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Home , Isobaric process

Introduction 1. Foundations of • Thermodynamics: phenomenological description of 1.1. Fundamental thermodynamical equilibrium bulk properties of matter in terms of only concepts a few “state variables” and thermodynamical laws. System : macroscopic entity under consideration. • Statistical physics: microscopic foundation of thermodynamics Environment : world outside of the system (infinite).

• ∼ 1023 degrees of freedom → 2–3 state variables! Open system : can exchange matter and with the environment. • “Everything should be made as simple as possible, but no simpler” (A. Einstein) Closed system : can exchange heat with the environment while keeping the number of particles fixed. Summary of contents: : can exchange neither matter nor heat with the environment. Can (possibly) still do • Review of thermodynamics by e.g. expanding. • Thermodynamical potentials Thermodynamical equilibrium: • space and probability • No macroscopic changes. • Quantum mechanical ensembles • Uniquely described by (a few) external variables of • Equilibrium ensembles state. • Ideal fluids • System forgets its past: no memory effects, no hysteresis. • Bosonic systems • Often the term global equilibrium is used, as opposed • Fermionic systems to local equilibrium, which is not full equilibrium at all (next page)! • Interacting systems Nonequilibrium: • Phase transitions and critical phenomena • Generally much more complicated than equilibrium state. • Simplest case: isolated systems each in an equilibrium state. • In a local thermodynamical equilibrium small regions are locally in equilibrium, but neighbour regions in different equilibria ⇒ particles, heat etc. will flow. Example: fluid (water) with non-homogeneous . • Stronger nonequilibrium systems usually relax to a local equilibrium.

Degrees of freedom (d.o.f.) is the number of quantities needed for the exact description of the microscopic state. Example: classical with N particles: 3N coordinates (x,y,z), 3N momenta (px,py,pz). State variables are parameters characterizing the macroscopic thermodynamical state. These are all extensive or intensive: Extensive variable: change value when the size (spatial and the number of degrees of freedom) is changed: volume V , N, internal U, S, total magnetic moment d3r M. R 1 Intensive variable: independent of the size of the system, and can be determined for every 1 = 2 semimicroscopical volume element: e.g. temperature T , p, µ, magnetic field H, ratios of extensive variables like ρ = N/V , s = S/N,.... Conjugated variables: A and B appear in pairs in dp = dU = · · · =0. expressions for the differential of the energy (or I1→2 I1→2 more generally, some state variable), i.e. in (B) The total change of an exact differential is forms ±A dB or ±B dA; one is always extensive independent on the path of integration: and the other intensive. 1 b Example: pressure p and volume V ; change in U when V is changed (adiabatically, at constant S) is dU = −pdV . a 2 Process is a change in the state.

Reversible process: advances via states dU − dU =0, a b infinitesimally close to equilibrium, Z Z so that we can write quasistatically (“slow process”). The direction of a reversible process can be reversed, obtaining 2 the initial state (for system + environment!) U(2) = U(1) + dU Z1 : T constant. Isobaric process : p constant. Exact differentials : V constant. Let us denote by dF¯ a differential which is not necessarily Isentropic or : S constant. exact (i.e. integrals can depend on the path). Assuming is a sudden or spontaneous it depends on 2 variables x, y, the differential change during which the system is far from equilibrium. In the intermediate steps global dF¯ = F1(x, y) dx + F2(x, y) dy state variables (p, T , . . .) are usually not well is exact differential if defined. ∂F ∂F Cyclic process consists of cycles which take the 1 = 2 . system every time to its initial state. ∂y ∂x

∂F (x,y) Then ∃F (x, y) so that F1(x, y)= ∂x and ∂F (x,y) 1.2. State variables and exact F2(x, y)= and differentials ∂y Let us suppose that, for example, the state of the system 2 can be uniquely described by state variables T , V ja N. dF¯ = F (2) − F (1) 1 Other state variables are then their unique functions: Z is independent on the path, and integrable. In this case p = p(T,V,N) (x, F1) and (y, F2) are pairs of conjugated variables with U = U(T,V,N) respect to F . Examples: are the following differentials exact? S = S(T,V,N) . . . dF¯ = y dx + x dy By applying differential calculus, the differential of p, for example, is dF¯ = x dx + x dy ∂p ∂p ∂p All physical state variables are exact differentials! This dp = dT + dV + dN ∂T ∂V ∂N will enable us to derive various identities between state  V,N  T,N  T,V . variables. . Integrating factor The differentials of state variables, If dF¯ = F1dx + F2dy is not exact, there exists an dp, dT , dV , . . ., are exact differentials. These have the integrating factor λ(x, y) so that in the neighbourhood of following properties the point (x, y) (A) Their total change evaluated over a closed path vanishes: λdF¯ = λF1dx + λF2dy = df

2 is an exact differential. λ and f are state variables. 1.3. Equations of state Example: find λ for the differential Encodes (some of the) physical properties of the equilibrium system. Usually these relate “mechanical” dF¯ = x dx + x dy . readily observable variables, like p, T , N, V ; not “internal” variables like S, internal energy U etc. A typical example: pressure of some gas as a function of T Legendre transformations and ρ. Legendre transformations can be used to make changes in Some examples: the set of the indepependent state variables. For example, let us look at the function f(x, y) of two variables. We Classical ideal gas denote ∂f(x, y) z = fy = ∂y pV = NkB T and define the function where N = number of molecules T = absolute temperature g = f − yfy = f − yz. −23 kB =1.3807 · 10 J/K = Boltzmann constant. Chemists use often the form (Note: z,y is a congjugated pair with respect to f!) Now pV = nRT dg = df − y dz − z dy = fxdx + fydy − y dz − z dy n = N/N0 = number of moles = fxdx − y dz. R = kB N0 =8.315J/K mol Thus we can take x and z as independent variables of the = function g, i.e. g = g(x, z). Obviously 23 N0 = 6.0221 · 10 = Avogadro’s number. ∂g(x, z) y = − . If the gas is composed of m different species of molecules ∂z the is still Corresponding to the Legendre transformation f → g there is the inverse transformation g → f pV = NkBT,

