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by: Jeremy Dao

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# PHYS 224 Final Exam Study Guide PHYS 224 A

Marketplace > University of Washington > PHYS 224 A > PHYS 224 Final Exam Study Guide
Jeremy Dao
UW
GPA 3.72

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Study Guide for the final exam: goes over all the material covered since midterm 2, including heat engines and refrigerators, free energy, phase equilibria and diagrams, vapor pressure, Boltzmann f...
COURSE
THERMAL PHYSICS (NW)
PROF.
COBDEN,DAVID
TYPE
Study Guide
PAGES
11
WORDS
CONCEPTS
thermal physics, Heat Engines, Refrigerators, free energy, phase diagrams, vapor pressure, Boltzmann factor, Partitions Function
KARMA
50 ?

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This 11 page Study Guide was uploaded by Jeremy Dao on Monday June 6, 2016. The Study Guide belongs to PHYS 224 A at University of Washington taught by COBDEN,DAVID in Spring 2016. Since its upload, it has received 70 views.

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Date Created: 06/06/16
Physics 224 Final Exam Study Guide ▯ Heat Engines: A "heat engine" converts heat (from eg. combustion, nuclear, solar, geothermal, etc.) – Most engines work in cycles (ex. 2 turns of the camshaft for the internal com- bustion engine in a car) →Q ;Qh;W lre per cycle – First Law for one cycle: ▯U = 0 = Q ▯ Qh▯ W l ) W = Q ▯ Qh l Ql i.e. a fraction Qh of heat from hot reservoir is dumped in the cold one 1 2 ▯ Efﬁciency – Deﬁne eﬃciency e as: e = W = Q h Q l= 1 ▯ Q l Qh Qh Qh fraction of h converted to useful work – Second Law →▯S totalfor one cycle) ▯ 0 ▯S engine 0 since the engine goes back to the same state ) ▯S h ▯S ▯l0 Q Q Q T T ) ▯ h + l▯ 0 ! l ▯ l ! e ▯ 1 ▯ l 0 < e < 1 Th Tl Q h Th Th Q ▯ Minimum change in entropy happens when quasistatiT,is lower bound on ▯S ▯ This places a fundamental limit on eﬃciency ▯ Higher the ratio oh T tl T , the better ▯ For ﬁnite temperatures, no engine can completely convert heat to work ▯ Example: An engine run between boiling and freezing water, generate elec- trical power P-1W. What is the min. rate X that heat must be generated (in hot water)? Let period of engine be ▯t W = P▯t Q h X▯t W P 273K = e = ▯ 1 ▯ = 0:27 Q h X 373K ) X = P ▯ 1W ▯ 3:7W e 0:27 This engine is very ineﬃcient, putting in 3.7W of heat to get out 1W of work. Can do better using higher T (eg. hot steam) h ▯ Refrigerators: (aka heat pump) – A refrigerator is basically a heat engine run backwards 3 – By the First Law: W = Qh▯ Q L By the Second Law: ▯S univ▯ 0 Q Q ) h▯ L ▯ 0 Th TL Q h Th ) Q ▯ T opposite to engine L L ▯ Coefﬁcient of Performance: (C.O.P) eﬃciency for refrigerator – QL QL 1 1 C:O:P = = = Qh ▯ Th ▯ C:O:P W Q h Q L QL ▯ 1 TL ▯ 1 T – Smaller the ratTo, the better L – Example: If we pump heat from freezing to boiling water: 1 COP = 373▯ 1 ▯ 3:7 273 therefore 1W from outlet produces ▯ 3.7 W of cooling ▯ Model heat engines: – Heat engines/fridges