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7: Statistical Physics

Last updated

Aug 14, 2020
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- 6: Optics
- 8: Thermodynamics

Johan Wevers

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Degrees of freedom

A molecule consisting of $n$ atoms has $s=3n$ degrees of freedom. There are 3 translational degrees of freedom, a linear molecule has $s=3n-5$ vibrational degrees of freedom and a non-linear molecule $s=3n-6$ . A linear molecule has 2 rotational degrees of freedom and a non-linear molecule 3.

Because vibrational degrees of freedom account for both kinetic and potential energy they count double. So, for linear molecules this results in a total of $s=6n-5$ . For non-linear molecules this gives $s=6n-6$ . The average energy of a molecule in thermodynamic equilibrium is $\left\langle E_{\rm tot} \right\rangle=\frac{1}{2}skT$ . Each degree of freedom of a molecule has in principle the same energy: the principle of equipartition.

The rotational and vibrational energy of a molecule modeled as a rigid harmonic oscillator are:

$W_{\rm rot}=\frac{\hbar^2}{2I}l(l+1)=Bl(l+1)~,~~W_{\rm vib}=(v+\frac{1}{2})\hbar\omega_0$

The vibrational levels are excited if $kT\approx\hbar\omega$ , the rotational levels of a hetronuclear molecule are excited if $kT\approx2B$ . For homonuclear molecules additional selection rules apply so the rotational levels are well coupled if $kT\approx6B$ .

The energy distribution function

The general form of the equilibrium velocity distribution function is $P(v_x,v_y,v_z)dv_xdv_ydv_z=P(v_x)dv_x\cdot P(v_y)dv_y\cdot P(v_z)dv_z$ with

$P(v_i)dv_i=\frac{1}{\alpha\sqrt{\pi}}\exp\left(-\frac{v_i^2}{\alpha^2}\right)dv_i$

where $\alpha=\sqrt{2kT/m}$ is the of a particle. The average velocity is given by $\left\langle v \right\rangle=2\alpha/\sqrt{\pi}$ , and $\left\langle v^2 \right\rangle=\frac{3}{2}\alpha^2$ . The distribution as a function of the absolute value of the velocity is given by:

$\frac{dN}{dv}=\frac{4N}{\alpha^3\sqrt{\pi}}~v^2\exp\left(-\frac{mv^2}{2kT}\right)$

The general form of the energy distribution function then becomes:

$P(E)dE=\frac{c(s)}{kT}\left(\frac{E}{kT}\right)^{\frac{1}{2}s-1}\exp\left(-\frac{E}{kT}\right)dE$

where $c(s)$ is a normalization constant, given by:

Even $s$ : $s=2l$ : $\displaystyle c(s)=\frac{1}{(l-1)!}$
Odd $s$ : $s=2l+1$ : $\displaystyle c(s)=\frac{2^l}{\sqrt{\pi}(2l-1)!!}$

Pressure on a wall

The number of molecules that collides with a wall with surface $A$ within a time $\tau$ is given by:

$\int\hspace{-1.5ex}\int\hspace{-1.5ex}\int d^3N=\int\limits_0^\infty \int\limits_0^\pi \int\limits_0^{2\pi} nAv\tau\cos(\theta)P(v,\theta,\varphi)dvd\theta d\varphi$

From this follows for the particle flux on the wall: $\Phi= \frac{1}{4} n\left\langle v \right\rangle$ . For the pressure on the wall it then follows that:

$d^3p=\frac{2mv\cos(\theta)d^3N}{A\tau}~,~~\mbox{so}~~p=\frac{2}{3}n\left\langle E \right\rangle$

The equation of state

If intermolecular forces and the volume of the molecules can be neglected then for gases from $p=\frac{2}{3}n\left\langle E \right\rangle$ and $\left\langle E \right\rangle=\frac{3}{2}kT$ it can be derived that:

$pV=n_sRT=\frac{1}{3}Nm\left\langle v^2 \right\rangle$

Here, $n_s$ is the number of of molecules and $N$ is the total number of molecules within volume $V$ . If the molecular volume and the intermolecular forces cannot be neglected the Van der Waals equation can be derived:

$\left(p+\frac{an_s^2}{V^2}\right)(V-bn_s)=n_sRT$

There is an isotherm with a horizontal point of inflection. In the Van der Waals equation this corresponds with the and of the gas. This is the upper limit of the area of coexistence between liquid and vapor. From $dp/dV=0$ and $d^2p/dV^2=0$ follows:

$T_{\rm cr}=\frac{8a}{27bR}~,~~p_{\rm cr}=\frac{a}{27b^2}~,~~V_{\rm cr}=3bn_s$

At the critical point: $p_{\rm cr}V_{m,\rm cr}/RT_{\rm cr}=\frac{3}{8}$ , which differs from the value of 1 which follows from the ideal gas law.

Scaled using the critical quantities, with $p^*:=p/p_{\rm cr}$ , $T^*=T/T_{\rm cr}$ and $V_m^*=V_m/V_{m,\rm cr}$ with $V_m:=V/n_s$ one obtains:

$\left(p^*+\frac{3}{(V_m^*)^2}\right)\left(V_m^*-\mbox{$\frac{1}{3}$}\right)= \mbox{$\frac{8}{3}$}T^*$

Gases behave the same for equal values of the reduced quantities: the law of the corresponding states. A virial expansion can be used for an even more accurate picture:

$p(T,V_m)=RT\left(\frac{1}{V_m}+\frac{B(T)}{V_m^2}+\frac{C(T)}{V_m^3}+\cdots\right)$

The $T_{\rm B}$ is the temperature for which the 2nd virial coefficient is 0. In a Van der Waals gas, this happens at $T_{\rm B}=a/Rb$ . The $T_{\rm i}=2T_{\rm B}$ .

