Second Year Thermodynamics. 1.1 Basic Concepts

Transcription

1 Second Year hermodynamics M. Coppins Jargon 1.1 Basic Concepts Microscopic: on the atomic scale; involving very few or single particles (atoms, molecules, electrons, etc). Macroscopic: on the everyday scale or larger; involving very large numbers of particles. hermodynamics is a macrocscopic theory. System: thing whose properties we are interested in, e.g., gas in container. Surroundings: things outside the system; can interact with system. System + surroundings often called the universe in thermodynamics text books. Isolated system: amount of matter and total energy of system is fixed, e.g., gas in closed rigid, thermally insulating container. Closed system: amount of matter in system is fixed, but energy can enter or leave system, e.g., gas in closed rigid container with walls which conduct heat. Energy can enter as heat P (= pressure) increases. Open system: matter and energy can enter and leave it, e.g., gas in open container. If heated P stays constant (e.g., 1 atm) but number of molecules in container decreases Equilibrium = state which things settle into if left alone. In equilibrium the macroscopic properties don t change with time. Even when macroscopic properties have settled into their equilibrium values the microscopic properties (e.g., velocity of a given molecule) are not constant. Could have mixture with several chemically different components. N i = number of molecules of component i. Chemical reactions N i s change. Eventually chemical reactions stop, and the system settles into chemical equilibrium: N i s are constant. Could have unbalanced forces, e.g., push piston into a container of gas V (= volume) gets smaller. Eventually settles into mechanical equilibrium: V is constant. Could put system into thermal contact with something at different (= temperature). Eventually settles into thermal equilibrium: is constant. Chemical + mechanical + thermal = thermodynamic equilibrium.

2 1.1.3 State variables In equilibrium macroscopic properties are constant in time (if left alone), uniform in space (if no external force field present 1 ). If an external force field is acting the macroscopic properties can vary spatially, e.g., in an isothermal atmosphere ( 1st Year SoM course, Lecture 5) the density falls with height due to gravity. State variable = a macroscopic property of a system in equilibrium (e.g., for a gas the state variables include: P, V,, N = number of molecules). Extensive state variables are proportional to the amount of material. Intensive state variables are independent of amount of material. Combine 2 identical volumes of gas. Produces a volume of gas with 2N, 2V, P,, i.e., N and V double (extensive), but P and are unchanged (intensive). Specific value: extensive variable per unit mass. Denoted with lower case letters. Chemists use moles instead of kilograms. 1 mole is amount of stuff that contains as many particles as there are atoms in 12 g of 12 C. Chemists specific values are therefore molar values. We won t use moles in the hermodynamics course. he most familiar state variables are P, V, N, and. here are many others, e.g., U = internal energy, β = thermal expansion coefficient Equation of state...is a relationship between state variables (usually P, V, N, and ). It might be derived theoretically or found experimentally. Ideal gas: (where k B = Boltzmann s constant = J K 1 ). PV = Nk B ( ) Chemists write the ideal gas equation of state in terms of moles: PV = N m R, where N m = the number of moles and R is the universal gas constant. Van der Waals gas: ( ) P +a N2 (V Nb) = Nk V 2 B ( ) (where a and b are constants; 1st Year SoM course, Lecture 9) 1 his will assumed throughout the hermodynamics course

3 hermodynamics applies to other macroscopic systems than gases, in which case the state variables P and V can be replaced with other more appropriate variables, e.g., Stretched metal wire: In this case P is replaced with J = tension, and V is replaced with L = length. Under certain conditions the equation of state has the form: { L = L 0 1+ J } YA +α( 0) ( ) where Y = Young s modulus, A = cross-sectional area, α = coefficient of linear expansion, 0 = reference temperature, and L 0 = length when = 0 and J = Independent state variables Of the four state variables P, V, N, and, we can specify any three; the equation of state then gives the value of the fourth one, e.g., ideal gas with P = Pa (1 atm), N = molecules, = 298 K V = 0.41 m 3 In thermodynamics we are usually interested in what happens when things change, e.g., change volume by V, what is? How many state variables can we vary independently? (1) Closed system: N = given constant. Can vary 2 of P, V,, equation of state gives 3rd one. Equation of State defines PV surface, e.g. ideal gas Any point on this surface represents an equilibrium. Real substances have more complicated P V surface, e.g.,

4 [Picture from University Physics (9th edition) by Young and Freedman.] (2) Open system: N = can change 3 independent state variables. (3) N fixed, but two different phases (e.g., solid and liquid). Phase diagram is the projection of the PV surface on the P plane ( 1st Year SoM course, Lecture 4). wo phases can only coexist in equilibrium along the lines on this diagram extra constraint only 1 independent state variable. [We consider the situation of two coexisting phases in detail in Lecture 3.5.] (4) Several chemically distinct components present more independent state variables chemistry. Most of the hermodynamics course (Lectures ) deals with a single component, single phase, closed system 2 independent state variables. We will mostly concentrate on a fixed mass of gas in a container, partly because this particular system is intrinsically important, and partly because it is easy to visualize what is happening. Remember, however, that thermodynamics doesn t just apply to gases, but also to other macroscopic systems.

5 Second Year hermodynamics M. Coppins Definitions 1.2, U, S he most important state variables in thermodynamics are: = temperature, U = interal energy, S = entropy. Like other state variables (e.g., P) these are measurable macroscopic properties of things in equilibrium. is in the equation of state ( Lecture 1.1). U and S can be found from the other state variables. Inthermodynamicsweareusuallyconcernedwithchangesinthings, e.g., = final init. First law deals with U ( Lecture 1.4). Second law deals with S ( Lecture 2.3). Although thermodynamics is a macroscopic theory, this lecture will include a brief and very superficial look at, U and S from a microscopic viewpoint. he microscopic theory will be covered in detail next term in the Statistical Physics course emperature Microscopic view: describes how particles of a system are distributed with respect to energy. Boltzmann law ( 1st Year SoM course, Lecture 5): probability of particle having energy close to E, in equilibrium, is proportional to e βe, where β = 1/k B. From a theoretical point of view it would be better to replace the state variable with β, but we are too used to temperature in everyday life. Example: gas molecules in Maxwell-Boltzmann velocity distribution function f(v x ) = ( ) 1/2 ) m exp( mv2 x 2πk B 2k B he probability that the x component of any given molecule s velocity is between v x and v x +dv x is f(v x )dv x.. Increase particles spread out more with respect to energy. We always use absolute temperatures. If β was used instead of then absolute zero would correspond to β =. Looked at this way the inaccessibility of absolute zero seems very natural.

6 Macroscopic view: wo systems are in thermal equilibrium if they have the same. is sometimes approximately constant during some process. he theoretical idealization of this is an isothermal process, in which is kept constant by placing the system in contact with a heat reservoir (i.e., an object so large that heat flow in or out of it does not change its temperature). Example: isothermal compression. gas stays equal to res during the compression because heat flows from the the gas into the reservoir Internal energy Microscopic view: U = total (kinetic + potential) energy of all the particles in the system. Ideal gas: particles do not interact (i.e., no potential energy of interaction) can use theorem of equipartition of energy: U = N n d 2 k B (n d = number of degrees of freedom: 1st Year SoM course, Lecture 3) For a fixed mass of ideal gas it can be proved thermodynamically that the internal energy depends only on its temperature ( Lecture 3.2). Microscopic theory required to obtain the functional form. Equipartition predicts U = n d 2 Nk B ( ) (n d = constant for give material, e.g., n d = 3 for a monatomic gas). Ideal gas: U = constant = constant. Van der Waals gas: U = 3 2 Nk B a N2 V Real gas: U depends on both and V. (depends on both and V) Entropy Microscopic view: S = k B lnw, where W = number of microstates corresponding to given macrostate. Increasing W increasing disorder at microscopic level.

7 Consider a system consting of N gas molecules in a box of volume V. Divide the box into a number of equal sized cells. Microscopic view: we know which cell each individual molecule is in. Macroscopic view: we know how many molecules there are in each cell, but not which ones they are. In this simple model we describe the macrostate by the density in each cell: ρ i = number of molecules in cell i/volume of cell. Example: 4 molecules, 2 cells. Macrostate I: all 4 molecules in cell 1: ρ 1 = 4, ρ 2 = 0. Only one microstate corresponds to this macrostate (i.e. W = 1). Macrostate II: 3 molecules in cell 1, 1 molecule in cell 2: ρ 1 = 3, ρ 2 = 1. here are 4 microstates corresponding to this macrostate (i.e., W = 4). Macrostate III: 2 molecules in cell 1, 2 molecule in cell 2: ρ 1 = 2, ρ 2 = 2. here are 6 microstates corresponding to this macrostate (i.e., W = 6).

8 All the microstates are equally probable probability of given macrostate occuring W. We could initialize the system in macrostate I by placing all four molecules in left hand side of box. After the molecules have spread out it is 6 more likely that the system is in macrosate III than back in I. In macrostate III ρ = uniform = N/V, i.e., unlike the other macrostates (in which ρ is non-uniform) this macrostate can be completely specified by the values of the state variables N and V. Real systems have N >> 4 (e.g., 1 mm 3 of air at atmospheric pressure contains molecules). Macrostate with uniform density is overwhelmingly more probable (much higher W) than macrostate with highly non-uniform density. If left to itself the system will settle into the macrostate with maximum W. his is equilibrium state ( Lecture 2.6) Macroscopic view: increasing S particles and/or energy become more dispersed. Once things become dispersed it is hard to get them back together again increasing S associated with irreversible changes ( Lecture 2.1). Ideal gas: S = S 0 + n d 2 Nk Bln +Nk B lnv ( ) where S 0 is a constant ( Lecture 3.2) S increases if V increases particles spread out more in space. increases particles spread out more with respect to energy. Place hot and cold objects in thermal contact. initial state: energy more concentrated in hot object. final state: thermal equilibrium, same everywhere, energy more spread out. Everyone is aware of the energy crisis, and that we should try to conserve energy. But, of course, energy is conserved. We are not runing out of energy, we are running out of usable energy. he energy in a barrel of oil does not vanish when the oil is burnt, but it becomes very much harder to use because it has been dispersed entropy increases. Rather than an energy crisis, it would be more true to say we have an entropy crisis.

