Lecture 8. Thermodynamic Identities (Ch. 3) We have been considering the entropy changes in the processes where two interacting systems exchanged the thermal energy but the volume and the number of particles in these systems were fixed. In general, however, we need more than just one parameter to specify a macrostate: Today we will explore what happens if we let the other two parameters (V and N) vary, and analyze the physical meaning of two partial derivatives of the entropy: We are familiar with only one partial derivative of entropy:

General Approach:  if monotonic as a function of U (“quadratic” degrees of freedom!), may be inverted to give When all macroscopic quantities S,V,N,U are allowed to vary: So far, we have abbreviated one of these partial derivatives: The other partial derivatives are: pressurechemical potential

Thermodynamic Identities  is an intensive variable, independent of the size of the system (like P, T, density). Extensive variables (U, N, S,V...) have a magnitude proportional to the size of the system. If two identical systems are combined into one, each extensive variable is doubled in value. With these abbreviations:  shows how much the system’s energy changes when one particle is added to the system at fixed S and V. The chemical potential units – J. The coefficients may be identified as: This identity holds for small changes  S provided T and P are well defined. - the so-called thermodynamic identity The 1 st Law for quasi-static processes (N = const): The thermodynamic identity holds for the quasi-static processes (T, P,  are well- define throughout the system)

Quasi-static Adiabatic Processes Quasistatic adiabatic (Q = 0) processes: (holds for all processes) In this case, - the change in internal energy is due to a “purely mechanical” compression (or expansion) that involves no change in entropy. (No entropy change = no heat flow). Let’s compare two forms of the 1 st Law : (quasi-static processes, N is fixed, and P is uniform throughout the system) Thus, for quasi-static processes : reduces to  isentropic processes

Pr A non-quasistatic compression. A cylinder with air (V = m 3, T = 300K, P =10 5 Pa) is compressed (very fast, non-quasistatic) by a piston (S = 0.01 m 2, F = 2000N,  x = m). Calculate  W,  Q,  U, and  S. holds for all processes, energy conservation quasistatic, T and P are well- defined for any intermediate state quasistatic adiabatic  isentropic non-quasistatic adiabatic The non-quasistatic process results in a higher T and a greater entropy of the final state. S = const along the isentropic line P V ViVi VfVf 1 2 2*2*  Q = 0 for both Caution: for non-quasistatic adiabatic processes,  S might be non-zero!!! An example of a non-quasistatic adiabatic process

Direct approach: adiabatic quasistatic  isentropic adiabatic non-quasistatic

The entropy is created because it is an irreversible, non-quasistatic compression. P V ViVi VfVf To calculate  S, we can consider any quasistatic process that would bring the gas into the final state (S is a state function). For example, along the red line that coincides with the adiabata and then shoots straight up. Let’s neglect small variations of T along this path (  U << U, so it won’t be a big mistake to assume T  const):  U = Q = 1J For any quasi-static path from 1 to 2, we must have the same  S. Let’s take another path – along the isotherm and then straight up: 1 2 P V ViVi VfVf 1 2  U = Q = 2J isotherm: “straight up”: Total gain of entropy:

The inverse process, sudden expansion of an ideal gas (2 – 3) also generates entropy (adiabatic but not quasistatic). Neither heat nor work is transferred: W = Q = 0 (we assume the whole process occurs rapidly enough so that no heat flows in through the walls). The work done on the gas is negative, the gas does positive work on the piston in an amount equal to the heat transfer into the system P V ViVi VfVf Because U is unchanged, T of the ideal gas is unchanged. The final state is identical with the state that results from a reversible isothermal expansion with the gas in thermal equilibrium with a reservoir. The work done on the gas in the reversible expansion from volume V f to V i : Thus, by going 1  2  3, we will increase the gas entropy by

Comment on State Functions U, S and V are the state functions, Q and W are not. Specifying an initial and final states of a system does not fix the values of  Q and  W, we need to know the whole process (the intermediate states). In math terms,  Q and  W are not exact differentials and we need to use  instead of d for their increments. Analogy: in classical mechanics, if a force is not conservative (e.g., friction), the initial and final positions do not determine the work, the entire path must be specified. x y z(x 1,y 1 ) z(x 2,y 2 ) - it is an exact differential if it is the difference between the values of some (state) function z(x,y) at these points: A necessary and sufficient condition for this: If this condition holds: e.g., for an ideal gas: - cross derivatives are not equal is an exact differential (S is a state function). Thus, the factor 1/T converts  Q into an exact differential.