f = g − zgz = g + yz. where now m N = Ni i=1 Often needed identities X and Let F = F (x, y), x = x(y,z), y = y(x, z) and z = z(x, y). If we want to give F in terms of (x, z), we can write p = pi, pi = NikBT/V, i X F (x, y)= F (x, y(x, z)). where pi is the partial pressure of the i:th component Applying differential rules we obtain identities Virial expansion of real gases ∂F ∂F ∂F ∂y When the interactions between gas molecules are taken = + ∂x ∂x ∂y ∂x into account, the receives corrections which  z  y  x  z are suppressed by powers of density ρ = N/V : ∂F ∂F ∂y = 2 3 ∂z ∂y ∂z p = kB T ρ + ρ B2(T )+ ρ B3(T )+ · · ·  x  x  x One can show that   Here Bn is the n:th virial coefficient. ∂x 1 = ∂y ∂y z Van der Waals equation   ∂x z The molecules of real gases interact   ∂x ∂y ∂z = −1 • repulsively at short distances; every particle needs at ∂y ∂z ∂x >  z  x  y least the volume b ⇒ V ∼Nb. and 6 ∂F • attractively (potential ∼ (r/r0) ) at large distances ∂x ∂y due to the induced dipole momenta. The pressure = z . ∂y  ∂F  decreases when two particles are separated by the  z ∂x z attraction distance. The probability of this is  ∝ (N/V )2.

3 We improve the ideal gas state equation where

′ ′ p V = NkBT P = electric polarization = atomic total dipole momentum/volume so that D = electric flux density ′ −12 V = V − Nb ǫ0 = 8.8542 · 10 As/Vm p = p′ − aρ2 = true pressure. = vacuum permeability. then In homogenous dielectric material one has (p + aρ2)(V − Nb)= Nk T. B b P = a + E, T   Solid substances where a and b are almost constant and a,b ≥ 0. The coefficient Curie’s law 1 ∂V When a piece of paramagnetic material is in magnetic α = p V ∂T field H we write  p,N and the isothermal B = µ0(H + M),

1 ∂V where κ = − T V ∂p  T,N M = magnetic polarization of solid materials are very small, so the Taylor series = atomic total magnetic moment/volume B = magnetic flux density V = V0(1 + αpT − κT p) −7 µ0 = 4π · 10 Vs/Am = vacuum permeability. is a good approximation. Polarization obeys roughly Curie’s law Typically ρC −10 M = H, κT ≈ 10 /Pa T −4 αp ≈ 10 /K. where ρ is the number density of paramagnetic atoms and C an experimental constant related to the individual atom. Stretched wire Note: Use as a thermometer: measure the quantity Tension [N/m2] M/H.

σ = E(T )(L − L0)/L0, where L0 is the length of the wire when σ = 0 and E(T ) is the temperature dependent elasticity coefficient.

Surface tension

t n σ = σ 1 − 0 t′   t = temperature ◦C t′ and n experimental constants, < < 1∼n∼2 ◦ σ0 = surface tension when t =0 C.

Electric polarization When a piece of material is in an external electric field E, we define D = ǫ0E + P,

4 If the system can exchange heat and particles and do Thermodynamics is based upon 4 laws (these can be work, the energy conservation law gives the change of the derived from statistical physics, but in thermodynamics internal energy these are considered “fundamental”). dU =dQ¯ −dW¯ + µdN, 1.4. 0th law of thermodynamics If each of two bodies is separately in thermal equilibrium where µ is the chemical potential. More generally, with a third body then they are also in thermal dU =dQ − f · dX + µ dN . equilibrium with each other. ⇒ there exists a property ¯ i i i called temperature and thermometer which can be used to X measure it. U is a state variable, i.e. dU is exact.

1.5. Work Cyclic process Before we discuss the 1st law, let us introduce the In a cyclic process the system returns to the original state. concept of Work. Work is exchange of such ”noble” Now dU =0, so ∆W = ∆Q (no change in thermal energy (as opposed to exchange of heat or matter) that energy). Now ∆Q+ is the heat absorbed by the system can be completely transformed to some other noble form duringH one cycle and −∆Q− (> 0) is the heat released. of energy; e.g. mechanical and electromagnetic energy. Sign convention: work ∆W is the work done by the p system to its environment. Example: pVT system 6 ∆Q+ ∆W = p ∆V. £ - £ Note: dW¯ is not an exact differential: the work done by ££ ' $ the system is not a function of the final state of the -∆W =Area = p dV = in pV T -system system (need to know the history!). Instead Q H 1  Q dW¯ = dV Qs − p & %∆Q - is exact, i.e, 1/p is the integrating factor for work. V Example:

