Thermodynamics Basic concepts
• Thermodynamics is a phenomenological description of macroscopic systems with many degrees of free- dom. Its microscopic justi cation 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. A macroscopic system left to itself will spontaneously approach its equilibrium state.
• An 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.
• The coupling to the environment allows us to control only a very small number of the system's degrees of freedom. The 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 23 (whose number can be of the order of Avogadro's number NA ∼ 10 ) 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 su ciently detailed to enable quantitative predictions of the macroscopic sys- tem dynamics. Historically, the development of thermodynamics led to applications such as engines and refrigerators.
• The most studied system in thermodynamics is gas. An 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 (insigni cant 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. Thermodynamics studies both classical and quantum systems.
Thermal equilibrium and temperature
• 0th law of thermodynamics: If two systems are separately in equilibrium with a third system, then they are also in equilibrium with each other.
• The 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 A and B are separately in equilibrium with a third system C, then they have something in common: being in equilibrium with C. This common property can be expressed as a state variable that separately characterizes the states of each system A and B.
• The universal state variable that characterizes the equilibrium of any system is temperature. Two systems in equilibrium with each other (A and B) have the same common temperature T (this is an alternative statement of the 0th law). Dierent kinds of equilibrium (mechanical, chemical, etc.) may be characterized by additional state variables (pressure, chemical potential, etc. respectively).
• We still have no de nition 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. The 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 de ne temperature.
1 • 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. This provides one convenient temperature scale, Celsius: water and ice in equilibrium have a reproducible temperature T = 0o C, while water and steam in equilibrium have a reproducible temperature T = 100o C (by de nition). • The modern scale for temperature has been established at the 10th General Conference on Weights and Measures in 1954. The SI unit of temperature is Kelvin. The reproducible temperature of the triple point of ice-water-gas system is de ned to be 273.16 K, while 0 K (absolute zero) is the lower bound for temperature of any system. The 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. The reference of 273.16 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 de nition). For this purpose, one uses an ideal gas. It has been experimentally observed that the quantity pV/T (p is pressure, V is volume and T is temperature in Kelvins measured by a typical accurate thermometer) is a constant in any equilibrium state of any su ciently dilute (approximately ideal) gas. Therefore, for two equilibrium states at the same pressure:
V1 V2 V2 = ⇒ T2 = T1 T1 T2 V1
Since one temperature (T1) 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. This observed linear relationship for an ideal gas is then promoted to the de nition of temperature, so that, by de nition, accurate thermometers measure temperature and not some arbitrary function of temperature.
The conservation of energy
• A 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.
• A system can exchange energy with its environment in dierent ways. The 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.
• The amount of work dW done on the system by external forces in an in nitesimal time interval is: X dW = JidXi 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, Ji are quantities that represent forces and Xi 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 dA due to a surface tension S: dW = SdA Volume change by dV due to internal pressure p: dW = −pdV
2 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 The number of particles change by dN at a chemical potential µ: dW = µdN
• There 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). The 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). This 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. The change of the system is microscopic and hidden from our view (the distribution of particle velocities changes).
• A system is said to evolve adiabatically if it does not exchange heat with its environment. • 1st law of thermodynamics: Energy is conserved in any process. If the system takes heat dQ from its environment, and work dW is done on it by external forces, then its internal energy changes by dE according to: dQ + dW = dE The 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) dQ does it take to change the temperature of the system by dT under various conditions... dQ CV = ··· measured at given volume V = const dT V dQ Cp = ··· measured at given pressure p = const dT p
• Force constant: a thermodynamic generalization of the spring constant 1 dX RT = ··· the change of displacement due to applied force, per unit volume V dJ T 1 dV κT = − ··· compressibility (pressure changes volume) V dp T 1 dM χT = ··· magnetic susceptibility (magnetic eld changes magnetization) V dB T • Thermal responses: 1 dV αp = ··· expansivity of a gas at constant pressure V dT p Ideal gas
• 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 + dQ → 0 in this case, so internal energy E does not change even though the volume of the gas changes. Therefore, internal energy of an ideal gas does not depend on its volume.
