MP203 Statistical and Thermal Physics Jon-Ivar Skullerud and James Smith October 3, 2017
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Contents 1 Introduction 3 1.1 Temperature and thermal equilibrium.................... 4 1.1.1 The zeroth law of thermodynamics................. 5 2 The ideal gas law 6 2.1 The ideal gas law and absolute temperature................ 6 2.1.1 What is an ideal gas?......................... 8 2.2 Microscopic model of an ideal gas...................... 8 2.3 Equipartition of energy............................ 10 2.4 Mean Free Path................................ 12 2
Chapter 1 Introduction We all think we know what temperature is, but it is notoriously difficult to define. In this course, we will arrive at a rigorous definition of temperature, but this definition does not appear to bear any direct or obvious relation to our commonsense understanding of the concept. This is one example of the paradoxical nature of the topic of statistical and thermal physics: on the one hand, it is a very down-to-earth subject we will for example study the performance of heat engines and refrigerators, and heat loss through walls, ceilings and windows of houses; but on the other hand it is full of riddles, not least how irreversible processes can arise from the motion of atoms and molecules which is described by time-reversible laws. Thermal physics has a history dating back to the seventeenth, with the discovery of Boyle s Law, providing a relation between the pressure and volume of a gas. It involves familiar concepts such as temperature, energy, pressure, heat flow and heat capacity; but also less familiar concepts such as entropy and latent heat. Most of the quantities can be directly measured and are macroscopic. Statistical physics dates back to the second half of the nineteenth century, with the work of Maxwell, Boltzmann and Gibbs. It describes the average behaviour of an extremely large number of particles, and the fluctuations around these averages, and how this gives rise to the thermal properties we can observe, including different states of matter, heat and the ideal gas law. Overview of topics Thermal equilibrium and temperature. The ideal gas law. Heat, work and the first law of thermodynamics Adiabatic, isothermal and cyclic processes Heat capacity; phase transitions and latent heat Heat transport: Fourier s law of heat conduction Entropy and the second law of thermodynamics; heat engines and refrigerators 3
Microscopic description of thermal systems, derivation of the ideal gas law Counting of quantum states; paramagnetism The relation between entropy and temperature Systems in contact with a heat bath. The Boltzmann distribution. Learning outcomes Define and distinguish the concepts of thermal equilibrium, temperature, heat, energy and entropy Calculate or estimate properties of an ideal gas from a particle model State the equipartition theorem and apply it to solids and gases Determine the heat exchanged and work performed in various thermal processes State the laws of thermodynamics and discuss some of their macroscopic implications, eg heat engines Calculate thermal averages using the Boltzmann distribution 1.1 Temperature and thermal equilibrium We start with some concepts and definitions: Thermodynamic system: Thermodynamic variable: Isolated system: Intensive quantity: Extensive quantity: a certain amount of stuff limited by a closed surface a macroscopic (measurable) quantity, such as energy or pressure no energy or matter exchange with the environment independent of the total mass of the system proportional to the total mass of the system Examples of thermodynamic variables: Extensive: Internal energy U Entropy S Volume V Particle number N Intensive: Temperature T Pressure p Chemical potential µ 4
