Answers to Physics 176 One-Minute Questionnaires March 20, 23 and April 3, 2009

Transcription

1 Answers to Physics 176 One-Minute Questionnaires March 20, 23 and April 3, 2009 Can we have class outside? A popular question on a nice spring day. I would be glad to hold class outside if I had some way to write on a blackboard and if I had a reasonable chance of keeping your attention. But with only six lectures left in the course and much interesting material to cover, I would like every lecture to count. Your questions do raise a point, that Duke should create some outside classrooms for just this purpose. I have been at some institutes in California and in Colorado where there were outdoor classrooms, with blackboards mounted on a wall and an awning to keep the Sun from blinding people. It was enjoyable to be outside while attending a lecture or discussing some science. How will we apply V q, the quantum volume, in a problem? I m just seeing a few formulas related to V q but I don t really understand how I would apply them. Problem 6.48 in Schroeder, which was part of your recent Assignment 12, hopefully gave you a sense of how to apply the quantum volume V Q. The key places where this quantity appears are in the equations on page 255 of Schroeder, especially the chemical potential µ for an ideal gas, Eq. (6.93). There is nothing tricky about V Q = lq 3 as shown in Eq. (6.83), it is a quantity with units of volume that depends on the temperature T of a system and on the mass m (in kg!) of the particles in the ideal gas, and it is straightforward to calculate. Perhaps the most important scientific point to keep in mind is that, when a gas becomes so cold or so dense that the volume per particle V/N becomes comparable to V Q in magnitude (say their ratio is less than 10). The assumption that the partition function Z N for N weakly interacting particles is given by Z1 N /N! then breaks down and one has to explicitly take into account the way particles occupy single-particle quantum states. Sections , which we are just starting to discuss, explain how to do this via Fermi-Dirac and Bose-Einstein distributions. 1

2 So the the egg-carton model is good for rough surfaces? temperature? Low There was a related question, When can we assume a surface to follow the egg-carton model? I am not a surface scientist so don t have a good or first-hand knowledge of when an egg-carton model is accurate for representative adsorbates and surfaces. I do know that Irving Langmuir and many scientists after him found that the surface coverage θ = N A /N S (where N A is the number of adsorbed molecules and N S is the total number of sites that could be occupied) for many kinds of molecules binding to many kinds of surface is related to the pressure P of the surrounding gas by Langmuir isotherms to excellent accuracy, and the temperature dependences of the pressures P 0 (T ) one obtains by fitting his isotherm θ = P/(P 0 + P ) to experimental data often make good sense chemically. But there are also cases where the eggcarton model breaks down, usually because the adsorbed atoms can move easily around on the surface. The answer to the first question is yes the egg-carton model is good for rough surfaces provided that one understands what rough means. All surfaces are rough at the atomic scale in that the electron cloud (by which I mean the quantum mechanical probability Ψ 2 to find electrons in a region of space) varies substantially near the surface, being larger in the location of a surface atom and smaller in between the surface atoms. If you shoot a low energy beam of electrons at any surface (which typically does not distort the surface electron cloud) and study the resulting scattered electrons, one can deduce directly the egg-carton-like structure of the surface. Alternatively, you can slide the atom-sized tip of an atomic force microscope (ATM) just above the surface of a solid, and detect the corrugated surface structure by the tiny variations in the electrical current that arises as electrons tunnel between the tip and the solid. The ATM tip is maintained a distance corresponding to many atoms away from the surface atoms and so does not distort the electron cloud during such a measurement. But you can t just treat the electron cloud as some passive rigid geometric egg carton, it can change its structure actively depending on the chemical properties of the molecule that adsorbs and on the chemical properties of the surface atoms. Some atoms and molecules are highly reactive and can form a strong covalent bond (or bonds for molecules) that lock the molecule in place on the surface, in which case the egg carton model is a good description. Other atoms and molecules, or surfaces, are such that only weak van der Waals bonds form and then the adsorbed object can often glide around 2

