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Fall 2018: Lecture 31 - Ideal Gases and Kinetic Theory

Boyle's Law

At constant temperature, it is found that the product of the pressure and volume of an ideal gas are constant

$PV=\mathrm{constant}$

This is named Boyle's Law, after Robert Boyle who formulated it in 1662.

Charles' Laws

Joesph Louis Gay-Lussac published Charles' Law in 1802, attributing it to unpublished work of Jacques Charles in the 1780's (Gay-Lussac has his own law..though it's not clear he should!).

Charles' Law states that at constant pressure the volume of a gas is proportional to the temperature.

$V\propto T$

Gay-Lussac's law

Gay Lussac's Law states that for a fixed volume the pressure is proportional to the temperature

$P\propto T$

Ideal Gas Law

The combination of the previous 3 laws implies that

$PV\propto T$

Our previous laws were for systems of constant mass, but we can see that the amount of mass should effect the volume (at a given pressure) or the pressure (at a given volume).

$PV\propto mT$

Measuring the amount of mass in moles will allow us to write the ideal gas law in terms of a universal constant. A mole of gas is a given number of molecules, Avagadro's number, $N_{A}=6.02\times 10^{23}$. If we have a certain mass $m$ of a gas which has a certain molecular mass (measured in atomic mass units, $\mathrm{u}$, which are also the number of grams per mole.), the the number of moles $n$ is given by

$n=\frac{m[\mathrm{g}]}{\textrm{molecular mass}[\mathrm{g/mol}]}$

and

$PV=nRT$ where $R=8.314\mathrm{J/(mol.K)}$

This equation is the ideal gas law

Ideal Gas Law for a number of molecules

The ideal gas law can also be written in terms of the number of molecules $N$

$PV=nRT=\frac{N}{N_{A}}RT=NkT$

where $k=\frac{R}{N_{A}}=\frac{8.314\mathrm{J/(mol.K)}}{6.02\times 10^{23}}=1.38\times 10^{-23}\mathrm{J/K}$ is the Boltzmann Constant.

Using the Ideal Gas Law to determine Absolute Zero

If $PV=nRT$ the absolute zero temperature occurs when $P=0$. In practice most gases will liquefy before this point, but we can measure the pressure of fixed volume of gas at a couple of reference points and extrapolate down to zero pressure to get an estimate for absolute zero.

What makes a gas ideal?

There are a number of conditions which must be satisfied for a gas to be considered ideal

  1. There must be a large number of molecules and they should move in random directions with a range of different speeds.
  2. The spacing between molecules should be much greater than the size of the molecules.
  3. Molecules are assumed to interact only through collisions.
  4. The collisions are assumed to be elastic.

Ideal gas law from a molecular perspective

We can work out the pressure exerted by an ideal gas on it's container by starting from the change in momentum of a molecule when it strikes the container wall

The average force due to one molecule is then

$F=\frac{\Delta(mv)}{\Delta t}=\frac{2mv_{x}}{2l/v_{x}}=\frac{mv_{x}^{2}}{l}$

The net force on the wall will be the sum of the forces from all $N$ molecules

$F=\frac{m}{l}\Sigma_{i=1..N} v_{xi}^{2}$

$\frac{\Sigma_{i=1..N} v_{xi}^{2}}{N}=\bar{v_{x}^{2}}$ → $F=\frac{m}{l}N\bar{v_{x}^{2}}$

$v^{2}=v_{x}^{2}+v_{y}^{2}+v_{z}^{2}$ → $\bar{v^{2}}=\bar{v_{x}^{2}}+\bar{v_{y}^{2}}+\bar{v_{z}^{2}}=\bar{3v_{x}^{2}}$

$F=\frac{m}{l}N\frac{\bar{v^{2}}}{3}$

$P=\frac{F}{A}=\frac{1}{3}\frac{Nm\bar{v^{2}}}{Al}=\frac{1}{3}\frac{Nm\bar{v^{2}}}{V}$

$PV=\frac{2}{3}N(\frac{1}{2}m\bar{v^{2}})=NkT$

$\bar{KE}=\frac{1}{2}m\bar{v^{2}}=\frac{3}{2}kT$

Maxwell Boltzmann Distribution

If we look at a simulation of particles moving according to the kinetic theory we can get some idea of the distribution of the speeds of the particles.

