}{k} \frac{d(k \ln W)}{d U}\\ & =\frac{1}{k} \frac{d S}{d U} \\ \Rightarrow \beta &=\frac{1}{k}\left(\frac{1}{T}\right)\\ &=\frac{1}{k T} \end{aligned}$$

Prove that the Partition function of a monatomic ideal gas molecule is :

$$ Z=\frac{V}{h^{3}}(2 \pi m k T)^{3 / 2} $$

Proof :

$$ \begin{aligned} & Z=\sum_{i} g_{i} e^{-E_{i} / k T} \\ & =\sum_{i} g(E) e^{-E / k T} \\ & =\int_{0}^{\infty} \frac{2 \pi V}{h^{3}}(2 m)^{3 / 2} E^{1 / 2} d E e^{-E / k T} \\ & =\frac{2 \pi V}{h^{3}}(2 m)^{3 / 2} \int_{0}^{\infty} E^{1 / 2} e^{-E / k T} d E \\ & =\frac{2 \pi V}{h^{3}}(2 m)^{3 / 2} \frac{\Gamma(3 / 2)}{(1 / k T)^{1+1 / 2}} \\ & =\frac{2 \pi V}{h^{3}}(2 m)^{3 / 2}(k T)^{3 / 2} \sqrt{\pi} / 2 \\ & Z=\frac{V}{h^{3}}(2 \pi m k T)^{3 / 2} \end{aligned} $$

( Partition function of the system containing $\mathbf{N}$ distinct ideal gas molecules $\left.Z_{N}=\frac{V^{N}}{h^{3 N}}(2 \pi m k T)^{3 N / 2}\right)$