MP203 Statistical and Thermal Physics. Jon-Ivar Skullerud and James Smith

Transcription

1 MP203 Statistical and Thermal Physics Jon-Ivar Skullerud and James Smith December 20, 2017

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3 Contents 1 Introduction Temperature and thermal equilibrium The zeroth law of thermodynamics The ideal gas law The ideal gas law and absolute temperature What is an ideal gas? Microscopic model of an ideal gas Equipartition of energy Mean Free Path The First Law of Thermodynamics Heat and Work Compression Work Isothermal and Adiabatic Compression Heat Capacity Molar Heat Capacities Calculating Heat Capacities Phase Transitions and Latent Heat Heat Transport Heat Conduction Fourier Relaxation Heat Conductivity of an Ideal Gas The Heat Equation Solutions to the Heat Equation Entropy and the Second Law of Thermodynamics 36 2

4 5.1 Two-State Systems and Combinatorics The Two-State Paramagnet Interacting Systems

5 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 4

6 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 5

7 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. 6

8 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 = 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 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. 7

9 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 = 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 = Pa = Pa. N = pv k B T = (N/m m (J/K 293 K = (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 = 22atm. 8

10 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 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 9

11 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 qq 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 = kg, where u is the atomic mass unit. Inserting this in (2.15 gives us J/K 293 K v rms = = 511 m/s. ( kg 10

12 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 mv2 y mv2 z = k BT ( mv2 x = 1 2 mv2 y = 1 2 mv2 z = 1 2 k BT. (

13 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 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 I yω 2 y 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 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. ( 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 12

14 Figure 2.2: The motion of a single molecule with diameter 2d pictured as a cylinder. 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 (2.25 where n = number density (not the molar number and σ = cross-section. 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 Then, by using the ideal gas law we get V N = k bt p λ mfp = k bt 2πd2 p (

15 Note: We assumed here that all molecules are perfect hard spheres, which is clearly not true for real atoms and molecules. However, the cross section σ can be defined (and measured/computed for all real molecules - it is a measure of the strength and range of the interactions between particles. Therefore, we can still use these expressions to determine the mean free path. Example 2.4 What is (a the mean free path and (b the average collision rate for nitrogen at 293K and 101kP a? A nitrogen molecule has an effective diameter d = m. Answer: (a λ mfp = k bt 2πd2 p = ( J/K (293K ( 2π ( m 2 ( N/m 2 = m This works out to be 300d, which means that the ideal gas condition is fulfilled. (b From example 2.3, we have that v rms = 511 m/s (v rms is not exactly equal to v avg but it s close. The collision rate is 1 time between collisions = v avg λ mfp s 1 which is billions of collisions per seconds! 14

16 Chapter 3 The First Law of Thermodynamics 3.1 Heat and Work Until the mid-19th century, heat was seen as a substance (caloric fluid that was transferred in the process of heating or cooling objects. It was measured in calories. 1cal = the amount of heat required to increase temperature of 1g of water from 145 C to 155 C at 1atm pressure. Note: 1 calorie in food is actually 1kcal = 1000cal. James Joule showed that you could achieve the same temperature increase by doing mechanical work, without any heat. This established that heat is not a substance, but a form of energy, or rather a form of energy transfer (like work. In thermodynamics, we define heat and work as Heat: Spontaneous flow of energy from one abject to another, caused by a temperature difference between them. Work: All other types of energy transferred to or from a system (mechanical, electrical, chemical, etc There are three types of heat: 1. Conduction - Heat transfer between nearby molecules 2. Convection - Bulk motion of a fluid 3. Radiation - Emission of electromagnetic waves The first law of thermodynamics states that the total energy is conserved in any process and can be represented mathematically as The First Law of Thermodynamics U = Q + W (3.1 15

17 Figure 3.1: Work done by a piston compressing a gas. In this equation, U = change in internal energy of a system Q = heat transferred to the system W = work done on the system Comments (a W is often taken to be the amount of work done by the system. In that case, the first 1st law reads U = Q W. (b According to the definitions above, it makes no sense to talk about the amount of heat in a system. We can talk about the amount of energy in the system, but it is never stored at heat or work. This is clearly not the same as the everyday meaning of heat, just as work also has a different meaning in physics. 3.2 Compression Work The most important type of work done on a system(typically a gas is compressing it, e.g. pushing a piston. Unless you slam the piston really fast (faster than the speed of sound, the force will equal the pressure force of the gas. F = pa (3.2 Where A equals the area of the piston. The work done pushing the piston a small distance x is W = F X = pa x (3.3 But A x is just the reduction in the volume of the gas, V = A x = W = p V (3.4 16

