THERMODYNAMICS. Contents

Transcription

1 THERMODYAMICS EREST YEUG Contents Part 1. otes and Solutions for Thermal Physics by Ralh Baierlein 1 1. Background Heating and Temerature Some dilute gas relationshis The First Law of Thermodynamics Heat caacity 2 Problems 3 2. Phase Equilibrium Latent heat Conditions for coexistence 4 Part 2. otes and Solutions on Fundamentals of Thermodynamics, 8th Edition, by Claus Borgnakke, Richard E. Sonntag Proerties of a Pure Substance 4 Part 3. Thermodynamics Revisited 4 5. Heat Caacity 4 6. Phase Equilibrium 5 References 6 7. Code listings 7 8. Automatic generation of documentation 8 Abstract. Everything about thermodynamics. I also look at thermodynamics for engineers from a theoretical and mathematical hysicists oint of view. I would like to seek more cross-olination between hysicists and mathematicians and engineers in thermodynamics. Part 1. otes and Solutions for Thermal Physics by Ralh Baierlein [1] 1. Background 1.1. Heating and Temerature. Heating: kee in mind 3 different tyes of heating for energy exchange between two systems: 1 Heating by conduction - literal contact, molecules jiggle faster from molecules jiggling faster by bouncing off them 2 Heating by radiation - em waves from hot source strike and excite target 3 heating by convection - energy transort by flow erhas a fluid This all relates to Q 1.2. Some dilute gas relationshis. Pressure according to kinetic theory. i.e. some kinetic theory F force on area A due to molecules momentum transferred to wall er collision n number of collisions in time t Thus F = n t ow = 2mv x since = mv x mv x = 2mv x elastic collision with momentum conversation Date: 18 octobre Key words and hrases. Thermodynamics. 1

2 2 EREST YEUG v x ta is a volume, inside of which gas molecules can be within distance v x t toward the wall. number density of molecules Assume equal distribution of velocities: Thus 1 2 n = v x ta 1 2 P = F A = 2mv xv x ta 1 2 A t Suose v 2 = vx 2 + vy 2 + vz 2 = 3 vx 2 = d vx. 2 An emirical gas law. ow 1.3. The First Law of Thermodynamics. Consider W = Q du. Consider ath in M, γ, γ : R M = U, γt = Ut, t γ XM, γ = U U + Suose Q γ = Qtdt γ t = Qt γ. = mvx 2 = P = m d v2 = 2 1 d 2 m v2 P = emirical = 1 2 m v2 = d 2 du = Q W or Q = du + W W γ = d γ = = Q γ du γ = Q γ U dt = Qt γdt = = Q U interreted as work done by gas. Q is heat transferred to gas system. U is the dro in internal energy of gas system as it does work Heat caacity. So define C Q. or interret C In this case, U For the case of a monatomic gas, U = d 2 number of molecules. ow Q = Q, = Q d + Udt Q d as energy inut by heating at constant volume over ensuing change in temerature., U = d 2. C = d 2 Q = Λ d + C d Q = Λ d + C d = Q d = C d d Q d = Λ d d + C d d ow from the thermodynamic identity, Q = W + du = d + du, Q d = d d + du d and from emirical ideal gas law, = which defines a hyersurface on M, so then d + d = d = d d = d d Q d = d d + du d In the case of the monatomic gas, U = d 2, and so du = d 2 d = C d and so comaring all the equations above, one recovers C = + C = 2 + d 2 EY: I m curious to know how this all generalizes for C, C P heat caacities, regardless of the tye of molecule we consider. The adiabatic relation for a classical ideal gas. Consider the adiabatic exansion or contraction! of a classical ideal gas. This means that Q = 0; there is no heat exchange to or from the gas system. Recall Q = du + W. If Q = 0, and suosing W = d, then 0 = du + d. EY : Either by definition, or the thermodynamic identity, dσ = du + d, then C := U. My question is this: for manifold of thermodynamic states M, M = U,, i.e. U is a global coordinate and is a local coordinate. One can make a Legendre transformation such that M is arametrized by,, where is the temerature. In general, one should say that U = U, C M, and so du = U U d + d, du Ω1 M. However, for this adiabatic rocess, we want Q = 0 = du + W = du + d = C d + d which imlies that du = C d. What haened to the U d? Is it that in this adiabatic rocess, the internal energy of the gas system goes to either doing work exansion or increases due to work being done on it contraction, and is characterized comletely by a dro or increase in its temerature, resectively? And so du = C, and d comletely describes what s going on with work done or work done on it? evertheless, using the emirical ideal gas law, =, Consider a ath γ : R M Thus, γt = t, t 0 = C d + d = C d + d in M, so that γt = + XM. 0 = C v + so + C dt ln f i + C ln f i = 0 or ln C = const. ow = C P C = γ 1 C C which is true, assuming the emirical ideal gas law, the thermodynamic identity, and, surely, for the case of a monatomic gas. Thus f γ 1 f = i γ 1 i

