* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Chapter 1 Thermodynamics
Conservation of energy wikipedia , lookup
Equipartition theorem wikipedia , lookup
Black-body radiation wikipedia , lookup
Heat transfer wikipedia , lookup
Thermal radiation wikipedia , lookup
Van der Waals equation wikipedia , lookup
Calorimetry wikipedia , lookup
Thermoregulation wikipedia , lookup
Maximum entropy thermodynamics wikipedia , lookup
Internal energy wikipedia , lookup
Entropy in thermodynamics and information theory wikipedia , lookup
First law of thermodynamics wikipedia , lookup
Equation of state wikipedia , lookup
Heat transfer physics wikipedia , lookup
Heat equation wikipedia , lookup
Thermal conduction wikipedia , lookup
Temperature wikipedia , lookup
Extremal principles in non-equilibrium thermodynamics wikipedia , lookup
Non-equilibrium thermodynamics wikipedia , lookup
Adiabatic process wikipedia , lookup
Chemical thermodynamics wikipedia , lookup
Thermodynamic system wikipedia , lookup
Chapter 1 Thermodynamics 1.1 Preliminaries Degree of freedom (d.o.f) is a number of parameters necessary to formulate the initial value problem divided by two. For a harmonic oscillator the initial value problem is specified with only two parameters: position coordinate (or angle) and velocity coordinate (or angular velocity) which together correspond to only a single d.o.f. In Hamiltonian mechanics the position q and momentum p coordinates come in conjugate pairs (q, p). Thus the number of d.o.f. is the number of 3 CHAPTER 1. THERMODYNAMICS 4 such pairs or one half of the total dimensionality of the phase space. Problem: How many d.o.f. in a problem of binary collision of protons in the LHC (Large Hadron Collider)? Solution: Binary collisions involve collisions of only two particles. Each particle is specified by position and velocity vector coordinates. This makes the number of d.o.f. (2 · 3 + 2 · 3)/2 = 6. Our main objective in this course is to understand the behavior of systems with a very large number of d.o.f. N ∫ 1 (e.g. NA = 6.02214 ◊ 1023 molecules in a box). The task is (in some sense) much more ambitious than the problems we are used to in physics courses where the number of d.o.f. is usually small. For N . 10 analytical methods may be useful; for N . 1010 computer may work; for N ≥ 1googol = 10100 statistical physics may be the only tool. There are two standard ways to study the large N limit: • phenomenological (e.g. thermodynamics) and • fundamental (e.g. statistical mechanics). We start with phenomenological approach and later will use a kinetic theory to justify a more fundamental (?) approach. Thermodynamics is a phenomenological theory of systems with many d.o.f. The main idea is that only a small number of measurable parameters (e.g. volume V , pressure P , temperature T , etc.) should be sufficient for describing the so-called equilibrium states. Equilibrium state is the state whose thermodynamic parameters do not change with time. It is an important experimental fact that in equilibrium all of the parameters are either extensive or intensive. For example, volume is extensive, but pressure is intensive. Extensive parameter is proportional to the amount of substance and intensive parameter is independent on the amount of substance. Problem: Give an example of neither extensive nor intensive quantity. Explain. CHAPTER 1. THERMODYNAMICS 5 Solution: When two wires with electrical resistances R1 and R2 are connected in series than the total resistance is an extensive quantity Rtotal = R1 + R2 , but when the wires are connected in parallel the total resistance is not extensive 1/Rtotal = 1/R1 + 1/R2 . Therefore, electrical resistance is not an extensive nor intensive quantity. Not all of the thermodynamic parameters are independent of each other. Equation of state, f (P, V, T ) = 0, (1.1) describes the dependence and reduces the number of independent parameters. For example, the equation of state for an ideal gas (sufficiently diluted gas) is given by, P V = N kT, (1.2) where N is the number of molecules and k is the Boltzmann’s constant. Because of a universal character of ideal gases (1.2) can be used to set a relative temperature scale (but to set an absolute scale and to