Outline Review Example Problem 1. Thermodynamics. Review and Example Problems: Part-2. X Bai. SDSMT, Physics. Fall 2014

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1 Review and Example Problems: Part- SDSMT, Physics Fall 014

2 1 Review Example Problem 1 Exponents of phase transformation

3 : contents 1 Basic Concepts: Temperature, Work, Energy, Thermal systems, Ideal Gas, etc. The Second Law of 3 Interacting systems: The connection between the microscopic and the macroscopic 4 Heat Engines and Refrigerators 5 Chemical

4 1. Equilibrium Equilibrium Thermal Mechanical Diffusive Chemical Exchanges Thermal energy Volume Particles Chemical reactions H O H + + OH H SO 4 H + + SO 4. Thermal systems Isolated system: No interaction with environment, no matter and energy exchange. Closed system: No matter exchange, has energy exchange. Open system: Has matter and energy exchange.

5 Ideal Gas, Carnot cycle, etc. 3. Ideal Gas PV = nrt PV = NkT n: number of moles of gas. R constant R = Carnot cycle J mol K Step 1: isothermal expansion, (V 1, P 1, T h ) (V, P, T h ) Step : adiabatic expansion, (V, P, T h ) (V 3, P 3, T c) Step 3: isothermal compression, (V 3, P 3, T c) (V 4, P 4, T c) Step 4: adiabatic compression, (V 4, P 4, T c) (V 1, P 1, T h ) 5. Equipartition theorem For a system with N particles, each with f DoF, and there is NO other non-quadratic temperature-dependent forms of energy, the total thermal energy in the system is U thermal = Nf 1 kt

6 The Second Law of 1. Macrostate, Microstate, Multiplicity, Applications: Einstein Solid, Ideal Gas,. Entropy: S = klnω The famous [ ( Sackur-Tetrode Equation for Ideal Gas: ( V S kn ln 4πmU ) ) ] 3/ + 5 N 3Nh 3. The Second Law of Any large system in equilibrium will be found in the macrostate with the greatest muitiplicity (aside from fluctuations that are normally too small to measure). If we start from a non-equilibrium system, it means: Multiplicity tends to increase as time goes. Isolated systems evolve from a organized system to a more randomized system!

Interacting systems: The connection between the

systems To understand heat flow and the evolution of a

consider two systems that interact with each other with

7 Interacting systems: The connection between the microscopic and the macroscopic Interacting thermal systems To understand heat flow and the evolution of a thermal system (reversible or irreversible), we need to consider two systems that interact with each other with heat and/or matter exchanges. Solid A N A, q A, U A, Ω A Solid B N B, q B, Ω B, U B

8 Interacting systems: The connection between the microscopic and the macroscopic - cnt. 1. Interacting thermal systems: Staring point Ω combined = Ω 1 Ω S combined = S 1 + S Away from equilibrium P a+b P a P b At equilibrium P a+b P a P b

9 Interacting systems: The connection between the microscopic and the macroscopic - cnt.. What governs a non-isolated system? The typical method: Starting from the total entropy of the entire system. S total = S + S environment ds total = ds + ds environment Using conditions: T e = T du + du e = 0 One gets: ds total = ds 1 T du = 1 T (du TdS) = 1 T df

10 Interacting systems: The connection between the microscopic and the macroscopic - cnt.

11 Heat Engines and Refrigerators Q h W Q c The efficiency of an engine: ɛ 1 Tc T h

12 Heat Engines and Refrigerators - cnt. hot reservoir, T H Q ΔS H = T H H Q H heat entropy heat work W Q ΔS C = T C C Q C cold reservoir, T C The coefficient of performance (COP): COP 1 = Tc T h /T c 1 T h T c

13 Chemical : Thermal quantities and potentials Table: Quantities that govern the thermal processes Process What governs the process constant E and V Entropy S constant T and V Helmholtz Free Energy F constant T and P Gibbs Free Energy G Table: Thermodynamic quantities Extensive quantities Intensive quantities Do change when the Do NOT change when the amount of matter changes amount of matter changes V, N, S, U, H, F, G, mass T, P, ρ, µ (chemical potential)

14 Chemical : Thermal quantities and potentials Potential variables identity U(S, V, N) S, V, N du = TdS PdV + µdn H(S, P, N) S, P, N dh = TdS + VdP + µdn F (T, V, N) V, T, N df = SdT PdV + µdn G(T, P, N) P, T, N dg = SdT + VdP + µdn

15 Chemical : Thermal quantities and potentials P T

Clausius-Claypeyron Relation What determines the phase boundary

(18 1888) German physicist & mathemabcian, one of the central

(1799 1864) French engineer and physicist, one of the founders

16 Clausius-Claypeyron Relation What determines the phase boundary is the condition? - The Clausius-Clapeyron Relation. ( ) German physicist & mathemabcian, one of the central founders of the science of thermodynamics. ( ) French engineer and physicist, one of the founders of thermodynamics. ΔT 1 ΔP 1 ΔP ΔT On the red line: ΔP 1 = ΔP ΔT 1 = ΔT dp dt = δs δv dp dt = L T δv

van der Waals Model short-distance repulsion U(r) 4 3 1 0-1 - -3 1.5.0.5 3.0 3.5 4.

17 van der Waals Model short-distance repulsion U(r) r distance Johannes van der Waals, 1873 long-distance attraction Using critical variables V c, T c, P c to normalize P, V, T : t = T /T c, p = P/P c, v = V /V c, the vdw equation becomes p = 8t 3 3v 1 v The vdw equation in terms of reduced variables.