3 Problems. Problem 4. Adiabatic comression. A diesel engine doesn t have a sark lug to ignite and exlode the fuel. Instead, the air in the cylinder is comressed so highly that the fuel ignites sontaneously when srayed into the cylinder. a f γ 1 f i γ 1 i f = = P f γ f P i γ i f i γ 1 i Run the Python scrit thermo.y to do the calculations. Here is some of the code from thermo.y for doing so one still needs to imort the necessary libraries: = 1 roomtem K = KCconv. subs T C, 2 0. rhs # room temerature in Kelvin Prob0104ans = adia t. subs gamma, subs f, 1. subs i, 1 5. subs t a u i, roomtem K # answer to Problem 4 o f Chater 1 Prob0104ans = Prob0104ans. l h s # K Prob0104ansC = s o l v e KCconv. subs T K, Prob0104ans, T C [ 0 ] # C s o l v e FCconv. subs T C, Prob0104ansC, T F [ 0 ] # F The final temerature is K or C or F b ow from the ideal gas law, which is obeyed at all thermodynamic states and so P f P i = P f P i = i f f i 2. Phase Equilibrium 2.1. Latent heat., T const., liquid gas, e.g. = 1 atm T = 373 K latent heat of vaorization L va by def. amount of energy sulied by heating. Q = du + W ɛ := U/ = average internal energy er molecule v := / = volume er molecule L va = dɛ + d dɛ γ = ɛ va ɛ liq dv γ = v va v liq > 0 if const., L va = dɛ + = dh, h := H/ = U+ EY : A better way to think about it is this: recall that Consider a ath γ M s.t. d γ = 0 d γ = 0 Q = du + W = du + d = dσ H = U + so dh = du + d + d = Q + d constant, Q γ = dh γ d γ = dh γ 0 = dh γ Q γ = dσ γ Q γ = dσ γ = σ g σ l THERMODYAMICS 3 Thus Q γ = s g s l = H g H l Q γ L s g s l = = 1 H g H l Latent heat versus heat caacity. Take slow, reversible rocess. ow σ C := σ C P := It s stated in Kittel and Kroemer 1980,. 166, Equation 37, Chater 6: Ideal Gas, Subsection Heat caacity [2], that σ U 1 C P = = + Suose σ = σ,. With dσ = du + W = dσ = du + d, P dσ = σ σ d + d = du + d σ dσ γ = + 0 = 1 du γ = 1 Then it s clear that σ ow suose σ = σ,. Consider also the enthaly, H = U +, and so ow for σ = σ,, So ow σ = σ, dσ = σ σ d + d = dh = P P U = C dh = du + d + d = dσ + d = dσ γ = d C := σ = U σ + 0 = 1 H U dh γ = du γ + d γ + d γ = σ U C = = + Kittel and Kroemer 1980 [2] argues that for ideal gas, U since U = U. = U = H 0 H so

4 4 EREST YEUG 2.2. Conditions for coexistence. Sec of Baierlein 1999 [1]. Recall G = F + = U + σ = G,, ow G = G,, va, liq = dg = G d va + G d liq = µ va 1 + µ liq 1 = 0 va liq = µ va, = µ liq, From Kittel and Kroemer 1980 [2], Examle : atoms in a box, Chater 3: Boltzmann Distribution and Helmholtz Free Energy, state of energy ɛ α 1 + ɛ β ɛ ξ, α, β,... ξ denote orbital indces of atoms in successive boxes. each entry occurs! times in Z1 EY: ! ways to fill α, β... ξ orbitals with distinguishable articles. Thus, Z = 1! Z 1 = 1! n Q Part 2. otes and Solutions on Fundamentals of Thermodynamics, 8th Edition, by Claus Borgnakke, Richard E. Sonntag [4] Proerties of a Pure Substance EY : Is the word vaor the same as gas? vaour, gaz, gas, vaore For coexistence equilibrium, µ g 0, 0 = µ l 0, 0 and and so 2 µ g 0 + d, 0 + d = µ l 0 + d, 0 + d d d = L v where v, the so-called vaor ressure equation or Clausius-Claeyron equation. For 2 aroximations, v = v g v l v g = g g Second, if L constant, and idealize vaor as ideal gas, = so d d = 3 = 0 ex L 0 / or ln 0 = L 0 cf. Ch. 10 Phase Transformations, , Derivation of the Coexistence Curve, ersus of Kittel and Kroemer 1980 [2]. Eq. 3 exlains the shae of the coexistence curve between solid and gas vaor sublimation and liquid and gas vaor vaorization; vaorization curve. Saturation is this = coexistence curve. Isotherms, Isothermals Recall G = F + F = U σ So then G = G, G = G,, and so dg = df + d + d = du dσ σd + d + d = σd + d dg = σd + d L 2 /. where the latter statement is when we include article transfer, so that For the ideal gas: For the an der Waals gas nq G,, = ln n dg = σd + d + µd for µ = µ, so then F, = F = where n Q = nq ln n 3/2 M 2π n / = nq = ln n G M = µ = ln, 2π 2 F vdw = ln F = =, G,, = b 2 2 a ln nq b evertheless, consider, when considering isotherms, isothermals, dg = σd + d + µd M = ln 2π 2 3/ a b 2 a 2 nq b Consider a ath γ on constant, constant total number of articles, and so dg γ = d γ = G g G l = d Part 3. Thermodynamics Revisited 5. Heat Caacity γ + 1 From Kittel and Kroemer 1980 [2], , Chater 6: Ideal Gas, Heat Caacity, Let c Σ s.t. dċ = 0 constant ressure. And so for c =, 0, ċ = T Σ. Q = Q, = Λ d + C d = du + W = du + d Qċ = 0 + C dċ = dσċ = duċ + d ċ = C = for heat caacity at constant ressure is larger than C exand volume of gas against constant ressure. ow recall for enthaly 3/2 σ U = + because additional heat must be added to erform the work needed to H = U + H = Hσ,