give a meaning to T = 0 it is necessary to postulate the Third Law of thermodynamics to be discussed later in the course). For example, we can measure PNVk when water boils and denote this temperature by T = 0, and then we can measure PNVk when water freezes and denote this temperature by T = 100. Then we can use linear interpolation and extrapolation to assign temperature to arbitrary values of PNVk . This is the Celsius scale. CHAPTER 1. THERMODYNAMICS 6 It is convenient to think of f (P, V, T ) as a 3D function which describes a given thermodynamical system and of (P, V, T ) as a point in 3D which describes a given state of the system. The projection of the surface of the equation of state on P ≠ V plane is known as the P ≠ V diagram. Reversible transformation of different types such as • 1-2. isothermal (i.e. constant temperature) • 2-3. isobaric (i.e. constant pressure) • 3-1. isochoric (i.e. constant volume) are described by different paths on the P ≠ V diagram. 1.2 Zeroth Law Every theory is based on a number of statements which cannot be proved (not that we can prove anything in physics). In mathematics such statements are called axioms, but in physics they go by different names: principles, laws, assumptions, etc. With this respect thermodynamics is not an exception as it is based on four laws of thermodynamics numbered from 0 to 3 (for purely historical reasons). Zeroth Law in words: If systems A and C are each in thermal equilibrium with system B, then A is in thermal equilibrium with system C. Zeroth Law in symbols: A ≥ B and C ≥ B ∆ A ≥ C (1.3) where ≥ is a relation between systems such that “A ≥ B” reads as “A is in a thermal equilibrium with B”. Problem: Under assumption that every system is in equilibrium with itself, prove that ≥ is an equivalence relation. Solution: Equivalence relation must satisfy three properties: 1) Reflexivity: By assumption, A ≥ A. 2) Symmetry: Let A ≥ B. From Reflexivity B ≥ B. Using Zeroth Law, B ≥ B and A ≥ B ∆ B ≥ A. 3) Transitivity: Let A ≥ B and B ≥ C. From Symmetry C ≥ B. From Zeroth Law, A ≥ B and C ≥ B ∆ A ≥ C. CHAPTER 1. THERMODYNAMICS 7 An important consequence of the equivalence relation “~” is that it divides all of the system into equivalence classes which can be labeled by temperature. Of course, this does not tell us why the temperature should be a real number as opposed to, for example, integers, complex or even more exotic p-adic number. 1.3 First Law The Zeroth Law does not tell us why temperature is a real number. Thus additional laws must be postulated before this fact can be established not only experimentally (e.g. using equation of state for ideal gases), but also more theoretically. The next assumption is the First Law which should be viewed as a statement about conservation of energy. First Law in words: The increment in the internal energy U of a system is equal to the difference between the increment of heat Q accumulated by the system and the increment of work W done by it. First Law in symbols: dU = dQ ≠ dW, where dU is an exact differential. (1.4) s Exact differential dX is a differential whose integral dX depends only on the limits of integration, but not on the path. The First law is used to define the state function U which is an extensive quantity: doubling the mass doubles the internal energy. For quasi-static (or sufficiently slowly varying) processes the work done by the system dW = P dV such that dU = dQ ≠ P dV. (1.5) Problem: Are the differentials dQ and dW exact or not? s Solution: W = P dV is not a full derivative and therefore depends not only on the initial and final points but also on the path. This can be shown explicitly by integrating along different path from (P1 , V1 ) to (P2 , V2 ) assuming that P1 < P2 and V1 < V2 : ⁄ (P1 ,V2 ) (P1 ,V1 ) ⁄ (P2 ,V2 ) P dV + (P1 ,V2 ) P dV = P1 (V2 ≠V1 ) ”= P2 (V2 ≠V1 ) = ⁄ (P2 ,V1 ) (P1 ,V1 ) ⁄ (P2 ,V2 ) P dV + (P2, V1 ) P dV. CHAPTER 1. THERMODYNAMICS 8 Therefore, dW is not an exact differential and since dU is exact due to the First Law dQ = dU ≠ dW is also not exact. It is convenient to think of the conjugate pair P and V as a generalized force (intensive quantity) and generalized displacement (extensive quantity) respectively. Other commonly used conjugate pairs include temperature T and entropy S, chemical potential µi and number of particles Ni of type i, etc. Then the total change in internal energy can be expressed as the sum of the products of generalized forces and generalized displacements: dU = T dS ≠ P dV + ÿ µi dNi . (1.6) i In other words, when a given generalized force is not balanced it causes a generalized displacement whose product equals to the energy transfer in or out of the system. 