3 on the surface, at least if the temperature is not too low. An example might be an argon or helium atom adsorbing onto the surface of a crystal of solidified xenon. These are nobel elements with closed valence shells and so are chemically inert, forming only extremely weak bonds at low temperatures. We can do a simple calculation to see if just the geometric dimples in the surface electron cloud play a role in whether the surface acts like an egg carton. For example, let s assume that a surface acts like a frictionless geometric egg carton, with dimples whose height are about half a chemical bond in size, which would be about 100 picometers or about m = 0.1 nm. Then we might expect the egg carton model to fail if the average kinetic energy (1/2)mv 2 = 2(kT/2) = kt of a surface molecule of mass m becomes comparable to or exceeds the potential energy mgh needed to lift a molecule over one of the hills surrounding a dimple: kt = mgh. (1) For a helium atom, m = 4 g 10 3 (kg/g)/( ) kg and g = 9.8 m/s 2 at sea level. So the temperature would have to exceed about T > mgh k = ( ) J/K K, (2) an exceedingly low temperature. So gravity is extremely weak compared to electronic bonds: any experiment, especially at room temperature, is so hot that adsorbed atoms have plenty of kinetic energy to easily jump over the tiny geometric barriers of a naive egg carton model. This means one has to take into account the details of the electron cloud and chemistry of the reagents, which means measuring the details experimentally or solving the Schrodinger equation for the adsorbed atom on the surface (which is fairly routine these days). The result is that it is not easy to deduce a priori whether an egg carton model gives an accurate description. A rough rule of thumb is that surfaces of atoms that bind together covalently or ionically such as diamond, silicon, and sodium chloride (salt) are well described by an egg carton model, atoms don t move around once they adsorb. But surfaces of atoms that bind together as metals or by van der Waals interactions (e.g., He on a xenon crystal) are better described as continuous areas, in which an atom can move about freely. 3

4 How is the Helmholtz free energy F measured experimentally? For the Gibbs free energy G, which is more widely used than F, your question is mainly answered at the bottom of page 151 and top of page 152 Schroeder: use a calorimeter to measure the heat lost or gained during some change in a system at constant pressure (change in enthalpy H), measure the entropy change S from the initial to final state of the system, e.g., using Eq. (3.50) on page 112 of Schroeder, then compute G = H T S. For the free energy, follow the same steps except use constant volume processes rather than constant pressure processes. If F is useful to know when a system is in equilibrium or other properties, what is G useful for? Why do chemists use G and not F? This is partially explained in the Schroeder book. Different thermodynamic potentials like U = U(S, V, N), enthalpy H = U + P V = H(S, P, N), Helmholtz free energy F = U T S = F (T, V, N), Gibbs free energy G = U T S +P V = G(T, P, N), and grand potential Ω = U T S µn = Ω(T, V, µ) are appropriate to use under different experimental conditions, namely when the variables on which they depend might be held constant. The free energy F = F (T, V, N) is useful to use when the variables T, V, and N are held constant, and it is fairly common that scientists and engineers work with an equilibrium system with a rigid container of fixed volume, fixed temperature, and fixed number of particles. The Gibb s free energy G = G(T, P, N) is appropriate to use when the temperature and pressure are held constant. This is the case for systems open to the atmosphere, in which case the surrounding air acts like a reservoir with constant temperature and pressure. The condition for N to be constant might not be consistent with having a vessel open to the atmosphere since particles can then be exchanged between the system of interest and the surrounding air (and then we want to use a thermodynamic potential for which the chemical potential µ is assumed constant, not N). But if the system of interest involves molecules that are not exchanged with the atmosphere, then G is ok to use. For example, many solids and liquids and magnetic systems like a paramagnet will not exchange particles with the air and so G can be used to analyze their properties. You can tell which variables a thermodynamic potential depends on by looking at the differentials in the corresponding thermodynamic identity. 4