9 Second Year hermodynamics M. Coppins 1.3 Mathematical methods hermodynamics has a reputation for being a hard subject. here are basically three reasons for this. hermodynamics is hard, reason 1: It is inherently strange because we don t use the reductionist world-view. It is quite unlike anything else in physics. We will encounter the other two reasons in this lecture Changes in state variables Closed system: N = fixed. But V can vary, and energy can enter or leave. U,, P, S can all change. Example: gas in cylinder with piston. Piston moves out V = V final V init > 0. Can we find U,, P...? We need to specify 2 independent variables. Example: ideal gas, P init = 10 5 Pa, V init = 1.0 m 3, init = 298K ( N = molecules). Change V and : V = m 3, = +1.0 K. What is P? V final = 0.99 m 3, final = 299 K. P final = Nk B final /V final P = P final P init = Pa. [We could also calculate the changes in other state variables, e.g., U = +503 J, S = 1.69 J K 1 ( Problem Sheet 1).] We started in equilibrium with state variables P init, V init, init... and ended in equilibrium with state variables P init + P, V init + V, init +... We didn t specify what caused this change. Given: initial state variables, and any 2 of P, V,, U, S... we can find the final state. hermodynamics is hard, reason 2: here are usually many different ways of solving a given problem (different choices of the pair of independent state variables). It often difficult to tell which way is best Small changes Reminder of some basic maths: y = y(x) = function of 1 variable. Make a small change x in x. Corresponding change in y is: y ( ) dy x. dx Make an infinitesimal change dx in x. Corresponing infinitesimal change in y is: dy = ( ) dy dx. dx

10 If z = z(x,y) is a function of 2 variables then making small changes x and y in x and y results in a small change in z given by ( ) ( ) z z z x+ y x y y x ( ) z where = the partial derivative of z with respect to x, keeping y constant. x y his suggests an alternative approach to find P given and V. Choose and V as the two independent state variables. ( ) ( ) P P We then have: P + V. V V For an ideal gas: P = Nk ( ) B P = Nk ( ) B P, V V V V herefore: P Nk B V Nk B V 2 V. Putting in the values we find P Pa ( 1% error). = Nk B V 2. his approach is more accurate for smaller changes. In the limit of infinitesimal changes in state variables we have ( ) ( ) P P dp = d + dv. V V Whatever we choose for our two independent state variables, we can always express the infinitesimal change in some other state variable by an equation like this. We do this a lot in thermodynamics. hermodynamics is hard, reason 3: Lots of pertial derivatives are involved. his makes it look mathemtically abstruse, although, in fact, once you get used to it the required maths is very straightforward Mathematical summary o do all the manipulations involving partial derivatives in thermodynamics we use just four mathematical results. hese are the boxed results below. Given a function z = z(x,y) of two variables: dz = ( ) z dx+ x y ( ) z dy ( ) y x We could rewrite the functional dependence ( ) as( y = ) y(x,z) ( or) x = ( x(y,z). ) ( hus ) there ( are ) 6 z z y y x x possible 1st order partial derivatives: x y x z y z hese partial derivatives are related to each other by two identities., y, x, z, x, z. y

11 he reciprocal rule: he cyclic rule: ( ) x y z Finally, mixed 2nd derivatives obey: ( where 2 z x y means x ( ) ) z. y x y ( ) x = y z ( ) y z x 1 ( ) y x 2 z x y = 2 z y x z ( ) ( ) z = 1 ( ) x y ( ) For the hermodynamics course you should be familiar with the use of the four boxed relationships, but you don t need to be able to derive them. For those interested, a non-examinable proof of the reciprocal and cyclic rules is given here. For a function z = z(x,y): Rewriting this as y = y(x,z) we have dz = dy = ( ) z dx+ x y ( ) y dx+ x z ( ) z dy ( ) y x ( ) y dz ( ) z x Substituting Eq into Eq and rearranging we obtain: { ( z ) ( ) ( ) } {( ) ( ) } z y z y + dx+ 1 dz = 0 x y y x x z y x z x Since dx and dz are independent the coefficients of dx and dz in this equation must both be zero. Setting the coefficient of dz to zero gives the reciprocal rule, and setting the coefficient of dx to zero (and using the reciprocal rule) gives the cyclic rule Path independence If we make successive very small changes dp the total change in P is P = dp But if we start in a state A and end in state B P = P B P A i.e., P is the same for any process between A and B. We say it is path independent. he change of any state variable between two equilibria is path independent.

12 We can re-express the condition for path independence in the following way. If we start in state A and do some process at the end of which we are back in state A then dp = dp = P A P A = 0. A A For any state variable, e.g. entropy S where the integral is round any closed path. ds = 0 ( ) If the concept of path independence seems a bit abstract consider the following analogy. Suppose there are various paths (actual paths, which you can walk along) by which we could climb a hill. he sketch shows two of them, starting at point A at the bottom (altitude: h A = 10 m), and ending at point B at the top (altitude: h B = 300 m). he change in altitude from A to B is path independent: dh = dh = h B h A = 290m. A B, path 1 A B, path 2 If we start from A, go up one path and then come back to A along the other one then dh = total change in altitude for the round trip = 0. But the distance travelled is not path independent.: dl A B, path 1 A B, path 2 If we start from A, go up one path and then come back to A along the other one then dl = total distance travelled for the round trip 0. h is an intrinsic property of the location, e.g., point B has an altitude h = 300 m. In thermodynamic-speak h is a state variable l is not a property of the location (not a state variable ). dh = very small change in altitude. dl = very small distance (not a change in something). In thermodynamics we have very small changes in state variables, e.g., d, ds,... very small quantities which are not changes in state variables. hese are denoted by a crossed d. For instance dq = a very small amount of heat going into our system ( Lecture 1.4). dl.

13 Second Year hermodynamics M. Coppins Statement of the 1st law 1.4 he first law U = Q+W ( ) where U = change in internal energy, Q = heat into system, W = work done on system.!! ake care!! Some books (e.g., Young and Freedman) define W = work done by system, so the 1st law becomes U = Q W. Microscopic view: Q is energy transferred by random particle motion, W is energy transferred by ordered particle motion. Macroscopically, heat is energy transferred as a result of a temperature difference. Energy conservation is built into the 1st law. U = U final U init is path independent ( Lecture 1.3). W is path dependent ( Section 1.4.2). herefore Q = U W is also path dependent. An infinitesimal change in the internal energy is given by where dq and dw are infinitesimal amounts of heat and work. du = dq+ dw ( ) We denote them with a d because they are not changes in things. hey are path dependent. Mathematically, du, dp, d, etc are examples of exact or perfect differentials. hey are changes in some property (in thermodynamics, changes in state variables). dq and dw are inexact or imperfect differentials Work Work is done compressing a gas. Assume friction is negligible the compression is done very slowly (these conditions will be relaxed in the next lecture). Compress the gas by applying a force F to a piston.

14 In equilibrium: F = PA. Increase force by small amount: Force = F +δf. piston moves small distance δx. δw = Small amount of work done = (F +δf)δx = PAδx+δFδx }{{}}{{} = PδV where δv = change in volume = Aδx ( sign because V decreases), 2 = 2nd order in small quantities negligible, i.e., infinitesimal work done on gas is : herefore On a PV diagram ( 1st Year SoM course, Lecture 3) dw = PdV. ( ) Vfinal W = PdV. ( ) V init. Area under curve = Vfinal V init PdV = W = work done by the gas. he PV diagram was originally invented by James Watt in the 1790 s. He considered it so useful that he regarded it as a trade sectret which helped to give his steam engines a competitive edge. It only entered mainstream thermodynamics after being re-invented by Émile Clapeyron in W is path dependent. Can do work on systems other than gases, e.g.:

15 Work done stretching metal wire: dw = JdL (J = tension, L = length). [Note opposite sign from the gas case. his is because W > 0 if length of wire in increased or if volume of gas is decreased.] Work done charging capacitor: dw = φdq (φ = voltage, q = charge) Adiabatic process = process in which dq = 0, e.g., adiabatic compression of a gas For an ideal gas: P = Nk B V, and U = n d 2 Nk B du = n d 2 Nk Bd ( Lecture 1.2). he 1st law with dq = 0 therefore gives: n d 2 Nk Bd = Nk B dv V d = 2 dv ln = lnv 2/n d +const V 2/n d = const n d V i.e., an ideal gas undergoing an adiabatic process satisfies the following equation 1 where γ = (n d +2)/n d = constant. V γ 1 = constant, or PV γ = constant. ( ) On a P V diagram adiabatic processes are steeper than isothermal processes, e.g., monatomic ideal gas (γ = 5/3). 1 his statement will be qualified in Lecture 2.3.

16 1.4.4 Heat capacities Heat capacity = C is defined by equation: dq = Cd Specific heat = c = heat capacity per unit mass (intensive variable). dq is path dependent so is C e.g., C V = heat capacity at constant volume, C P = heat capacity at constant pressure. We can derive a general equation linking C V and C P.!! his is the first bit of serious thermodynamic maths in the course!! ( ) ( ) U U Choose V and as independent state variables du = dv + V 1st law: dq = du +PdV dq = { ( ) } U P + dv + V V d ( ) U d. ( ) V ( ) U At constant volume dv = 0 dq = d But at constant volume: dq = C V d. V herefore ( ) U C V =. ( ) At constant pressure: dq = C P d. o get an expression for C P we want an equation like Eq but with a term multiplying dp instead of dv on the right hand side. ( ) ( ) V V Write V = V(,P) dv = d P + dp (since dp = 0). P { ( ) }( ) ( ) U V U Substitute into Eq dq = P + d + d. V P V But dq = C P d and ( ) U V = C V. herefore { ( ) U C P = P + V V }( V ) P +C V. ( ) For an ideal gas: C V = n d 2 Nk B, C P = C V +Nk B, γ = C P C V ( Problem Sheet 2).

17 Second Year hermodynamics M. Coppins Reversible process 2.1 Reversibility = a process in which the system and its surroundings can be restored to their original state without any changes anywhere else. It is a theoretical concept (cf frictionless plane in mechanics, or a point charge in electromagnetism). All real processes are irreversible, but there are processes which are reversible to a good approximation Dissipation Energy in ordered motion is converted into heat (i.e., internal energy) temperature rise. Examples Pendulum. Energy falls with time due to friction at pivot and air resistance. Joule heating. Energy in electric current lost due to electrical resistance. Dissipation energy is dispersed entropy increases ( Lecture 1.2). Processes involving dissipation are irreversible. For instance, we could set the pendulum swinging at its initial amplitude again by giving it a push periodically with a motor, and cool the surrounding air to restore it to its initial temperature. BU this would involve changes in the motor and cooling system (e.g., using up fuel). We cannot get the pendulum and its surroundings back to their original state without changing something else.

18 2.1.3 Qusai-static process = a process which is so slow that the system stays in equilibrium at all times. Gas in container. Initially in state (P init,v init ). Compress it with piston to final state (P final,v final ). (1) Very slow (qusai-static) compression: at any stage the gas has well defined values of state variables. Can plot this process on a PV diagram. (2)VeryfastcompressiontoV final,andthenallowedtosettleintoequilibrium: statevariables not well defined during compression (e.g., pressure is not uniform). Can plot initial and final states on a PV diagram, but not the process joining them Reversible work o identify the conditions for work to be done reversibly we again consider our standard example: gas in a container with a piston. Apply force F piston moves small distance δx = δv/a small amount of work done δw = Fδx = FδV/A. (1) Finite friction. F must be larger than PA by a finite amount in order to overcome friction δw > PδV. Dissipation involved process is not reversible. (Can restore the gas to its original volume but doing so does not recover the energy dissipated by the friction.) (2) Zero friction. Apply force F larger than P A piston accelerates. hen either: (a) piston oscillates in and out for ever does not reach an equilibrium, or (b) oscillations are damped out by dissipation in gas not reversible. While piston is oscillating the system is not in equilibrium not qusai-static cannot be plotted on PV diagram.

