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+ | ~~SLIDESHOW~~ | ||

+ | |||

+ | ====== Fall 2018: Lecture 31 - Ideal Gases and Kinetic Theory ====== | ||

+ | |||

+ | /* | ||

+ | ---- | ||

+ | If you need a pdf version of these notes you can get it [[http://www.ic.sunysb.edu/class/phy141md/lecturepdfs/141lecture32F11.pdf|here]] | ||

+ | |||

+ | ===== Video of lecture ==== | ||

+ | |||

+ | The video should play in any browser, but works best in anything that isn't Internet Explorer. If you are having trouble watching the video within the page you can [[http://www.ic.sunysb.edu/class/phy141md/lecturevids/phy141lecture32f2011.mp4|download the video]] and play it in [[http://www.apple.com/quicktime/download/|Quicktime]]. | ||

+ | |||

+ | <html> | ||

+ | <video id="video" width="640" height="360" controls="true"/> | ||

+ | <source src="lecturevids/phy141lecture32f2011.mp4" type="video/mp4"></source> | ||

+ | <source src="lecturevids/phy141lecture32f2011.ogv" type="video/ogg"></source> | ||

+ | |||

+ | |||

+ | <object width="640" height="384" type="application/x-shockwave-flash" data="player.swf"> | ||

+ | <param name="movie" value="player.swf" /> | ||

+ | <param name="flashvars" value="file=lecturevids/phy141lecture32f2011.mp4" /> | ||

+ | |||

+ | </object> | ||

+ | |||

+ | |||

+ | </video> | ||

+ | </html> | ||

+ | |||

+ | Note: In the messing around with the hotplate I forgot to start the screencapture. Thankfully I did remember to put the flip camera on.. | ||

+ | |||

+ | */ | ||

+ | ===== 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 [[wp>Boyle%27s_law|Boyle's Law]], after Robert Boyle who formulated it in 1662. | ||

+ | |||

+ | {{Boyles_Law_animated.gif}} | ||

+ | |||

+ | ===== Charles' Laws ===== | ||

+ | |||

+ | Joesph Louis Gay-Lussac published [[wp>Charles%27s_law|Charles' Law]] in 1802, attributing it to unpublished work of Jacques Charles in the 1780's (Gay-Lussac has his own law..though it's not clear he should!).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. | ||

+ | |||

+ | $V\propto T$ | ||

+ | |||

+ | {{Charles_and_Gay-Lussac's_Law_animated.gif}} | ||

+ | |||

+ | ===== Gay-Lussac's law ===== | ||

+ | |||

+ | [[wp>Gay-Lussac%27s_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 [[wp>Molecular_mass|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 [[wp>Ideal_gas_law|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 [[wp>Boltzmann_constant|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 [[wp>Absolute_zero|absolute zero]]. | ||

+ | |||

+ | /* | ||

+ | Through laser cooling and molecular trapping techniques it is now possible (but difficult!) for temperatures on the order of a $\mathrm{nK}$ to be achieved. Prof. [[http://ultracold.physics.sunysb.edu/index.html|Dominik Schneble]] produces ultra-cold ($\mu K$) Bose-Einstein condensates in the basement of this building! Prof. [[http://www.stonybrook.edu/metcalf/hmetcalf.html|Hal Metcalf]] was one of the key players in the original development of laser cooling. | ||

+ | */ | ||

+ | |||

+ | ===== What makes a gas ideal?===== | ||

+ | |||

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

+ | |||

+ | - There must be a large number of molecules and they should move in random directions with a range of different speeds. | ||

+ | - The spacing between molecules should be much greater than the size of the molecules. | ||

+ | - Molecules are assumed to interact only through collisions. | ||

+ | - 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 | ||

+ | |||

+ | {{kinetictheory.png}} | ||

+ | |||

+ | 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 | ||

+ | |||

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

+ | |||

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

+ | |||

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

+ | |||

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

+ | |||

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

+ | |||

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

+ | |||

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

+ | |||

+ | |||

+ | ===== Maxwell Boltzmann Distribution ===== | ||

+ | |||

+ | If we look at a [[http://www.chm.davidson.edu/vce/kineticmoleculartheory/Maxwell.html|simulation]] of particles moving according to the kinetic theory we can get some idea of the distribution of the speeds of the particles. | ||

+ | |||

+ | The [[wp>Maxwell%E2%80%93Boltzmann_distribution|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 [[wp>James_Clerk_Maxwell|James Clerk Maxwell]] based on symmetry arguments, later [[wp>Ludwig_Boltzmann|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 [[wp>Gaussian_function|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 [[wp>Gaussian_integral|standard integral of a gaussian function]]. | ||

+ | |||

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

+ | |||

+ | This is explained quite nicely on [[http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/ExpIntegrals.htm|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 [[wp>Integration_by_parts|integration by parts]]. | ||

+ | |||

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

+ | |||

+ | ===== Finding A ===== | ||

+ | |||

+ | |||

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

+ | |||

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

+ | |||

+ | |||

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

+ | |||

+ | $u=v$<html>        </html> $dv=ve^{-Bv^{2}}dv$ | ||

+ | |||

+ | $du=dv$<html>        </html>$v=-\frac{1}{2B}e^{-Bv^{2}}$ | ||

+ | |||

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

+ | |||

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

+ | |||

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

+ | |||

+ | ===== Finding B ===== | ||

+ | |||

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

+ | |||

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

+ | |||

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

+ | |||

+ | $u=v^{3}$<html>        </html> $dv=ve^{-Bv^{2}}dv$ | ||

+ | |||

+ | $du=3v^{2}dv$<html>        </html> $v=-\frac{1}{2B}e^{-Bv^{2}}$ | ||

+ | |||

+ | |||

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

+ | |||

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

+ | |||

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

+ | |||

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

+ | |||

+ | ===== 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 | ||

+ | |||

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