18 In general, the pressure (and hence the force can change as volume is reduced, and we have W = x2 x2 F dx = padx = x 1 x 1 = W = V2 V2 V 1 pdv (3.5 V 1 p(v dv (3.6 If we know the pressure as a function of volume, we can use this to calculate the work. Example 3.1 1mol of ideal gas is compressed from an initial volume of 20.0L at atmospheric pressure, while keeping the temperature constant, to a final volume of 15.0L. Find the amount of work done. Answer:We know that Hence, pv = nrt = p(v = nrt V V2 nrt V2 W = dv = nrt V 1 V = nrt [ln(v 2 ln(v 1 ] = nrt ln for constant T. (3.7 V 1 ( V1 = W = p 1 V 1 ln V 2 dv V ( V1 V 2 (3.8 (3.9 (3.10 ( 20.0m = ( N/m 2 ( m 3 3 ln 10.0m 3 (3.11 = 2020J 0.288= 581J (3.12 This is an example of isothermal compression. For an ideal gas, we also know that the total energy is given by the equipartition theorem, U = f 2 Nk bt where f is the number of degrees of freedom. So if N and T do not change, the gas loses 581J of heat to the environment. 17

19 Example 3.2 An ideal gas is made to do undergo the cyclic process shown on the right. For each step A, B, C, determine whether the following are positive, negative or zero: (i the work done on the gas, (ii the change in the energy content of the gas, (iii the heat added to the gas. Finally, determine the sign of each of these for the whole cycle. Answer: A. (i Here we have expansion at constant pressure. The work done is then since V 2 > V 1 V2 W = pdv = p(v 2 V 1 < 0 (3.13 V 1 (ii From the Ideal Gas Law, pv = Nk b T, if V increases while p, N are constant, the temperature T must increase. Since the energy U Nk b T for an ideal gas, this implies U > 0. (iii The gas loses energy through work, but gains energy overall, so heat must be added i.e Q > 0. B. (i We are now at constant volume (V 3 = V 2 so no work is done, i.e. W = 0. (ii For the same reason as in step 1., T must increase and therefore U > 0. (iii U = Q + W < 0 but W > 0, so Q > 0. C. (i Now we have compression along a line where p V, say p = bv. Then V1 V1 [ 1 ] W = pdv = bv dv = V 3 V 3 2 bv 2 V1 = 1 V 3 2 b(v 3 2 V1 2 > 0. (3.14 (ii In this case, p and V both decrease, so pv = Nk b T decreases, and U > 0. (iii U = Q + W < 0 but W > 0, so Q < 0. 18

20 Total: (i We have W tot = W A + W C = p 1 (V 2 V b(v 2 2 V 2 1 (3.15 since V 3 = V 2. But we also have p 1 = bv 1, so W = 1 2 b(v 2 2 V 2 1 bv 1 (V 2 V 1 (3.16 = 1 2 bv 2 2 bv 1 V bv 2 1 (3.17 = 1 2 (V 2 V 1 2 > 0 (3.18 (ii Since we get back to where we started, U is the same, i.e. U = 0 (iii U = Q + W = 0 = Q = W, i.e. heat escapes from the gas. The net result result of this process is that work is done on the gas and is converted into heat. 3.3 Isothermal and Adiabatic Compression We will consider two important special cases of compression work: Isothermal: The temperature does not change (slow Adiabatic: No heat flow in or out (faster We already looked at isothermal compression in Example 3.1. For an ideal gas, pv = Nk b T, so if T (and N is constant, we have p(v = Nk bt V. V2 W = V 1 = Nk b T p(v dv = Nk b T [ ] V2 ln V = Nk b T ln V 1 V2 V 1 ( V1 V 2 dv V (3.19 (3.20 Also, for an ideal gas, U Nk b T = constant U = 0. Therefore ( V1 Q = U W = W = Nk b T ln V 2 (3.21 If V 1 > V 2 (compression, W > 0 (work done on the gas and Q < 0 (heat leaves the gas Now consider adiabatic compression, i.e. Q = 0. In this case, U = Q + W = W. According to the equipartition theorem, U = f 2 Nk bt (f is the number of degrees of freedom. 19