19 Now consider a situation in which there is a very small amount of friction, and a force is applied which is larger than PA by a very small amount. he piston is displaced a very small distance and the gas reaches a new equilibrium with a slightly larger pressure. A very small amount of work is done δw = PδV ( Lecture 1.4). Large number of very small displacements large total displacement. his process would be very nearly reversible. In the limit of very small infinitesimal we have: no friction, gas is in equilibrium throughtout process (i.e., quasi-static), dw = PdV, process is reversible. his is a theoretical concept, but it can be closely approximated by reducing friction and pushing the piston slowly. Of course, if the friction was exactly zero and we applied a force exactly equal to PA the piston would not move. Reversible compression is an idealization of the situation with very small friction, in which the force is very slowly increased, so that at any time it is only very slightly larger than PA. In practical terms this means that the time over which the compression takes place must be much longer than the time over which the gas settles into equilibrium Reversible heat Heat flows between objects at different temperatures. Consider a fixed volume of gas in thermal contact with a reservoir at 1 : gas = 1. Now place it in thermal contact with a reservoir at 2 > 1. Heat Q flows from the reservoir into the gas and eventually equilibrium is reached in which gas = 2. his process is not quasi-static. When first placed in contact with Reservoir 2 the gas is not in equilibrium (e.g., temperature is not uniform, convection currents are probably occuring). not reversible. We could restore the gas to its original state by putting it back in contact with Reservoir 1. Heat Q would flow from the gas to the reservoir and gas 1.

20 But the reservoirs have changed: a net amount of heat Q has been transferred from Reservoir 2 to Reservoir 1. We could move Q back to Reservoir 2, but that would require a change somewhere else (e.g., a heat pump, Lecture 2.2.) Now place the gas successively in contact with a sequence of reservoirs, each of which has a temperture larger than the last one by a very small amount δ. A very small amount of heat δq flows into the gas from each reservoir, and the gas very quickly reaches equilibrium. Largenumberofreservoirs largetotaltemperaturedifference = δ andlargetotal heat flow Q = δq. Goingbackthroughthesequenceofreservoirsinreverseorder gasisrestoredtoitsoriginal state, and most reservoirs are restored to their original states. he only irreversible aspect is a very small net amount of heat δq transferred from the hottest reservoir to the coolest. In the limit of very small infinitesimal we have an infinite number of reservoirs, and: heat flow takes place between objects with infinitesimal temperature difference, gas is in equilibrium throughtout process (i.e., quasi-static), process is reversible. ( Lecture 2.6) Isothermal work Allow a gas to expand while in thermal contact with a reservoir at res. Heat Q flows from the reservoir to the gas, to keep gas = res. he gas does work W. If the expansion is quasi-static and there is no friction, this process is reversible (heat flows between objects at the same temperature).

21 Second Year hermodynamics M. Coppins Heat engines 2.2 Engines, fridges and heat pumps...convert heat into work. hey are the technological foundation of our society (e.g., power stations, transportation). In a heat engine internal energy of a gas is increased by heat and work, some of that added internal energy is used to do work (e.g., turn turbine), the rest is thrown away. Example: jet engine. his is a type of gas turbine. 1: Intake. Low temperature gas (air) is sucked in. 2: Compressor. Gas is compressed (work is done on it). 3: Combustion chamber. Fuel is added and ignited. he gas is rapidly heated at constant pressure. 4: urbine. he expanding hot gas does work on turbine blades. his drives the compreesor. 5: Exhaust. he hot gas is expelled (energy is lost). [Picture from Wikipedia] Cycles heoretical model: heat engine operates in a cycle with a fixed mass of gas (= the system ). Represent 1 cycle schematically: Q in = heat into system in 1 cycle, Q out = heat out of system in 1 cycle. Both of these quantities are defined to be positive.

22 W E = W out W in = net work done by engine in 1 cycle > 0. After 1 cycle the system returns to its initial state, i.e., after 1 cycle all state variables return to their initial values. U = change in internal energy in 1 cycle = 0. 1st law: U = W E +Q in Q out W E = Q in Q out. Q out = energy thrown away. A perfect engine would completely convert Q in to work, i.e., Q out = 0 and W E = Q in. his is not possible ( Lecture 2.3). Define efficiency of heat engine: η = fraction of Q in which is converted into work = WE Q in. Using 1st law: Brayton cycle = idealized gas turbine/jet engine. η = 1 Q out Q in. ( ) A B: low temperature gas compressed adiabatically B C: gas heated at constant pressure (Q in during this stage). C D: hot gas expands adiabatically and does work. D A: gas is cooled at constant pressure (Q out during this stage). he jet engine is an open system: hot gas is expelled and a fresh supply of low temperature gas is taken in. Some fraction of the energy put into the gas is lost in exhaust. If we suppose that the hot gas is cooled and recirculated we can account for the lost energy, he sections of the Brayton cycle correspond to the sections of the jet engine: A: air intake (1), A B: compressor (2), B C: combustion chamber (3), C D: turbine (4), D: exhaust (5).

23 Area under B D = W out = work done by gas. Area under D B = W in = work done on gas. Area enclosed by cycle = W E = W out W in net work done by gas (this is true for any cycle) wo reservoir engine = theoretical heat engine in which the heat source is a hot reservoir at H and the heat sink is a cold reservoir at C. Define: Q H = heat in from hot reservoir in 1 cycle (= Q in ), Q C = heat out to cold reservoir in 1 cycle (= Q out ). Example: coal fired power station: system: water/steam, hot reservoir: furnace, H 800 K, cold reservoir: outside environment (e.g., cooling towers), C 300 K. Heat Q H is transferred from hot reservoir, work W E is done by system, and heat Q C is transferred to cold reservoir. Schematic diagram of 1 cycle of a 2-reservoir engine: 1st law: W E = Q H Q C.

24 2.2.5 Refrigerator = heat engine run in reverse. Work W F is done on system, heat Q C is transferred from cold reservoir, and heat Q H is transferred to hot reservoir. Cold reservoir = contents of fridge, hot reservoir = air around fridge. Schematic diagram of 1 cycle of a fridge: In this case Q H = Q out and Q C = Q in. [Remember: in and out refer to the system.] 1st law: W F = net work done on system in 1 cycle = Q H Q C. A perfect fridge would transfer Q C from cold reservoir to hot reservoir without any work being done, i.e., W F = 0 and Q C = Q H. his is not possible ( Lecture 2.3). Define coefficient of performance of a fridge: ω F = fraction of W F which is used to cool cold reservoir = Q C W F. Note: Q C can be larger than W F, i.e., ω F can be larger than Heat pump = fridge used to heat a building. Cold reservoir = ground outside building, hot reservoir = interior of building. he schematic diagram of 1 cycle of a fridge also applies to a heat pump. Define coefficient of performance of a heat pump: ω P = fraction of W F which is used to heat hot reservoir = Q H W F. Note: 1st law Q H > W F ω P > 1. η, ω F and ω P all have the form What you get out What you put in We don t use the the word efficiency for ω F or ω P because they can exceed 100%.

25 Second Year hermodynamics M. Coppins Alternative statements 2.3 he second law he 2nd law concerns a fundamental asymmetry in the universe. his asymmetry manifests itself in many different ways, and there are, therefore, many different ways of expressing the 2nd law. Kelvin statement. No process is possible in which the sole effect is the absorption of heat from a reservoir and its complete conversion into work (i.e., a perfect engine is impossible, Lecture 2.2). Clausius statement. No process is possible in which the sole effect is the transfer of heat from a colder reservoir to a hotter reservoir (i.e., a perfect fridge is impossible, Lecture 2.2). Mathematical form. = dq, in a reversible process ds > dq ( ), in an irreversible process he phrase sole effect in the Kelvin and Clausius statements is very important. Various processes might appear to violate the 2nd law, but there is always something else happening ( Sec ) Perpetual motion machines are usually classified into: perpetual motion machines of the first kind: these violate the 1st law of thermodynamics. perpetual motion machines of the second kind: these violate the 2nd law of thermodynamics Equivalence of Kelvin and Clausius statements We can show that the Kelvin and Clausius statements of the 2nd law are equivalent by (1) showing that if the Kelvin statement is violated, then so is the Clausius statement, and (2) showing that if the Clausius statement is violated, then so is the Kelvin statement. Here we carry out stage (1), and relegate stage (2) to Problem Sheet 3. Suppose a perfect heat engine exists which completely converts heat from a hot reservoir (Q E H ) into work (WE ), with no other effect, i.e., it violate the Kelvin statement.

26 We use the work from this engine to run a fridge which discharges heat into the same hot reservoir. We adjust the engine so that it produces exactly the amount of work required to run the fridge, i.e., W E = W F. he net effect of the engine + fridge combination is: From the 1st law we have Q F C = QF H QE H. he sole effect is thus the transfer of heat from the cold reservoir reservoir to the hot one, i.e., it violates the Clausius statement Reversible, adiabatic process Reversible process: ds = dq, adiabatic process: dq = 0. Reversible, adiabatic process is isentropic (i.e., S = constant, ds = 0). In Lecture 1.4 we derived the ideal gas adiabatic law: PV γ = constant. he derivation assumed: (1) state variables well defined at all times during the process (i.e., quasi-static), (2) dw = PdV. i.e., the process is reversible. he ideal gas adiabatic law describes an isentropic process.