dW¯ = p dV − σAdL − E · dP − H · dM. The total change of heat is

In general ∆Q = ∆W = ∆Q+ + ∆Q−, dW¯ = fidXi = f · dX, i The efficiency of a (∆W > 0) is work/(heat X where fi is a component of a generalized force and Xi a taken): component of a generalized displacement. ∆W ∆Q+ + ∆Q− |∆Q−| η = = =1 − . 1.6. 1st law of thermodynamics ∆Q+ ∆Q+ |∆Q+| Total energy is conserved In addition to work a system can exchange heat (thermal energy) or chemical energy, associated with the exchange 1.7. 2nd law of thermodynamics of matter, with its environment. Thermal energy is 2nd law can be stated in various (historical) ways: related to the energy of the thermal stochastic motion of (a) Heat flows spontaneously from high to microscopic particles. low temperatures. The total energy of a system is called internal energy U. (b) Heat cannot be transferred from a cooler heat Sign conventions: reservoir to a warmer one without other changes. (c) In a cyclic process it is not possible to convert all ∆Q  Environmentheat taken from the hotter heat reservoir into work.  ' $ + (d) It is not possible to reverse the evolution of a system towards thermodynamical equilibrium without System - ∆W converting work to heat. Qk Q Q (e) The change of the total entropy of the system and its Q Q µ ∆N environment is positive and can be zero only in chemical energy reversible processes. & % 5 (f) Of all the engines working between the temperatures We define the efficiency as T1 and T2 the Carnot engine has the highest ∆W ∆Q efficiency. η = =1 − 1 . ∆Q2 ∆Q2 We consider the infinitesimal process Because the processes are reversible the cycle C can be reversed and C works as a heat pump. Let us consider 6y two Carnot cycles A and B, for which U(2) ∆WA = ∆WB = ∆W. C 2 'CW t A is an enegine and B a heat pump. The efficiences are U(1) dQ¯ 1 correspondingly t ∆W ∆W ηA = and ηB = . ∆QA ∆QB - x T 2 D Q D Q A W B Now AB T 2 > T 1 dQ¯ = dU +dW¯ = dU + f · dX, D Q A -W D Q B -W so there exists an integrating factor 1/T so that T 1 1 Let us suppose that dQ¯ = dS T η > η , is exact. The state variable S is entropy and T turns out A B to be temperature (on an absolute scale) The second law so that ∆QB > ∆QA or ∆QB − ∆QA > 0. The heat (e) can now be written as would transfer from the cooler reservoir to the warmer one without any other changes, which is in contradiction dS tot ≥ 0, with the second law (form b). So we must have dt η ≤ η . where Stot is the entropy of the system + environment. A B For the entropy of the system only we have By running the engines backwards one can show that 1 dS ≥ dQ,¯ T ηB ≤ ηA, where the equality holds only for reversible processes. so that ηA = ηB, i.e. all Carnot engines have the same The entropy of the system can decrease, but the total efficiency. entropy always increases (or stays constant). Similarly, it follows that the Carnot engine has the For reversible processes the first law can be rewritten as highest efficiency among all engines (also irreversible) dU =dQ¯ −dW¯ + µ dN = T dS − p dV + µ dN. working between given temperatures: assume that engine A in previous figure is some other engine than Carnot. Then the argument above implies that ηA ≤ ηB. If A is 1.8. not reversible, efficiency is not necessarily the same while Illustrates the concept of entropy. The Carnot engine C running the system backwards and the inequality remains consists of reversible processes in force; if it is reversible, reversing the process gives ηA = ηB. a) isothermal T2 ∆Q2 > 0 Note: The efficiency does not depend on the realization b) adiabatic T2 → T1 ∆Q =0 of the cycle (e.g. the working substance). Only c) isothermal T1 ∆Q1 > 0 reversibility is essential! ⇒ The efficiency depends only d) adiabatic T1 → T2 ∆Q =0 on the temperatures of the heat reservoirs.

Now ∆U =0, so ∆W = ∆Q2 − ∆Q1 (wrong sign here, Defining temperature scale via Carnot engines: Let us for simplicity). consider Carnot’s cycle between temperatures T3 and T1. Now a D Q 2 η =1 − f(T3,T1), d b where we have the identity c ∆Q1 D Q 1 f(T3,T1)= . ∆Q3

6 is exact and the entropy S and the temperature T are T 3 D Q 3 D Q state variables. Because the Carnot cycle has the highest D W 2 3 3 D Q 2 D W 1 3 efficiency, a cycle containing irreversible processes satisfies T 2 D Q 2 D Q D W 1 2 1 ∆Q T D Q 1 1 1 ηirr =1 − < ηCarnot =1 − T 1 ∆Q2 T2 Here or ∆Q2 ∆Q1 ∆Q − < 0. 2 T2 T1 f(T3,T2) = ∆Q3 Thus for an arbitrary cycle we have ∆Q1 f(T2,T1) = dQ¯ ∆Q2 ≤ 0, (∗) ∆Q1 T f(T3,T1) = I ∆Q3 where the equality holds only for reversible processes. so Because entropy S is a state variable, it cannot depend on the integration path. Thus, for an arbitrary process f(T3,T1)= f(T3,T2)f(T2,T1). 1 → 2 the change of the entropy can be obtained from the The simplest solution is formula dQ¯ ∆S = dS = . T1 rev rev T f(T2,T1)= . Z Z T2 i r r 2 We define the absolute temperature so that

T η =1 − 1 . T 1 2 r e v This (theoretical) definition was first used by Kelvin, and According to the formula (∗) we have it gives us our familiar absolute temperature scale, up to dQ¯ dQ¯ a scale factor. − < 0, T T (This is by no means unique; one could also use Zirr Zrev a f(t2,t1) = [t2/t1] , with any a 6= 0, this would just give us or 1/a dQ¯ a different temperature scale t = const. × T .) ∆S > . T The Carnot cycle satisfies Zirr This is usually written as dQ¯ =0, T dQ¯ I dS ≥ , T since, during the isothermal part a, and the equality is valid only for reversible processes. For dQ¯ ∆Q = 2 an isolated system (∆Q = 0, e.g. system + environment!) T T Za 2 we have ∆S ≥ 0. and part c dQ¯ ∆Q1 ∆Q2 Thus, for an irreversible process, = − = − . c T T1 T2 Z ∆Qirr ≤ ∆Qrev = T ∆S, This is valid also for an arbitrary reversible cycle ∆Wirr = ∆Qirr − ∆U ≤ ∆Wrev = f · dX C where “rev” is a hypothetical reversible process