21 Clearly, the temperature will have to change, as will the pressure. Consider an infinitesimal compression dv, with temperature change dt. The change in internal energy is thus du = f 2 Nk bdt = W = pdv (3.22 f 2 Nk bdt = Nk bt dv V f dt 2 T = dv V (3.23 (3.24 using the fact that p = Nk bt. Integrating on both sides, we get V f T2 dt 2 T 1 T = f ( 2 ln T2 V2 dv = T 1 V 1 V = ln ( V2 V 1 (3.25 Now, exponentiate both sides, using e ln(x = x, and a ln(x = ln(x a ( T1 T 2 f/2 = V 2 V 1 V 1 T f/2 1 = V 2 T f/2 2 or V T f/2 = constant (3.26 For any given adiabatic compression, from V 1 to V 2, this gives us the change in temperature and therefore the energy, U = f 2 Nk b T = W. We may also find an expression for the pressure pv γ = constant, γ = f + 2 f = adiabatic exponent (3.27 Deriving this is left as an exercise. Example mol of Oxygen gas (O 2 has temperature 310K and volume 12l. (a What is the final temperature if the gas expands adiabatically? (b How much work is done by the gas in the process? Oxygen gas at this temperature has rotations, but no oscillation/vibrations. Answer: (a The number of degrees of freedom is f = 3(translation + 2(rotation = 5. For adiabatic expansion, we have ( V 1 T f/2 1 = V 2 T f/2 V1 2/f 2 T 2 = T1 = V 2 ( 12 2/5 310K= 258K (

22 (b Here we can use that W = U for an adiabatic process, and U = f 2 Nk bt = f 2 n mrt (3.29 U 1 = 5 2 n mrt 1 = 5 1mol 8.31J/(mol K 310K = 6440J 2 (3.30 U 2 = 5 2 n mrt 1 = 5 1mol 8.31J/(mol K 258K = 5360J 2 (3.31 W = U 2 U 1 = 1080J (3.32 This is the work done on the gas. The work done by the gas is 1080J. We can also consider work done at constant pressure. In this case the expression for the work is quite simple: W = pdv = p dv = p(v 2 V 1 = p V (3.33 Summary First law of thermodynamics: the change in internal energy = heat transferred to the system + work done on the system. U = Q + W Compression work: W = pdv Isothermal compression/expansion: (T =constant p(v = Nk bt V Constant volume: ( V1 W = Nk b T ln, U = 0, Q = W V 2 W = 0 Q = U = f 2 Nk b T Constant pressure: W = pdv Adiabatic expansion/compression: Q = 0 V T f/2 = constant, pv γ = constant, γ = f + 2 f W = U = f 2 Nk b(t 2 T 1 Underlined expressions hold for ideal gases. but not in general. 21

23 3.4 Heat Capacity The heat capacity of an object is the amount of heat required to increase its temperature, per unit of temperature increase. C Q (3.34 T We will also be looking at the specific heat, which is heat capacity per unit mass c C m (3.35 C is an extensive quantity. c is an intensive quantity - does not depend on the amount of stuff. But out definitions are ambiguous! We know that energy can be added as either heat or work, depending on the specific process. Even if the internal energy only depends on T, we may be doing work, so the expression C = U(T W T is ambiguous. There are two important special cases: 1. Constant Volume Here, no work is done, so C v = lim T 0 Q T = lim T 0 U T Where C v is heat capacity at constant volume. = V = const ( U T V (3.36 Note: 2. Constant Pressure This is more common in everyday life: objects expand and do work on the surrounding aire (for example, which remains at the same pressure. Here we have W = p V (3.37 Q C p = lim T 0 T = lim U + p V ( U ( V = + p (3.38 T 0 T p= const T p T p Where C p is heat capacity at constant pressure The partial derivatives ( U T V ( U T p in general since U may depend on pressure or volume. It matters which of them is kept constant! Example 3.4 Water has a specific heat capacity c p = 1.00cal/(g K = 4187J/(kg K. How much energy is needed to 22

