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Thermodynamic Laws/Definition of Entropy Carnot Cycle This is a review of Thermodynamics and Statistical Mechanics. Thermodynamic Laws/Definition of Entropy • 1st law of thermodynamics, the conservation of energy: dU = dq − dw ∆ = Q − W, (1) where dq is heat entering the system and dw is work done by the system. Note the convention: (+) for energy entering the system and (-) for energy leaving the system. • 2nd law of thermodynamics, entropy: In any spontaneous transition, the entropy of the universe increases. There are many equivalent statements of the second law, such as: heat never flows from cold to hot, or, there is no such thing as a perpetual motion machine. In a reversible transformation, the entropy of the universe does not change. Note: this does not mean that the entropy of some sub-universal system will not increase or decrease. It is important to consider all parts of your system. To find the change in entropy of a system between state A and state B, connect A and B by a reversible path. Then Z B dq S(B) − S(A) = . (2) A T Using Equation 2, we can illustrate the point about reversible transformations above: Consider a reversible isothermal expansion of an ideal gas in contact with a thermal reservoir 3 at temperature T . For an ideal gas, U = 2 kT , for an isothermal expansion dU = 0, and dq = P dV , so Q Z dq Z P dV = = ∆S = . (3) T T T nRT P = V , so Z P Z VB nR V dV = dV = nR ln B > 0, (4) T VA V VA −Q but heat leaves the reservoir, thus ∆Sres = T and ∆Suniverse = 0. Carnot Cycle The Carnot Cycle is usually discussed with an ideal gas as the working substance, but in reality any thermodynamic system (a paramagnet, an electrochemical cell, etc.) can be used. A Carnot Cycle is a cycle involving two reversible isothermal transitions and two reversible adiabatic transitions. If the working substance is an ideal gas, the P − V diagram of the cycle looks like Fig. 1, and the S − T diagram looks like Fig. 2, The efficiency of the Carnot Cycle is given by W Q − Q T − T T η = = 2 1 = 2 1 = 1 − 1 , (5) Q2 Q2 T2 T2 1 Figure 1: Carnot Cycle: P − V diagram Figure 2: Carnot Cycle: S − T diagram which is work done by the system divided by the heat absorbed by the system. For a “Carnot Refrigerator,” run the cycle backwards: Q1 Q1 1 1 T1 ηref = = = = = , (6) Q2 T2 W Q2 − Q1 − 1 − 1 T2 − T1 Q1 T1 heat extracted from the system divided by work required to do so. Note that as T1 → T2, ηref → ∞. In general, the efficiency is what you get out η = . (7) what you need to put in Typical Phase Diagrams Figs. 3 and 4 show the phase diagrams for a single component system. 2 Figure 3: P − v diagram: Pressure vs. specific volume Thermodynamic Potentials Let’s begin with Eq. 1: dU = dq − dw. (8) If we connect states by reversible processes we get T dS = dq and dW = P dV for a gas system. So, dU = T dS − P dV, (9) and setting one of ∂s or ∂v equal to zero yields, ∂U ∂U = T, = −P. (10) ∂S V ∂V S Define ∂F ∂F F ≡ U − T S, dF = −SdT − P dV, ∂T = ∂V = −S. | {z } V T (11) Helmholtz free energy Define ∂H ∂H H ≡ U + P V, dH = T dS + V dρ, ∂S = T, ∂ρ = V. | {z } ρ S (12) enthalpy Define ∂G ∂G G ≡ H − T S, dG = −SdT + V dρ, ∂T = −S, ∂ρ = V | {z } ρ T (13) Gibbs’ free energy 3 Figure 4: P − T diagram: Pressure vs. Temperature. The dotted line characterizes the anamolous behavior of water (and any substance that expands when frozen). The differential form tells us the natural variables to use with each function. A, for example (Fig. 11), is best expressed in terms of T and V : A(T,V ). In physics we often do problems at constant (V, T ) and work with A. Chemists are prone to a constant (T, ρ) and work with G. Note that we can get Maxwell’s relations from the fact that each of these is an exact differential. For example, ∂F ∂F dF = dT + dV, (14) ∂T V ∂V T thus ∂F ∂F ∂T V = −S and ∂V T = −ρ, (15) but ∂2F ∂2F = , (16) ∂T ∂V ∂V ∂T so ∂S ∂ρ = , (17) ∂V T ∂T V which is one of Maxwell’s relations. Note from Dr. Collins: “I have not seen Maxwell’s relations on any GRE example test (yet!)”