Thermodynamics. Basic concepts. Thermal equilibrium and temperature

Transcription

1 hermodynamics Basic concepts hermodynamics is a phenomenological description of macroscopic systems with many degrees of freedom. Its microscopic justication is provided by statistical mechanics. Equilibrium is a state of an undisturbed system that does not change over time in any observable way - at least within time periods that are much longer than the time intervals of observation. macroscopic system left to itself will spontaneously approach its equilibrium state. n isolated system cannot exchange energy or anything else with its environment. However, a system can be coupled to the environment in various ways. Dierent types of coupling enable dierent types of equilibrium. he coupling to the environment allows us to control only a very small number of the system's degrees of freedom. he controllable and measurable properties of a system are various state variables, quantities such as temperature, pressure, volume, energy, charge, magnetization, etc. Most degrees of freedom (whose number can be of the order of vogadro's number N ) are far beyond our ability to control, measure or even mathematically handle. Nevertheless, thermodynamics, formulated by mathematical relationships between state variables (laws of thermodynamics), is suciently detailed to enable quantitative predictions of the macroscopic system dynamics. Historically, the development of thermodynamics led to applications such as engines and refrigerators. he most studied system in thermodynamics is gas. n ideal gas is a collection of many non-interacting identical particles, which can't collide with one another but bounce from the walls of a container. Realistic gases at very low densities (rare collisions), or with very weak interactions (insignicant collisions) can be approximated by an ideal gas. Other systems of interest are uids (interacting gases and liquids), solids (crystals and amorphous), electrons in materials, magnets, plasmas, nuclear matter, biological systems, etc. hermodynamics studies both classical and quantum systems. hermal equilibrium and temperature 0 th law of thermodynamics: If two systems are separately in equilibrium with a third system, then they are also in equilibrium with each other. he rst implication is that two systems can be in equilibrium with respect to each other, meaning that if such two systems are in contact with each other (and isolated from everything else) then their properties will not change over time. Second, if two systems and B are separately in equilibrium with a third system C, then they have something in common: being in equilibrium with C. his common property can be expressed as a state variable that separately characterizes the states of each system and B. he universal state variable that characterizes the equilibrium of any system is temperature. wo systems in equilibrium with each other ( and B) have the same common temperature (this is an alternative statement of the 0 th law). Dierent kinds of equilibrium (mechanical, chemical, etc.) may be characterized by additional state variables (pressure, chemical potential, etc. respectively). We still have no denition of temperature. From our everyday experience, we regard temperature as a measure of how hot something is. More practically, we can measure temperature of a system by putting some uid in contact with it, waiting until the uid and the system equilibrate, and measuring the volume by which the uid expands or contracts. he change of uid's volume is roughly proportional to the change of its temperature, but depends also on the choice of uid. We must pick a particular uid and decide what reference points and scale to use in order to rigorously dene temperature.