24 (a boil enough water to make a cup of tea; and (b heat the water for a bath? Answer: We will assume the mains water in (a starts out with a temperature of 7 C. The water in (b might come from your attic tank with a temperature of 15 C. (a A cup is approx 0.20l, so its mass is approx 0.20kg. The temperature is increased by T = 100 C-7 C = 93K The heat required is Q = C p T = c p m T = 4187J/(kg K = 0.20kg 93K 78kJ (3.39 (b Now we heat water from 15 C to body temperature, 35 C T = 22K The volume of a bath tub is 1.5m 0.5m 0.5m = 0.38m 3 = 380l m 380kg Q = c p m T = ( J= 35MJ (3.40 If we like, we may convert this to kwh: 1W = 1J/s 1kWh = 1000W 1hr = 1000W 3600s = 3.6MJ (3.41 Heating the bath water requires approx 10kWh. Boiling a cup of tea requires approx 0.02kWh Note: For liquids and solids, the heat capacity nearly always means constant pressure (pressure of the surrounding medium. That means we allow the object to expand thermally (preventing this by applying pressure would be difficult. The difference between c p and c v is very small in practice Molar Heat Capacities Another useful quantity is the molar heat capacity [molar specific heat], which is heat capacity per mol: c m = C n m. Here are the specific heats and molar specific heats of some substances: 23

25 Substance c v [J/kg K] c m [J/mol K] Lead, P b Silver, Ag Copper, Cu Aluminium, Al Tungsten, W Is there a pattern here? Helium, He Argon, Ar Oxygen, O Nitrogen, N Hydrogen, H Ammonia, NH Carbon Dioxide, CO Methane, CH Calculating Heat Capacities Assume the equipartition theorem holds where f is the number of active degrees of freedom. Then, U = f 2 Nk bt C v = assuming f does not depend on T. The molar specific heat is ( U = ( f T v T 2 Nk bt = f 2 Nk b = f 2 n mr (3.42 C m v = C v n m = f 2 R (3.43 For monatomic gases: f = 3 c m v = 3 2 R = 12.47J/mol K diatomic gases: f = 5 c m v = 5 2 R = 20.79J/mol K polyatomic gases: f = 6 c m v = 3R = 24.95J/mol K metals/solids: f = 6 c m v = 3R = 24.95J/mol K The actual values for polyatomic gases such as NH 4, CH 4, CO 2 are closer to 7 2 R = 29.10J/mol K suggesting that a vibrational degree of freedom may be active. The number of active degress of freedom depends on the temperature - at low temperature all degrees of freedom freeze out (so c v 0 as T 0 for solids, while rotations and then vibrations (in gases become active at higher temperatures. Figure 3.2 shows a typical plot of c m v vs T for a diatomic gas. Note that at very low T the substance will become liquid, while at very high T the atoms will dissociate. In both cases the ideal gas description is no longer valid. What about c p (constant pressure? We found that C p = ( U ( V + p T p T p 24 (3.44

26 Figure 3.2: A plot showing how the specific heat of a diatomic gas increases with temperature. For an ideal gas: Putting this together: U = f 2 Nk bt C v = pv = Nk b T V = Nk bt p For specific heat capacities we would have ( U T p ( U = T ( V T p V = Nk b p = f 2 Nk b (3.45 (3.46 C p = f ( 2 Nk Nkb b + p = C v + Nk b (3.47 p C p C V = Nk b = nr (3.48 c p = c v + Nk b m (3.49 It is important to remember that these equation hold for ideal gases only. The relationship between c v and c p for solids and liquids is much more complicated than for ideal gases and is beyond the scope of this course. But since liquids and solids can practically be assumed to be incompressible, c p and c v have almost the same value. Hence, it is common to just have one value for their heat capacity. Example 3.5 We transfer 1000J as heat to a diatomic gas, allowing the gas to expand with p help constant. The molecules can rotate but not vibrate. (a How much of the 1000J goes into increasing the internal energy of the gas? (b How much of this goes into translational kinetic energy, and how much into rotational motion? (c If there is 7.60mol of the gas, by how much does its temperature increase? 25

27 Answer: (a Since the pressure is kept constant we can use C p = C v + n m R = 5 2 n mr + n m R = 7 2 n mr (3.50 T = Q = Q 7 C p n 2 mr (3.51 But U = 5 2 n mrt (3.52 U = 5 2 n mr T = 5 ( Q 2 n mr 7 n = 5 Q= 714J ( mr 7 (b We have k trans = 3 2 n mrt ; k rot = 2 2 n mrt = 2 5 U k trans = 3 2 U= 429J; k rot = 2 U= 286J ( (c The temperature increase is T = Q 7 n 2 mr = 1000J = 4.52K ( mol 8.31J/mol K Note that (a and (b could be answered without knowing n m and T. 3.5 Phase Transitions and Latent Heat When a system undergoes a phase transition - e.g. ice melting to become water, or water boiling to become vapour - it can absorb heat without increasing its temperature. Essentially, this energy goes into breaking the bonds between the molecules in the lower temperature phase. This energy is called latent heat, defined as the heat per unit mass required to completely melt/boil a substance. L Q m By convention, this is always taken to be at constant pressure. (3.56 Example 3.6 The latent heat for melting ice is 333J/g. How much heat is required to melt 6.0 litres of ice at 0 C and heat it to 45 C to wash dishes? 26