An equation of state for solids and liquids is given by:

$\frac{V}{V_0}=1+\gamma_p\Delta T-\kappa_T\Delta p= 1+\frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_{p}\Delta T+\frac{1}{V}\left(\frac{\partial V}{\partial p}\right)_{T}\Delta p$

Collisions between molecules

The collision probability of a particle in a gas that is moving over a distance $dx$ is given by $n\sigma dx$ , where $\sigma$ is the . The mean free path is given by $\displaystyle\ell=\frac{v_1}{nu\sigma}$ with $u=\sqrt{v_1^2+v_2^2}$ the relative velocity between the particles. If $m_1\ll m_2$ then: $\displaystyle\frac{u}{v_1}=\sqrt{1+\frac{m_1}{m_2}}$ , so $\displaystyle\ell=\frac{1}{n\sigma}$ . If $m_1=m_2$ then: $\displaystyle\ell=\frac{1}{n\sigma\sqrt{2}}$ . This means that the average time between two collisions is given by $\displaystyle\tau=\frac{1}{n\sigma v}$ . If the molecules are approximated by hard spheres the cross section is: $\sigma= \frac{1}{4} \pi(D_1^2+D_2^2)$ . The average distance between two molecules is $0.55n^{-1/3}$ . Collisions between molecules and small particles in a solution result in . For the average motion of a particle with radius $R$ it can be derived that: $\left\langle x_i^2 \right\rangle=\frac{1}{3}\left\langle r^2 \right\rangle=kTt/3\pi\eta R$ .

A gas is called a if $\ell\gg$ the volume of the gas, something that can easily occur at low pressures. The equilibrium condition for a vessel which has a hole with surface area $A$ in it if $\ell\gg\sqrt{A/\pi}$ is: $n_1\sqrt{T_1}=n_2\sqrt{T_2}$ . Together with the general gas law follows: $p_1/\sqrt{T_1}=p_2/\sqrt{T_2}$ .

If two plates slide along each other at a distance $d$ with velocity $w_x$ the $\eta$ is given by: $\displaystyle F_x=\eta\frac{Aw_x}{d}$ . The velocity profile between the plates is in that case given by $w(z)=zw_x/d$ . It can be derived that $\eta=\frac{1}{3}\varrho\ell\left\langle v \right\rangle$ where $v$ is the thermal velocity.

The heat conductance in a stationary gas is described by: $\displaystyle\frac{dQ}{dt}=\kappa A\left(\frac{T_2-T_1}{d}\right)$ , which results in a temperature profile $T(z)=T_1+z(T_2-T_1)/d$ . It can be derived that $\kappa=\frac{1}{3}C_{mV}n\ell\left\langle v \right\rangle/N_{\rm A}$ . Also: $\kappa=C_V\eta$ . A better expression for $\kappa$ can be obtained with the : $\kappa=(1+9R/4c_{mV})C_V\cdot\eta$ with an error $<$ 5%.

Interaction between molecules

For dipole interaction between molecules it can be derived that $U\sim-1/r^6$ . If the distance between two molecules approaches the molecular diameter $D$ a repulsive force between the electron clouds appears. This force can be described by $U_{\rm rep}\sim\exp(-\gamma r)$ or $V_{\rm rep}=+C_s/r^s$ with $12\leq s\leq20$ . This results in the Lennard-Jones potential for intermolecular forces:

$U_{\rm LJ}=4\epsilon\left[\left(\frac{D}{r}\right)^{12}-\left(\frac{D}{r}\right)^6\right]$

with a minimum $\epsilon$ at $r=r_{\rm m}$ the following holds: $D\approx0.89r_{\rm m}$ . For the Van der Waals coefficients $a$ and $b$ and using critical quantities: $a=5.275 N_{\rm A}^2D^3\epsilon$ , $b=1.3N_{\rm A}D^3$ , $kT_{\rm kr}=1.2\epsilon$ and $V_{\rm m,kr}=3.9N_{\rm A}D^3$ .

A simpler model for intermolecular forces assumes a potential $U(r)=\infty$ for $r<D$ , $U(r)=U_{\rm LJ}$ for $D\leq r\leq3D$ and $U(r)=0$ for $r\geq3D$ . This gives for the potential energy of one molecule: $\displaystyle E_{\rm pot}=\int_D^{3D}U(r)F(r)dr$ with $F(r)$ the spatial distribution function in spherical coordinates, which for a homogeneous distribution is given by: $F(r)dr=4n\pi r^2dr$ .

Some useful mathematical relations are:

$\int\limits_0^\infty x^n{\rm e}^{-x}dx=n!~~,~~ \int\limits_0^\infty x^{2n}{\rm e}^{-x^2}dx=\frac{(2n)!\sqrt{\pi}}{n!2^{2n+1}}~~,~~ \int\limits_0^\infty x^{2n+1}{\rm e}^{-x^2}dx=\mbox{$\frac{1}{2}$}n!$