5 Thus the most fundamental identity du = T ds P dv + µ dn, (3) tells us that U = U(S, P, N) since, if we make a small change du in the energy, it arises from small changes in the variables S, V, and N. Similarly, the thermodynamic identity for the grand potential Ω: dω = d(u ST µn) = S dt P dv N dµ, (4) tells us that Ω = Ω(T, V, µ) and so is appropriate to use for an experiment in which a reservoir has fixed temperature T and fixed chemical potential µ. Note that there is a somewhat paradoxical implication of any given thermodynamic identity. When we write Ω = Ω(T, V, µ), we think of fixing the variables T, V, and µ in some experiment. If these are quantities fixed, then the complementary quantities S, P, and N that multiply each differential respectively are not fixed and so can fluctuate. However, for macroscopic systems with of order Avogadro number of elements, it is extremely unlikely to observe any deviation from the mean (the deviation will be of order 1/ N and so tiny) so the coefficients will be observed to take on their average values S, P, and N. These average values can in turn be computed via derivatives of the partition function. How does rotational and electronic energy affect the pressure? Thanks for catching this. My brief comment about this in lecture was wrong: in general, the properties associated with internal degrees of freedom of a molecule, as summarized in the internal partition function Z int, do not depend on the volume of the ideal gas and so the pressure P = ( F/ V ) T,N does not depend on details of the molecule and you get the ideal gas law result, P = NkT/V. (See Eq. (6.91) on page 255 and the related discussion.) This can change if the temperature becomes so low or the gas so dense that the wave functions of the molecules start to overlap, in which case the quantum mechanical properties of the molecule (rotational, vibrational, electronic structure) can influence the volume of the system, but the system is then no longer an ideal gas so one needs to use the Fermi-Dirac or Bose- Einstein distributions to describe what is going on. How does entropy and other thermodynamic identities tie into heat conductivity? Or does it? Thermal conductivity, a number that describes how rapidly heat spreads from a region of higher temperature to a region of lower temperature, de- 5

6 scribes a nonequilibrium property of a system (what happens when the temperature is not spatially uniform) and so there is no way to deduce the conductivity from thermodynamic quantities like the entropy S, free energy F, or partition function Z, which characterize systems in equilibrium. A calculation, part of which is described in Section 1.7 of Schroeder, leads to an explicit formula for the thermal diffusivity κ of an ideal gas (this is the quantity that gives us the time scale L 2 /κ to relax to equilibrium): κ = 2 3 π c σ 0 kt m, (5) where c is the intensive specific heat per molecule. This formula contains some thermodynamic quantities like c and the thermal energy kt but also contains a quantity σ 0 known as the scattering cross section, which is basically the effective cross sectional area of the molecule when another molecule collides with it. It is σ 0 that can not be deduced from thermodynamics nor from the quantum levels of the isolated molecule (and so not from the partition function). The cross section σ 0 can also vary with the relative speed of the molecules in a complicated quantum mechanical way. In the carbon monoxide problem (competing with oxygen to bind to hemoglobin), shouldn t there be a degeneracy or at least two different cases in blocking O 2 binding since the CO can bind to the hemoglobin with the carbon end or oxygen end? The carbon monoxide molecule CO is not symmetric in its properties. As you might remember from your introductory chemistry course, atomic oxygen is somewhat more electronegative (stronger tendency to attract electrons from another atom since O is closer to having a closed valence shell) than carbon. This means that the outermost molecular valence electrons responsible for binding carbon and oxygen together spend more time on average in the vicinity of the oxygen atom, which means that oxygen end of the CO molecule is slightly negative compared to the carbon end of the molecule; CO is an electrically polarized molecule. Now biological molecules like hemoglobin consist of long folded polymer chains of amino acids (of which there are 20 different kinds found for all life on Earth) and some of the amino acids become electrically charged in water, either positively or negatively. (The Wikipedia article Hemoglobin shows a schematic picture of the hemoglobin molecule, which consists of four identical subunits bound together, with each subunit containing one oxygen binding site.) The binding sites of biological molecules like hemoglobin 6