28 Answer: Q melt = L m = J/kg 6.0kg = J (3.57 Q heat = c water m T = 4.186J/gK 6000g 45K = J (3.58 Q tot = 3.1MJ = 0.87kWh (

29 Chapter 4 Heat Transport 4.1 Heat Conduction So far we have only considered systems in thermal equilibrium or the beginning and end of the equilibrium porcess (e.g. two bodies with different temperatures are brought into contact. But we may also want to know how fast the equilibrium state is reached - this can be the difference between mild discomfort and a visit to A&E, or between a small heating bill and fuel poverty. Specifically, we will look at how fast heat flows from a hot object to a cold object when they are put in contact: heat conduction (Heat can also be transferred through radiation and convection. The radiation law requires more advanced physics, while convection is a lot more complicated so we will not deal with it here. Imagine a slab of material with thickness x and area A. On the left side of the slab, the temperature is T = T 2 and on the right side T = T 1. If T 2 > T 1, heat will flow from left to right. On general grounds, we can expect that 1. Q A: the bigger the area, the higher the heat flow. 2. Q t: if T 1, T 2 are kept constant, the heat flows at a constant rate per time unit. 3. The thicker the slab, the smaller the heat flow. 4. The bigger the temperature difference T = T 2 T 1, the bigger the flow. In particular, Q = 0 if T 1 = T 2. The simplest expression consistent with 1-4 is Figure 4.1: Q t A T x or Q t = κadt dx (4.1 Note the - sign: dt dx < 0 if T 2 > T 1 28

30 We can write this as J Q = Q A T = κdt dx Fouriers Law of Heat Conduction (4.2 The quantity J Q is called the heat flux = heat per area per time. κ [kappa] is the thermal conductivity - it is a property of the material. Example 4.1 Calculate the rate at which heat would be lost on a cold winters day through a 62m x 38m brick wall 32cm thick, if the temperature inside is 22 C and outside is 2 C. The thermal conductivity of the brick is 0.74W/(m K. Answer: The heat loss is dq dt = κ A T x = (0.74W/(m K 62m 3.8m 24K 13000W ( m Fourier Relaxation Fourier s law of heat conduction, for a system where the temperature T varies only in the x-direction, is J Q = dq Adt = κ T x. (4.4 Here, J Q is the heat flux, which is the heat flow per unit time and area, A is the area of the surface through which the heat flows and κ is the thermal conductivity of the system. Imagine now that we have an amount of matter with heat capacity C V in a container with area A, thickness d and thermal conductivity κ, and that the temperature outside the container is kept constant at T o. The temperature T in on the inside of the container will now vary with time as heat flows out of (or into the container. Across the walls of the container itself, we have T x = T in T o. (4.5 d Substituting this into (4.4 and using dq = C V dt, we get for the temperature T in = T (t of the matter inside the container, dq Adt = C V dt i Adt = C V d(t T o Adt d(t T o T T o = κ d (T T o (4.6 = κa dc V dt (4.7 ln(t T o + C = κa dc V dt (4.8 T (t = T o + Be κa dc V t. (4.9 29

31 The constant B = e C can be determined if we know the temperature T i at the start of the process, t = 0: T (0 = T o + B = T i B = T i T o. (4.10 Note that we made some assumptions here which may not be very well satisfied in real life cases. In particular, we assume that the temperature does not vary across the piece of material we are considering, which would imply that the heat transport inside the material is perfect (or that the piece of matter is small enough that it always remains in internal thermal equilibrium. We also assume that there are no temperature gradients in the medium outside, so that any heat coming out from the container is immediately carried away Heat Conductivity of an Ideal Gas Let us consider two boxes of ideal gas, each of width l, but with slightly different temperatures T and T. This means that the molecules in box on the right are moving slightly faster on average. 1 2 m v2 x left = 1 2 k bt ( m v2 x right = 1 2 k b(t + T (4.12 In a time t, some of the molecules on the right will have moved into the left box, and visa versa. Specifically, take l λ mfp (mean free path. Figure 4.2: t = average time between collisions = λ mfp v In this time, the molecules will (on average have moved freely (without collisions. Half the molecules on the left (those with v x > 0 will have moved into the right hand box, and half of those on the right will have moved to the left. [We are cheating a bit here. A careful calculation shows that molecules move l = 2 3 λ mfp in the x-direction]. This implies a transfer of heat from right to left, Q = 1 2 U R 1 2 U L = 1 f 2( 2 N Rk b (T + T f 2 N Lk b T (4.13 = 1 2 (f 2 Nk b T = 1 2 Cbox v T (4.14 [We take the density to be the same on both sides] If we write Cv box unit volume. = V c vol = l A c vol, where c vol = Cv V = f 2 nk b is the heat capacity per 30