C i connecting the initial and final states of “irr”-trajectory. Note the sign conventions: ∆W work done by the system; ∆Q heat absorbed by the system. ∆Wirr ≤ ∆Wrev means less work done by “irr” (∆W > 0) or more work done to because it (∆W < 0). dQ¯ dQ¯ = =0. T T IC i ICi 1.9. 3rd law of thermodynamics X Nernst’s law (1906): So dQ¯ dS = lim S =0. T T →0

7 A less strong form can be stated as: 2. Thermodynamic potentials When the maximum temperature occuring in the process from a state a to a state b approaches zero, also the 2.1. Fundamental equation entropy change ∆Sa→b → 0. According to the first law (in a TVN-system, generalizes Note: There are systems whose entropy at low easily), for reversible processes temperatures is larger than true equilibria would allow. This is due to very slow relaxation time dU = T dS − p dV + µ dN . (∗) S, V and N can be considered to be natural variables of the internal energy U, i.e. U = U(S,V,N). Furthermore, from the law (∗) one can read the relations ∂U = T ∂S  V,N ∂U = −p (∗∗) ∂V  S,N ∂U = µ. ∂N  S,V Scaling law of extensive variables: all extensive variables must be linear functions of system size V (and each other). Now U, S, V and N are extensive so we have U(λS, λV, λN)= λU(S,V,N) ∀λ. (∗∗∗) Taking a derivative of (∗∗∗) wrt. λ, we obtain the Euler equation for homogenous functions ∂U ∂U ∂U U = S + V + N . ∂S ∂V ∂N  V,N  S,N  S,V Substituting the partial derivatives in (∗∗) this takes the form U = TS − pV + µN or 1 S = (U + pV − µN). T This is called the fundamental equation. 2.2. Internal energy and Because ∂U T = ∂S  V,N and ∂U p = − , ∂V  S,N so ∂T ∂ ∂U ∂ ∂U ∂p = = = − . ∂V ∂V ∂S ∂S ∂V ∂S Similar relations can be derived also for other partial derivatives of U and we get so called Maxwell’s relations ∂T ∂p = − ∂V ∂S  S,N  V,N ∂T ∂µ = ∂N ∂S  S,V  V,N ∂p ∂µ = − . ∂N ∂V  S,V  S,N

8 In an irreversible process is called . Now T ∆S > ∆Q = ∆U + ∆W, dH = dU + p dV + V dp so ∆U

∆Wfree = ∆W1 + ∆W2 = p1∆V1 + p2∆V2 In an irreversible process one has = (p − p )∆V = (p − p )A ∆L 1 2 1 1 2 ∆Q = ∆U + ∆W − µ ∆N

H = U + pV ∆H ≤−∆Wfree

9 i.e. ∆Wfree is the minimum work required for the change and the partial derivative relation ∆H. ∂T ∂V ∂S Note: Another name of enthalpy is heat function (in = p ∂S ∂T constant pressure).  p ∂V p we can write  Joule-Thomson phenomenon T T ∂V Flow of gas through a porous wall (“choke”): dT = dS + dp. C C ∂T D Q = 0 p p  p Substituting into this the differential dS in constant p 1 enthalpy (∗) we get so called Joule-Thomson coefficients V 1 ∂T T ∂V V = − . c h o k e ∂p C ∂T T  H p " p # p1 and p2 are constant (in time), p1 >p2 and the process irreversible. When a differential amount of matter passes Defining the heat expansion coefficient αp so that through the choke the work done by the system is 1 ∂V αp = , V ∂T p dW¯ = p2dV2 + p1dV1.   we can rewrite the Joule-Thomson coefficient as V1 V2 ∂T V Initial state Vinit 0 = (T αp − 1). ∂p H Cp Final state 0 Vfinal   The work done by the system is We see that when the pressure decreases the gas

• cools down, if T αp > 1. ∆W = dW¯ = p2Vfinal − p1Vinit. • warms up, if T α < 1. Z p According to the first law we have For ideal gases ∂T = 0 holds. For real gases ∂T ∂p H ∂p H is below the inversion temperature positive, so the gas ∆U = U − U = ∆Q − ∆W = −∆W,     final init cools down. so that 2.4. Free energy Uinit + p1Vinit = Ufinal + p2Vfinal. The Legendre transform Thus in this process the enthalpy H = U + pV is ∂U U → F = U − S constant, i.e. the process is isenthalpic, ∂S  V,N ∆H = Hfinal − Hinitial =0. or F = U − TS We consider now a reversible isenthalpic (and dN = 0) defines the (Helmholtz) free energy. process init→final. Here Now dF = −S dT − p dV + µ dN, dH = T dS + V dp =0, so the natural variables of F are T , V and N. We can so read the partial derivateves V dS = − dp. (∗) ∂F T S = − ∂T Now T = T (S,p), so that  V,N ∂F ∂T ∂T p = − dT = dS + dp. ∂V T,N ∂S ∂p    p  S ∂F µ = . ∂N On the other hand  T,V ∂T T From these we obtain the Maxwell relations = , ∂S ∂p ∂S p Cp =   ∂V ∂T  T,N  V,N where Cp is the isobaric (see ∂S ∂µ thermodynamical responses). Using the Maxwell relation = − ∂N ∂T  T,V  V,N ∂T ∂V ∂p ∂µ = = − . ∂p ∂S ∂N ∂V  S  p  T,V  T,N

10 In an irreversible change we have 2.6. Grand potential The Legendre transform ∆F < −S ∆T − p ∆V + µ ∆N, ∂U ∂U i.e. when the variables T , V and N are constant the U → Ω= U − S − N ∂S V,N ∂N S,V system drifts to the minimum of the free energy.     Correspondingly defines the grand potential ∆Wfree ≤−∆F, Ω= U − TS − µN. when (T,V,N) is constant. Free energy is suitable for systems where the exchange of Its differential is heat is allowed; i.e. the control variables are T and V (typically at constant N). Very useful quantity in physics! dΩ= −S dT − p dV − N dµ,