The Maxwell-Boltzmann Distribution gives the probability $f(v)$ that a particular particle in an ideal gas has a given speed $v$. It was originally derived by James Clerk Maxwell based on symmetry arguments, later Ludwig Boltzmann derived it on a more general basis. We will follow Maxwell's reasoning.

Velocity distribution

Suppose we have a gas of $N$ particles. The number of particles which have a velocity in the x direction between $v_{x}$ and $v_{x}+dv_{x}$ will be $Nf(v_{x})dv_{x}$

$\int_{-\infty}^{\infty}f(v_{x})dv_{x}=1$

The same function $f$ should describe the $y$ and $z$ direction as well, so

$Nf(v_{x})f(v_{y})f(v_{z})dv_{x}dv_{y}dv_{z}$

gives the number of particles with velocity between $v_{x}$ and $v_{x}+dv_{x}$,$v_{y}$ and $v_{y}+dv_{y}$, and, $v_{z}$ and $v_{z}+dv_{z}$

Maxwell realized that all directions are equivalent the distribution function should depend only on the total speed of the particle

$f(v_{x})f(v_{y})f(v_{z})=F(v_{x}^2+v_{y}^2+v_{z}^2)$

This implies that the form of $f$ must be

$f(v_{x})=Ae^{-Bv_{x}^2}$

Which is a Gaussian Function.

Mean velocity

The fact that the velocity in a given direction is given by the function

$f(v_{x})=Ae^{-Bv_{x}^2}$

implies that the mean velocity in any given direction is zero. Does this also mean the mean particle speed is zero?

From velocity to speed

We now know that

$f(v_{x})=Ae^{-Bv_{x}^2}$

But we need to find out what A and B are, and we also would like to express the function in terms of the speed of the molecules rather than the velocity in a given direction.

Starting from the number of particles with velocity between $v_{x}$ and $v_{x}+dv_{x}$,$v_{y}$ and $v_{y}+dv_{y}$, and, $v_{z}$ and $v_{z}+dv_{z}$

$Nf(v_{x})f(v_{y})f(v_{z})dv_{x}dv_{y}dv_{z}=NA^{3}e^{-B(v_{x}^2+v_{y}^2+v_{z}^2)}dv_{x}dv_{y}dv_{z}$ $=NA^{3}e^{-Bv^2}dv_{x}dv_{y}dv_{z}$

To transform $dv_{x}dv_{y}dv_{z}$ to an integral in the speed we consider a sphere in velocity space and can reason that the equivalent volume to $dv_{x}dv_{y}dv_{z}$ is $4\pi v^{2}dv$

So the probability distribution as a function of speed is

$f(v)dv=4\pi v^2A^{3}e^{-Bv^{2}}dv$

Finding A and B

We now have

$f(v)dv=4\pi v^2A^{3}e^{-Bv^{2}}dv$

We can use some of the information about the system to find out what $A$ and $B$ have to be. All particles have to have a velocity between 0 and $\infty$, so

$\int_{0}^{\infty}f(v)dv=1$

We saw before that

$\bar{K}=\frac{1}{2}m\bar{v^{2}}=\frac{3}{2}kT$

so

$\int_{0}^{\infty}\frac{1}{2}mv^{2}f(v)dv=\frac{3}{2}kT$

Gaussian Intergral

We will need to use the standard integral of a gaussian function.

$\int_{-\infty}^{\infty}e^{-ax^{2}}=\sqrt{\frac{\pi}{a}}$

This is explained quite nicely on this page

The integrals we need to do

$\int_{0}^{\infty}f(v)dv$

and

$\int_{0}^{\infty}\frac{1}{2}mv^{2}f(v)dv$

can be found using integration by parts.