1.4 Second Law Second Law in Kelvin words: No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work. Second Law in Clausius words: No process is possible whose sole result is the transfer of heat from a body of lower temperature to a body of higher temperature. To show the equality of the two statements it will be useful to define a heat engine. Heat engine is a system which undergoes a cyclic transformation that takes heat Q1 from a warmer reservoir, converts some of it to work W and rejects the rest Q2 = Q1 ≠ W to a colder reservoir. In contrast, refrigerator, is a heat engine running backwards in time: use work to extract heat from a colder reservoir and to reject it to a warmer reservoir. Problem: Prove that the Kelvin and the Clausius statements of the Second Law are equivalent. Solution: Assume that Kelvin statement is false ∆ Extract Q of heat from a reservoir at temperature T2 and convert it entirely to work W = Q. Then convert this work back into heat Q = W and transfer it to reservoir at temperature T1 > T2 ∆Clausius statement is false. Assume that Clausius statement is false ∆ Extract Q of heat from a reservoir at temperature T2 and transfer it to reservoir at temperature T1 > T2 ∆ Operate an engine between temperatures T1 and T2 designed such that W = Q ∆Kelvin statement is false. CHAPTER 1. THERMODYNAMICS 1.5 9 Carnot engine Carnot engine is an engine consisting of two isothermal paths (i.e. T = 0) operating at two different temperatures and two adiabatic paths (i.e. Q = 0) connected in a cyclic transformation. The efficiency of engines is defined as ÷= W , Q1 (1.7) and for a Carnot engine it is given by Q2 . (1.8) Q1 Carnot’s theorem: a) All irreversible engines operating between temperatures T1 and T2 < T1 are less efficient than a Carnot engine operating between the same temperatures. b) All reversible engines operating between temperatures T1 and T2 < T1 are equally efficient as a Carnot engine operating between the same temperatures. ÷ =1≠ CHAPTER 1. THERMODYNAMICS 10 Proof: a) Combine an arbitrary heat engine whose efficiency is ÷ with a reversed Carnot engine whose efficiency is ÷ Õ so that there is no work done by the combined system. If ÷ = ÷ Õ then the process does nothing in conflict with irreversibility assumption. If ÷ > ÷ Õ then the process is in the conflict with Clausius statement of the Second Law, i.e. W = Wc ∆ ÷Q = ÷ Õ QÕ ∆ Q= ÷ Q ≠ Q > 0. ÷Õ Thus, ÷ Æ ÷c . Together ÷ Æ ÷c and ÷ ”= ÷c give us ÷ < ÷c . b) We have already proved that ÷ Æ ÷ Õ regardless of reversibility. Now we can reverse the process by combining a reversed heat engine with a Carnot engine which leads to conclusion ÷ Õ Æ ÷. Together ÷ Æ ÷ Õ and ÷ Õ Æ ÷ give us ÷ = ÷Õ. According to Carnot’s theorem the efficiency of reversible process between any two temperatures is a universal number, i.e. ÷(T1 , T2 ). This allows us to define not only relative, but also absolute temperature scale. Consider three Carnot cycles 1-2, 2-3 and 1-3 operating between different temperatures T1 and T2 , T2 and T3 , T1 and T3 respectively, where without loss of generality we assume that T1 > T2 > T3 . We can now construct a combine cycle such CHAPTER 1. THERMODYNAMICS 11 that the heat Q2 rejected by 1-2 is absorbed by 2-3 which is a reversible and thus a cycle 1-3 by Carnot’s theorem. Then the heat absorbed by reservoir at T3 must satisfy both Q3 = Q1 ≠ W13 = Q1 (1 ≠ ÷(T1 , T3 )) and Q3 = Q2 ≠W23 = Q2 (1≠÷(T2 , T3 )) = (Q1 ≠W12 )(1≠÷(T2 , T3 )) = Q1 (1≠÷(T1 , T2 ))(1≠÷(T2 , T3 )). Therefore and (1 ≠ ÷(T1 , T3 )) = (1 ≠ ÷(T1 , T2 ))(1 ≠ ÷(T2 , T3 )) 1 ≠ ÷(T1 , T2 ) = f (T2 ) f (T1 ) (1.9) for an arbitrary function f (T ) which is by convention is set to be a linear function, i.e. T2 T1 (1.10) Q2 T2 = . Q1 T1 (1.11) ÷ =1≠ and from the definition of