2 2 It has been noticed that heated water mixed with ice maintains its cold temperature until all ice is melted. Similarly, heated boiling water maintains its hot temperature until all water evaporates. his provides one convenient temperature scale, Celsius: water and ice in equilibrium have a reproducible temperature 0 o C, while water and steam in equilibrium have a reproducible temperature 100 o C (by denition). he modern scale for temperature has been established at the 10 th General Conference on Weights and Measures in he SI unit of temperature is Kelvin. he reproducible temperature of the triple point of ice-water-gas system is dened to be K, while 0 K (absolute zero) is the lower bound for temperature of any system. he water triple point is a much better reference than freezing or boiling temperatures, because the former is an equilibrium between all three phases of water that occurs only at a particular pressure, while freezing/boiling critical temperatures actually depend on the ambient pressure. he reference of K is chosen in order to make the temperature change of one Kelvin (K) identical to the temperature change of one Celsius (determined at standard atmospheric pressure). In addition to a scale, one must also specify a concrete reference procedure for measuring temperature, which will provide a linear scale (by denition). For this purpose, one uses an ideal gas. It has been experimentally observed that the quantity pv/ (p is pressure, V is volume and is temperature in Kelvins measured by a typical accurate thermometer) is a constant in any equilibrium state of any suciently dilute (approximately ideal) gas. herefore, for two equilibrium states at the same pressure: V 1 1 V V 2 V 1 1 Since one temperature ( 1 ) can be determined by achieving equilibrium with a reference system (water at its triple point), any other temperature can be found by measuring volume in an ideal gas expansion experiment. his observed linear relationship for an ideal gas is then promoted to the denition of temperature, so that, by denition, accurate thermometers measure temperature and not some arbitrary function of temperature. he conservation of energy fundamental property of any system is energy. Internal energy E is the part of the system's total energy that is stored in its own degrees of freedom and depends only on the state (and not environment) variables. For any system of particles, internal energy is the total kinetic energy of all particles (measured in the center-of-mass frame) together with the potential energy of interactions between the particles. system can exchange energy with its environment in dierent ways. he general exchange of energy is work W done on the system by the environment (or work W done on the environment by the system). Normally, we will not be concerned with work that accelerates the center of mass or changes the potential energy of the system due to external forces. Hence, all work will go toward internal energy. he amount of work dw done on the system by external forces in an innitesimal time interval is: dw i J i dx i as long as the system is not thrown far from equilibrium (so its state can still be faithfully represented by state variables such as temperature, pressure, etc.). Here, J i are quantities that represent forces and X i are the corresponding state variables that represent responses (displacements) of the system to the applied force. For example: Length change dl due to an external (tension) force F : dw F dl Surface area change by d due to a surface tension S: dw Sd Volume change by dv due to internal pressure p: dw pdv

3 Magnetization change by dm in an external magnetic eld B: dw BdM Electric polarization change by dp in an external electric eld E: dw EdP he number of particles change by dn at a chemical potential µ: dw µdn here are processes in which energy is exchanged but we cannot identify the detailed corresponding change of the system's state (typically due to its microscopic complexity). he energy exchanged in such processes is called heat. For example, if two solid objects at dierent temperatures are placed in contact with each other, the hotter object will transfer heat to the colder one until both objects reach the same temperature (equilibrium). his need not lead to any change of volume, the number of particles, etc, which we would need to observe in order to recognize this energy exchange as work according to one of the above formulas. he change of the system is microscopic and hidden from our view (the distribution of particle velocities changes). system is said to evolve adiabatically if it does not exchange heat with its environment. 1 st law of thermodynamics: Energy is conserved in any process. If the system takes heat from its environment, and work dw is done on it by external forces, then its internal energy changes by de according to: + dw de he sign convention is such that a positive value indicates energy added to the system. Response functions Response functions are easy-to-measure quantities that depend on state variables and thus characterize the state of a system. Heat capacity: how much energy (heat) does it take to change the temperature of the system by d under various conditions... C V measured at given volume V const d V C p measured at given pressure p const d p Force constant: a thermodynamic generalization of the spring constant R 1 dx the change of displacement due to applied force, per unit volume V dj κ 1 dv compressibility (pressure changes volume) V dp χ 1 dm magnetic susceptibility (magnetic eld changes magnetization) V db hermal responses: Ideal gas α p 1 V dv expansivity of a gas at constant pressure d p Joule's free expansion experiment: If an ideal gas expands through empty space (without having any environment to exchange heat or work with), then its initial and nal temperature are found to be the same. But, according to the 1st law of thermodynamics: de dw + 0 in this case, so internal energy E does not change even though the volume of the gas changes. herefore, internal energy of an ideal gas does not depend on its volume.

4 4 In general, internal energy of any system depends on its temperature. It must also be an extensive quantity, i.e. proportional to the number of particles N (a larger system has more energy). However, it is natural for an ideal gas to have energy independent on volume. In fact, this implies that E of an ideal gas does not depend on the concentration of particles n N/V (N is xed but V changes in the Joule's free expansion experiment), which merely reects the fact that particles do not interact with one another. Since particles do not interact, the internal energy of an ideal gas is just the kinetic energy of its particles. nd, we learned that it depends only on temperature. herefore, temperature is a measure of the kinetic energy of particles in an ideal gas. Heat capacities: C V de + pdv de d V d V d V C p de + pdv de + p dv d p d p d p d p Note that dv (the change of volume) at xed volume in the rst line is zero. Since internal energy of an ideal gas depends only on temperature, we have: Hence: de d hermal response of an ideal gas is simply: de de V d p d C p C V p dv pv α p d p α p 1 V dv d p 1 V his is because temperature is dened in such a way that: for an ideal gas. V p 1 pv const (in equilibrium) dv V d p Equation of state for an ideal gas: Note that C p and C V are dierent, and both are proportional to the total number of particles N. he latter is obvious if one recalls that internal energy E must be proportional to the number of particles, and heat capacity expresses the change of energy with a change of temperature under some conditions. hus, we know that: C p C V pv α p pv N We also know that the denition of temperature implies pv const. for an ideal gas. herefore, we only need to provide a proportionality constant to match the units on the left and right hand sides of the above proportionality. his constant, known as Boltzmann's constant (k B J/K), is merely a factor that converts between the units and scales of energy (Joule for pv ) and temperature (Kelvin for ). We conclude that an ideal gas obeys the following equation of state: pv k BN pv Nk B