3 • 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 re ects 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. And, we learned that it depends only on temperature. Therefore, temperature is a measure of the kinetic energy of particles in an ideal gas.
• Heat capacities: dQ dE + pdV dE CV = = = dT V dT V dT V dQ dE + pdV dE dV Cp = = = + p dT p dT p dT p dT 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:
dE dE dE = = dT V dT p dT Hence: dV Cp − CV = p = pV αp dT p • Thermal response of an ideal gas is simply: 1 dV 1 V 1 αp = = = V dT p V T p T This is because temperature is de ned in such a way that:
pV dV V = const ··· (in equilibrium) ⇒ = T dT p T for an ideal gas.
• Equation of state for an ideal gas: Note that Cp and CV are dierent, and both are proportional to the total number of particles N. The 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. Thus, we know that:
pV C − C = pV α = ∝ N p V p T We also know that the de nition of temperature implies
pV = const. T for an ideal gas. Therefore, we only need to provide a proportionality constant to match the units on the left and right hand sides of the above proportionality. This constant, known as Boltzmann's −23 constant (kB = 1.38064852 × 10 J/K), is merely a factor that converts between the units and scales of energy (Joule for pV ) and temperature (Kelvin for T ). We conclude that an ideal gas obeys the following equation of state: pV = k N ⇒ pV = Nk T T B B
4 • Internal energy of an ideal gas: Consider a gas con ned to a xed volume V , whose temperature increases from T to T + dT . No work is done, but the increase in temperature requires heat that goes to internal energy:
dQ = dE = CV dT Here we used the de nition of heat capacity at xed volume. We will integrate this equation to obtain
E(T ). By de nition, CV is proportional to the number of particles N. We may write CV = xkBN, where x is dimensionless given that Boltzmann constant kB 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 T only as a function of a dimensionless ratio T/Tc, where Tc 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). Therefore, there
is no special temperature scale Tc that can characterize an ideal gas: the quantity x and hence CV are independent of temperature. We may safely integrate:
E(T ) = CV T
where we chose the reference for energy E(T = 0) = 0. Empirically, or in statistical mechanics, one nds: 3 5 3 C = Nk ,C = Nk ,E(T ) = Nk T V 2 B p 2 B 2 B for a monoatomic ideal gas.
Heat engines
• An idealized heat engine takes heat QH from a hot heat source and dumps heat QC to a cold heat sink in every cycle of operation, thus doing useful work W = QH − QC in each cycle. The engine e ciency is de ned as: W Q − Q η = = H C (0 ≤ η ≤ 1) QH QH • An idealized refrigerator is a heat engine working backward in time. It uses externally provided work W to extract some heat QC from a cooled system and dump heat QH = W + QC to a hot heat reservoir. A gure of merit for the performance of a refrigerator is given by:
Q Q ω = C = C (0 ≤ ω < ∞) W QH − QC
• A 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 (1st law of thermodynamics).
• A perpetual motion machine of the second kind is a hypothetical engine that uses heat as a source of energy (fuel). A 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. The 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.
• 2nd law of thermodynamics: States the observed fact that perpetual motion machines of the second kind are not possible. There 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 = ∞).
5 • 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 QC from the same cold reservoir and heat output QH = QC + W to a hot reservoir. The combined machine takes no work, but transfers heat Q + QC 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 QH and heat output Q, thus producing work W = QH − 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 QH − Q from the hot reservoir and converts all of it to work, in violation of Kelvin's law.
• Physically, 2nd 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). The opposite cannot be achieved without providing external work (Clausius).