. Also, all of the above assume N is fixed. If N can vary each gets another term and we must worry about chemical potentials: ∂U dU = T dS − P dV + µdN where ∂N S,V = µ = chemical potential (18) The rest of the potentials also get this term, and we get a new potential, the grand potential - Φ: Φ = A − µN where dΦ = −SdT − ρdV − NdU (19) 4 Heat Capacity Heat capacity is specific heat if you divide by volume or mass ∆Q c = holding volume constant (20) V ∆T ∂U = where U is internal energy (21) ∂T V ∆U c = holding pressure constant (22) P ∆T cP > cV because the volume expands at constant pressure and the system does work which means it stores less energy. Note that for non PVT systems analogous heat capacities can be defined. In a magnetic system, P → H and V → M and we can define cH and cM . A statistical approach can be used to obtain the classical heat capacity of common sys- tems. If the Hamiltonian of a particle in the system can be written as N1 N2 X 2 X 2 H = AjPj + BjQj (23) j j where Pj and Qj are generalized momenta of coordinates, then 1 U = NkT f Equipartition Theorem (24) 2 where N is the total number of particles and f = N1 + N2, the number of quadratic coordi- nates and momenta. Thus, 1 c = fNk (25) V 2 1 1 and the specific heat is 2 fk. You get 2 k per degree of freedom (momentum of coordinate degrees of freedom count equally). For this expression to work, the temperature must be large compared to the quantum of energy associated with a particular motion. For example, if a diatomic molecule has a vibrational frequency ω, then the quantum of energy is ~ω 1 1 2 1 (E = (n + 2 )~ω). So kT hω to get NkT contributions to U ( 2 NλT for x and 2 NkT for ρ2). Consider the example of a diatomic molecule like N2. See Fig. 5. We can write the Hamiltonian as 1 1 1 1 1 1 1 H = Mv2 + Mv2 + Mv2 + µr˙2 + kr2 + Iω2 + Iω2 (26) 2 xcm 2 ycm 2 zcm 2 2 2 z 2 x Note that Iy = 0 for quantum mechanical reasons, so there is no rotational term about the y-axis. We immediately know that at high temperatures 7 7 cV 7 H = 2 kT N, cV = 2 Nk, and cV = N = 2 k. (27) 3 3 One final thing: For and ideal gas, cV = 2 Nk = 2 nR where N is the number of molecules, n is the number of moles, k is Boltzman’s constant, and R is the universal gas constant. ∂H 5 Also cP = ∂T P = 2 Nk. In general, for gases with ideal-like behavior, but internal degrees of freedom (rotations, stretches), cP = cV + Nk. 5 Figure 5: Diatomic molecule Statistical Mechanics of systems at constant T and V The probability a state of the system is occupied is proportional to e−Es/kT where E is the energy of the state. The expectation value of the energy (which gives us U, the internal energy, is P −Es/kT s Ese P −Es/kT s e If the system is made up of identical particles or a set of identical smaller systems (a bunch of harmonic oscillators), we can just find the expected energy of the particle of a smaller system, and multiply by N. A couple of examples. Consider the two level system, a bunch of particles can have energy O or E. The expectation value of the energy of N particles is N Oe−O/kT + Ee−E/kT < E >= e−O/kT + e−E/kT NEe−E/kT E U = 1+e−E/kT ; as T → ∞,U → N 2 . The average particle energy is halfway between the ground state and excited state. If U E > , (28) N 2 then e−E/kT 1 > , (29) 1 + e−E/kT 2 and 1 1 > −→ eE/kT < 1 and T < 0 (30) eE/kT + 1 2 6 Figure 6: number microstates vs. U The two level problem is a classic problem which leads to negative temperatures. To see why, let’s look at it a different way. Let’s plot the number of configurations of the system that will give a particular energy. If the total energy is 0, there is only one configuration, and in the ground state. Same for U = NE so the plot looks like Fig. 6. Here, S = k ln(Ω) ∂S 1 and dU = T dS − P dV , so ∂U = T , where Ω is the multiplicity. Refer to Fig. 6. When S is increasing, T is positive. When S is decreasing, T is negative. This a characteristic of any system with a maximum total energy: the plot of the number of microstates vs. energy will have a maximum, and thus negative temperatures.
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