5 Internal energy of an ideal gas: Consider a gas conned to a xed volume V, whose temperature increases from to + d. No work is done, but the increase in temperature requires heat that goes to internal energy: de C V d Here we used the denition of heat capacity at xed volume. We will integrate this equation to obtain E( ). By denition, C V is proportional to the number of particles N. We may write C V xk B N, where x is dimensionless given that Boltzmann constant k B supplies the correct units. Since internal energy E of an ideal gas depends only on temperature, x cannot depend on any state variable except perhaps temperature. However, being dimensionless, x could depend on temperature only as a function of a dimensionless ratio / c, where c is some special temperature scale that supposedly characterizes an ideal gas. Since the particles of an ideal gas do not interact with one another, it is not possible for an ideal gas to undergo any phase transition or other qualitative change at any critical temperature. Similarly, the particles do not change their character at any temperature (not exactly true in quantum mechanics if the particles have internal structure, such as molecules). herefore, there is no special temperature scale c that can characterize an ideal gas: the quantity x and hence C V are independent of temperature. We may safely integrate: E( ) C V where we chose the reference for energy E( 0) 0. Empirically, or in statistical mechanics, one nds: C V 3 2 Nk B, C p 5 2 Nk B, E( ) 3 2 Nk B for a monoatomic ideal gas. Heat engines n idealized heat engine takes heat from a hot heat source and dumps heat Q C to a cold heat sink in every cycle of operation, thus doing useful work W Q C in each cycle. he engine eciency is dened as: η W Q C (0 η 1) n idealized refrigerator is a heat engine working backward in time. It uses externally provided work W to extract some heat Q C from a cooled system and dump heat W + Q C to a hot heat reservoir. gure of merit for the performance of a refrigerator is given by: ω Q C W Q C (0 ω < ) Q C perpetual motion machine of the rst kind is a hypothetical engine that produces useful work without any source of energy (fuel). It is ruled out by the conservation of energy (1 st law of thermodynamics). perpetual motion machine of the second kind is a hypothetical engine that uses heat as a source of energy (fuel). machine like this would be extremely productive because it could extract heat even from a cold source, without the help of a colder heat sink. he conservation of energy only limits the maximum energy that such a machine could extract (given by the total internal energy of the heat source). But in reality, it is not possible to obtain energy by refrigerating the environment: energy wants to ow from a hotter to a colder body. 2 nd law of thermodynamics: States the observed fact that perpetual motion machines of the second kind are not possible. here are multiple equivalent formulations: Kelvin's statement: It is not possible to convert heat completely into work (perfect engine, η 1). Clausius' statement: It is not possible to transfer heat from a cold to a hot system without doing any work (perfect refrigerator, ω ).