27 2.3.4 Calculating entropy changes dq Reversible process. S = (example Sec ). Irreversible process. ds > dq. So how do we calculate S? S is a state variable S = S final S init depends only on initial and final states (i.e., path independent). hermodynamic trick: Choose a reversible path between the initial and final states and calculate S along that (example Sec ) Reversible isothermal volume change Reversible volume change of ideal gas in thermal contact with reservoir at temperature res. Heat Q flows into the gas from the reservoir, to keep gas = res. he gas does work W. Ideal gas: = constant U = constant ( Lecture 1.2) Q = W, i.e., heat from the reservoir is completely converted into work violates Kelvin statement of 2nd law? NO!! his is not the sole effect. he volume of the gas changes. U = constant dq = dw (1st law) = PdV (reversible process) = Nk B gas dv. V If the gas expands from volume V 0 to V 1 the entropy change is: dq V1 ( ) Nk B S gas = = gas V dv = Nk V1 Bln. V 1 > V 0 S > 0. V 0 V 0

28 2.3.6 Adiabatic free expansion Consider a gas in a thermally insulated container. he container is divided into two parts by a removable partition. Initially the left hand part is filled with gas, while there is a vacuum in the right hand part. We remove the partition and allow the gas to fill the entire container. he process is adiabatic (thermally insulated container) d Q = 0 S = 0? NO!! he process is irreversible ds > 0. Use thermodynamic trick to find S. Assume: ideal gas, partition can be removed without doing any work on the gas (i.e., dw = 0). dq = dw = 0 U = 0 (1st law). Ideal gas: U = constant = constant ( Lecture 1.2). GasexpandsfrominitialstatewithvolumeV 0 (=volumeoflefthandpart)tofinalstatewith volume V 1 (= volume of whole container). Initial and final states have same temperature. Same initial and final states as in reversible isothermal ( ) expansion ( Sec ). V1 same entropy change, i.e., S gas = Nk B ln (> 0). Suppose partition divided the container in half, i.e., V 1 = 2V 0 S gas = Nk B ln2 Microscopic view of entropy ( Lecture 1.2): S = k B lnw where W = number of microstates corresponding to given macrostate. S = k B (lnw final lnw init ) = k B ln(w final /W init ) Probability of given macrostate occuring W. Probability that any 1 gas molecule is in left hand half of container = 1 2. Probability that any 2 gas molecules are in left hand half of container = = ( 1 2) 2. W init /W final = probability that N gas molecules are in left hand half of container = herefore S gas = k B ln2 N = Nk B ln2 V 0 ( ) N 1. 2

29 Second Year hermodynamics M. Coppins Reversible, two-reservoir engine 2.4 he Carnot cycle he most basic theoretical heat engine: is reversible (works just as well as a fridge), has only two reservoirs ( Lecture 2.2). [Remember: the system here is the working substance, e.g., the gas in a gas turbine, the steam in a steam engine, etc.] During reversible heat flow the temperature of the system stays constant (isothermal) and heat flows to/from a reservoir at the same temperature, or the temperature of the system changes by a finite amount, requiring an infinite number of infinitesimally different reservoirs. his is because reversible heat flow only takes place between objects at the same temperature or at infinitesimally different temperatures ( Lecture 2.1) In our theoretical engine the temperature of the system must stay constant when heat flows, otherwise we would need an infinite number of reservoirs. During Q in (= Q H ) system stays on = H isotherm. During Q out (= Q C ) system stays on = C isotherm. hese two isotherms must be connected to make a closed cycle. During the connecting processes changes (from H to C or vice versa), Q = 0 during these processes because we only have 2 reservoirs. the connecting processes must be adiabats. he complete cycle, formed by 2 isotherms and 2 adiabats is a...

30 2.4.2 Carnot cycle Assuming the system is a gas, for which P and V are appropriate state variables, the PV diagram of a Carnot cycle has the basic form shown. A B: isothermal expansion, heat Q in = Q H from hot reservoir at H. B C: adiabatic expansion. C D: isothermal compression, heat Q out = Q C to cold reservoir at C. D A: adiabatic compression.!! ake care!! Here the subscript C can refer to the cold reservoir (e.g., C, Q C ) or to point C on the cycle (e.g., V C would be the volume of the system at that point in the cycle). By a lucky coincidence the temperature of the system at point C is C = temperature of the cold reservoir. he Carnot cycle shares the basic properties of all two-reservoir engines ( Lecture 2.2): the net work done in one cycle (W E ) is the area enclosed on the PV diagram, the efficiency is given by η = 1 Q C Q H Efficiency of a Carnot engine with an ideal gas Ideal gas: = constant U = constant ( Lecture 1.2) herefore U A B = 0. Q H Q H = B A B A PdV = 0 (1st law). ( NkB B )dv = Nk B H V Similarly U C D = 0. Q C D C Q C = Nk B C ln A dv V = Nk B H ln ( VB PdV = 0 (1st law; note that Q C is heat out of system, hence the sign). ( VD V C ) ( ) VC = Nk B C ln. V D V A ).

31 Q C Q H = C ln(v C /V D ) H ln(v B /V A ). ( ) V γ 1 = constant along a reversible adiabat in an ideal gas ( Lecture 1.4). B C: H V γ 1 B D A: H V γ 1 Divide: ( VB V A A ) γ 1 = = C V γ 1 C = C V γ 1 D ( VC V D herefore Eq reduces to: ) γ 1 V B V A = V C V D. and the efficiency (η = 1 Q C /Q H ) is given by S diagram Q C Q H = C H ( ) η Carnot = 1 C H. ( ) A reversible process must be quasi-static. herefore it can be plotted on a PV diagram. Alternatively it can be plotted on a S diagram. Isotherms are just lines of constant on a S diagram. In a reversible process dq = ds ( Lecture 2.3). Lines of constant S (isentropes) on a S diagram are therefore adiabats. he area under a curve on a PV diagram is the work done ( Lecture 1.4). he area under a curve on a S diagram also has a physical meaning.

32 Sfinal Area under curve = ds = dq = Q = heat into the system. S init Plotted on a S diagram a Carnot cycle is a rectangle Area under A B = Q in = Q H. Area under C D = Q out = Q C. Area enclosed by cycle = Q H Q C = W E ( Lecture 2.2) = net work done by system (true for any cycle).

33 Second Year hermodynamics M. Coppins Proof of the theorem, part Carnot s theorem his proof is similar to the proof of the equivalence of the Kelvin and Clausius statements of the 2nd law ( Lecture 2.3). Carnot engine ( Lecture 2.4): 2 reservoir reversible engine. Assume our Carnot engine has an efficiency η Carnot, and in 1 cycle takes in heat Q H from the hot reservoir, expels heat Q C to the cold reservoir, and does net work W = η Carnot Q H (definition of efficiency, Lecture 2.2). he Carnot engine is reversible. Run in reverse it is a Carnot fridge. In 1 cycle: heat Q H out, heat Q C in, and net work W in. We now run our Carnot fridge with the work produced by a second engine operating between the same 2 reservoirs. hisnewenginehasanefficiencyη, andin1cycleittakesinheatq H fromthehotreservoir, expels heat Q C to the cold reservoir, and does net work W. We connect it to our Carnot fridge and adjust it so that W = W.

34 he net effect of the Carnot fridge + engine combination is: i.e., a device which in 1 cycle transfers heat Q net = Q H Q H from the hot reservoir to the cold one. But Q H = W η = W ( ) η = Q η Carnot ηcarnot H Q η net = Q H 1. η η > η Carnot Q net < 0 sole effect is heat from cold reservoir to hot one violates 2nd law impossible. η = η Carnot Q net = 0 after 1 cycle no net effect, everthing is back in its initial state reversible. η < η Carnot Q net > 0 sole effect is heat from hot reservoir to cold one 2nd law OK, but to restore reservoirs to their initial state would need external device to transfer Q net back irreversible. hus we see that the efficiency of the second engine must satisfy the condition η η Carnot, no heat engine operating between two reservoirs can be more efficient than a Carnot engine. his is the 1st part of Carnot s theorem.

35 2.5.2 Proof of the theorem, part 2 Suppose the second engine used in the proof of Carnot s theorem was, in fact, another Carnot engine. It too would be reversible, and we could run either engine as the fridge. Reversing engine 1 η 2 η 1. Reversing engine 2 η 1 η 2. hese two conditions can only be satisfied if η 1 = η 2 all Carnot engines operating between the same two reservoirs have the same efficiency. his is the 2nd part of Carnot s theorem. Carnot engine with ideal gas as working substance: η Carnot = 1 C H ( Lecture 2.4). his equation therefore must apply to any Carnot engine. here is a very simple alternative proof of this result using entropy. he heat in/out (Q H and Q C ) in 1 cycle of any Carnot engine can be represented as areas on S diagrams ( Lecture 2.4). Comparing these areas we see Q C Q H = C H. he efficiency of any heat engine is given by η = 1 Q out Q in ( Lecture 2.2). Since Q in = Q H and Q out = Q C, we find that the efficiency of any Carnot engine is η Carnot = 1 C H. ( ) Our first proof of this result, involving a thought experiment with engines and fridges, did not use entropy. We could therefore ask if, having proved the result without entropy, we can use it to deduce that entropy must exist. We return to this interesting question in Lecture Implications of Carnot s theorem We have shown that the 2nd law imposes an upper limit on the efficiency of any 2-reservoir engine, given by Eq his is a remarkable result. It is completely independent of the mechanical details or choice of working substance in any particular engine.

36 For instance, consider a power station in which the hot reservoir is a furnace at H = 800 K and the cold reservoir is the outside environment at C = 300 K. No matter what technological improvements we make, or what new working substances we use, the 2nd law tells us that the efficiency can never exceed 62.5%. Carnot s theorem also allows us to define an absolute thermodynamic temperature scale. raditionally a temperature scale is defined in terms of a property of some given material which changes with temperature. For instance, it is found experimentally that the pressure of a given fixed volume of dilute gas is always the same if it is placed in thermal contact with objects at the same temperature (how we detect if two objects are at the same temperature before we have defined a temperature scale will be discussed in Lecture 3.1). We could therefore define the temperature of a given object as the ratio of the measured pressure, P, of the gas when placed in thermal contact with the object, and the measured pressure when placed in thermal contact with some reference object. A convenient choice of reference object is water at its triple point. If we denote the measured pressure of the fixed volume of gas when in thermal contact with this reference object by P P then we can define the temperature of any other object as = (P/P P ). he constant volume gas thermometer, which employs this approach, is one of the devices used in establishing the international temperature calibration standard IS-90. Notice that we have actually used two reference points: the triple point of water and a zero-point where P = 0. As we know, very dilute gases behave as ideal gases, so P 0 at absolute zero. Alternatively, we can define a theoretical thermodynamic temperature in terms of the heat in and out of a Carnot engine. We use as our 2 reservoirs (1) water at its triple point, and, (2) the object whose temperature we wish to find. Any Carnot engine has the property that in 1 cycle ( C / H ) = (Q C /Q H ). hus if we measure the heat to/from the two reservoirs in one cycle of our Carnot engine, then we can define the temperature of the object to be = ( Q 2 / Q 1 ), where we have defined the temperature of the triple point of water to be exactly K. he advantage of this as a theroretical definition of absolute temperature is that it is equally valid for any Carnot engine, i.e., it is independent of the working substance.