18 Maxwell construction plots P/P c V/V c

19 Vapor pressure and critical point Understand phase transition using van der Waals Model. (1) vapor pressure P v for each T: at which the liquid-gas transformation takes place () critical pressure P c, critical temperature T c, and critical volume V c: the critical point. P c P v V c T c

20 Final Exam Time: Start at 9:00 AM MT/10:00 AM CT, Tuesday, October 14, 014. Due at 1:00 PM MT/:00 PM CT (4 hours), Tuesday, October 14, 014. Location: Not required to be in class.

21 Problem 1 Review Exercise 01: This series of problems are dedicated to the study of the behavior of van der Waals fluid near the critical point. (1) Expand the vdw equation in a Taylor series in (V V c), keeping terms through order (V V c) 3. Prove that, for T close enough to T c, the quadratic term in (V V c) becomes negligible compared to the others and may be dropped. Answer: To be simple, we take the vdw equation in terms of reduced variables p = 8t 3. When we use reduced quantities, v = V /V 3v 1 v c, the expansion is in terms of (V /V c V c/v c) = (v 1). Taylor expansion is: f (x 0 + h) = f (x 0) + hf (x 0) + h f (x! 0) +... The derivatives we need: p = 4t(3v v 1) + 6v 3 p = 144t(3v 1) 3 18v 4 v 3 p v 3 = 196t(3v 1) 4 + 7v 5

22 Problem 1 (1) - cnt. p(v) p(1) + p v 1 (v 1) + 1 p v 1 (v 1) p v 3 1 (v 1) 3 p(v) (4t 3) 6(t 1)(v 1) + 9(t 1)(v 1) 3 (9t 8)(v 1)3 T being close enough to T c means T T c 0, or in reduced variables, t 1 0, let s note this as t 1 O(0). When v 1 t 1: p(v) (4t 3) 6O (0) + 9O 3 (0) 3 (9t 8)O3 (0) When (v 1) t 1: p(v) (4t 3) 6O(0)(v 1) + 9O (0) 3 (9t 8)(v 1)O(0) So, 9(t 1)(v 1) is always smallest, which we can drop. Then we have: p(v) (4t 3) 6(t 1)(v 1) 3 (9t 8)(v 1)3

23 Problem 1 - cnt. () The expression for P(V ) is antisymmetric about the point V = V c. Use this fact to find an approximate formula for the vapor pressure as a function of temperature. Answer: p(v) (4t 3) 6(t 1)(v 1) 3 (9t 8)(v 1)3 When (v 1) > 0, p(v) (4t 3) 6(t 1) (v 1) 3 (9t 8) (v 1) 3 When (v 1) < 0, p(v) (4t 3) + 6(t 1) (v 1) + 3 (9t 8) (v 1) 3 So, the 6(t 1)(v 1) 3 (9t 8)(v 1)3 is antisymmetric. The pressure of the phase transition is p(v) = (4t 3) - This line split the enclosed area into two parts with the same area.

24 Problem 1 - cnt. (3) Find an expression for the difference between volume between the gas and the liquid phases at the vapor pressure. You should find (V g V l ) (T c T ) β, where β is a critical exponent. Experiments show that β has a universal value of about 1/3, but vdw model predicts a larger number. Answer: The volumes of the liquid and gas at the transition pressure are those at the transition pressure p = 4t 3. That is, (4t 3) = (4t 3) 6(t 1)(v 1) 3 (9t 8)(v 1)3 6(t 1) 3 (9t 8)(v 1) = 0 At T T c, or t 1, 6(t 1) 3 (v 1) = 0 v 1 = 1 t or, v = 1 1 t v g v l = 4 1 t So, β is 1.

25 Problem 1 - cnt. (4) Use the previous result to calculate the latent heat of the transformation as a function of T, and sketch this function. Answer: The Clausius-Clapeyron relation: L = T (V g V l ) dp dt L = tp cv c 4 1 t 4 L = 16P cv ct 1 t L = kntct 1 t L knt c 6t 1 t dp = L dt T δv = tpcvc(vg v l) dp dt So, the latent heat is

26 Problem 1 (4) - cnt. L knt c L knt c 6t 1 t by red line. 6 1 t by black line. 6.0*sqrt(1-x) L/kNT c t

27 Problem 1 - cnt. (5) The shape of the T = T c isotherm defines another critical exponent, called δ : (P P c) (V V c) δ. calculate δ in vdw model. Answer: When t = 1, from the Taylor expansion of the vdw equation: p(v) (4t 3) 6(t 1)(v 1) 3 (9t 8)(v 1)3 p(v) 1 3 (v 1)3 p(v) 1 3 (v 1)3 P P c 3 (V Vc)3 So, δ = 3 Note: The experimental value of δ is typically around 4 and 5.

Outline Review Example Problem 1 Example Problem 2. Thermodynamics. Review and Example Problems. X Bai. SDSMT, Physics. Fall 2013

Outline Review Example Problem 1 Example Problem 2. Thermodynamics. Review and Example Problems. X Bai. SDSMT, Physics. Fall 2013 Review and Example Problems SDSMT, Physics Fall 013 1 Review Example Problem 1 Exponents of phase transformation 3 Example Problem Application of Thermodynamic Identity : contents 1 Basic Concepts: Temperature,