6 6 If Kelvin's statement is violated, so is Clausius': Suppose we have a machine that violates Kelvin's law by extracting heat Q from a cold reservoir and converting all of it to work W Q. Use the obtained work to run a normal refrigerator with heat intake Q C from the same cold reservoir and heat output Q C + W to a hot reservoir. he combined machine takes no work, but transfers heat Q + Q C from the cold to the hot reservoir, in violation of Clausius' law. If Clausius' statement is violated, so is Kelvin's: Suppose we have a machine that violates Clausius' law by transferring heat Q from a cold to a hot reservoir without requiring any work. Run a normal engine next to this machine between the same two reservoirs. Suppose the engine is tuned to have heat intake and heat output Q, thus producing work W Q. In every cycle, the engine dumps the same amount of heat Q into the cold reservoir that the refrigerating machine takes from there. Hence, the combined machine only takes heat Q from the hot reservoir and converts all of it to work, in violation of Kelvin's law. Physically, 2 nd law of thermodynamics implies that any thermodynamic system (with many degrees of freedom) spontaneously evolves toward its equilibrium state. Heat naturally ows from a hot to a cold body, spontaneously leading to an equilibrium in which both bodies have the same temperature (this does not happen only if the two bodies are isolated). he opposite cannot be achieved without providing external work (Clausius). Carnot engines Carnot engine is any heat engine that has reversible cycles, and exchanges heat only with two heat reservoirs at xed temperatures, C (cold) and H (hot). In order to operate reversibly, a Carnot engine must be in an equilibrium at every point of its cycle. Otherwise, 2 nd law of thermodynamics implies the existence of some irreversible processes that tend to bring the engine to an equilibrium state. Carnot engine utilizes some working medium, for example a gas. By denition, the cycle of a Carnot engine has for stages: 1) isothermal heat intake from the hot reservoir at H 2) adiabatic expansion (by inertia) with temperature drop from H to C 3) isothermal heat output Q C to the cold reservoir at C 4) adiabatic compression (by inertia) with temperature increase from C to H he total work W per cycle must be equal to the heat balance W Q C according to the 1 st law of thermodynamics. here is no violation of the 2 nd law. Carnot engine is the most ecient engine possible. Proof: Use an arbitrary engine with heat intake and heat output Q C to run a Carnot engine as a refrigerator (which is possible because the Carnot engine is reversible). he engine's work output is W Q C per cycle, and eciency η W/. he Carnot refrigerator uses W to extract heat Q C from the cold reservoir and dump heat Q H Q C + W in the hot reservoir, so it would have eciency η W/ if it ran as an engine. he total heat transfer from the hot to the cold reservoir Q C Q C cannot be negative according to 2nd law of thermodynamics (Clausius). herefore, W W η η Consequently, all Carnot engines have a universal maximum eciency that does not depend on the system that realizes the engine. he only parameters that characterize a Carnot engine are the heat reservoir temperatures H and C, so the universal maximum eciency should be a function of H and C only.

7 Universal maximum heat engine eciency: Consider two Carnot engines operating between three heat reservoirs at temperatures 1 > 2 > 3. he rst engine operates between reservoirs 1 and 2, has heat intake, heat output Q C and eciency η( 1, 2 ) Q C 1 Q C he second engine operates between reservoirs 2 and 3, has heat intake, heat output Q C and eciency η( 2, 3 ) Q H Q C 1 Q C If the engines are tuned so that Q C, then there is no net heat exchange at the reservoir 2 and the two engines together function as a single Carnot engine that runs between the reservoirs 1 and 3. he eciency of the combined Carnot engine is: η( 1, 3 ) Q C 1 Q C Using Q C, we nd: 1 η( 1, 3 ) Q C Q C [ ( 1 1 Q C [ 1 η( 2, 3 ) Q C )] [ 1 Q C ( 1 Q C ][ ] 1 η( 1, 2 ) he nal expression cannot depend on 2, since the initial expression doesn't. We may assume the following single formula: 1 η( 1, 2 ) 2 1, 1 η( 2, 3 ) 3 2, 1 η( 1, 3 ) 3 1 which cancels out 2 in the last expression on the right-hand-side and makes everything work. his amounts to: η( H, C ) 1 C H here is freedom to choice a dierent formula that satises the above relationship. his freedom can be used to construct a denition of temperature, based solely on the fundamental principles of thermodynamics and independent of the uid (e.g. ideal gas) used by the thermometer standard. emperature dened by the above formula is actually identical to the temperature dened earlier using an ideal gas thermometer, which we will show next by considering a Carnot engine with an ideal gas as a working uid. Carnot engine based on an ideal gas: Knowing that the engine is always in an equilibrium state, we may use the equation of state pv Nk B for an ideal gas. We also know that internal energy of an ideal gas is E( ) xnk B, where x a dimensionless constant (x 3 2 for an ideal monoatomic gas). Stage 1: internal energy E( H ) does not change, but the gas expands from the initial volume V H to V H and does work W 1 equal to the heat intake: ˆ ˆ ( ) dv V W 1 pdv Nk B H V Nk B H log H )] V H