Carnot engines
• A Carnot engine is any heat engine that has reversible cycles, and exchanges heat only with two heat reservoirs at xed temperatures, TC (cold) and TH (hot). • In order to operate reversibly, a Carnot engine must be in an equilibrium at every point of its cycle. Otherwise, 2nd law of thermodynamics implies the existence of some irreversible processes that tend to bring the engine to an equilibrium state.
• A Carnot engine utilizes some working medium, for example a gas. By de nition, the cycle of a Carnot engine has for stages:
1) isothermal heat intake QH from the hot reservoir at TH
2) adiabatic expansion (by inertia) with temperature drop from TH to TC
3) isothermal heat output QC to the cold reservoir at TC
4) adiabatic compression (by inertia) with temperature increase from TC to TH
st • The total work W per cycle must be equal to the heat balance W = QH − QC according to the 1 law of thermodynamics. There is no violation of the 2nd law.
• Carnot engine is the most e cient engine possible.
Proof: Use an arbitrary engine with heat intake QH and heat output QC to run a Carnot engine as a refrigerator (which is possible because the Carnot engine is reversible). The engine's work
output is W = QH − QC per cycle, and e ciency η = W/QH . The Carnot refrigerator uses W to extract heat 0 from the cold reservoir and dump heat 0 0 in the hot reservoir, so it QC QH = QC + W would have e ciency 0 0 if it ran as an engine. The total heat transfer from the hot to the η = W/QH cold reservoir 0 0 cannot be negative according to 2nd law of thermodynamics QH − QH = QC − QC (Clausius). Therefore,
0 W W 0 QH ≥ QH ⇒ ≤ 0 ⇔ η ≤ η QH QH • Consequently, all Carnot engines have a universal maximum e ciency that does not depend on the system that realizes the engine. The only parameters that characterize a Carnot engine are the heat
reservoir temperatures TH and TC , so the universal maximum e ciency should be a function of TH and TC only.
6 • Universal maximum heat engine e ciency: Consider two Carnot engines operating between three heat reservoirs at temperatures T1 > T2 > T3. The rst engine operates between reservoirs 1 and 2, has heat intake QH , heat output QC and e ciency
QH − QC QC η(T1,T2) = = 1 − QH QH The second engine operates between reservoirs 2 and 3, has heat intake 0 , heat output 0 and QH QC e ciency 0 0 0 QH − QC QC η(T2,T3) = 0 = 1 − 0 QH QH If the engines are tuned so that 0 , then there is no net heat exchange at the reservoir and QC = QH T2 the two engines together function as a single Carnot engine that runs between the reservoirs T1 and T3. The e ciency of the combined Carnot engine is:
0 0 QH − QC QC η(T1,T3) = = 1 − QH QH Using 0 , we nd: QC = QH
0 0 0 0 QC QC QH QC QC 1 − η(T1,T3) = = 0 = 0 QH QH QH QH QH 0 QC QC = 1 − 1 − 0 1 − 1 − 0 QH QH h ih i = 1 − η(T2,T3) 1 − η(T1,T2)
The nal expression cannot depend on T2, since the initial expression doesn't. We may assume the following single formula:
T2 T3 T3 1 − η(T1,T2) = , 1 − η(T2,T3) = , 1 − η(T1,T3) = T1 T2 T1
which cancels out T2 in the last expression on the right-hand-side and makes everything work. This amounts to: TC η(TH ,TC ) = 1 − TH There is freedom to choice a dierent formula that satis es the above relationship. This freedom can be used to construct a de nition of temperature, based solely on the fundamental principles of thermodynamics and independent of the uid (e.g. ideal gas) used by the thermometer standard. Temperature de ned by the above formula is actually identical to the temperature de ned 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 = NkBT for an ideal gas. We also know that internal energy of an ideal gas is , where a dimensionless constant ( 3 for an ideal monoatomic E(T ) = xNkBT x x = 2 gas).