32 We can confirm Fouriers law, J Q = Q 1 A T = l/ v 2 lc vol T = 1 2 v lc vol T l = κ T l, (4.15 with κ = 1 2 v lc vol 1 2 v2 λ mfp c vol (4.16 We can now substitute the expressions for v 2, λ mfp, c vol ; κ 1 v2 λ mfp c vol = 1 3kB T 2 2 2m 3 = 4σ f 2 k b 1 2nσ f ( kb T m 2 nk b (4.17 1/2= 3 ( kb T 1/2 4σ cmolecule v (4.18 m The precise prefactor here is not important since we have made a lot of approximations. The important lessons are: 1. κ T - This is experimentally well confirmed in practice. 2. κ is independent of the pressure and density. 3. At constant T, κ 1 mσ - experimentally well confirmed Note that σ (cross-section is a measure of the strength of the interaction between the molecules. So if there are no interactions, the thermal conductivity is infinite, while it is small if the interactions are strong. This observation holds also for other transport properties such as viscosity. - the stronger the interaction, the smaller the viscosity/conductivity etc. 4. κ c molecule v f = the number of degrees of freedom. 4.2 The Heat Equation Figure 4.3: The heat flowing in from the right is Look at a uniform rod, with T varying along the x-direction. Look at heat flowing into a segment of length x. The heat flowing in from the left in a time t is T (x Q L = A t κ x (4.19 T (x + x Q R = A t κ x (

33 The net heat flow equals the change in energy, [ T (x T (x + Deltax ] Q L + Q R = A tκ x x (4.21 = U = cm x T = cρa x T, (4.22 where T is the change in temperature in time t. This gives us or in the limit t 0, x 0, T t = κ [ T (x/ x T (x + x/ x ] cρ x (4.23 T t = K 2 T x 2 with K = κ cρ The Heat Equation ( Solutions to the Heat Equation The heat equation is a partial differential equation (depending on x and t. Solving such equations is in general very complicated, and you do not have the mathematical or numerical tools for this. We can looks at some special cases though: 1. Stationary (time-independent solution Assume the temperatures T 1, T 2 at either end of a heat conducting material of length L are help constant in time. In that case, there is no reason why the temperature within the material should change with time. Specifically, T t x=0 = T t Then = 0 ; we can take T x=l t = 0 x (4.25 T t = 0 = K 2 T x 2 2 T x 2 = 0 (4.26 This now looks like an ordinary differential equation, T (x = 0 Which has the solution T (x = Ax + B (4.27 The constants A, B are determined from the boundary conditions T (x = 0 = T 1, T (x = L = T 2 B = T 1, A = T 2 T 1 L ( Heat flow from a small region Let us assume that at some time t 0 the temperature profile of a thin rod is T (x, t 0 = T 1 e x2 /a + T 0 (

34 If a is small this will describe a system with temperature T 1 T 0 in a very small region, and T 0 elsewhere. Consider the following expression: T (x, t = T 0 + A t e x2 /4Kt (4.30 This gives T x = A ( 2x e x2 /4Kt = Ax t 4Kt 2 T x = A /4Kt 2 2Kt 3/2 e x2 Ax ( 2Kt 3/2 = A ( 1 x2 e x2 /4Kt 2Kt 3/2 2Kt 2Kt 3/2 e x2 /4Kt 2x 4Kt e x2 /4Kt T t = A /4Kt 2t 3/2 e x2 + e x2 /4Kt + A ( + x2 e x2 /4Kt t 4Kt 2 e x2 /4Kt = A 2t 3/2 ( 1 x2 2Kt We can see that this satisfies the heat equation. (4.31 (4.32 (4.33 (4.34 (4.35 This solution describes a very narrow, very tall peak at small t, and a broad, flat distribution at late t. This is typical of how temperatures get evened out with time. Example 4.2 A certain solid material has heat capacity C = 150J/(kgK, thermal conductivity κ = 300W/(Km and density 10kg/m 3. We consider a rod of this material with length 1m and cross-section 1cm 2. assume the temperature only varies along the length direction of the rod. (a Assume the rod is attached to a tank of ice water at 0 C on one end and a tank of boiling water at 100 c ircc at the other end.if the temperature of the rod does not change with time, calculate the amount of heat flowing through the rod, from the boiling water to the ice water, per second. (b How long does it take for the heat flowing through the rod to melt a single gram of ice in the ice water tank? Answer: We (a if T does not change with time, the heat equation tells us that T (x = T 0 + T 1 T 0 x (4.36 L 33