1.1.1 The zeroth law of thermodynamics Let us start with a fundamental fact, which is often called the zeroth law of thermodynamics: If two systems are both in thermal equilibrium with a third system, they are in thermal equilibrium with each other. This fact is what allows us to define temperature as an objective property of a system, namely the property (whatever it is) that systems in thermal equilibrium with each other have in common. We can therefore say: If two systems are in thermal equilibrium with each other, they have the same temperature. Another fundamental fact of thermal equilibrium is: If two systems have been in contact for long enough, they are in thermal equilibrium with each other. Here, two objects or systems are deemed to be in contact if they are able to exchange energy. This allows us to achieve thermal equilibrium in practice. Combining these facts, we can come up with a provisional definition of temperature: Temperature is the thing that is the same for two objects if they have been in contact for long enough. In general, a system with higher temperature will transfer energy to a system with lower temperature until they achieve equilibrium. We will get back to this. For now, we will leave these rather abstract definitions and turn to some more concrete relations between thermodynamic variables. 5
Chapter 2 The ideal gas law 2.1 The ideal gas law and absolute temperature The first relation discovered between thermodynamic quantities (and, indeed, the first law of physics written as an equation relating two quantities) was Boyle s Law. It was first discovered by Henry Power and Richard Towneley in the mid-17th century, and then confirmed and published by the 14th child of the 1st Earl of Cork, Robert Boyle. It states that at constant temperature, the pressure of a quantity of gas is inversely proportional to its volume, pv = c, (2.1) where c is a constant. This law holds for most gases at constant temperature, and can be verified in air pumps. Much later, Jacques Charles (in the 1780s), John Dalton (1801) and Joseph Louis Gay- Lussac (1802) discovered a relation between temperature and volume at fixed (atmospheric) pressure, namely that when a gas heats up, it expands in proportion to the temperature, V 1 V 2 = kv 1 (T 1 T 2 ), (2.2) where k is another constant. If we take T 1 to be the freezing point of water, and measure temperatures in degrees Celsius, we find that k = 1 273.15 C We can see from Charles s Law (2.2) that if we lower the temperature, the volume of the gas will get smaller, and at some temperature T 0 it will become zero. Below that temperature, the volume will be negative, which probably does not make any physical sense. Setting V 2 = 0 in (2.2) we find V 1 = kv 1 (T 1 T 0 ) = T 0 = T 1 1 k. (2.3) This was realised by William Thomson, who later became Lord Kelvin, in 1848. Since the notion of a negative volume makes no sense, we can postulate that T 0 is the lowest temperature that can be attained, and define this as absolute zero. We can then define absolute temperature relative to this, ie T abs T T 0. (2.4) From now on, temperature will always mean absolute temperature. 6
Absolute temperature is usually measured in Kelvin (K) [not K] Boyle s Law (2.1) and Charles s Law (2.2) are both special cases of The ideal gas law pv = nrt Nk B T (2.5) In this equation, n = the molar weight of the gas, R = the gas constant = 8.31 J/(mol K), N = the number of molecules in the gas, k B = Boltzmann s constant = 1.381 10 23 J/K. Example 2.1 How many molecules are there in 1 m 3 of air at room temperature and atmospheric pressure? Take the air to be an ideal gas. Answer: We must first define what we mean by room temperature and atmospheric pressure. The standard values for these are With these values, we find T room = 20 C = 293 K, p atm = 1013 hpa = 1013 10 2 Pa = 1.013 10 5 Pa. N = pv k B T = 1.01013 105 (N/m 2 ) 1.0 m 3 1.381 10 23 (J/K) 293 K = 2.5 1025. (2.6) Example 2.2 A cylinder contains 12 litres of oxygen at 20 C and 15 atmospheres pressure. The temperature is increased to 35 C and the volume reduced to 8.5 litres. What is the resulting pressure? Answer: We have T 1 = 20 C = 293 K, T 2 = 35 C = 308 K. The ideal gas law gives us p 1 V 1 = Nk B T 1 = Nk B = p 1V 1 T 1 p 2 V 2 = Nk B T 2 = p 2 = Nk BT 2 = p 1V 1 T 2 V 1 T 2 = p 1 = 15 atm 12 V 2 T 1 V 2 V 2 T 1 8.5 308 293 = 22atm. 7