37 Second Year hermodynamics M. Coppins Entropy and equilibrium dq = 0 for an isolated system. ds = dq/ in a reversible process. 2.6 he arrow of time he entropy of an isolated system is constant in a reversible process. ds > dq/ in an irreversible process. All real processesd are irreversible. he entropy of an isolated system increases in any real process. he system can have many different parts which interact with each other. Even though the total entropy (= the sum of the entropies of all the parts) must increase in any real process, the entropy of some parts might decrease while the entropy of other parts increases. Much of this lecture involves a detailed thermodynamic analysis of a simple isolated system which is initially not in equilibrium: a fixed volume of gas initially at temperature 0, placed in thermal contact with reservoir at R > 0. System = gas + reservoir An irreversible process takes place (the temperature of the gas increases). Nothing else is involved in this process: the gas + reservoir form an isolated system. During the process the total entropy S tot = S gas +S res increases. Eventually S tot reaches its maximum value process stops. But we know that the process stops when the system reaches equilibrium. For an isolated system equilibrium corresponds to a state of maximum entropy Heat flow across a finite temperature difference o analyze this situation we need to be able to find the entropy changes involved when heat flows from the reservoir to the gas. he reservoir and gas have different temperatures irreversible process ( Lecture 2.1).

38 1 HeatQintogas rises. Suppose reachesavalue 1 thenq = C V d = C V ( 1 0 ) 0 (assuming C V = constant volume heat capacity, is independent of ). Irreversible use thermodynamic trick ( Lecture 2.3) to calculate S gas. Reversible process between initial state of gas (V 0, 0 ) and final state of gas (V 0, 1 ): use an infinite number of infinitesimally different reservoirs ( Lecture 2.1) At each reservoir dq = heat into gas from reservoir = C V d. ds gas = dq S gas = 1 0 (where = temperature of that particular reservoir). ( ) C V d 1 = C V ln. 0 he entropy of the reservoir also changes. Initial state of reservoir ( R, U res 0 ). Final state of reservoir ( R, U res 0 +Q res ), where Q res = heat into the reservoir. Foranyreversibleprocessbetweenthesetwostatestheentropychangewouldbe S res = Q res R. hermodynamic trick same S res if the process is irreversible. For our situation the 1st law (i.e., energy conservation) Q res = Q = heat into gas. herefore S res = C V( 1 0 ) R hermal equilibrium ( ) 1 S tot = S gas + S res = C V ln C V( 1 0 ) 0 R What temperature does the gas reach at equilibrium? i.e., what 1 maximizes entropy? d S tot d 1 = C V 1 C V R = 0 when 1 = R. d 2 S tot d 2 1 = C V 2 1 < 0, i.e., 1 = R at maximum of S tot. We have deduced that at equilibrium the gas reaches the same temperature as the reservoir simply by imposing the conditions energy is conserved, entropy is maximized Entropy changes Define δ = R 0 = initial temperature difference between the gas and the reservoir. ( ) ( ) 1 R S gas = C V ln = C V ln = C V ln(1+x), where x = δ S res = C V( 1 0 ) C V( R 0 ) R 0 +δ = C V δ( 0 +δ) 1 = C V x(1+x) 1.

39 Now suppose δ << 0. i.e., x << 1. ( ) S gas C V x x2 2 + > 0 S res C V x(1 x+ ) < 0 ( ) S tot C V x x2 x 2 2 x+x2 = C V 2 > 0 Entropy of the gas increases. Entropy of the reservoir decreases. otal entropy increases. [Can show that this is true for arbitrary δ Problem Sheet 4.] δ 0: S gas C V x, S res C V x S tot 0 process is reversible heat flow between objects with infinitesimal temperature difference is reversible Energy degradation Consider two possible ways in which an ideal gas can go from an initial state (V 0, 0 ) to a final state (V 1, 0 ), where V 1 > V 0, i.e., the gas expands but its final temperature is the same as its initial temperature. 1. Reversible, isothermal expansion ( Lecture 2.1) Gas is in thermal contact with reservoir. otal entropy change S tot = S gas + S res = 0 (reversible process). ( ) V1 Work done by gas = Nk B 0 ln ( Assessed Problem Sheet 2). V 0 We could use this energy for some useful purpose (generate electricity, move a plane, etc). 2. Adiabatic free expansion ( Lecture 2.3) he gas is the only thing involved in this process ( ) V1 S tot = S gas = Nk B ln ( Lecture 2.3). V 0 S tot > 0 (irreversible process). he gas expanded without doing any work. Compared to the reversible, isothermal expansion we wasted the opportunity for extracting an amount of useful work = Nk B 0 ( ) V1 ln Energy has been degraded. o recover the opportunity for the gas to do some work, we would need to compress it back to volume V 0, which would require at least the same amount of work to be done on the gas. In an irreversible process S tot energy degradation. V 0

40 2.6.6 he arrow of time In any real (irreversible) process ds things > dq things, where dq things = heat into all the things involved in the process from the rest of the universe = 0 since the rest of the universe is not involved in the process. he total change in entropy of the things involved S things = ds things > 0. he entropy of the universe always increases in real processes. his defines the arrow of time. Events occur in the direction of increasing entropy.

41 Second Year hermodynamics M. Coppins Scope of this lecture 3.1 Classical thermodynamics hermodynamics is a purely macroscopic theory. Nothing illustrates this point more vividly than the fact that the whole thermodynamic formalism was developed in the nineteenth century, before people were even sure that atoms existed. It is hard to believe that a sophisticated theory involving concepts like internal energy can be derived without knowing anything about the microscopic world. Yet that is exactly what Carnot, Kelvin, Clausius and others managed to do, and it represents one of the most impressive achievements of classical physics. Anyone studying thermodynamics should have some awareness how this was done, partly because of its historical importance, and partly because it demonstrates that physics need not be constrained by the fetters of the reductionist world-view. In this lecture we therefore start again and go back over what we have already done, but follow a purely macroscopic approach, carefully avoiding any reference to the underlying mechanism. Done this way the subject is usually referred to as classical thermodynamics. he main purpose of this lecture is to illuminate the subject and provide extra insight. he only thing in it which forms part of the examinable, core course material is the Clausius inequality (Sec , below). he lecture will therefore depart from the usual format, and be given on Powerpoint. For this reason the notes go over the usual 4 page limit Preliminaries o start we assume that we can measure mechanical properties: mass, length and force. Example: for a fixed mass of gas we can measure V = volume and P = pressure. Alternative properties could be measured for other systems, e.g., in the case of a stretched metal wire we could measure L = length and J = tension. We can also measure W = work done on system, e.g., do the work with a mechanism driven by a mass m which falls through a distance h; work done = mgh. Initially we can measure V, P, W he zeroth law and temperature Zeroth law of thermodynamics: if object A is in thermal equilibrium with object B, and object B is in thermal equilibrium with object C, then A is in thermal equilibrium with C. How do we tell if two systems are in thermal equilibrium? Example: object B is a fixed volume of gas in a container with a pressure gauge. Place it in contact with object A. If they are not in thermal equilibrium, P B will change. Eventually P B settles down to a constant value A and B are in thermal equilibrium.

42 Now place B in contact with object C. If P B does not change A and C are in thermal equilibrium. Zeroth law can define a new (macroscopic) property called temperature such that, if two objects are in thermal equilibrium they have the same value of this property. We could use P B to ascribe a value to this new property; = P B /P P B, where P P B = pressure measured when in thermal equilibrium with water at its triple point Note that this purely macroscopic approach provides a way of measuring, but doesn t provide any insight into what is. he microscopic approach provides good insight ( describes way particles are distributed with respect to energy), but doesn t provide a way of measuring. We can now measure V, P, W, he first law and internal energy Certain materials act as thermal insulators. Consider two identical volumes of gas, A and B, at different temperatures (i.e., P A P B ). A thermally insulating material prevents them reaching thermal equilibrium. Any process which takes place on a system enclosed in thermal insulation is called adiabatic. Consider an object initially in state P 0, V 0, 0 [remember: we can measure these properties]. We enclose the object in thermal insulation and do some work [remember: we can also measure work] as a result of which the object ends up in state P 1, V 1, 1. Experimentally we find the following: First law of thermodynamics: to take a given object from a given initial state to a given final state adiabatically always requires the same amount of work. Example: assume the object is a thermally insulated container of gas, and do work either by pushing in a piston, or by stirring the gas with a paddle wheel. First law we can define a new (macroscopic) property, internal energy, U, such that the change in this property is the adiabatic work done. Note that this purely macroscopic approach provides a way of measuring U, but doesn t provide any insight into what U is. he microscopic approach provides good insight (U = total energy of all the particles), but doesn t provide a way of measuring U. We can now measure V, P, W,, U.

43 3.1.5 Heat Now suppose we measure the work required to take the system from the given initial state to the given final state (1) with thermal insulation (i.e., adiabatically): work required = W 1. (2) without thermal insulation: work required = W 2. In general W 1 W 2. Define: Q = heat into system = W 1 W 2. Note that this purely macroscopic approach provides a way of measuring Q, but doesn t provide any insight into what Q is. he microscopic approach provides good insight (Q = energy transferred by random particle motion), but doesn t provide a way of measuring Q. We can now measure V, P, W,, U, Q Carnot cycle We are now in a position to be able to construct a heat engine with two reservoirs, and which runs in a Carnot cycle, i.e. a single cycle has the following 4 stages: (1) Place the working substance in thermal contact with the hot reservoir (at temperature H ), and allow it to undergo a frictionless, quasi-static expansion. Heat Q H enters the working substance from the reservoir. (2) hermally insulate the working substance and allow it to undergo a further frictionless, quasi-static expansion (this part of the cycle is adiabatic, by definition). (3) Place the working substance in thermal contact with the cold reservoir (at temperature C ), and allow it to undergo a frictionless, quasi-static compression. Heat Q C leaves the working substance and goes into the reservoir. (4) hermally insulate the working substance and allow it to undergo a further frictionless, quasi-static compression (this part of the cycle is adiabatic, by definition). After 1 cycle the working substance has returned to its original state (but the reservoirs haven t). he 1st law the net work done by the working substance is W E = Q H Q C. Note: W E could be negative, i.e., net work is done on the working substance. In this case we would be running the engine in reverse, as a fridge, and in the above description of the cycle the words expansion and compression should be interchanged. If we use an ideal gas as the working substance ( Lecture 2.4) Q C Q H = C H. his equation is valid for Q H and Q C either positive (engine) or negative (fridge). [Of course we know that this equation is valid for any Carnot cycle, with any working substance. However, the proof of that result follows from the 2nd law, which we haven t yet encountered in our development of classical thermodynamics, in this lecture.]