8 8 Stage 2: the gas expands adiabatically from the volume V H to V C and cools down from H to C. here is no heat exchange, so the work W 2 done by the gas is only at the expense of its internal energy: W 2 E( H ) E( C ) xnk B ( H C ) During any small step of the adiabatic process, the changes of internal energy de, temperature d, volume dv and pressure dp are related by: de xnk B d xd(pv ) x(pdv + V dp) x( dw + V dp) x( de + V dp) We rst used the expression for internal energy, then the equation of state, and nally the adiabatic condition that the work dw pdv done on the gas (by the environment) equals the change of internal energy de. Focusing again on de dw, we nd: de pdv x 1 + x V dp ˆ dp p 1 + x ˆ ˆ dv dv x V γ V ( ) ( ) p V log γ log p 0 p p 0 ( V V 0 ) γ V 0 pv γ p 0 V γ 0 const where γ (1 + x)/x equals 5 3 for a monoatomic ideal gas. he last formula relates volume and pressure of an ideal gas that evolves adiabatically: the product pv γ stays constant as long as the ideal gas does not exchange heat with its environment. If we combine this with the equation of state to eliminate pressure, we get a relationship between the volumes at the beginning and the end of the adiabatic expansion: Nk B H V H H V γ 1 H V γ H Nk B C V γ C V C V H V C ( C H C V γ 1 C ) 1 γ 1 Stage 3: internal energy E( C ) does not change, but the gas shrinks from the initial volume V C to V C and does (negative) work W 3 (at xed temperature) equal to the heat output: ( ) V Q C W 3 Nk B C log C Stage 4: the gas shrinks adiabatically from the volume V C to its original volume V H, and heats up from C back to H. here is no heat exchange with the environment, so the (negative) work W 4 done by the gas is only at the expense of its internal energy: V C W 4 E( C ) E( H ) xnk B ( C H ) he starting and ending volumes (V C, V H ) are related to the starting and ending temperatures ( C, H ) by the same adiabatic condition that we derived above: C V γ 1 C H V γ 1 H ( ) 1 V C H γ 1 V H C

9 he engine's eciency is: η Q C 1 Q C 1 + C log H log ( ) V C V C ( V H V H ) 1 + [ ( ) 1 ] γ 1 V C log H H VC C H log [ V C V H ( ) 1 γ 1 C H ] 1 C H We rst substituted the expressions for and Q C obtained from stages 1 and 3 (recall that Q C was negative, thus the plus sign in front of the fraction with logarithms), and then eliminated V C and V H using the adiabatic conditions from the stages 2 and 4. he nal two logarithms have inverted arguments, like log(z) and log(1/z), hence they dier only by sign. Once they are canceled, the nal expression for Carnot engine eciency is identical to the one we constructed previously from completely abstract considerations. Entropy Entropy is a state variable related to temperature the same way volume (as a state variable) is related to pressure (as an external force). wo systems in mechanical contact with each other, but at dierent pressures, change their volumes such that the system at the higher pressure expands and the system at the lower pressure contracts. Similarly, two systems in thermal contact at dierent temperatures change so the one at the higher temperature loses entropy (as it cools) and the one at the lower temperature gains entropy (as it warms up). he existence of entropy as a state variable follows from Clausius' theorem. Clausius' theorem: ny system that undergoes a cyclic change, and receives heat increments at progressively changing temperatures, obeys: 0 Proof: Connect a Carnot engine to a heat reservoir at a xed temperature 0 and the considered system acting as the other heat reservoir. Ensure that all of the system's heat exchange is directed to the Carnot engine, by running an entire Carnot cycle for every innitesimal step of the system (during which its temperature can be considered constant). hen, the total Carnot engine's heat intake 0 and output (which goes to the system) are related by the universal eciency: η We derived the formula on the right by assuming 0 >, but exactly the same holds for 0 < when the Carnot engine runs a refrigerator cycle (both, 0 and, 0 exchange their roles). Now integrate out the last formula over the entire system's cycle (which is many Carnot engine's cycles): 1 0 Q We used the fact that 0 is constant, and labeled by Q 0 the total heat intake of the Carnot's engine from the 0 reservoir in one system's cycle. By conservation of energy, the total amount of work done by the Carnot engine and system together is W Q 0 in one system's cycle (the combined system takes energy only from the 0 heat source, and outputs energy only through work because it is not connected to any external heat sink). However, Kelvin's statement of the 2 nd law of thermodynamics prohibits converting all of heat to work under any circumstances. he combined system can operate only if W Q 0 < 0, which means that work ( W ) is actually done on the system and converted to heat ( Q 0 ) that is dumped (instead of extracted) at the 0 reservoir. his proves the theorem.