35 Where x = 0 at the ice end, x = L = 1m at the boiling end, T 0 = 0 C and T 1 = 100 C Therefore, T x = T 1 T 0, and (4.37 L dq T = κ A dt x = 300W/Km (10 2m 2 100K = 3.0W (4.38 1m The minus sign is because heat flows from right to left, into the ice water. (b The latent heat of melting ice is 333J/g. Therefore, to melt 1g of ice requires Q = 333J. This takes t = Q dq/dt = 333J 3.0J/s = 111s (= 1min 51s to achieve. Summary Fouriers law of heat conduction: J Q = Q A t = κdt dx Heat conductivity for ideal gas: κ 1 ( 2 < v > λ mfpc vol cmolecule v kb T 1/2 f = σ m 2 k ( b kb T 1/2 σ m Heat Equation: T t = κ 2 T x 2 ; K = κ cρ Stationary Heat Flow ( T/ t = 0: T (x = Ax + B = T 0 + T 1 T 0 x L Figure 4.4: A stationary solution 34

36 Figure 4.5: Heat flow fram a small region 35

37 Chapter 5 Entropy and the Second Law of Thermodynamics Recall the first law, U = Q + W (5.1 Basically this says that you cannot get more energy out of a system than you put in. Specifically for a cyclical process (which can repeat itself over and over, U = 0, Q net + W net = 0, or Q in + W in = Q out + W out This means that you cannot make an engine that does more work than the energy you put into it (perpetual motion of the first kind But there is more! You could imagine a perfect engine (perpetual motion of the second kind that takes heat in and produces the same amount of work. THIS IS NOT POSSIBLE There are also processes that are irreversible - they only proceed in one direction, and not in the other: A stone falls to the ground and stays there. It does not jump back up on its own accord. A cup of coffee cools down in a cold room. it does not spontaneously heat up, taking heat from the room. Red wine is spilled on a table cloth. It does not by itself gather together and unstain the cloth. From the point of view of conservation of energy (or momentum, or angular momentum, ect there is nothing wrong with these processes going in reverse, or with a perfect engine. These facts of life are related to the second law of thermodynamics. There are several formulations of this law: 36

38 1. Kelvin Postulate No series of processes is possible whose sole result is the absorption of heat from a thermal reservoir and the complete conservation of this energy to work. 2. Clausius Postulate No process is possible whose sole result is the transfer of heat from a reservoir at one temperature to another reservoir at a higher temperature. 3. Entropy Formulation In a closed system, the entropy of the system will either increase or remain constant. It will never decrease The third formulation is the most general, and may be shown to entail the two others, but to understand it, we need to know what entropy is. Clausius introduced entropy S as S = f We will come back to this later. A more fundamental definition was found by Boltzmann, where Ω is the multiplicity of states. i dq T S = k b ln(ω (5.2 To understand what this means, we need to take a step back and introduce ideas of microstates, macrostates and combinatorics. A microstate is a complete enumeration of the states (eg. position, momentum, energy levels, spin/angular momentum of each and every one of the atoms/molecules in a system. A macrostate is described and distinguished by the values of bulk quantities such as total energy, pressure, magnetisation, ect. The fundamental assumption of statistical mechanics: In an isolated system in thermal equilibrium, all microstates are equally probable Let us consider a pair of dice. We throw the dice once. The microstates are given by the values shown on each die. There are 36 of them: 37