44 3.1.7 he second law and the Clausius inequality We now construct a theoretical device for converting heat into work. his device includes a small Carnot engine as one of its components. We can derive a very important thermodynamic result, the Clausius inequality, by insisting that our device does not violate: Second law of thermodynamics (Kelvin statement): no process is possible in which the sole effect is the absorption of heat from a reservoir and its complete conversion into work. he main part of our device is some arbitrary system which undergoes a cyclic process (main cycle) comprising many separate stages; Stage 1 Placethesysteminthermalcontactwithareservoirattemperature 1. hesystem takes in heat δq 1 from the reservoir, and does some work δw 1. As a result it moves to another equilibrium. Stage 2 Placethesysteminthermalcontactwithareservoirattemperature 2. hesystem takes in heat δq 2 from the reservoir, and does some work δw 2. As a result it moves to another equilibrium. and so on. Note: some of the δq i s and δw i s could be negative. A 3-stage cycle is shown schematically below, although, of course, we can envisage a cycle with many more than 3 stages. At the end of the main cycle the system has returned to its initial state, and U sys tot = (δq i δw i ) = 0, i i.e., the total work done by the system is W sys = i δw i = i δq i. If W sys > 0 we might appear to have violated the 2nd law, but this is not neccessarily the case. he Kelvin statement specifically forbids the conversion of heat from a single reservoir into work. If we have more than one reservoir the heat flow from some of them can be negative (δq i < 0). In this case we can have W sys > 0 and still satisfy Kelvin (e.g., in any

45 two-reservoir engine). hus before making use of the 2nd law we top-up all the reservoirs with heat from a single external reservoir. his is where the Carnot engine comes in. We use it to transfer heat from the external reservoir (at temperature ) to each of reservoirs 1, 2, in turn, and restore them to their original states. his process is shown schematically for the i th reservoir below. For each reservoir we run the Carnot engine so that it deposits exactly the same amount of heat, δq i, which had previously been transferred to the arbitrary system. his will require an amount of heat δq i from the external reservoir, and will result in the Carnot engine doing some work δwi = δq i δq i. For some reservoirs this will mean running the Carnot engine in reverse ( δq i < 0 and δwi < 0), but that is not a problem. Each time we use our Carnot engine we know δq i = i δq i, i.e., δq i = δq i i At the end of the whole process (main cycle plus reservoir top-ups) the arbitrary system and all the reservoirs apart from the external one are back in their original states. he total work done by the system is W sys = δq i, while the total work done by the Carnot engine i is W E = i δw i = i (δq i δq i ). hus the total work done by the composite device is W tot = W sys +W E = i δq i + i (δq i δq i ) = i δq i = i δq i i. Now we can make use of the 2nd law: W tot 0, i.e., i δq i i 0. If δq i = 0 then W tot = δq i = 0, i.e., the external reservoir (along with everthing i i i else involved) has returned to its original state process was reversible. If δq i < 0 then W tot = δq i < 0, i.e., net heat flow into the external reservoir i i i process was not reversible (to get everthing involved back to its original state we need to extract this heat from the external reservoir involve changes in some other device). In limit of infinite number of infinitesimally different reservoirs Clausius inequality: { dq = 0, for a reversible cycle < 0, for an irreversible cycle

46 3.1.8 Entropy We made no assumptions about the form of the main cycle which our arbitrary system was dq taken round. hus any cycle which is reversible will satisfy the condition = 0. In the Clausius inequality = reservoir temperature. But if the infinitesimal heat flow d Q is reversible the system and reservoir must be at the same (or infintesimally different) temperature ( Lecture 2.1). hus can also be taken as the system temperature. Allthisimpliesthat dq rev / (where dq rev isareversibleinfinitesimalheatflow)mustbethe infinitesimal change in some new state variable of the system, S, which we will call entropy. Clausius inequality round any reversible cycle ds = 0. Consider an irreversible cycle consisiting of an irreversible process from state A to state B, followed by a reversible process from B back to A. Clausius inequality But A B ) ( dq rev = A B A A B dq B < 0 ds = S A S B A B A ) ( dq ) ( dq irrev irrev + A B ) ( dq < S B S A. Applying this result to an infinitesimal irreversible process we obtain: i.e., the Clausius inequality can be re-expressed as: { = dq/, in a reversible process ds > dq/, in an irreversible process rev < 0. dq < ds ( ) his is the mathematical form of the 2nd law. Note that this purely macroscopic approach provides a way of measuring S: in any reversible process S =. However, it doesn t provide any insight into what S is. dq he microscopic approach provides good insight (S = k B lnw Lecture 1.2), but doesn t provide a way of measuring S. We can now measure V, P, W,, U, Q, S.

47 Second Year hermodynamics M. Coppins he first law revisited 3.2 he fundamental equation For a reversible process: dq = ds ( Lecture 3.1) and dw = PdV ( Lecture 2.1). So, for a reversible process the 1st law becomes: du = ds PdV. BU U,, S, P and V are all state variables. If we change S and V by given amounts ds and dv, then the corresponding change in internal energy, du, will be the same whether the process is reversible or irreversible, this form of the 1st law must be valid even for an irreversible process. du = ds P dv Fundamental equation of thermodynamics ( ) For an irreversible process: dq < ds ( Lecture 3.1) and dw > PdV ( Lecture 2.1). When they are added the differences cancel. he fundamental equation is always true for a closed system (N = constant) for which P and V are the appropriate basic state variables. We will use this equation a great deal in the rest of the course Entropy of an ideal gas Fundamental equation ds = 1 (du +PdV). Ideal gas: du = n d 2 Nk Bd ( Lecture 1.4), P = Nk B V ds = n d 2 Nk d B +Nk dv B V S = S 0 + n d 2 Nk Bln +Nk B lnv where S 0 is an integration constant he fundamental equation in action ( ) ( ) U U Write U = U(S,V) du = ds + dv. S V V S ( ) ( ) U U Compare with fundamental equation = and P =. S V V S Note: in this case we choose S and V as the 2 independent state variables because we want to compare with the fundamental equation, which has terms with ds and dv on the right hand side.

48 Use relationship for mixed 2nd derivatives ( Lecture 1.3): ( ) ( ) P = V S S. V 2 U V S = 2 U S V his is the first Maxwell relation (there are 3 more). ( ) ( ) ( ) ( ) U U P he 3 equations we have just derived, namely =, P = and =, S V V S V S S V follow directly from the fundamental equation. hey are therefore always true. It would be very hard, if not impossible, to derive them, and demonstrate their general validity, using a microscopic theory. But once we know the fundamental equation the thermodynamic proof is simplicity itself hermodynamic potentials Internal energy is one of four thermodynamic potentials. hese are four extensive state variables, all of which have the dimensions of energy. hey will be discussed more fully in the next lecture. Here we state their definitions and derive some mathematical results. he thermodynamic potentials are: U = internal energy, H = enthalpy = U +PV, F = Helmholtz function = U S, G = Gibbs function = U +PV S. Corresponding to each of the thermodynamic potentials there is a different form of the fundamental equation: du = ds PdV, dh = du +PdV +VdP dh = ds +VdP, df = du ds Sd df = Sd PdV, dg = du +PdV +VdP ds Sd dg = Sd +VdP. In Sec , above, we derived equations for and P, together with the first Maxwell relation, from the form of the fundamental equation involving du. We can apply the same procedure to the other forms of the fundamental equation. he form of the fundamental equation for dh has terms involving ds and dp on the right hand side. Choose S and P as our 2 independent state variables: ( ) ( ) H H H = H(S,P) dh = ds + dp. S P P S ( ) ( ) H H Compare with dh = ds +VdP = and V =. S P P S ( ) ( ) 2 H P S = 2 H V (Maxwell relation 2): =. S P P S S P

49 he form of the fundamental equation for df has terms involving d and dv on the right hand side. Choose and V as our 2 independent state variables: ( ) ( ) F F F = F(,V) df = d + dv. V ( ) F Compare with df = Sd PdV S = V ( ) 2 F V = 2 F S (Maxwell relation 3): = V V V ( F and P = V ( ) P Using the same method with the equation for dg we find: S =. V ). ( ) G, V = P ( ) ( ) S V and (Maxwell relation 4): =. P P [You will have an opportunity to derive these 3 equations on Problem Sheet 5.] Energy equation Using the equation for P obtained in the previous section, we find ( ) { } ( ) ( ) F U S P = = (U S) = + V V V V Use the 3rd Maxwell relation to obtain: ( ) U = V his is the thermodynamic energy equation. ( ) P V Apply this equation to a system for which PV (i.e., an ideal gas). ( ) P In this case = P ( ) U, and the energy equation = 0. V V. ( ) G, P P. ( ) Integrating U = U(),i.e.,internalenergyofanidealgasdependsonlyonitstemperature. his is a classic example the thermodynamic approach. Although the energy equation is completely general and accurate, on its own it doesn t tell us very much. However, if we give it some small scrap of information about something (e.g., we have a gas for which PV ) it can process the information to yield other, unexpected, facts about the thing. his is why I use the phrase thermodynamic machine : the mathematical formalism of thermodynamics allows us to discover new properties of things once we know a small amount about them. Notice that the machine doen t care about the origin of the information which was fed into it. In our example, we could have obtained the relationship P V from kinetic theory, or from experiments with dilute gases, or it might just be a made-up property of a hypothetical substance.

50 3.2.6 Other systems he infinitesimal heat flow in a reversible process is always dq rev = ds. For a gas, in which the appropriate basic state variables are P and V, the infinitesimal work done in a reversible process is dw rev = PdV. In other systems we have other basic state variables, and different expressions for dw. For instance, for a stretched metal wire or elastic band the infinitesimal work done in a reversible process is dw rev = JdL, where J = tension and L = length. In this case, therefore, the fundamental equation becomes du = ds +JdL. All the equations derived from the fundamental equation for a gas can be applied directly to a stretched metal wire/elastic band simply by making the substitutions: P J and V L. For instance, using ( the) fundamental ( equation ) ( in the ) form du = ds PdV we obtained the U P equations P = and = ( Section 3.2.3, above). V V S S S Without doing any more maths we can immediately ( ) write ( down ) the ( following ) equations for U J a stretched metal wire/elastic band: J = and = L L S rying to derive these equations from a microscopic theory of solid objects under tension would be an extremely formidable challenge. S V S. L

51 Second Year hermodynamics M. Coppins What do they mean? 3.3 hermodynamic potentials he thermodynamic potentials were introduced in Lecture 3.2. hey are defined as follows: U = internal energy, H = enthalpy = U +PV, F = Helmholtz function = U S, G = Gibbs function = U +PV S. What do these state variables mean? From a microscopic perspective the internal energy is the total energy of all the particles of the system. he other three, however, do not have simple microscopic interpretations. It is best to regard H, F, and G as highly useful, purely macroscopic variables. It is often more convenient to formulate the analysis of a given situation in terms of one of these functions instead of the microscopically more meaningful internal energy. In this lecture we briefly consider a few situations of this kind he Joule-homson process his process is used to cool and liquify gases. he gas is forced through a porous plug (e.g., a barrier with a small hole) by a pressure difference, i.e., the pressure is lower after the plug (P 2 ) than before it (P 1 ). he internal energy of the gas changes because work is done on the gas pushing it through the plug, and then the gas does work on the gas ahead of it, pushing it away. he internal energy is increased on one side of the plug (work done on the gas) and then reduced on the other (work done by the gas). So it is not obvious that this process produces a decrease in temperature rather than an increase. In fact, it is only at low temperatures that the temperature is reduced. Above a threshold value, called the inversion temperature, the Joule-homson process produces a temperature rise. Consider a parcel of gas, of fixed mass, which is forced through the plug by a pressure difference produced by pistons on either side. 1 Initially the gas is on the left hand side. It has a pressure P 1, volume V 1 and internal energy U 1. he right hand piston is in contact with the plug. here. 1 In fact the pressure difference is usually produced by a pump, but this does not affect the argument