2.5. so the natural variables are T , p andµ. The Legendre transformation The partial derivatives are now

∂U ∂U ∂Ω U → G = U − S − V S = − ∂S ∂V ∂T  V,N  S,N  p,µ ∂Ω defines the Gibbs function or the Gibbs free energy p = − ∂V  T,µ G = U − TS + pV. ∂Ω N = − . ∂µ Its differential is  T,V We get the Maxwell relations dG = −S dT + V dp + µ dN, ∂S ∂p so the natural variables are T , p and N. For the partial = ∂V ∂T derivatives we can read the expressions  T,µ  V,µ ∂S ∂N ∂G = S = − ∂µ T,V ∂T V,µ ∂T      p,N ∂p ∂N ∂G = . V = ∂µ T,V ∂V T,µ ∂p      T,N ∂G In an irreversible process µ = . ∂N  T,p ∆Ω < −S ∆T − p ∆V − N ∆µ,

From these we obtain the Maxwell relations holds, i.e. when the variables T , V andµ are kept ∂S ∂V constant the system moves to the minimum of Ω. = − ∂p ∂T Correspondingly  T,N  p,N ∂S ∂µ = − ∆Wfree ≤−∆Ω, ∂N T,p ∂T p,N     when (T,V,µ) is constant. ∂V ∂µ = . The grand potential is suitable for systems that are ∂N ∂p  T,p  T,N allowed to exchange heat and particles. In an irreversible process Bath ∆G< −S ∆T + V ∆p + µ ∆N, A bath is an equilibrium system, much larger than the system under consideration, which can exchange given holds, i.e. when the variables T , p and N stay constant extensive property with our system. the system drifts to the minimum of G. Pressure bath Correspondingly F D V ∆Wfree ≤−∆G, when (T,p,N) is constant. The exchanged property is the volume or a corresponding The Gibbs function is suitable for systems which are generalized displacement; for example magnetization in a allowed to exchange mechanical energy and heat in heat- magnetic field. and pressure baths. Heat bath

11 T The heat capacity C is defined so that T D Q ∆Q ∆S D S C = = T . ∆T ∆T Particle bath D S m Keeping volume and N constant, we define

m DN ∂S C = T . V ∂T  V,N Baths can also be combined; for example a suitable potential for a pressure and heat bath is the Gibbs Now, according to the first law at constant N,V function G. dU = T dS − p dV + µ dN = T dS. 2.7. Thermodynamic response functions Response functions are thermodynamic quantities most Using S = −(∂F/∂T )V,N, we can write accessible to experiment. They give us information about how a specific state variable changes as other independent ∂U ∂2F CV = = −T state variables are changed. They can be classfied as ∂T ∂T 2  V,N  V,N mechanical (compressbility, susceptibility) and thermal (heat capacity) responses.

1) Heat expansion coefficient 5) Isobaric heat capacity ∂S 1 ∂V Cp = T αp = V ∂T ∂T p,N  p,N   or Because 1 ∂ρ dH = T dS + V dp + µ dN, αp = − , ρ ∂T p,N   and using S = −(∂G/∂T )p,N , one can write where ρ = N/V . ∂H ∂2G Cp = = −T . 2) Isothermal compressibility ∂T ∂T 2  p,N  p,N 1 ∂V 1 ∂ρ κ = − = T V ∂p ρ ∂p  T,N  T,N Relating response functions Now 3) Adiabatic compressibility ∂S ∂S (V (p,T ),T ) = ∂T ∂T 1 ∂V 1 ∂ρ  p  p κS = − = . ∂S ∂S ∂V V ∂p S,N ρ ∂p S,N = +     ∂T ∂V ∂T The velocity of sound depends on the adiabatic  V  T  p compressibility like and (a Maxwell relation)

1 ∂S ∂p cS = , = , mρκS ∂V ∂T r  T  V where m the particle mass. so ∂p ∂V One can show that C = C + T . p V ∂T ∂T 2  V  p αp κT = κS + VT . Since Cp ∂p ∂T ∂V = −1 ∂T ∂V ∂p  V  p  T or 4) Isochoric heat capacity ∂V ∂p ∂T α Heat capacity C is a measure of the amount of heat = − p = p , ∂V ∂T V κT needed to raise the temperature of a system by a given   ∂p  amount. T so   In a reversible process we have 2 αp Cp = CV + VT . ∆Q = T ∆S. κT

12 Thus, Cp > CV . 2.9. Stability conditions of matter In a steady equilibrium the entropy has the true 2.8. Thermodynamical equilibrium state maximum so that small variations can only reduce the According to the 2nd law, the entropy of an isolated entropy. equilibrium system must be at maximum. Thus, any local We again consider an isolated system divided into parts, fluctuation must cause the entropy to decrease; if it were and denote the equilibrium values common for all not so, the system could move to a new higher entropy fictitious parts by the symbols T , p and {µj } and the 0 state, which cannot happen in an equilibrium system by equilibrium values of other variables by the superscript . definition. In order to study the maximality of the entropy, the We divide the system into fictitious parts: entropy of the partial system α, Sα, needs to be expanded into second order in {∆Uα, ∆Vα, ∆Njα}. We write Sα close to an equilibrium as the Taylor series a p a , T a ,