efficiency 1.6 Entropy The Second Law suggests a new thermodynamical quantity, called entropy and usually denoted by S. It is conveniently introduced using Clausius’s theorem. Clausius’s Theorem: For any cyclic transformation j dQ Æ 0. T (1.12) The equality holds for reversible transformations. Proof: Subdivide the cycle into infinitesimal transformations where the temperature T remains roughly constant. During each transformation the system receives dQ of heat and does dWS of work. Arrange a series of Carnot cycles operating between CHAPTER 1. THERMODYNAMICS 12 temperatures T and TR where is the temperature of an arbitrary reservoir. The whole purpose of each Carnot cycle is to take dQR of heat from reservoir and deliver to the system dQ of heat and to do dWC of work. Then according to the absolute definition of temperature dQR dQ + dW TR = = . (1.13) dQ dQ T By the Kelvin’s statement of the Second Law the total work must i i s be non-positive dW = (dWC + dWS ) = (dQ + dWC ) Æ 0 and thus, j (dQ + dWC ) = TR j j dQ dQ Æ0 ∆ Æ0 T T (1.14) since TR > 0. For a reversible cycle we can run the process in i dQ opposite direction to show that T Ø 0 which together with i dQ i Æ 0 implies dQ = 0. T T An immediate consequence of the Clausius’s Theorem is sthat for reversible transformations between two states A and B the integral AB dQ does not deT pend on the path. Indeed if we take two distinct reversible paths parametrizes by dQ and dQÕ then together they would form a reversible cycle, i.e. ⁄ B A and thus, dQ ⁄ A dQÕ + =0 T T B (1.15) ⁄ B dQ ⁄ B dQÕ = . (1.16) T T A A This suggests that for a reversible transformation we can define and exact differential dQ . (1.17) T whose integral defines relative entropy up to an arbitrary constant of integration. This also produces a new pair of conjugate thermodynamic variables generalized force T and generalized displacement S, i.e. dS © dU = T dS ≠ P dV. (1.18) Although the entropy was defined using only reversible processes (as a reference), the Clausius’s Theorem implies that for all systems CHAPTER 1. THERMODYNAMICS 13 ⁄ B dQ . (1.19) T (This can be seen by connecting the irreversible process from A to sB with i s B dQ s A dQ B dQ any reversible process from B to A such that dQ = + A T B T = A T + T S(B) ≠ S(A) Æ 0.) Moreover, for thermally isolated systems (i.e. dQ = 0) S(B) ≠ S(A) Ø A S(B) ≠ S(A) Ø 0 (1.20) where the equalities hold for only reversible transformations. This shows that in a thermal equilibrium (when the state of the system does not change) the system must be in a state of maximal entropy. Second Law in symbols: dS Ø0 (1.21) dt Not all of the processes we observe in nature are reversible. Many processes are irreversible, which is a rather surprising fact given that the fundamental laws of physics (as we know them) are usually time symmetric. Then, why does the nature always picks some initial conditions and not other? Why don’t we see a thermal state? Why is there an asymmetry between past and future? Why do we see arrow of time? 1.7 dQ equations Consider a system specified by three parameters P, V, T any two of which are independent variables due to equation of state. Then, the differential of internal energy can be expressed in three possible forms dU (P, V ) = dU (T, V ) = dU (P, T ) = A ˆU ˆP B A ˆU ˆT B ˆU ˆP B A V A B dV (1.22) A B dV (1.23) A B dT. (1.24) ˆU dP + ˆV V ˆU dT + ˆV T ˆU dP + ˆT P T P Using the First Law given by Eq. (1.5) we obtain can the so-called ”Q equations: dQ(P, V ) = dU (P, V ) + P dV = A ˆU ˆP B V dP + AA ˆU ˆV B P B + P dV (1.25) CHAPTER 1. THERMODYNAMICS A dQ(T, V ) = dU (T, V ) + P dV = AA dQ(P, T ) = dU (P, T )+P dV = B ˆU ˆT ˆU ˆP 14 AA .dT + V B +P T A ˆU ˆV B B ˆV ˆP B B + P dV T AA dP + T ˆU ˆT (1.26) B +P P A (1.27) where the heat capacities at constant pressure and volume are defined by CP © A ˆQ ˆT B = P CV © A ˆQ ˆT A ˆU ˆT B B +P P = V A A ˆU ˆT B ˆV ˆT B (1.28) P (1.29) . V Roughly speaking the heat capacities describe how much heat can be stored in a system as its temperature is increased. Problem: Under assumption that U (T ) is a function of temperature alone and CV does not depend on temperature, prove that for an ideal gas it is more efficient (requires less energy) to heat