7 typically consists of several charged and neutral amino acids that are close to one another because of the specific way the long polymer chains has folded up. (In the case of the hemoglobin, a metallic iron atom is also positioned at the binding site.) The electrical charges of the binding site will then favor particular geometric orientations of ligands (molecules that bind to the site) and exclude other orientations. So one would expect that the polarized CO molecule will bind one end preferentially compared to the other end and there should not be a degeneracy. Note: my understanding of chemistry would have led to the opposite prediction of what is shown in Figure 7.2 on page 259, in which the carbon end of CO is binding to the Fe ion. I would have expected the slightly more negative oxygen end of CO to bind preferentially to the positively charged Fe 2+ ion that dominates the hemoglobin binding site. I checked on the Internet and found that, generally, carbon monoxide does bind to positively charged metallic ions via the carbon end so this is not special to hemoglobin. So my chemistry understanding is incorrect, perhaps there is a student in the class with a better knowledge of chemistry who can explain why the carbon end binds preferentially. What happens to the Maxwell distribution in the high temperature limit with relativity? Or does it not matter with the change to a plasma? As I showed by some calculations in class, for an ideal gas at room temperature that satisfies the Maxwell speed distribution, there are essentially no molecules in a large room moving faster than about eight times the speed of sound. Even one hundred times the speed of sound (about m/s) is so much slower than the speed of light ( m/s) that special relativity is irrelevant for nearly all gases that are cold enough not to be ionized. (An exception would be a gas of neutrinos, which are particles with small masses and that interact extremely weakly.) Relativity becomes important when the speeds of molecules become comparable to the rest mass mc 2 of the molecules, i.e., when the ratio kt/(mc 2 ) is no longer tiny. Then things start to happen in the gas that make the entire discussion complicated, namely collisions between particles can lead to the creation of new particles or the annihilation of existing particles, i.e., particle number and type are no longer conserved. In order to have an equilibrium gas with a significant number of relativistic elements, the temperature would have to be so high that the molecules would ionize and one would obtain a plasma that consists of freely moving 7

8 negative electrons and positively charged ions. (Actually, the rest energy of electrons and protons is so large compared to bonding energies, of order one giga-electronvolt compared to one ev, that all molecules in a relativistic gas would disintegrate into atomic ions and electrons.) For thermal energies kt substantially below the rest mass of the particles, the charged particles in the plasma will also eventually be described by a Maxwell speed distribution (when equilibrium is attained. But plasmas are more complicated than ideal gases of neutral molecules because the cross section for two charged particles to scatter off one another decreases with increasing relative energy so relativistic particles scatter weakly off other particles which means that so it can take a long time for a relativistic plasma to reach thermodynamic equilibrium. For a given experiment with a sufficiently hot plasma (like the 10 8 K thermonuclear plasmas generated at the Princeton Plasma Physics Lab, where I worked some years ago as a theoretical physicist), the plasma may never reach equilibrium during an experiment and so one can not use a Maxwell distribution to describe the plasma. If a magnetic field is present, it is even more complicated because particles will typically have one temperature along the magnetic field lines (along which particles move easily) and a different temperature perpendicular to the field lines (since diffusion of a charged particle across field lines is slow). You should appreciate that the derivation of the Boltzmann factor is completely general and is valid for special relativity. Thus the relativistic energy of a neutral particle of rest mass m with velocity v is γmc 2 where γ = 1/ 1 (v/c) 2 is the usual Lorentz relativistic factor. So the Boltzmann factor to observe this particle with velocity v in an equilibrium gas is the usual expression [ p(v) exp E ] [ ] γmc 2 exp. (6) kt kt This gives the velocity distribution. To get the speed distribution, one has to multiply this by the number of vectors with speed v and here the calculation is more complicated, because of relativity we can t use the expression 4πv 2 since that would allow speeds faster than the speed of light. The correct distribution is called the Maxwell-Juttner distribution but I ll let you investigate the details further on your own. The relativistic formula has been used to describe electrons in insterstellar clouds that have been heated by X-rays emitted from nearby stars, I don t know of any applications on Earth except possibly to fusion plasmas. 8