V a , . . . 0 0 0 0 Sα(Uα, Vα, {Njα})= Sα(Uα, Vα , {Njα}) ∂S 0 ∂S 0 + ∆Uα + ∆Vα D U = D V = D N i = 0 ∂U V,N ∂V U,N Extensive variables satisfy     ∂S 0 + ∆N ∂N jα S = Sα j j U,V X   α 0 0 X 1 ∂Sα ∂Sα V = Vα + ∆ ∆Uα + ∆ ∆Vα 2 ∂Uα ∂Vα α (  V,N  U,N X 0 U = Uα ∂Sα + ∆ ∆Njα α ∂N X j jα U,V ) X   Nj = Njα. + · · · . α X Here ∆U = U − U 0 and correspondingly for other Since each element is in equlibrium the state variables are α α α quantities. The variations of partial derivatives stand for defined in each element, e.g. 0 ∂Sα Sα = Sα(Uα, Vα, {Njα}) ∆ = ∂U  α V,N 0 and 2 0 1 pα µjα ∂ S ∂ ∂S ∆Sα = ∆Uα + ∆Vα − ∆Njα. 2 ∆Uα + ∆Vα Tα Tα Tα ∂U ∂V ∂U  V,N "  V,N #U,N Let us assume that the system is composed of two parts: 0 ∂ ∂S α ∈{A, B}. Then + ∆Njα ∂Nj ∂U j "  V,N #U,V ∆UB = −∆UA, ∆VB = −∆VA and ∆NjB = −∆NjA X and similarly for other partial derivatives. so The 1st order differentials drop out, because

α ∆Uα = 0. Thus, ∆S = ∆Sα α P X ∆Sα = 1 1 pA pB 0 0 = − ∆UA + − ∆VA 1 ∂Sα ∂Sα TA TB TA TB ∆ ∆Uα + ∆ ∆Vα     2( ∂Uα V,N ∂Vα U,N µjA µjB     − − ∆NjA. 0 T T ∂Sα j  A B  + ∆ ∆N . X ∂N jα j  jα U,V ) In an equilibrium S is at its maximum, so ∆S = 0 and X This can be rewritten as TA = TB

pA = pB ∆Sα =

µjA = µjB . 1 1 pα ∆ ∆Uα + ∆ ∆Vα 2( Tα Tα This is valid also when the system consists of several     phases. µjα − ∆ ∆N . T jα j α ) X  

13 Using the first law we get 3. Applications of thermodynamics

1 3.1. Classical ideal gas ∆S = −∆T ∆S + ∆p ∆V 2T α α α α For a full description of the thermodynamics of a system α ( X we need to know both the equation of state and some thermodynamic potential. EOS gives us mechanical − ∆µjα∆Njα . response functions, but for thermal response functions we j ) X need also some potential. This can be further written as From the ideal gas equation of state

1 C 1 pV = NkB T ∆S = − V (∆T )2 + [(∆V )2 ] 2T T α κ V α Nα α ( T we obtain directly the mechanical response functions X 0 ∂µ 2 1 ∂V NkB 1 + (∆Nα) , α = = = ∂N p V ∂T Vp T  p,T )  p,N 1 ∂V Nk T 1 where κ = − = B = . T V ∂p Vp2 p  T,N ∂V 0 ∂V 0 (∆V ) = ∆T + ∆p . Thermal response functions cannot be derived from the α Nα ∂T α ∂p α  N,p  N,T equation of state. Empirically it has been observed Since ∆S ≤ 0, we must have 1 C = fk N. V 2 B ∂µ C ≥ 0, κ ≥ 0, ≥ 0. Here 1 fk is the specific heat capacity/molecule and f is V T ∂N 2 B the number of degrees of freedom of the molecule. Atoms/molecule f translations rotations The condition CV ≥ 0 is a condition for thermal stability: if a small excess of heat energy is added to a volume 1 33 0 element of fluid, the temperature of the volume element 2 53 2 must increase. polyatomic 6 3 3 For real gases f = f(T,p), which is different from ideal The condition κT ≥ 0 is a condition for mechanical stability: if a volume of a small fluid element (fixed N) gas because of internal degrees of freedom (vibrations), increases, the pressure must go down, so that the larger interactions between the molecules and quantum pressure from the environment halts and reverses the mechanical effects. growth. Entropy Likewise ∂µ ≥ 0 is a condition for chemical stability. ∂N Entropy can be obtained from thermal and mechanical Recall that response functions by integrating along the trajectory ∂S ∂2F V CV = T = −T > 0 ∂T ∂T 2  V,N  V,N V S Thus, F is a concave function of T . S 0 V 0 Similarly, T T 0 T 2 1 ∂p ∂ F The differential is = −V = V 2 κT ∂V ∂V  T,N  T,N ∂S ∂S dS = dT + dV ∂T ∂V and F is a convex function of V .  V  T 1 ∂p = C dT + dV, T V ∂T  V since according to Maxwell relations ∂S ∂p = . ∂V ∂T  T  V Integrating we get T C V Nk S = S + dT V + dV B 0 T V ZT0 ZV0 T V = S0 + CV ln + NkB ln T0 V0

14 or In the process V1 → V2 and ∆Q = ∆W =0, so ∆U = 0. T f/2 V Process is irreversible. S = S + Nk ln . 0 B T V " 0  0 # a) Ideal gas Now Note: A contradiction with the 3rd law: S → −∞, when 1 U = fk T N, T → 0. 3rd law relies on the quantum nature of real 2 B matter! so T1 = T2, because U1 = U2. The cange in the entropy is thus Internal energy V2 ∆S = NkB ln . We substitute into the first law (N const.) V1 dU = T dS − p dV b) equation of state the differential The internal energy and the number of particles are constant: ∂S ∂S dS = dT + dV, ∂U ∂U ∂T ∂V dU = dT + dV =0 .  V  T ∂T ∂V  V  T and get ∂T The Joule coefficient ∂V U,N characterizes the behaviour ∂S of the gas during free expansion (cf. Joule-Thompson dU = C dT + T − p dV.  V ∂V coefficient):   T  ∂T ∂U According to Maxwell relations and to the equation of = − ∂V T state we have ∂V ∂U  U,N ∂T V ∂S ∂p NkB p 1  αp = = = , = p − T . ∂V ∂T V T CV κT  T  V   so Note that this is differential form (∂V ); with a finite dU = CV dT process one must integrate over differential changes. and 3.3. Mixing entropy 1 Conside different gases A and B, separated by a partition: U = U0 + CV (T − T0)= U0 + fkBN(T − T0). 2 A B