Stage 1: internal energy does not change, but the gas expands from the initial volume E(TH ) VH to 0 and does work equal to the heat intake: VH W1
0 dV VH QH = W1 = pdV = NkBTH = NkBTH log ˆ ˆ V VH
7 Stage 2: the gas expands adiabatically from the volume 0 to and cools down from to . VH VC TH TC There is no heat exchange, so the work W2 done by the gas is only at the expense of its internal energy:
W2 = E(TH ) − E(TC ) = xNkB(TH − TC ) During any small step of the adiabatic process, the changes of internal energy dE, temperature dT , volume dV and pressure dp are related by:
dE = xNkBdT = 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: x dE = −pdV = V dp 1 + x dp 1 + x dV dV = − ≡ −γ ˆ p x ˆ V ˆ V p V log = −γ log p0 V0 p V −γ = p0 V0 γ γ const pV = p0 V0 = where equals 5 for a monoatomic ideal gas. The last formula relates volume and γ = (1 + x)/x 3 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:
NkBTH 0γ NkBTC γ 0 V H = VC VH VC
0γ−1 γ−1 TH VH = TC VC 1 γ−1 0 TC VH = VC TH
Stage 3: internal energy E(TC ) does not change, but the gas shrinks from the initial volume VC to 0 and does (negative) work (at xed temperature) equal to the heat output: VC W3 0 VC QC = −W3 = ··· = −NkBTC log VC
Stage 4: the gas shrinks adiabatically from the volume 0 to its original volume , and heats VC VH up from TC back to TH . There is no heat exchange with the environment, so the (negative) work W4 done by the gas is only at the expense of its internal energy:
W4 = E(TC ) − E(TH ) = xNkB(TC − TH ) The starting and ending volumes ( 0 , ) are related to the starting and ending temperatures VC VH (TC ,TH ) by the same adiabatic condition that we derived above:
0γ−1 γ−1 TC VC = TH VH
1 γ−1 0 TH VC = VH TC
8 The engine's e ciency is:
1 γ−1 V 0 V H TH C TC log TC log VC TC Q − Q Q VC T η = H C = 1 − C = 1 + = 1 + = 1 − C V 0 1 QH QH H γ−1 TH TH log V C TC VH TH log VH TH
We rst substituted the expressions for QH and QC obtained from stages 1 and 3 (recall that QC was negative, thus the plus sign in front of the fraction with logarithms), and then eliminated 0 and VC 0 using the adiabatic conditions from the stages 2 and 4. The nal two logarithms have inverted VH arguments, like log(z) and log(1/z), hence they dier only by sign. Once they are canceled, the nal expression for Carnot engine e ciency 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). Two 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). The existence of entropy as a state variable follows from Clausius' theorem.
• Clausius' theorem: Any system that undergoes a cyclic change, and receives heat increments dQ at progressively changing temperatures T , obeys: dQ ≤ 0 ˛ T
Proof: Connect a Carnot engine to a heat reservoir at a xed temperature T0 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 in nitesimal step of the system (during which its temperature T can be considered constant). Then, the total Carnot engine's heat intake dQ0 and output dQ (which goes to the system) are related by the universal e ciency: dQ T dQ dQ η = 1 − = 1 − ⇒ = 0 dQ0 T0 T T0
We derived the formula on the right by assuming T0 > T , but exactly the same holds for T0 < T when the Carnot engine runs a refrigerator cycle (both T,T0 and dQ, dQ0 exchange their roles). Now integrate out the last formula over the entire system's cycle (which is many Carnot engine's cycles): dQ 1 Q0 = dQ0 = ˛ T T0 ˛ T0
We used the fact that T0 is constant, and labeled by Q0 the total heat intake of the Carnot's engine from the T0 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 = Q0 in one system's cycle (the combined system takes energy only from the T0 heat source, and outputs energy only through work because it is not connected to any external heat sink). However, Kelvin's statement of the 2nd law of thermodynamics prohibits converting all of heat to work under any circumstances.