always have a working substance (usually a ﬂuid) that ex- pands/contracts, does work, absorbs heat – Max eﬃciency or COP is achieved if the engine is reversible, meaning quasistatic, and not heat ﬂow through a ﬁnite temperature diﬀerence – the simplest (classic) model of a reversible engine is a Carnot engine, which uses a portion of ideal gas as a working substance – The engine is placed in contact with the hot reservoir and allowed to expand (isothermal). Then it is placed between the reservoirs and expands until it cools down from h to TL(adiabatic). Then it is placed into contact with the cold reservoir and compressed (isothermal). Finally, the engine is removed and compresses until it gets backhto T (adiabatic). – These expansions and compressions can be plotted on a PV diagram: 4 – Recall Q = T▯S. Therefore, we can calculate that: (watch signs) ▯ ▯ ▯ ▯ ▯W = ▯Q = NkT ln V2 W = Q = ▯NkT ln V4 a h h V1 c L L V3 From the adiabatic expansion/compression: ▯1 V1 V3 PV = constant ) TV = const: ! = V2 V4 QL TL Therefore: e = 1 ▯ = 1 ▯ Q h Th – This is the upper limit on eﬃciency for any engine. Real engines are irreversible and fast, so their eﬃciency will be less ▯Real engines: – Internal Combustion Engines: ▯ ▯ ▯1 W V1 e = Q = 1 ▯ V 2 where 1 is the compression ratio 2 – Real Refrigerators: ▯ Circulated ﬂuid coolant, relies on throttling (ﬂuid moves from high pres- sure to low pressure) ▯ ▯ @T ▯ ▯H = 0 ! ▯(U+PV ) = 0, therefore, cooling occurs when > 0 @P H @T ▯ For ideal gase@P is zero, but real gases will sometimes give cooling 5 ▯ Enthalpy – Deﬁne Enthalpy as: H = U + PdV – Useful for: gives heat ﬂow for process (at constant P), cooling in fridge – ▯H = ▯U + P▯V = Q + (▯P▯V ) + W other = Q + W other – H is an example of a thermodynamic potential, is a function of state ▯ Free Energy – Notes that "free" means "available", how much useful energy can you get out – Helmholtz’s free energyF = U ▯ TS – Gibb’s free energyG = U ▯ TS + PV – For a process at constant P and T: ▯G = ▯U ▯ T▯S + P▯V = Q + W other P▯V ▯ T▯S + P▯V = Q ▯ T▯S + W other By the second law: ▯S univ 0 ▯S + ▯S ▯ 0 system surroundings Q ▯S ▯ ▯ 0 T ) Q ▯ T▯S ▯ 0 ) ▯G ▯ W other – ▯G is the minimum amount of work needed to induce the process (not counting the "free work" from the atmosphere) – If ▯G < 0;W otheran be < 0. Therefore the process can be made to do useful workW useful ▯W other ▯▯G – ▯▯G = G ▯ G is the "available energy", the max amount of useful work you i f can get out ▯ Thermodynamic potentials – Natural Variables U dU = TdS ▯ PdV + ▯dN U(S;V;N) H = U + PV dH = TdS + V dP + ▯dN H(S;P;N) G = U + PV ▯ TS dG = ▯SdT + V dP + ▯dN G(T;P;N) F = U ▯ TS dF = ▯SdT ▯ PdV + ▯dN F(T;V;N) – Each potential has diﬀerent natural independent variables. In calculations, it’s easies to work with the potential whose natural variables are given. 