52 Both pistons now move slowly to the right so as to maintain a steady flow. During this process the pressures P 1 and P 2 remain constant, with P 2 < P 1. Eventually all the gas is in the right hand side, with pressure P 2, volume V 2 and internal energy U 2, and the left hand piston is in contact with the plug. 0 he work done on the gas while on the left hand side is P 1 dv = P 1 V 1. V 1 V2 he work done on the gas after it reaches the right hand side is P 2 dv = P 2 V 2. 0 Assuming the process is adiabatic (Q = 0) the 1st law U = U 2 U 1 = P 1 V 1 P 2 V 2, where U 1 = internal energy before passing through the plug, and U 2 = internal energy after passing through the plug. herefore U 1 +P 1 V 1 = U 2 +P 2 V 2, i.e., H 1 = H 2, the enthalpy of the gas remains constant in the Joule-homson process. he Joule-homson process is an example of a situation in which a fluid moves as a result of a force exerted on it by a pressure difference. In general fluids experience a force whenever there is a pressure gradient. he motion of a fluid which is inviscid and incompressible (liquids and gases at low flow speeds) can be described by Bernoulli s principle: the speed, v, of a small parcel of the fluid satisfies the equation v 2 2 +gz + P ρ = constant, where g = acceleration due to gravity, z = height, and ρ = density of the fluid. For a compressible fluid Bernoulli s principle becomes: v 2 +gz +h = constant, 2 where h = specific enthalpy (= enthalpy per unit mass). In the Joule-homson process there is no change in height, and it is sufficiently slow that v 0 enthalpy of fixed mass of gas = constant Enthalpy and heat dh = ds +VdP ( Lecture 3.2). In reversible process: dq = ds in reversible, constant P process: dq = dh. But dh = change in state variable. Can calculate heat in/out in reversible, constant P process from H init and H final. Q out = heat out of system = dq = ds = H But H = H init H final = decrease in H. Can visualise the system s enthalpy being extracted as heat. For this reason H is sometimes called the heat content.

53 But don t forget that enthalpy is a state variable and heat isn t. wo hundred years ago people believed that heat was a fluid (called caloric) and that one could meaningfully talk about the amount of heat contained in something. We now know that the caloric theory of heat is rubbish. So when using the expression heat content bear in mind it is a metaphor, only applicable for a reversible, constant P process. Chemical reactions are usually reversible, at constant P. H reaction = heat absorbed in reaction, where H reaction = H final H init, and H final = enthalpy of products, and H init = enthalpy of reactants, i.e., H > 0 endothermic reaction, H < 0 exothermic reaction Free energy df = Sd PdV ( Lecture 3.2). In reversible process: dw = PdV in reversible, isothermal process: dw = df. But df = change in state variable. Can calculate work done in reversible, isothermal process from F init and F final. W out = work done by system = dw = PdV = F But F = F init F final = decrease in F. Can visualise the system s Helmholtz function being extracted as work. For this reason F is sometimes called the free energy. In irreversible process: dw > PdV ( Lecture 2.1). W out = dw < PdV W out < F, i.e., decrease in F = maximum work that can be extracted he link to the microscopic world Statistical physics describes the macroscopic properties of matter in terms of the underlying microscopic properties. he link between microscopic and macroscopic is provided by the partition function, Z ( Statistical Physics course, next term). Once Z has been evaluated the macroscopic properties are found by: calculating the Helmholtz function using F = Nk B lnz ( Statistical Physics), using the equations derived in Lecture 3.2: ( ) ( ) F F S =, P = V V, U = F +S = F ( ) F. V

54 3.3.6 Specific Gibbs function Like the other thermodynamic potentials, G is an extensive state variable. It is often more useful to define an intensive variable: g = specific Gibbs function = Gibbs function per unit mass = G/M, where M = Nm p = mass of the system, m p = mass of 1 particle (atom, molecule, etc). In a given equilibrium state: value of g is independent of the amount of material, G = Mg. We can obtain intensive versions of the equations for V and S derived in Problem Sheet 5, by dividing throughout by M: ( ) ( ) G g S = s =, P P ( ) ( ) G g V = v = P P, where s = S/M = specific entropy, and v = V/M = specific volume = ρ 1 (ρ = density). Alternatively we can define the Gibbs function per particle: ĝ = G/N = m p g. [hese equations will be used in Lectures 3.4 and 3.5.]

55 Second Year hermodynamics M. Coppins N constant 3.4 Open systems So far we have considered closed systems in which N = constant. he thermodynamic formalism is slightly modified if N constant. Examples (1) Open container. N (= number of particles in container) can change. (2) Phase changes (e.g., liquid condensing from gas). wo different phases can coexist, and N i (= number of particles of part in phase i) can change. (3) Chemical reactions. N i (= number of particles of i th component) changes during reaction. In this course we consider situations (1) and (2) One component, one phase Closed system: N = constant 2 independent state variables. Fundamental equation: du = ds PdV Change S by ds and V by dv change in U is specified. Open system: N constant 3 independent state variables. Fundamental equation: du = ds PdV +µdn, µ = chemical potential = increase in internal energy per particle added if ds = dv = 0. WritefundamentalequationintermsofGibbsfunction: G = U +PV S ( Lecture3.2): dg = Sd +VdP +µdn ( ) G G = G(,P,N) dg = ( ) G Comparing µ = N,P P,N d + ( ) G dp + P,N ( ) G dn N,P Write G(,P,N) = Nĝ(,P), where ĝ = Gibbs function per particle { } µ = N (Nĝ) = ĝ = m p g,p where m p = mass of 1 particle, and g = specific Gibbs function ( Lecture 3.3).

56 Fermions are particles with half integer spin (e.g., electrons). hey obey the exclusion principle: 2 identical fermions cannot occupy the same quantum state. Consider a system which consists of a collection of fermions. At = 0 they fill a range of energy levels up to ε F = Fermi energy. his is still approximately true at finite if k B << ε F (e.g., for free electrons in a metal). Suppose we add one extra fermion but otherwise leave the system unchanged (ds = dv = 0). he levels are unchanged but the internal energy of the system increases by ε F since the extra fermion must have energy ε F (all the lower levels are filled). But µ = increase in internal energy per particle added under conditions of ds = dv = 0 for fermions µ = ε F Phase transitions We consider equilibrium phase transitions: system stays on the equilibrium P V surface. Everyday phase-transitions are usually close to this. Examples of exceptions to this condition: (1) placing an ice cube in boiling water, (2) a supercooled (or supersaturated) vapour. We consider an idealized phase transition from gas to liquid, reducing at fixed P (e.g., the condensation of water vapour at 1 atm as the temperature falls below 100 C). InordertoensurethatwestayonthePV surfacewewilldothisprocessreversibly: keeping the pressure of the gas/liquid constant with a piston, and slowly reducing by placing it successively in contact with an infinite sequence of infinitesimally different reservoirs. he process follows a constant P path on the PV surface. he projection of this path in the PV and the P planes is shown on the next page.

57 A B: single phase (gas). Heat from system to reservoir, dq = C P d. and V fall. B C: two phases coexist in equilibrium. Heat from system to reservoir, but dq = ldm, where l = latent heat and dm = mass of substance which has changed phase (i.e., extracting heat dq causes a mass of gas dm to condense to a liquid). V continues to fall as more gas condenses into liquid, but stays constant. C D: single phase (liquid). Heat from system to reservoir, dq = C P d. and V fall One component, two phases Focusing on the phase transition itself (B C), we see from the P diagram that specifying P (e.g., 1 atm) fixes (e.g., 100 C). hus if we are at a phase transition there is an extra constraint only 1 independent state variable, instead of the usual 2. But V changes during this stage specifying P does not fix V. So how many independent state variables are there? In situations involving more than one co-existing phase or component it is much more convenient to use only intensive variables. P and are intensive, but N, V, U, S, H, F and G are all extensive. For each extensive variable we define a corresponding specific value, by dividing by the mass M = Nm p, where m p = mass of single particle. he separate phases have separate values. Chemists define their intensive varaibles with respect to moles, e.g., the molar enthalpy. In this course we use specific values defined with respect to mass, e.g., the enthalpy per kg. In general there are two independent intensive state variables, e.g., we could write u = specific internal energy = u(,p). If 2 phases co-exist in equilibrium then specifying P gives only one independent intensive state variable. his is an example of Gibbs phase rule: the number of independent intensive state variables = n c n p +2, where n c = number of chemically independent components and n p = the number of phases present. he total value of any extensive property depends on the masses of each phase present, e.g., U = M 1 u 1 +M 2 u 2, where M 1 and u 1 are the mass and specific internal energy of phase 1, and M 2 and u 2 are the mass and specific internal energy of phase 2. Other specific values include: v = specific volume (= ρ 1 ), s = specific entropy, g = specific Gibbs function.

58 During a phase change in which phases 1 and 2 coexist in equilibrium: P and remain constant intensive variables remain constant. v 1 v 2, e.g., density of water vapour < density of liquid water. s 1 s 2, e.g., specific entropy of water vapour > specific entropy of liquid water (supply heat at constant to vaporize water). M 1 +M 2 remains constant dm 1 = dm 2. he change in internal energy can be written: du = d(m 1 u 1 +M 2 u 2 ) = u 1 dm 1 +u 2 dm 2 = (u 1 u 2 )dm 1. We assumed that the process was reversible: dq = ds = d(m 1 s 1 +M 2 s 2 ) = (s 1 dm 1 +s 2 dm 2 ) = (s 1 s 2 )dm 1, dw = PdV = Pd(M 1 v 1 +M 2 v 2 ) = P(v 1 dm 1 +v 2 dm 2 ) = P(v 1 v 2 )dm 1. 1st law (u 1 u 2 )dm 1 = (s 1 s 2 )dm 1 P(v 1 v 2 )dm 1 u 1 s 1 +Pv 1 = u 2 s 2 +Pv 2 g 1 = g 2.