2.1.1 What is an ideal gas? As the name implies, the ideal gas law is an idealised description of real gases, and is not exactly satisfied for all gases (not to mention liquids and solids). In practice, it holds for sufficiently dilute gases, and it is a good description for most normal gases at normal conditions. The theoretical definition of an ideal gas is An ideal gas is a gas of point particles whose only interactions are perfectly elastic collisions. If the average distance between molecules in the gas is much larger than their size, this is a good approximation. We will later express this in terms of the mean free path of a molecule. 2.2 Microscopic model of an ideal gas Let us now change perspective and consider a gas with N 10 23 molecules enclosed in a volume V, and try to work out the pressure it exerts on the container walls. Computing the motion of all the molecules from Newton s laws is clearly hopeless, even if they never collide with each other, only with the walls. Let us instead look at averages. We start by considering the pressure exerted on a single wall of the box by a single molecule. L v Figure 2.1: A single molecule bouncing off the wall of an enclosed box. and t is the average time interval between collisions. The molecule exerts a force on the wall only when it bounces off the wall. Let us assume that this collision is elastic, and that no momentum is imparted to the wall in the y or z directions this will be true on average. The average force is given by F x = (mv x), (2.7) t where (mv x ) is the change in the momentum of the molecule (in the x- direction) as it bounces off the wall, For an elastic collision with the wall, (mv x ) = 2mv x : the molecule just bounces back with the same speed, but with v x reversed. t is the time it takes for the molecule to get to the opposite wall, bounce off that wall, and come back again, ie t = 2L v x. (2.8) 8
Putting this together, we find that the average force is F x = 2mv x 2L/v x = mv2 x L. (2.9) The pressure p is defined as the force divided by the area A of the wall, ie where V is the volume of the box. p (x) = F x A = mv2 x AL = mv2 x V, (2.10) We now assume that the gas is isotropic, ie that the average velocities and pressure are the same in all directions. This means that v 2 x = v 2 y = v 2 z = 1 3 (v2 x + v 2 y + v 2 z) = 1 3 v2, (2.11) where X means the average of X. If we have N molecules, each one of them will contribute to the pressure according to (2.10). Adding all this up, we bet But the ideal gas law tells us pv = Nmv 2 x = 1 3 Nmv2. (2.12) pv = Nk B T = k B T = 1 3 mv2 (2.13) 1 2 mv2 = 3 2 k BT (2.14) We see that the temperature is directly related to the average kinetic energy of the molecules in the gas. This expression also gives us the root mean square (rms) average speed of a molecule in the gas 3kB T v rms = m. (2.15) Note that this is not the same as the average, the most common, or the median speed, although all of these have similar values. The rms average is often used as the best measure of the typical magnitude of a quantity which can be both positive and negative. Example 2.3 Find the rms average speed of a nitrogen molecule in air at room temperature. Answer: We need the mass of a nitrogen (N 2 ) molecule. We can look this up in a table, or remember that m N2 = 2m N = 2 14u = 28 1.661 10 27 kg, where u is the atomic mass unit. Inserting this in (2.15) gives us 3 1.381 10 23 J/K 293 K v rms = = 511 m/s. (2.16) 28 1.661 10 27kg 9
Comments When we derived this expression, we made several simplifying assumptions: 1. There are no forces between the molecules which would cause them to slow down or deviate from their paths. 2. The collisions between the molecules and the walls are elastic. 3. The gas is isotropic: no direction is preferred over any other. The third condition is innocent: most gases (and liquids) are isotropic. If not, we would have different pressures in different directions (which could happen in some circumstances, but we will not deal with those here). The second condition is not problematic either: (a) The wall can be introduced as a thought experiment, allowing us to compute the pressure exerted by the gas on whatever it borders (including other bits of gas). (b) If the collisions are inelastic, the kinetic energy that is lost will go into internal energy (eg, vibrations) in the wall which can be released in subsequent collisitons. If the gas and wall are in thermal equilibrium, the collisions will on average be elastic, ie on average as much energy is gained as lost from the collisions. The first condition is the crucial one, but it can be relaxed: (a) Elastic collisions between molecules do not