6 – Eg. if T, P, N are speciﬁed ! use G ▯ ▯ ▯ ▯ ▯ ▯ dG = @G dT + @G dP + @G dN @T @P @N P;N T;N P;T ▯ ▯ ▯ ▯ ▯ ▯ @G @G @G S = ▯ @T V = @P ▯ = @N P;N T;N T;P ▯ Rela▯ion▯between G and ▯ – @G = ▯(T;P) because ▯ can’t depend on N, since ▯ is intensive and N is @N T;P extensive – Integrate w.r.t N ! G = N▯(T;P) + f(T;P) Clearly for N = 0;G = 0 ) f(T;P) = 0 – Therefore G = N▯ , ▯ is the Gibbs free energy per particle ▯ Equilibrium and free energy – For a process at constant T and P, we saw tfat Gi▯ G = ▯G ▯other – For a spontaneous procesother 0, therefore ▯G ▯ 0 , therefore G can only decrease – In equilibrium, G must be minimized – Minimizing G corresponds to maximizing the entropy of the universe (this can also be seen from the derivation of ▯G, see Free Energy section) ▯ Maxwell Relations: Partial derivatives commute and only 3 independent variables – Example: ▯ ▯ ▯ ▯ @ ▯ @ ▯ @ ▯ @ ▯ ▯ ▯ G = ▯ ▯ G @P T;N @T P;N @T P;N @P T;N # # ▯S V ▯ @S▯ ▯ @V ▯ ) ▯ = @P T;N @T P;N and many, many, more equations ▯ @P▯ ▯ @T ▯ – Note that ﬂipping the derivatives upside is ok too: = @S T;N @V P;N ▯ Phase Equilibria and Phase Diagrams – All substances can exist in multiple phases: solids-liquids-vapor-plasma – What is the equilibrium phase at a give T and P: Eqm ! max S total minimize G for system G = N▯ therefore for a given amount of stuﬀ N ! minimize ▯ So if v < ▯Land ▯s, vapor will be stable 7 – Why are there phase transitions? Generally v (P;T);L (P;T);▯s(P;T) are diﬀerent functions, therefore at a give P, T, the phase with lowest ▯ will be stable – A phase diagram is a map showing which phase is stable as function of P, T Generic Phase Diagram (simple substance) Line on the phase diagram between phase A and B is the phase boundary along which ▯A= ▯ B Cross a phase boundary ! phase transition (i.e. melting, condensation, etc.) – Two kinds of special points: (1) Critical Point cT cP ) Liquid and vapor are both ﬂuids, the distinction is only by density. Liquid can be converted to vapor smoothly with no phase transition, by going around the critical point. (2) Triple Point (tr;tr) where 3 phase boundaries meet Here ▯s= ▯ L v Only at this T and P can all 3 phases be stable at the same time – dG = ▯SdT + V dP + ▯dN = d(▯N) S V ) d▯ = ▯ dT + dP N N ▯ ▯ ▯ ▯ @▯ S @▯ V = ▯ < 0 = > 0 @T P N @P T N So ▯ always decreases as T increases and ▯ always increases as P increases. We can use this to predict phase transitions as we change T or P. 8 For a ﬁxed pressure, ▯ decreases as T increases. Entropy S of the vapor is larger (more degrees of freedom in gas), so its slope is more negative. This is why as temperature is increased, the phase goes from liquid to vapor. For ﬁxed temperature, ▯ increases as P increases. For a gas, volume V is larger, so its slope is larger. Thus, as pressure is increased, phase goes from vapor to liquid. ▯ Shape of Phase Boundary:AB(T) between phases A;B – Consider moving along boundary. Change T by ▯T;P by ▯P so to stay on phase boundary. – Along boundary: ▯A= ▯ B▯▯ A ▯▯ B ) ▯S AT + V AP = ▯S BT + V BP dP ▯P AB = dT ▯T ▯ ▯ @P S A S B ▯S L = = = @T boundaryV A V B ▯V T▯V Where L is the latent