59 Second Year hermodynamics M. Coppins First order phase changes 3.5 Phase changes Consider a situation in which phases 1 and 2 co-exist in equilibrium. Convenient to use only intensive variables ( Lecture 3.4). Divide extensive variables V, S, and G by mass M (intensive) specific values v, s and g. In general the specific variables for a given phase of a given substance depend on 2 independent intensive variables (e.g., P and ). But at a given pressure two phases only co-exist in equilibrium at a unique temperature at a phase transition the intensive variables are functions of a single independent variable (e.g., P). During the phase transition P and remain constant. Heat in/out changes the relative proportions of the two phases: dq = ldm, where l = latent heat and dm = mass of substance which has changed phase. dq 0 but d = 0 c P = constant pressure specific heat = C P M = dq Md. At the given values of P and the specific volumes, v, and specific entropies, s, of the two phases differ. But the specific Gibbs functions, g, are the same ( Lecture 3.4), i.e., during a phase transition from phase 1 to phase 2 v 1 v 2, s 1 s 2, g 1 = g 2. ( ) ( ) g g But v = and s = ( Lecture 3.3) P P Solid/liquid, liquid/gas, solid/gas transitions are examples of first order phase transitions: = constant during transition, and two phases co-exist, involve latent heat c P, g is continuous, ( ) ( ) g g and P P are discontinuous. he behaviour of various specific variables as the temperature changes through a 1st order phase transition is shown schematically on the next page.

60 Second order phase transitions also occur, e.g., onset of superconductivity: g and its first derivatives are continuous, but second derivatives are discontinuous Clausius-Clapeyron equation his is an equation for the slope (i.e., dp ) of the phase boundary P = P(). d Consider part of the phase boundary between phase 1 and phase 2. A and B are two points on the phase boundary. g A 1 = specific Gibbs function of phase 1 at point A. g B 1 = specific Gibbs function of phase 1 at point B. g A 2 = specific Gibbs function of phase 2 at point A. g B 2 = specific Gibbs function of phase 2 at point B.

61 Write g = g(,p). hen( making ) small( changes ) d and dp the corresponding small change in g is g g dg = d + dp = sd +vdp, P P dg 1 = s 1 d +v 1 dp and dg 2 = s 2 d +v 2 dp, g B 1 = g A 1 +dg 1 = g A 1 s 1 d +v 1 dp g B 2 = g A 2 +dg 2 = g A 2 s 2 d +v 2 dp. Subtract: (g B 2 g B 1 ) = (g A 2 g A 1 ) (s 2 s 1 )d +(v 2 v 1 )dp. But g A 2 = g A 1 and g B 2 = g B 1, Clausius-Clapeyron equation: dp d = s v where s = s 2 s 1 and v = v 2 v 1 Can rewrite the Clausius-Clapeyron equation by noting that s = S M = 1 M where Q trans = heat absorbed/released during phase change, and = temperature during phase change = constant. [Remember: we assumed the process was reversible Lecture 3.4.] ( Qtrans Choosing phases 1 and 2 such that Q trans > 0 (e.g., 1 = liquid, 2 = gas) then Q trans = lm s = l dp d = l v Liquid/gas phase boundary v = v 2 v 1 = v gas v liquid. he specific volume of the gas phase is nearly always much larger than the specific volume of the liquid phase, v v gas. he condition that v gas >> v liquid is valid everywhere along the liquid/gas phase boundary, except very close to the critical point. ) Above the critical temperature there is no phase transition between liquid and gas. Increasing the pressure of the gas increases its density. Eventually it becomes so dense that it is like a liquid.

62 Assuming the gas can be treated as ideal: v gas = V M = V = k B Nm p m p P where m p = mass of 1 gas molecule. v 1 = m pp k B Clausius-Clapeyron equation becomes: dp d = l lgm p P k B, 2 where l lg = latent heat of vaporization. Rearranging: dp P = l lgm p k B d 2 and integrating lnp = C l lgm p k B where C = integration constant. Assume that some point (P 0, 0 ) on the phase boundary is known (e.g., for water, could use P 0 = 1 atm, 0 = 373 K) C = lnp 0 + l lgm p k B 0. hen for any other temperature ( ), the pressure at which the liquid/gas phase transition P takes place is given by: ln = l ( lgm p 1 1 ). P 0 k B 0

63 Second Year hermodynamics M. Coppins Photon gas 3.6 hermodynamics of radiation Any object at finite emits photons, an empty container with a vacuum inside is actually full of photons. We can regard this as a photon gas. his a relativistic, quantum system, but thermodynamics can still be applied to it. Photon gases and ideal gases have many properties in common. In both systems the particles: don t collide, have a distribution of energies, have an isotropic distribution of directions, exert pressure on walls of container (photon gas exerts radiation pressure). From a classical point of view radiation pressure is the force exerted by the magnetic field of an electromagnetic wave on currents flowing in a solid surface which have been induced by the wave itself. Photon gases and ideal gases also have significant differences, for instance: gas molecules have a distribution of speeds, photons all have the same speed (c), gas molecules are reflected from walls, photons are absorbed and emitted by walls, the relationship between particle energy (e) and momentum (p) is: e = p2 2m p for gas molecules, e = pc for photons. Cavity radiation: equilibrium state of a photon gas in a closed cavity. Unlike a normal gas N (= number of photons in a closed cavity) can vary. o see this, consider a reversible isothermal expansion of a photon gas. Suppose V = 0 initially (this is not possible for a normal gas) no photons (N = 0). As V increases the cavity fills with photons N increases. Q = heat into gas, W = work done by gas.

64 1st law internal energy of photon gas changes: U = Q+W, U must depend on and V (these are the things we can vary). Varying and V also changes N, i.e., U is a function of 2 independent state variables, even though dn 0, µdn = 0 µ (= chemical potential) = 0 for photon gas ( Lecture 3.4) Pressure and energy flux Define û = U V = internal energy density. But U = Ne where e = average energy/particle. û = ne where n = N V = number density of particles (photons). In a photon gas the photons have an isotropic distribution, i.e., there are photons moving in all directions. In this case the (radiation) pressure is: P = û 3. We can regard this as a form of the equation of state of the photon gas. As usual in thermodynamics we don t ask where it came from, merely accept it as an input. In next term s Electromagnetism course you will derive the expression P = û for the radiation pressure exerted by a plane electromagnetic wave which is completely absorbed at a surface. Since our surface is also emitting we might have expected an equal contribtion to the pressure from the emitted photons, giving P = 2û. Instead we have P = û 3. here are two effects to consider here: (1) emission from the outer surface of the container, and (2) the isotropy of the radiation in the cavity. (1) In equilibrium the walls of the container are at a uniform temperature, and emit photons from their inner and outer surfaces. Since the photons are emitted in all directions, the total momentum (vector sum) they impart to the walls of the container is zero. hus they do not contribute to the radiation pressure. he absorbed photons only impinge on the wall from one side. hey therefore do impart net momemntum. (2) In a plane electromagnetic wave all the photons are moving in a single direction. In our case, however, the photons have an isotropic distribution of directions, i.e., they can have any direction of motion. his introduces the factor of 3. he analagous situation for a monatomic ideal gas would be the comparison between n d = 1 (1 degree of freedom) and n d = 3 (3 degrees of freedom). n d = 1. Molecules are constrained to move in 1-dimension: U = N 1 2 k B û = Nk B P = 2û. 2V n d = 3. Molecules are free to move in any direction: U = N 3 2 k B û = 3Nk B P = 2 2V 3û. If the ideal gas molecules were absorbed by the walls of the container, rather than reflected from them, the pressure would be reduced by a factor of 2 P = û, just like the photon gas. 3 Γ = particle flux onto wall of cavity = number of photons/area/time. Double n double the number of photons reaching wall in any given interval of time. Γ = αn where α = constant (depends on the speed of the particles, i.e., c in our case). Γ E = energy flux onto wall (SI unit W m 2 ) = Γe.

65 In a normal gas Γ E Γe because the particles have different speeds. In any given interval more of the faster particles reach the wall. hese particles have more energy Γ E > Γe. For a Maxwellian distribution Γ E = Γ2k B, but e = 3 2 k B. hus we find Γ E = αne = αû Energy density Consider two closed cavities A and B. hey differ in most respects (size, shape, material, etc). However, each is in thermal contact with a reservoir at the same temperature. he internal energy densities of the photon gases in the 2 cavities are û A and û B. We place the two in contact and open a small hole (area A) between them. he rate at which energy flows from A to B is Γ A B E A = αû A A. he rate at which energy flows from B to A is Γ B A E A = αû B A. In equilibrium these rates must be equal, otherwise there would be net flow of energy from one reservoir to another reservoir at the same temperature. û A = û B. û depends only on, and is independent of size, shape, material... for a photon gas: U = û()v (i.e., internal energy is a function of both and V). his is another way in which ideal gases and photon gases differ: as we know the internal energy of an ideal gas is a function only of temperature. ( ) U We now apply the energy equation ( Lecture 3.2): V ( ) { (û )} V (ûv) = û 3 V 3 û = ( ) û 3 ( ) û But û is a function only of dû d. V V = û 3. dû 3 d = 4 3û dû û = 4d. Integrating lnû = 4ln +lnλ (where Λ = integration constant), energy density of a photon gas is given by: û = Λ 4, ( ) P V P, internal energy of a photon gas: U = Λ 4 V, pressure of a photon gas: P = Λ 3 4.

66 3.6.4 Entropy U = U(,V) du = ( ) ( ) U U d + dv = 4Λ 3 Vd +Λ 4 dv. V V Fundamental equation ( Lecture 3.2) ds = 1 (du +PdV). ds = 1 ( 4Λ 3 Vd +Λ 4 dv + Λ ) 3 4 dv = 4Λ 2 Vd Λ3 dv. But 2 d = 1 3 d(3 ) ds = 4 3 Λ{ Vd( 3 )+ 3 dv } = 4 3 Λd(V3 ), entropy of a photon gas is given by: S = 4 3 ΛV3. [Note that we set the integration constant to zero here Lecture 3.7.] Blackbody radiation Consider two objects at the same temperature. Both objects emit and absorb radiation. hus they can exchange energy. his is a form of heat flow. In equilibrium each object must emit just as much radiation as it absorbs (otherwise there could be a net heat flow between the two objects and one of them would spontaneously start getting hotter), good absorbers are good emitters; poor absorbers are poor emitters. A perfect absorber is called a black body (it absorbs all incident radiation). It must therefore also be the best possible emitter. We can determine the basic temperature scaling of the power radiated by a black body by placing one inside a cavity in which there is a photon gas in equilibrium at temperature. Heat (in the form of radiation) can flow between the reservoir and the black body. In equilibrium the black body must have: temperature (otherwise heat to/from reservoir until it is). Ψ = power/area out of black body = power/area in (from photon gas), Power/area into black body = energy flux = αû = αλ 4, Ψ = power/area emitted by black body at temperature = σ 4 (Stefan s law) where σ = constant. We cannot determine the value of σ (Stefan s constant) from purely thermodynamic considerations. It was originallyfoundexperimentally, andcanalsobeobtainedfromstatisticalmechanics: σ = Wm 2.