spoil the argument, since they do not change the total (or average) kinetic energy. (b) Inelastic collisions where energy is lost to internal motion inside molecules, or even to molecules breaking up do ruin the argument, but the result can be rewritten by taking these types of motion into account, as we shall see later. (c) The most important assumption is that molecules move in straight lines most of the time. This breaks down if the gas is very dense (so that collisions happen almost continuously), or if there are strong long-ranged forces between them. Electromagnetic forces might be examples of this. 2.3 Equipartition of energy Our result for the average kinetic energy can be written and (since we assume the gas is isotropic) 1 2 mv2 = 1 2 mv2 x + 1 2 mv2 y + 1 2 mv2 z = 3 1 2 k BT (2.17) 1 2 mv2 x = 1 2 mv2 y = 1 2 mv2 z = 1 2 k BT. (2.18) 10
This is an example of equipartion of energy: Motion in the three directions (x, y, z) corresponds to three different degrees of freedom, where each degree of freedom receives 1 2 k BT of energy. For a monatomic gas where the molecules are single atoms, such as helium (He), neon (Ne), argon (Ar) this is the only form of motion that can exist. Also, in this case, at normal temperatures, all collisions are elastic and the atoms are essentially spherical. However, for most molecules, other forms of motion are also possible. Rotation: A diatomic gas where the molecules consists of two atoms, such as nitrogen (N 2 ), oxygen (O 2 ), hydrogen (H 2 ) can rotate about two axes. If we call the axis joining the two molecules the z-axis, then the molecule can rotate about the x and y axes (but not the z axis). This is also the case for a linear molecule such as CO 2. We therefore have K rot = 1 2 Iω2 x + 1 2 Iω2 y = 2 degrees of freedom. (2.19) Most multiatomic gases (eg, H 2 O, CH 4 ) can rotate about all three axes, so K rot = 1 2 I xω 2 x + 1 2 I yω 2 y + 1 2 I zω 2 z = 3 degrees of freedom. (2.20) Note that in general, the moments of inertia I x, I y, I z about the three axes are different. Vibration: is given by A diatomic gas can also vibrate (oscillate). The energy in the oscillation E vib = 1 2 mv2 r + 1 2 kr2 r = 2 degrees of freedom, (2.21) where r r is the distance (or displacement from equilibrium) between the two atoms, and v r is their relative velocity. Inelastic collisions can transfer energy between linear motion, rotation and vibration. As a result, all active degrees of freedom receive 1 2 k BT of energy on average. There is a catch to this, which is that quantum mechanics sets a lower limit to allowable vibration (and rotaion) energies, so usually vibrations do not feature at normal temperatures, since the lowest allowable vibration energy is usually larger than 1 2 k BT. This means that for a diatomic gas we have 1 2 mv2 x = 1 2 mv2 y = 1 2 mv2 z = 1 2 Iω2 x = 1 2 Iω2 y = 1 2 k BT. (2.22) The total energy in a gas of N such molecules is U = N 5 ( 1 2 k BT ) = 5 2 Nk BT. (2.23) 11
Figure 2.2: The motion of a single molecule with diameter 2d pictured as a cylinder. 2.4 Mean Free Path Let us now look at how far a molecule travels between collisions. The average value of this is called the mean free path. We will assume all molecules are perfect spheres with diameter d. Two molecules collide when their centres come within a distance d of each other. Consider a single molecule (assume all others at rest). We can treat this molecule as having a diameter of 2d; all others are point particles. We can picture the motion of the molecule as a cylinder (series of cylinders), colliding with anything inside the cylinder. The number of molecules inside a cylinder with length l is N c = N V V cyl = N V πd2 l (2.24) This is the average number of collisions the molecule experiences over the distance l (time t = l ). The mean free path is v λ = l N c = V πd 2 N 1 nσ where n = number density (not the molar number) and σ = cross-section. (2.25) Here we assumed all other molecules were at rest. If we take their motion into account, it turns out that we get a factor of 2. λ mfp = 1 2nσ = V 2πd2 N (2.26) We can compare this with the size d of the molecules. The ideal gas approximation holds if λ mfp d (2.27) 12
Then, by using the ideal gas law we get V N = k bt p λ mfp = k bt 2πd2 p (2.28) 13