up the system at constant volume than at constant pressure. Solution: Since U is not a function of P nor V , A ˆU ˆT B = P A ˆU ˆT B = V dU dT and the difference between specific heat capacities (1.28) and (1.29) is given by CP ≠ CV = P A ˆV ˆT B P = A ˆ(P V ) ˆT B = N k. P where the last equality is obtained from the equation of state for an ideal gas (1.2). Since both the number of molecules and the Boltzmann’s constant are positive number we conclude that CP ≠ CV > 0 or that it is easier heat up the system at constant pressure rather than at constant volume. The dQ equation in their present form are not very useful. Using the Second Law we can rewrite equations (1.26) and (1.27) dQ 1 = dS = T T A ˆU ˆT B V 1 dT + T AA ˆU ˆV B T B + P dV (1.30) ˆV ˆT B B P dT CHAPTER 1. THERMODYNAMICS A dQ = dS = T 1 T A ˆU ˆP B T P + T A ˆV ˆP 15 B B A A 1 dP + T T ˆU ˆT B P + T A ˆV ˆT B B = A A ˆS ˆV B B P (1.31) and use the exactness of dS, i.e. dS = A ˆS ˆT B V A ˆS dT + ˆV B T A dV ∆ ˆ ˆV A A AA ˆU ˆV A 1 T A ˆU ˆP B = A ˆU ˆP B ˆS ˆT B B V T ˆ ˆT dT P T V to derive the following relations ˆ ˆP A 1 T A ˆ ˆV A ˆU ˆT B 1 T A ˆU ˆT P P + T A ˆP ˆT B B A V ˆV ˆT T ˆ = ˆT B B P T 1 T ˆ = ˆT B +P T BB (1.32) V T P + T A ˆV ˆP B B T or A ˆU ˆV B +P = T T B V and ≠T A ˆV ˆT B P T +P A ˆV ˆP B P (1.33) . (1.34) T Then one can use these expressions to simplify the dQ equations T dS = CV dT + T T dS = CP dT ≠ T 1.8 A ˆP ˆT B dV (1.35) A ˆV ˆT B dP. (1.36) V P Chemical potential Chemical potential, denoted byµ, is defined as work required to increase the number of particle by one. The name - chemical potential - is due to the fact that µ, describes the tendency of particles to move from higher densities to lower densities. Paired together with a number of particles N it forms another conjugate pair which appears in the most fundamental equation of thermodynamics: dU = T dS ≠ P dV + µdN (1.37) CHAPTER 1. THERMODYNAMICS 16 which combines in a single equations the First and Second laws. As was already noted T ,P and µ are intensive quantities and S, V and N are extensive. This can be expressed as a scaling law U (⁄S, ⁄V, ⁄N ) = ⁄U (S, V, N ) (1.38) which can differentiated with respect to ⁄ at ⁄ = 1, A ˆU ˆS B A ˆU S≠ ˆV V,N B A ˆU V + ˆN S,N B N = U (S, V, N ). (1.39) S,V By combined with (1.37) we obtain Euler equation U = ST ≠ P V + µN (1.40) SdT ≠ V dP + µdN = 0. (1.41) as well as the Gibbs-Duhem equation In a more general form it reads as SdT ≠ V dP + ÿ µi dNi = 0, (1.42) i where µi is a chemical potential and Ni is the number of particle of type i. 1.9 Perpetual motion machines If the First Law could be violated then it would open an exciting possibility of perpetual motion machine of the first kind: do work without input of energy. Such machines would be in a conflict with conservation of energy, but there is certainly no prove that they cannot exist. Moreover it is well known that for gravitational (more precisely general relativistic) systems the mass and energy is not always well defined. Although the perpetual motion machines of the first kind are not allowed by the First Law, there are perpetual motion machines of the second kind which are not in conflict with conservation of energy, yet they have never been constructed. It is an experimental fact that not all of the processes allowed by the conservation of energy are observed in nature. For example, we see ice cubes melting in warm water to cool it down, but we never see a time-reversed process where cool water splits into warm water with ice cubes. Such processes are called irreversible, in contrast to reversible processes which can run backwards in time if its initial and final conditions are interchanged. In thermodynamic such processes are prohibited by postulating the Second Law. If the Second Law could be violated it would allow perpetual motion machines of the second kind: do work using thermal energy. CHAPTER 1. THERMODYNAMICS 1.10 17 Thermodynamic potentials As we have seen the extrema (maximum) of entropy corresponds to an equilibrium state of an isolated system (i.e. dQ = dW = 0) dS = 0. (1.43) dt Can this concept be generalized to open systems? For this purpose in addition to U we define three new thermodynamic potentials: Enthalpy, Helmholtz free energy, Gibbs free energy and Landau free energy. • Enthalpy H = U + PV (1.44) It is convenient to express H in differential form as a function of S and P dH = dU + P dV + V dP = T dS + V dP. (1.45) When there is no heat exchange (i.e. dQ = 0) and the external force is constant (i.e. P = const), dH = dU + P dV + V dP = dW + P dV Æ 0. (1.46) where the equality corresponds to quasi-static processes (i.e. dW = ≠P dV ). • Helmholtz free energy A = U ≠ TS (1.47) It is convenient to express A in differential from as a function of V and T dA = dU ≠ T dS ≠ SdT = ≠P dV ≠ SdT. (1.48) When there is no external work (i.e. dW = 0) and the temperature remains constant (i.e. T = const), dA = dU ≠ SdT ≠ T dS = dQ ≠ T dS Æ 0. (1.49) where the equality corresponds to reversible processes (i.e. dQ = T dS). As an example of the variational principle consider gas at a constant temperature T in a box of volume V divided by a sliding piston into V1 and CHAPTER 1. THERMODYNAMICS 18 V2 . In the equilibrium state 0 = dA = = = or A A ˆA ˆV1 ˆA ˆV1 AA A B T ˆA dV1 + ˆV2 T ˆA dV1 + ˆV2 B B ˆA ˆV1 T B ˆA ˆV1 A ≠ = T A A A dV2 (1.50) T B T B B ˆA ˆV2 ˆA ˆV2 B d(V ≠ V1 ) dV1 T B T and using (1.48) we can conclude that in the equilibrium state the pressures on both sides of the piston must be equal. • Gibbs free energy G = U ≠ TS + PV (1.51) It is convenient to express G in differential form as a function of P and T since dG = dU ≠ SdT ≠ T dS + P dV + V dP = V dP ≠ SdT. (1.52) When the external force (i.e. P = const) and temperature remains constant (i.e. T = const) , dG = dU ≠SdT ≠T dS+P dV +V dP = dU +P dV ≠T dS = dQ≠dW +P dV ≠T dS Æ 0. (1.53) where the equality corresponds to reversible (i.e. dQ = T dS) and quasi-static (i.e. dW = ≠P dV ) processes. Note that the exactness of differential dH, dA, and dG, implies A ˆT ˆP B A ˆP ˆT B and A ˆV ˆT B S A ˆV ˆS B V A ˆS ˆV B P = = A . P ˆS =≠ ˆP . T B T . CHAPTER 1. THERMODYNAMICS 19 Other thermodynamic relations are conveniently summarized by the socalled Maxwell or thermodynamic square: For a thermodynamic potential of interest (in bold) whose arguments are placed in the neighboring corners (in italic) the derivative of the potential with respect to one of its argument with other argument held fixed is determined by going along a diagonal line either with or against the direction 1 2of ˆU the arrow. Going against the arrow yields a minus sign (e.g. P = ≠ ˆV , S=≠ 1.11 1 2 S ˆG ). ˆT P Third Law Entropy of a system is defined for all states using a reference state by connecting it with a reversible transformation. However, if the surface defined by the equation of state is disconnected then the reversible transformations might not exist. Moreover it is important to have definition of entropy for distinct systems which may later come into contact with each other. The role of the Third Law is to defines an absolute scale of entropy which defines uniquely the entropy of an arbitrary equilibrium state of any system. Third Law in words: The entropy of any system at absolute zero is zero. Third Law in symbols: lim S(T ) = 0 T æ0 (1.54) Consequences of the Third Law: 1. Partial derivative of S at T = 0 with respect to every thermodynamic CHAPTER 1. THERMODYNAMICS 20 parameter X vanishes: lim T æ0 A ˆS ˆX B = 0. (1.55) T 2. Heat capacity CX at T = 0 with fixed thermodynamic parameter X vanishes: lim CX = lim T æ0 T æ0 A ˆQ ˆT B X = lim T T æ0 A ˆS ˆT B = 0. (1.56) X 3. Absolute zero cannot be reached in a finite number of steps. This statement is often used as an alternative definition of the third law. We will come back to it at the end of the course when the quantum statistical mechanics is introduced. Physical theories are always based on assumptions which very often turn out to be false (sooner or later). Now, that you have seen all of the theoretical assumption (laws of thermodynamics) which go into a phenomenological theory of thermodynamics it is a good time to ask which of the assumption is likely to be wrong? Zeroth Law (Universality of Temperature): A ≥ B and C ≥ B ∆ A ≥ C (1.57) First Law (Conservation of Energy): dU = dQ ≠ dW. (1.58) Second Law (Arrow of Time): dS Ø0 dt (1.59) Third Law (Quantum Mechanics): lim S(T ) = 0 T æ0 (1.60)