heat, which equals T▯S. This is known as the Clasius-Clapeyron Equation – Notes: 9 ▯ ▯ (1) Normally, if V > V , then S > S . Therefore @P > 0, so T B A B A @T boundary boil increases with P. An exception to this is water V liq V solidice ﬂoats) S liq> S solid Thus for water, melting temperature decreases as P increases (2) If L = 0;▯S = 0;▯V = 0 we get a "second-order" phase transition e.g. magnetic ! non-magnetic, superconducting ! normal metal ▯ Vapor Pressure – Consider liquid water under air. Air contains some water vapor. – In equilibrium: ▯ L ▯ atvtemp T. – Partial pressure: P w P LV (T) = P satT) saturated vapor pressure Pw – Deﬁne the relative humidity as RH = Psat(is a percentage) If RH < 100% P < P ! ▯ < ▯ !water will evaporate w sat v w If RH > 100% ▯ > v ! wawer will condense. ▯ The Boltzmann factor – Consider small, well-deﬁned system (object), e.g. piece of solid, or molecule, etc. in thermal equilibrium with a large reservoir (surroundings) – From quantum mechanics, we know that the object has discrete allowed energies (its spectrum) quantum states: i = 0;1;2;::: E < 0 < E 1 ::: 2 – Deﬁne P =iprobability of ﬁnding object in state i Pi/ =imultiplicity of object and reservoir when object is state i = R imultiplicity of reservoir only b/c state of object is known = e SRik S R = k ln R i i 10 – For 2 state (i 6= j): Pj Rj SRj▯SRik ▯SR=k P = = e = e ▯ i ▯ Ri @S 1 1 ▯S R R ▯E R (▯▯E) = ▯ (E ▯ j ) i @E R T T V;N Pi Ej▯Ei Ei ) = e kT ! P i e ▯ kT Pj ▯ Absolute Probability and the Partition Function 1 Ei – Let P i e▯kT , where Z is a number, determined by normalization. Z X 1 X Ei X i Pi= 1 ! e▯ kT= 1 ! Z = e▯kT Z i i i – Z is the main calculational tool in statistical mechanics – All equilibrium properties can be derived from Z – For a ﬂuid, Z(N;V;T) ▯ Freeze-out: For low enough temperature,1E ▯ 0 >> kT P 0 1 object is frozen P i6=0 0 in ground state ▯Average value of any property X of object – X X E X =< x >= X P = 1 e▯kT i i Z i i 1 X 1X Covariace ▯ = so Z = e▯▯Ei; X = Xie▯▯Ei kT Z i i – Example: Internal Energy 1X @ U = E = Eie▯▯Ei= ▯ (lnZ) Z @▯ i @ dT @ 1 @ 2 @ = = ▯ 2 ! U = kT (lnZ) @▯ d▯ @T kT @T @T 11 ▯ Free Energy F = U ▯ TS ▯ ▯ ▯ ▯ @F @F dF = ▯SdT ▯ PdV + ▯dN ▯ S = @T F = U + T @T ▯ ▯ V;N V;N @F @ F ▯ T = U = kT 2 (lnZ) @T V;N @T The solution to this diﬀerential equation is F = ▯kT lnZ From this, we can get everything else: ▯ ▯ ▯ ▯ S = ▯ @F = @ ▯ (kT lnZ) G = N▯ = ▯N @ ▯ (kT lnZ) ▯@T ▯V;N @T ▯;N ▯ ▯ @▯ ▯;V ▯ P = ▯ @V = @V▯ (kT lnZ) C V @T = @T▯ kT 2@T(lnZ) ▯@F ▯T;N @ ▯;N V V ▯ = ▯ @N T;V= @N T;V(kT lnZ) ▯ Equipartition Theorem One quadratic d.o.f q E = cq = E(q) @ P ▯▯Ei – We want E = ▯ @▯lnZ Z = ie – From quantum mechanics, q has discrete, equally spaced (by ▯q) values E = E(q ) q = n▯q;n = integer i n X1 Z +1 1 ▯▯E(qn) 1 ▯▯E(q) Z = ▯q e ▯q = ▯q e dq for small ▯q ▯1 ▯1 Z 1 r = 1 e▯▯cqdq = 1 ▯ Gaussian integral ▯q ▯q c▯ ▯1 1 lnZ = const + ▯▯ 2 ▯ ▯ @ 1 1 kT E = ▯ ▯ ln▯ = = @▯ 2 2▯ 2

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