If we choose U0 = CV T0, we get for the internal energy T A T B p p 1 A B U = fNk T. 2 B

We assume that initially pA = pB = p and TA = TB = T . Now 2 The partition is removed and the gases mix. The process αp Cp = CV + VT is irreversible is adiabatic so ∆Q = 0. κT In a mixture of ideal gases every component satisfies the or state equation 1 C = Nk f +1 pj V = NjkB T. p B 2   or The concentration of the component j is Cp = γCV , Nj pj xj = = , where γ is the adiabatic constant N p where the total pressure p is γ = Cp/CV = (f + 2)/f. p = pj . j 3.2. Free expansion of gas X Method 1: Each constituent gas expands to volume V . Since V 1 D Q = D W = 0 pA = pB and TA = TB, we have Vj = V xj . The change in the entropy is (see the free expansion of a gas) V ∆S = N k ln j B V j j X

15 or Example: ∆Smix = −NkB xj ln xj . j j AB C D X M H SO H O SO Now ∆S ≥ 0, since 0 ≤ x ≤ 1. j 2 2 2 2 mix j ν −2 −3 2 2 Method 2: j For a process taking place in constant pressure and We define the degree of reaction ξ so that temperature the Gibbs function is the suitable potential: dNj = νj dξ. G = U − TS + pV 1 When ξ increments by one, one reaction of the reaction = fk TN − TS + pV = · · · 2 B formula from left to right takes place. = NkBT [φ(T ) + ln p]= Nµ(p,T ), Convention: When ξ = 0 the reaction is as far left as it can be. Then where ξ ≥ 0. µ0 f φ(T )= − ξ − ( + 1) ln T. Let us assume that p and T remain constant during the kB T 2 reaction. Then a suitable potential is the Gibbs function Before mixing G = µ N . G = N k T [φ (T ) + ln p] j j (b) j B j j j X X Its differential is and after mixing

dG = µj dNj = dξ νj µj . G(a) = Nj kBT [φj (T ) + ln pj ], j j j X X X We define so the change in the Gibbs function is ∂G p ∆ G ≡ = ν µ . ∆G = G − G = N k T ln j r ∂ξ j j (mix) (a) (b) j B p  p,T j j X X ∆rG is thus the change in the Gibbs function per one = Nj kBT ln xj . j reaction (often called the affinity). X Since (p,T ) is constant G has a minimum at equilibrium. Because The equilibrium condition is thus ∂G S = − , ∂T ∆ Geq = ν µeq =0.  P,{Nj } r j j j we get for the mixing entropy X

In a nonequilibrium dG/dt < 0, so if ∆rG> 0 we must ∆Smix = S(a) − S(b) = − Nj kB ln xj . have dξ/dt < 0, i.e. the reaction proceeds to left and vice j X versa. Let us assume that the components obey the ideal Gibbs’ paradox: If A ≡ B, i.e. the gases are identical no gas equation of state. Then changes take place in the process. However, according to µ = k T [φ (T ) + ln p + ln x ], the former discussion, ∆S > 0. The reason is that in j B j j classical ensemble the particles are distinguishable, and where pj = xj p is the partial pressure of component j and mixing of A and B really happens. In quantum mechanics 0 this apparent contradiction is removed by employing µj 1 quantum statitics of identical particles. φj (T )= − ηj − (1 + fj ) ln T. kBT 2 3.4. Chemical reactions So Consider for example the νj νj ∆rG = kBT νj φj (T )+ kBT ln p xj . → j 2 H2S+3O2←2 H2O+2SO2. X  P Y  In general the chemical reaction formula is written as The equilibrium condition can now be written as

ν − νj j j 0= νj Mj . xj = p K(T ), j j X Y P

Here νj ∈ I are the stochiometric coefficients and Mj where − νj φj (T ) stand for the molecular species. K(T )= e j P 16 is the equilibrium constant of the reaction, which depends an intensive quantity it depends only on relative only on T . The equilibrium condition is historically called fractions, so the law of mass action. For the reaction above µjα = µjα(p,T,x1α,...,xH−1,α),

2 2 xC xD and the conditions (∗) take the form 2 3 = pK(T ). xAxC µ1A(p,T,x1A,...,xH−1,A)=

The heat of reaction is the change of heat energy ∆rQ per µ1B(p,T,x1B,...,xH−1,B ) one reaction to right. A reaction is . . • Endothermic, if ∆rQ> 0 i.e. the reaction takes heat. µHA(p,T,x1A,...,xH−1,A)=