39 The macrostates are given by the total value of the dice. There are 11 of them (the numbers 2, 3,..., 12. The multiplicity is how many microstates there are for each macrostate. Here we have: Macrostate Multiplicity Using the fundamental assumption of statistical mechanics (which here is the same as saying the dice are not loaded we find that 7 is six times more likely to occur than either 2 or 12. According to the definition of entropy, we find that the state 7 has the highest entropy, specifically S(2 = S(12 = k b ln(1 = 0 (5.3 S(7 = k b ln(6 1.79k b (5.4 What this means is that for 2 and 12, the microstate is uniquely determined, whereas there are many ways of achieving a Two-State Systems and Combinatorics Let us now move from dice to coins, and consider a sequence of N coin tosses, which can each result in heads ( or tails (. This is actually a reasonable model of magnetisation, as we shall see soon. The microstate now records the sequence of individual tosses, eg. The macrostate counts the number of heads and tails, eg N = 5, N = 4 The question is now: How many ways are there of choosing N heads out of N (= N + N tosses? This will give us the multiplicity Ω(N, N [or Ω(N, N tot ]. Let us proceed by induction, assuming we do N coin tosses. N = 0: this means all tosses were tails, which is a unique microstate, i.e. Ω(0 = 1 N = 1: any one of the N tosses could be head, i.e. Ω(1 = N N = 2: we can choose the first head among among N tosses, and the second among the remaining N 1 N(N 1. However, it does not matter which one we call first or second, so the number of pairs is Ω(2 = N(N

40 N = 3: we can choose the first N ways: the second N 1 ways and the third N 2 ways N(N 1(N 2 ordered triplets. But all orderings are equivalent, and there are 3 2 ways or ordering 3 items, so Ω(3 = N(N 1(N N = 4: we now have N(N 1(N 2(N 3 ordered quadruplets, and ways or ordering each, i.e. N(N 1(N 2(N 3 Ω(4 = In general, for N coins and n heads: = N(N 1...(N n + 1 Ω(N, n = n(n ( N(N [(N n(n n 1...1]n! = N! N (N n!n! binomial coefficient n This result, Ω(N, n = N! n!(n n! ( N, (5.5 n holds not only for coin tosses, but for any system of N items which can be in one of two states (and not just in those cases either!. For example, distributing items (balls - or molecules! between two containers a system of magnetic dipoles with two possible directions The second example is called the two-state paramagnet, and is a good description of many real materials. We will come back to it soon. We can generalise our results for 2-state systems to systems with more possible states. The mathematics of the is called combinatorics. For example, if we have a 3-state system (let us call the three possible outcomes/states R,G,B with the number in each state labelled as n R, n G, n B, we can show that for a system of size N (N items where n R + n G + n B = N, The total number of microstates = 3 N The total number of macrostates = (N+1(N+2 2 The multiplicity of the macrostate Ω(n R, n G, n B = N! n R!n G!,n B! Exercise: Check this for youself! 39

41 5.1.1 The Two-State Paramagnet Consider a material consisting of N elementary magnets (magnetic dipoles. These can be electrons, or protons or nuclei or even protein molecules! In most solids they are usually unpaired electrons which carry a magnetic moment related to their spin (internal angular momentum. The magnetic moment µ characterises the strength of a magnet, defined by τ = µ B (torque created by external magnetic field (5.6 A magnetic dipole in an external magnetic field B has energy V = µ B. For a single electron, we have µ 2 = g e µ B s 2 where g e 2, µ B = e h 2m e = Bohr magneton, and s 2 = ± 1 is the spin quantum number. 2 We see that for a single electron there are two possible values (spin up, spin down for the magnetic moment, and if we have N atoms with one such electron each, we have a 2-state system similar to our system of coin tosses! Paramagnetism is when elementary magnets tend to align in the external magnetic field (like a compass needle since this minimises the potential energy V = µ B. This creates an increased magnetic field (magnetisation Ferromagnetism is when the magnets/spins interact with each other so that they line up in the same direction spontaneously. Diamagnetism is when the material creates a magnetic response opposite to the external field - this happens because the field induces a rotating current - most materials are diamagnetic to some extent. Consider now a two-state paramagnet with elementary magnetic moment µ, in a magnetic field B = B z. Each particle has energy U i = ±µb. The total energy of a system of size N, with N spin-up dipoles and N spin-down dipoles is U tot = µb(n N = µb(n 2N [ since N + N ] (5.7 The total magnetisation is M = µ(n N = U B (5.8 A macrostate is any state with a given (N, N, which gives a definite value for the macroscopic observables U, M. We have already found the multiplicity of the macrostates: Ω(N!, N = N! N!(N N! = N! N!N! (5.9 If N (and N, N is not very large, we can calculate this exactly, and use this to find the entropy as a function of energy or magnetism. But Ω(N, N quickly becomes huge: 40