The combined system can operate only if W = Q0 < 0, which means that work (−W ) is actually done on the system and converted to heat (−Q0) that is dumped (instead of extracted) at the T0 reservoir. This proves the theorem.
9 • If and only if a cycle is reversible, it is characterized by: dQ = 0 ˛ T
Proof: Since the direct cycle obeys dQ/T ≤ 0, then the opposite cycle has reversed heat exchanges at every step, dQ → −dQ, so¸ that (−dQ)/T ≤ 0. Both inequalities can hold only if dQ/T = 0. Likewise, dQ/T = 0 implies that¸ both inequalities hold, so that the reverse cycle is¸ possible. ¸
• For a reversible process between an initial state A and a nal state B, the integral
B dQ ˆ T A depends only on the initial and nal states, and not on the path of the system's evolution from A to B.
Proof: This follows from the previous result for reversible processes:
B A B B dQ dQ dQ dQ dQ = + = − = 0 ˛ T ˆ T 1 ˆ T 2 ˆ T 1 ˆ T 2 A B A A if we break a reversible cycle into arbitrary reversible segments A → B and B → A which go along dierent paths (1 and 2 respectively).
• Entropy S is a state variable de ned in every equilibrium state X by:
X dQ S(X) = S(R) + ˆ T rev. R
where R is some reference state (e.g. the state at T = 0), SR is a de ned 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 de ne 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 A and ends in a state B,
B dQ ≤ S(B) − S(A) ˆ T A Proof:
B B B A dQ h i dQ dQ dQ − S(B) − S(A) = − − ˆ T ˆ T ˆ T rev. ˆ T rev. A A R R B R A dQ dQ dQ = + + ˆ T ˆ T rev. ˆ T rev. A B R B A dQ dQ dQ = + = ≤ 0 ˆ T ˆ T rev. ˛ T A B
10 We substituted the de nition of entropy in the rst line (SR cancels out), then manipulated the integral limits and signs to connect individual reversible paths into longer paths. The dierence of entropies is merely an integral over some reversible path from B to A. The generic (irreversible) path from A to B and the reversible one from B to A are eventually joined into a single irreversible cycle to which we apply Clausius' theorem.
• An adiabatically isolated system spontaneously evolves to the state of maximum entropy, which is the equilibrium state. This is yet another equivalent formulation of the 2nd law of thermodynamics.
Proof: Suppose the system evolves from the initial state A to a state B some time later. Since the system is adiabatically isolated, it does not exchange heat with the environment, so that:
B dQ S(B) ≥ S(A) + = S(A)(dQ ≡ 0) ˆ T A
Therefore, at any later time, the entropy of the system S(B) is greater (or equal) than its initial entropy S(A). The only way to stop the entropy growth (without exchanging heat) is to reach equilibrium in which the state does not change any more.
• A system undergoing an in nitesimal reversible transformation receives energy from the environment through heat dQ and work dW done on it: X dQ = T dS , dW = JidXi i
(Note that dQ/T = dS is the dierential form of the integral dQ/T = ∆S, obtained when the initial and nal states are only in nitesimally dierent). Therefore,´ by the 1st law of thermodynamics, the change of system's internal energy is: X dE = dQ + dW = T dS + JidXi i This equation can be integrated to obtain relationships between the quantities that characterize equilib- rium (E, S, V, N . . . T, 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 T plays the role of a force parameter (like Ji ∈ {p, µ . . .}), and entropy S plays the role of a state response (proportional to the number of particles N).
Thermodynamic potentials
• Entropy is a thermodynamic potential that achieves its maximum value in equilibrium with the en- vironment at temperature T , under the condition that the system is adiabatically isolated. One can de ne other thermodynamic potentials that similarly achieve their extreme values in other kinds of equilibrium under various conditions.