5 Then dh = du + d + d = Q + d Thus, for c Σ, dċ = 0 constant ressure. Hence H dhċ = = Qċ + 0 = C Hence, H Q C = = : Phase Equilibrium : Phase Diagram : Phase Coexistence Curve : Antoine Equation arameters 6. Phase Equilibrium For coexistence equilibrium, µ g 0, 0 = µ l 0, 0 and µ g 0 + d, 0 + d = µ l 0 + d, 0 + d, and so d d = L v v. Make the aroximation that v v g v l v g = g g If L constant, 4 = 0 ex L 0 / or ln and idealize vaor as ideal gas, =, so d d = 0 = L 0 Consider the Antoine Equation Parameters given in the IST ational Institute of Standards and Technology Chemistry WebBook 1 L 2 /. THERMODYAMICS 5 with 5 log 10 P = A B/T + C Then P = 10 A= B T +C = 10 A 10 B T +C B = 10 A ex ln 10 T + C ow D T +C = D T DC T T +C, and so B ln 10C B ln 10 P = 10 A ex ex T T + C T Consider how much the ressure is changed due to this C arameter. Consider P 0, P 1, defined as such: B ln 10 B ln 10C B ln 10 P 0 = 10 A ex P 1 = 10 A ex ex = P 0 B ln 10C = ex T T T + C T P 1 T T + C So a deviation can be estimated from 1 P0 P 1. Oen u thermochem.y. The saturation curve or coexistence curve for, for examle, oxygen, from liquid to gas, can be reroduced. One needs to inut in the Antoine arameters from the IST website 2 >>> O2coec = CoexistCurve , , >>> l o t O2coec. c u r v e S I. rhs, T, , u t x := ux, t = Ua, t =: U t a U t >>> O2coec. c u r v e S I. r h s. subs T, >>> O2coec. c u r v e S I. r h s. subs T, >>> O2coec. c u r v e S I. r h s. subs T, a gt = g t g 1 t = g t xa, t = x 1 Phase change data for Oxygen 2 htt://webbook.nist.gov/cgi/cbook.cgi?id=c &mask=4#thermo-phase

6 6 EREST YEUG References [1] Ralh Baierlein. Thermal Physics Cambridge University Press July 28, 1999, ISB-13: [2] Charles Kittel, Herbert Kroemer, Thermal Physics, W. H. Freeman; Second Edition edition, ISB-13: [3] Bernard F. Schutz, Geometrical Methods of Mathematical Physics, Cambridge University Press, ISB-13: [4] Claus Borgnakke, Richard E. Sonntag. Fundamentals of Thermodynamics, 8th Edition, Wiley, December 26, ISB-13: There is a Third Edition of T. Frankel s The Geometry of Physics [?], but I don t have the funds to urchase the book about $ 71 US dollars, with sales tax. It would be nice to have the hardcoy text to see new udates and to use for research, as the second edition allowed me to formulate fluid mechanics and elasticity in a covariant manner. Please hel me out and donate at ernestyalumni.tilt.com.

7 THERMODYAMICS 7 7. Code listings

8 8 EREST YEUG 8. Automatic generation of documentation Demontration using eydoc: eydoc --df -o /home/fnielsen/tm/eydoc/ --name RBBase wikiedia/ai.y This examle does not use brede_str_nmf but another more well-documented module called ai.y that are used to download material from Wikiedia.