$\int u\,dv = uv - \int v\,du$

Finding A

$\int_{0}^{\infty}f(v)dv=1$

$4\pi A^{3}\int_{0}^{\infty}v^{2}e^{-Bv^{2}}dv=1$

$\int u\,dv = uv - \int v\,du$

$u=v$         $dv=ve^{-Bv^{2}}dv$

$du=dv$        $v=-\frac{1}{2B}e^{-Bv^{2}}$

$4\pi A^{3}\int_{0}^{\infty}v^{2}e^{-Bv^{2}}dv=4\pi A^{3}([(-\frac{v}{2B}e^{-Bv^{2}})]_{0}^{\infty}+\int_{0}^{\infty}\frac{1}{2B}e^{-Bv^{2}}dv)$

$=\frac{4\pi A^{3}}{2B}\int_{0}^{\infty}e^{-Bv^{2}}dv=\frac{4\pi A^{3}}{2B}\frac{1}{2}\sqrt{\frac{\pi}{B}}=1$

$4\pi A^{3}=\frac{4B^{3/2}}{\sqrt{\pi}}$

Finding B

$\bar{K}=\frac{1}{2}m\bar{v^{2}}=\frac{3}{2}kT$

$\int_{0}^{\infty}\frac{1}{2}mv^{2}f(v)dv=\frac{3}{2}kT$

$2m\pi A^{3}\int_{0}^{\infty}v^{4}e^{-Bv^{2}}dv=\frac{3}{2}kT$

$u=v^{3}$         $dv=ve^{-Bv^{2}}dv$

$du=3v^{2}dv$         $v=-\frac{1}{2B}e^{-Bv^{2}}$

$2m\pi A^{3}\int_{0}^{\infty}v^{4}e^{-Bv^{2}}dv=2m\pi A^{3}([(-\frac{v^3}{2B}e^{-Bv^{2}})])_{0}^{\infty}+\int_{0}^{\infty}\frac{3v^2}{2B}e^{-Bv^{2}}dv)$

$=\frac{3m}{4B}(4\pi A^3)\int_{0}^{\infty}v^{2} e^{-Bv^{2}}dv=\frac{3m}{4B}$

$\frac{3m}{4B}=\frac{3}{2}kT$

$B=\frac{m}{2kT}$

And finally .. the Maxwell-Boltzmann distribution

$4\pi A^{3}=\frac{4B^{3/2}}{\sqrt{\pi}}$

$B=\frac{m}{2kT}$

$f(v)dv=4\pi v^2A^{3}e^{-Bv^{2}}dv$

$f(v)=4\pi (\frac{m}{2\pi k T})^{\frac{3}{2}}v^{2}e^{-\frac{1}{2}\frac{mv^{2}}{kT}}$

Now that we have this we can find various useful quantities, such as the average velocity

$\bar{v}=\int_{0}^{\infty}vf(v)dv=2\sqrt{\frac{2}{\pi}}\sqrt{\frac{kT}{m}}\approx 1.6 \sqrt{\frac{kT}{m}}$

or the most probable speed of a particle, from the condition $\frac{df}{dv}=0$ which gives

$v=\sqrt{\frac{2kT}{m}}\approx 1.41 \sqrt{\frac{kT}{m}}$

A third measure of velocity that we could already obtain before deriving the Maxwell Boltzmann distribution, is the root-mean-square, or rms velocity. Here we take the square root of the average of the squared velocity

$\frac{1}{2}m\bar{v^{2}}=\frac{3}{2}kT$ → $\sqrt{\bar{v^{2}}}=\sqrt{\frac{3kT}{m}}\approx 1.73 \sqrt{\frac{kT}{m}}$

phy141kk/lectures/31-18.txt · Last modified: 2018/11/25 10:46 by kkumar
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