10 10 If and only if a cycle is reversible, it is characterized by: 0 Proof: Since the direct cycle obeys / 0, then the opposite cycle has reversed heat exchanges at every step,, so that ( )/ 0. Both inequalities can hold only if / 0. Likewise, / 0 implies that both inequalities hold, so that the reverse cycle is possible. For a reversible process between an initial state and a nal state B, the integral depends only on the initial and nal states, and not on the path of the system's evolution from to B. Proof: his follows from the previous result for reversible processes: + 1 ˆ B if we break a reversible cycle into arbitrary reversible segments B and B which go along dierent paths (1 and 2 respectively). Entropy S is a state variable dened in every equilibrium state X by: ˆX S(X) S(R) + where R is some reference state (e.g. the state at 0), S R is a dened value of entropy in the state R (e.g. zero), and the evolution from R to X in the integral is done reversibly (slowly) by moving the system through equilibrium states. It is possible to dene S(X) as a state variable only because the integral does not depend on the path. For an arbitrary (irreversible) process that starts in a state and ends in a state B, R rev. S(B) S() Proof: [ ] S(B) S() R ˆR + B ˆ + B rev. rev. rev. ˆ R ˆ + R rev. rev. 0

11 We substituted the denition of entropy in the rst line (S R cancels out), then manipulated the integral limits and signs to connect individual reversible paths into longer paths. he dierence of entropies is merely an integral over some reversible path from B to. he generic (irreversible) path from to B and the reversible one from B to are eventually joined into a single irreversible cycle to which we apply Clausius' theorem. n adiabatically isolated system spontaneously evolves to the state of maximum entropy, which is the equilibrium state. his is yet another equivalent formulation of the 2 nd law of thermodynamics. Proof: Suppose the system evolves from the initial state to a state B some time later. Since the system is adiabatically isolated, it does not exchange heat with the environment, so that: S(B) S() + S() ( 0) herefore, at any later time, the entropy of the system S(B) is greater (or equal) than its initial entropy S(). he only way to stop the entropy growth (without exchanging heat) is to reach equilibrium in which the state does not change any more. system undergoing an innitesimal reversible transformation receives energy from the environment through heat and work dw done on it: ds, dw i J i dx i (Note that / ds is the dierential form of the integral / S, obtained when the initial and nal states are only innitesimally dierent). herefore, by the 1 st law of thermodynamics, the change of system's internal energy is: de + dw ds + i J i dx i his equation can be integrated to obtain relationships between the quantities that characterize equilibrium (E, S, V, N..., p, µ...). In that sense, the equation is valid as a connection between equilibrium states, regardless of whether these states are reached by reversible or irreversible processes. Note that temperature plays the role of a force parameter (like J i {p, µ...}), and entropy S plays the role of a state response (proportional to the number of particles N). hermodynamic potentials Entropy is a thermodynamic potential that achieves its maximum value in equilibrium with the environment at temperature, under the condition that the system is adiabatically isolated. One can dene other thermodynamic potentials that similarly achieve their extreme values in other kinds of equilibrium under various conditions. Helmholtz free energy F E S is minimized in equilibrium, under the condition that the system is mechanically isolated (does not do or receive any work) and kept at a constant temperature. Proof: Consider a small change of state, with δe δq + δw δq since there is no work. Note that the change is not necessarily reversible, so it need not go through equilibrium. However, δq δs according to Clausius' theorem (equal only for equilibrium changes). herefore, at constant temperature, δf δe δ( S) δq δs 0. he Helmholtz free energy decreases in every state transformation under the specied conditions, and achieves its minimum only in equilibrium when the state evolution stops.