Entropy Change for Different Processes Type of interaction Exchanged quantity Governing variable Formula thermalenergytemperature mechanicalvolumepressure diffusiveparticles chemical potential The last column provides the connection between statistical physics and thermodynamics. The partial derivatives of S play very important roles because they determine how much the entropy is affected when U, V and N change:

Temperature U A, V A, N A U B, V B, N B To discuss the physical meaning of these partial derivatives of S, let’s consider an ideal gas isolated from the environment, and consider two sub-systems, A and B, separated by a membrane, which, initially, is insulating, impermeable for gas molecules, and fixed at a certain position. Thus, the sub-systems are isolated from each other. (We have already considered such a system when we discussed T, but we want to proceed with P and  ) Assume that the membrane becomes thermally conducting (allow exchange U while V and N remain fixed). The system will evolve to the state of equilibrium that is characterized by a maximum entropy. One of the sub-systems can increase its energy only at the expense of the other sub-system, and it’s crucial that the sub- system receiving energy increases its entropy more than the donor loses. Thus, the This was our logic when we defined the stat. phys. temperature as: - the energy should flow from higher T to lower T; in thermal equilibrium, T A and T B should be the same. sub-system with a larger  S/  U should receive energy, and this process will continue until  S A /  U A and  S B /  U B become the same:

Mechanical Equilibrium and Pressure Let’s now fix U A,N A and U B,N B, but allow V to vary (the membrane is insulating, impermeable for gas molecules, but its position is not fixed). Following the same logic, spontaneous “exchange of volume” between sub-systems will drive the system towards mechanical equilibrium (the membrane at rest). The equilibrium macropartition should have the largest (by far) multiplicity  (U, V) and entropy S (U, V). U A, V A, N A U B, V B, N B - the sub-system with a smaller volume-per-molecule (larger P at the same T) will have a larger  S/  V, it will expand at the expense of the other sub-system. In mechanical equilibrium: - the volume-per-molecule should be the same for both sub-systems, or, if T is the same, P must be the same on both sides of the membrane. S AB VAVA V A eq S A S B The stat. phys. definition of pressure:

The Equation(s) of State for an Ideal Gas Ideal gas: (fN degrees of freedom) The “energy” equation of state (U  T): The “pressure” equation of state (P  T): - we have finally derived the equation of state of an ideal gas from first principles! In other words, we can calculate the thermodynamic information for an isolated system by counting all the accessible microstates as a function of N, V, and U.

Problem: (a) Calculate the entropy increase of an ideal gas in an isothermal process. (b) Calculate the entropy increase of an ideal gas in an isochoric process. (c) What is the process in which the entropy remains constant? You should be able to do this using (a) Sackur-Tetrode eq. and (b) (a)(b) (c) an adiabatic process a quasistatic adiabatic process = an isentropic process (all the processes are quasi-static)

Diffusive Equilibrium and Chemical Potential Sign “-”: out of equilibrium, the system with the larger  S/  N will get more particles. In other words, particles will flow from from a high  /T to a low  /T. Let’s fix V A and V B (the membrane’s position is fixed), but assume that the membrane becomes permeable for gas molecules (exchange of both U and N between the sub-systems, the molecules in A and B are the same ). S UAUA NANA U A, V A, S A U B, V B, S B For sub-systems in diffusive equilibrium: In equilibrium, - the chemical potential

Chemical Potential: examples Einstein solid: consider a small one, with N = 3 and q = 3. let’s add one more oscillator:  To keep dS = 0, we need to decrease the energy, by subtracting one energy quantum. Thus, for this system Monatomic ideal gas: At normal T and P, ln(...) > 0, and  < 0 (e.g., for He,  ~ - 5· J ~ eV. Sign “-”: usually, by adding particles to the system, we increase its entropy. To keep dS = 0, we need to subtract some energy, thus  U is negative.

The Quantum Concentration At T=300K, P=10 5 Pa, n << n Q. When n  n Q, the quantum statistics comes into play. - the so-called quantum concentration (one particle per cube of side equal to the thermal de Broglie wavelength). When n Q >> n, the gas is in the classical regime. n=N/V – the concentration of molecules The chemical potential increases with the density of the gas or with its pressure. Thus, the molecules will flow from regions of high density to regions of lower density or from regions of high pressure to those of low pressure.   0 when n  n Q,   0 when n increases

Ideal Gas in a Gravitational Field Pr Consider a monatomic ideal gas at a height z above sea level, so each molecule has potential energy mgz in addition to its kinetic energy. Assuming that the atmosphere is isothermal (not quite right), find  and re-derive the barometric equation. In equilibrium, the two chemical potentials must be equal: note that the U that appears in the Sackur-Tetrode equation represents only the kinetic energy

S and C P - this is a useful result that removes the restriction of constant volume. For P = const processes, Q = C P dT  At T  0, the graph goes to 0 with zero slope. After initial fast increase of the slope, the rate of the S increase slows down (C P – almost const). When solid melts, there is a large  S at T = const, another jump – at liquid–gas phase transformation. “Curving down” in both liquid and gas phases – because C P  const.  S T ice water vapor Pr Sketch a graph of the entropy of H 2 0 as a function of T at P = const.

Future Directions Although the microcanonical ensemble is conceptually simple, it is not the most practical ensemble. The major problem is that we must specify U - isolated systems are very difficult to realize experimentally, and T rather than U is a more natural independent variable. EnsembleMacrostateProbabilityThermodynamics micro- canonical U, V, N (T fluctuates) canonical T, V, N (U fluctuates) grand canonical U, V,  (N fluctuates)