• Helmholtz free energy F = E − TS 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 ≤ T δS according to Clausius' theorem (equal only for equilibrium changes). Therefore, at constant temperature T , δF = δE −δ(TS) = δQ−T δS ≤ 0. The Helmholtz free energy decreases in every state transformation under the speci ed conditions, and achieves its minimum only in equilibrium when the state evolution stops.
11 Helmholtz free energy changes as dF = dE − d(TS) = JdX − SdT in equilibrium (reversible) state evolution, since dE = T dS + JdX. This implies: dF dF J = ,S = − dX T dT X ∂F ∂ F E = F + TS = F − T = −T 2 ∂T ∂T T
• 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 Jf , which reduces the work input of the system: δW = (J − Jf )δX ≤ JδX. Therefore, at constant force J, δH = δE − δ(JX) = δW − JδX ≤ 0. Enthalpy decreases in every state transformation under the speci ed conditions, and achieves its minimum only in equilibrium when the state evolution stops. Enthalpy changes as dH = dE − d(JX) = T dS − XdJ in equilibrium (reversible) state evolution, since dE = T dS + JdX. This implies: dH dH T = ,X = − dS J dJ S dH d(E − JX) dE − JdX dE − dW dQ = = = = = CJ (heat capacity at xed J) dT J dT J dT J dT J dT J • Gibbs free energy G = E − TS − 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 < T δS, δW < JδX. Since both T and J are kept constant, δG = δE − δ(TS) − δ(JX) = (δQ − T δS) + (δW − JδX) ≤ 0.
• Grand potential G = E − TS − 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
• 3rd law of thermodynamics: The entropy of all systems at absolute zero temperature is a universal constant, which can be de ned 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. The two dierent low- temperature phases (which can be either stable or metastable) are observed to have the same entropy in the limit T → 0. It is found in statistical mechanics that the law does not apply to model systems that have an in nite number of minimum-energy states in the T → 0 limit. Whether such systems exist in nature is an open question. Interacting particles quite generally tend to order at su ciently 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 3rd 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 T → 0. Note that the 3rd 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 T → 0 limit. This is, however, a mathematical model. Every realistic gas deviates from the ideal gas behavior more and more as T → 0.
12 • Regardless of the values of state variables Xi ∈ {V,N...}, the entropy vanishes at zero temperature: dS S(T = 0; Xi) = 0 ⇒ lim = 0 T →0 dXi T The last formula simply states that the change of entropy due to a change of any state variable is zero when we keep temperature at T = 0.
• Heat capacities measured at any xed state variable X must vanish at zero temperature.
Proof: Using dQ = CX (T )dT , we nd that the integral
T T dQ dT S(T,X) − S(0,X) = = CX (T ) ˆ T X ˆ T 0 0
diverges unless CX (T ) → 0 as T → 0. A nite entropy, thus, implies CX → 0. This does not hold for an ideal gas, whose CV = xNkB is independent of temperature. A realistic low-density gas can approximate the ideal gas behavior and have CV (T ) ≈ xNkB for T T0 > 0, but it generally exhibits CV (T ) → 0 for T T0, where T0 is the critical temperature of a phase transition to a low-temperature ordered phase.
• Thermal 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 de ned as: 1 dX αJ ≡ X dT J Consider the so called Gibbs free energy F = E − TS − JX, and take its derivatives:
dF = dE − d(TS) − d(JX) = (T dS + JdX) − (T dS + SdT ) − (JdX + XdJ) = −SdT − XdJ
dF ∂F dF ∂F ∂2F dS dX ≡ = −S, ≡ = −X ⇒ = − = − dT J ∂T dJ T ∂J ∂T ∂J dJ T dT J The result dS dX = dJ T dT J 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:
1 dX 1 dS 1 dS dX 0 dS as αJ = = = = RT → 0 , T → 0 X dT J X dJ T X dX T dJ T dX T because we found out earlier that dS/dX → 0 at xed T → 0. The generalized compressibility 0 (normalized to instead of system volume) generally does not diverge in the limit. RT X T → 0
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