12 Helmholtz free energy changes as df de d( S) JdX Sd in equilibrium (reversible) state evolution, since de ds + JdX. his implies: J df, S df dx d X E F + S F F 2 ( ) F Enthalpy H E JX is minimized in equilibrium, under the condition that the system is adiabatically isolated and subjected to a constant external force J. Proof: Consider a small change of state, with δe δq + δw δw since there is no heat transfer. Work done in a reversible change is dw JdX. However, here the change is not necessarily reversible due to the presence of friction J f, which reduces the work input of the system: δw (J J f )δx JδX. herefore, at constant force J, δh δe δ(jx) δw JδX 0. Enthalpy decreases in every state transformation under the specied conditions, and achieves its minimum only in equilibrium when the state evolution stops. Enthalpy changes as dh de d(jx) ds XdJ in equilibrium (reversible) state evolution, since de ds + JdX. his implies: dh d J d(e JX) d dh, X dh ds J dj S de JdX de dw C J (heat capacity at xed J) J d J d J d J Gibbs free energy G E S JX is minimized equilibrium, under the condition that the system is kept at a xed temperature and subjected to a constant external force J. Proof: Following the previous two arguments, δe δq + δw and δq < δs, δw < JδX. Since both and J are kept constant, δg δe δ( S) δ(jx) (δq δs) + (δw JδX) 0. Grand potential G E S µn is minimized equilibrium, under the condition that the system is kept at a xed temperature and a xed chemical potential µ. Zero temperature limit 3 rd law of thermodynamics: he entropy of all systems at absolute zero temperature is a universal constant, which can be dened as equal to zero. Deduced by measuring entropy in experiments that take a substance from one low-temperature phase to another, through a unique (disordered) high-temperature phase. he two dierent lowtemperature phases (which can be either stable or metastable) are observed to have the same entropy in the limit 0. It is found in statistical mechanics that the law does not apply to model systems that have an innite number of minimum-energy states in the 0 limit. Whether such systems exist in nature is an open question. Interacting particles quite generally tend to order at suciently low temperatures (into crystals, for examples), and there are always no more than one or a few ordered structures that minimize energy. Glassy systems are not ordered and experimentally seem to violate the 3 rd law, but it is never clear if they are the true stable minimum-energy states or far-from-equilibrium states that are just stuck due to a too slow dynamics at 0. Note that the 3 rd law of thermodynamics does not apply to the ideal gas. Particles of an ideal gas do not interact and can never order, even in the 0 limit. his is, however, a mathematical model. Every realistic gas deviates from the ideal gas behavior more and more as 0.

13 Regardless of the values of state variables X i {V, N...}, the entropy vanishes at zero temperature: ds S( 0; X i ) 0 lim 0 0 dx i he last formula simply states that the change of entropy due to a change of any state variable is zero when we keep temperature at 0. Heat capacities measured at any xed state variable X must vanish at zero temperature. Proof: Using C X ( )d, we nd that the integral S(, X) S(0, X) ˆ 0 X ˆ 0 C X ( ) d diverges unless C X ( ) 0 as 0. nite entropy, thus, implies C X 0. his does not hold for an ideal gas, whose C V xnk B is independent of temperature. realistic low-density gas can approximate the ideal gas behavior and have C V ( ) xnk B for 0 > 0, but it generally exhibits C V ( ) 0 for 0, where 0 is the critical temperature of a phase transition to a low-temperature ordered phase. hermal expansivities must vanish at zero temperature. Proof: Expansivity with respect to a state variable X (that changes in response to a generalized force J) is dened as: α J 1 dx X d J Consider the so called Gibbs free energy F E S JX, and take its derivatives: df de d( S) d(jx) ( ds + JdX) ( ds + Sd ) (JdX + XdJ) Sd XdJ df d he result F J S, df dj F J X 2 F ds dx dj d J J ds dx dj d is one of the so called Maxwell's relations (the other ones are obtained in the same manner by taking derivatives of other thermodynamic functions). Now we have: α J 1 X dx d 1 ds 1 ds dx R ds 0, as 0 J X dj X dx dj dx because we found out earlier that ds/dx 0 at xed 0. he generalized compressibility R (normalized to X instead of system volume) generally does not diverge in the 0 limit. J