• Exothermic, if ∆rQ< 0 i.e. the reaction releases µHB (p,T,x1B,...,xH−1,B ). heat. There are We write ∆rG as • M = (H − 1)F + 2 variables, ∆ G = −k T ln K(T )+ k T ν ln px . r B B j j • Y = H(F − 1) equations. j X The simultaneous equations can have a solution only if Now M ≥ Y or ∆Q = ∆U + ∆W = ∆U + p ∆V = ∆(U + pV ) F ≤ H +2. = ∆H, This condition is know as the Gibbs phase rule. For pure matter the equilibrium condition since ∆p = 0. When the total amount matter is constant µA(p,T )= µB(p,T ) dG = −S dT + V dp defines in the (p,T )-plane a coexistence curve. If the holds in a reversible process and phase B is in equilibrium with the phase C we get another curve G 1 G G S V µ (p,T )= µ (p,T ). d = dG − dT = − + dT + dp B C T T T 2 T 2 T T     The phases A, B can C can be simultaneously in H V = − dT + dp, equilibrium in a crossing point, so called triple point, of T 2 T these curves. because G = H − TS. We see that 3.6. Phase transitions ∂ G In a the chemical potential H = −T 2 . ∂T T   p,N ∂G µ = ∂N Now  p,T ∂ ∆ G d r = −k ln K(T ), ∂T T B dT is continuous. Instead   ∂G so that S = − 2 d ∆ H = k T ln K(T ). ∂T p r B dT   This expression is known as the heat of reaction. and ∂G V = ∂p 3.5. Phase equilibrium  T In a system consisting of several phases the equilibrium are not necessarily continuous. conditions for each pair (A and B) of phases are A transition is of first order, if the first order derivatives (S, V ) of G are discontinuous, and of second order, if the TA = TB = T 1st order derivatives are continuous but 2nd order pA = pB = p discontinuous. Otherwise the transition is continuous.

µjA = µjB , j =1,...,H, (∗) T m 1 = m 2 where H is the number of particle species in the system. 1 Let us assume that the number of phases is F , so for each 2 species there are F − 1 independent conditions (∗). Now µiα = µiα(p,T, {Njα}). Because the chemical potential is p

17 In a first order transition from a phase 1 to a phase 2 so that −S1dT + V1dp = −S2dT + V2dp ∂G (2) ∂G (1) ∆S = − + or on the curve ∂T ∂T  p  p −1 dp S2 − S1 ∆S T ∆H ∂G (2) ∂G (1) = = = ∆V = − . dT V2 − V1 ∆V ∆V ∂p ∂p  T  T and we end up with the Clausius-Clapeyron equation When we cross a coexistence curve p and T stay constant, dp 1 ∆H so = . dT coex T ∆V ∆Q = T ∆S = ∆U + p ∆V = ∆(U + pV )   = ∆H. Here ∆H = H2 − H1 and ∆V = V2 − V1.

∆Q is called the phase transition heat or the . Examples Note: First order transitions have non-zero latent heat a) Vapour pressure curve but not the higher order ones. p fusion curve Typical phase diagram of solid-liquid-gas -system: lines c r i t i c a l p o i n t are 1st order transitions, critical point is 2nd order. C (triple point is 1st order). s o l i d f l u i d ◦ v a p o r Water: pT = 610Pa, TT =0.01 C; sublim ation T ◦ c u r v e vapor pressure pc = 22 MPa, Tc = 374.15 C. triple point c u r v e p T We consider the transition solid critical liquid → vapour. point liquid Assuming ideal gas we have triple point Nk T ∆V = V = B , gas v T p because V ≪ V , and p l(iquid) v(apor)

dp ∆Hlvp critical = . point 2 solid dT coex NkBT liquid const T isotherm   If the vapourization heat (the latent heat) ∆Hlv is gas roughly constant on the vapour pressure curve we can

triple integrate point −∆H /NkB T V p = p0e lv .

coexistence (this assumption is not true near crit. point!) Because of phase coexistence, phase diagrams are simplest b) Fusion curve in “force”-type coordinates (T, p, µ, H,...~ ). Now ∆V = V − V 3.7. Phase coexistence ls l(iquid) s(olid) can be positive or negative (for example H2O). T coexistence According to the Clausius-Clapeyron equation c u r v e dp ∆Hls 1 = 2 dT T ∆Vls we have p dp On the coexistence curve dT > 0, if ∆Vls > 0 1)

dp G1(p,T,N)= G2(p,T,N) dT < 0, if ∆Vls < 0 2) . and p d p p d p > 0 < 0 dG = −S dT + V dp d T d T 1 ) s o l i d f l u i d 2 ) f l u i d when the number of particles N is constant. Along the s o l i d curve

G1(p + dp,T + dT,N)= G2(p + dp,T + dT,N), T T

18 We see that when the pressure is increased in constant p temperature the system i s o t h e r m 1) drifts ”deeper” into the solid phase, BAI II 2) can go from the solid phase to the liquid phase. V c) Sublimation curve In many equations of state the phase transition happens Now when there is apparent instability dp/dV > 0 (for

dH = T dS + V dp = CpdT + V (1 − T αp) dp, example, van der Waals). In this case, we can use Maxwell’s construction: The points A and B have to be because S = S(p,T ) and using Maxwell relations and chosen so that the area I = area II. definitions of thermodynamic response functions ∂S ∂S ∂V C dS = dp + dT = − dp + p dT. ∂p ∂T ∂T T  T  p  p The vapour pressure is small so dp ≈ 0, and T 0 s Hs = Hs + CpdT solid phase Z0 T 0 v Hv = Hv + Cp dT vapour (gas). Z0 Let us suppose that the vapour satisfies the ideal gas state equation. Then Nk T Nk T ∆V = B − V ≈ B , vs p s p so dp ∆Hvs p ∆Hvs = ≈ 2 , dT T ∆Vvs NkBT where ∆Hvs = Hs − Hv. For a monatomic ideal gas 5 Cp = 2 kBN, and T p ∆H0 5 1 CsdT ′ ln = − vs + ln T − 0 p dT +constant. p Nk T 2 k N T 2 0 B B Z R 0 Here ∆Hvs is the sublimation heat at vanishing temperature and pressure.

Coexistence range p i s o t h e r m s C undercooled

B A V o v e r h e a t e d dV Matter is mechanically stable only if dp < 0. Thus the range of stability lies outside of the points A and B. Overheated liquid and undercooled vapour are metastable (supercooling, -heating). According to the Gibbs-Duheim relation (consider dG!) S V dµ = − dT + dp N N we have along isotherms V dµ = dp. N Thus, when the phases A and B are in equilibrium, B V µ − µ = dp =0. A B N ZA

19