New User Special Price Expires in

Let's log you in.

Sign in with Facebook


Don't have a StudySoup account? Create one here!


Create a StudySoup account

Be part of our community, it's free to join!

Sign up with Facebook


Create your account
By creating an account you agree to StudySoup's terms and conditions and privacy policy

Already have a StudySoup account? Login here

PHYS224 Midterm 2 Study guide

by: Jeremy Dao

PHYS224 Midterm 2 Study guide PHYS 224 A

Marketplace > University of Washington > PHYS 224 A > PHYS224 Midterm 2 Study guide
Jeremy Dao
GPA 3.72

Preview These Notes for FREE

Get a free preview of these Notes, just enter your email below.

Unlock Preview
Unlock Preview

Preview these materials now for free

Why put in your email? Get access to more of this material and other relevant free materials for your school

View Preview

About this Document

Study guide for midterm 2: Covers all the material starting after midterm 1. Highlights important equations, including their basic derivations. Covers Entropy, the S-T equation, Mixtures, temperatu...
Study Guide
thermal physics, entropy, Heat Engines
50 ?





Popular in Department

This 10 page Study Guide was uploaded by Jeremy Dao on Saturday May 14, 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 94 views.


Reviews for PHYS224 Midterm 2 Study guide


Report this Material


What is Karma?


Karma is the currency of StudySoup.

You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 05/14/16
Physics 224 Midterm 2 Study Guide ▯ Reversibility – If a process increases total entropy S of the universe, it’s irreversible. THe reverse process would decrease entropy, which is not allowed by the 2nd law of thermo. – If S remains constant, it is reversible. – A reversible process for a gas must be: (1) Quasistatic, so P and T are well-defined throughout (2) Involve no heat flow across finite temp. difference (heat flow increases entropy) Note that these are only necessary conditions, and do NOT guarantee that the process is reversible. ▯ Sackur-Tetrode Equation: Entropy of an ideal monatomic gas. " # ! V ▯ 4▯m U ▯ 2 5 S = Nk ln + N 3h N 2 – Entropy S(N;U;V ) is a function of state – Expansion from V to V > V with T = T (therefore U = U ) i f i f i f i ▯ ▯ V f ▯S = S ▯fS =iNk ln V i which holds no matter the process of expansion. – Case 1: isothermal and quasistatic (reversible) Z Q = ▯U ▯ W = 0 + PdV Z Vi ▯ ▯ = NkT dV = NkT ln Vf = T▯S V P Vi i Q = T▯S 1 2 Therefore the entropy of the gas increases, so why is it reversible? Reversible means that ▯S of the universe = 0 for the whole process Here, ▯Suniv= ▯S gas ▯S reservoirS reservoir▯ Q because the entropy of T the reservoir decreases as heat flows out, so: Q Q ▯S univ= + T ▯ T = 0 – Case 2: "Free expansion" Gas expands into vacuum ! no work done: W = 0 ▯U = Q + W = 0 ) T f T =iT ▯ V ▯ ) ▯S = S f S =iNk ln f V i ▯ ▯ Vf ▯S univ Nk ln V > 0 ! IRREVERSIBLE i Can NOT plot on a PV-diagram: Process is not quasistatic, so P is not well- defined ▯ Mixing Consider two ideal gases A and B, again in a thermally isolated box separated by a partition: Since A and B are different ideal gases, they don’t interact and don’t collide. So the situation is the same as free expansion for A and B, so temperature doesn’t change. 3 Since the gases don’t interact, entropy is extensive: ▯S univ= ▯S A ▯S B 2V 2V = N A ln + N B ln = +2Nkln2 > 0 ! IRREVERSIBLE V V Even though quasistatic and no heat flow, still irreversible. Mixing increases entropy even if slow and isothermal ▯ Gibbs paradox: What if A and B are the same? – THen ▯S = 0 (if gases are same, no mixing) – Conclusion: Either gases are the same (indistinguishable) or they’re not! (dis- tinguishable) The two situations are fundamentally different due to quantum mechanics. (N! on bottom of multiplicity of ideal gas). ▯ Maxwell’s demon – Start with uniform mixture of gases A and B. A little demon selectively operates valve to let through gas A moving to the right, and gas B moving to the left. Eventually, the gases will unmix. – Q: Why does S univnot decrease? A: The demon must generate entropy. – Its thought to be because it gains info each time it measures, which eventually must be lost. Can show that each bit of info erased generates entropy k ln2. ▯ The true definition of temperature – Consider 2 systems in thermal contact. Allow small amount of energy dU to flow. U + U = constant dU = dU = ▯dU A B A B = A B S = S A S B 4 In eqm. and S are maximal as a function of U (since it is the only thing that A can vary) @S = 0 ) @ (S + S ) = 0 ) @SA = ▯ @SB = @SB @U A @UA A B @UA @UA @UB Therefor@U must correspond to T. Define 1▯ @S T @U – If S increases faster than U, then increases faster as system gains energy →corresponds to colder, lower T. – Colder system is more likely to gain entropy. – Heat flows b/c the system with lower T intotalaster than the one with higher T decreases it. ▯ Pressure – Now allow exchange of volume. When does the wall not move? (What is eqm. in this situation?) U + U = const. dV = dV = ▯dV A B A B In eqm. S = S + Sis maximal A B @S @SA @SB @SA @SB @SB 0 = = + ! ) = ▯ = @VA @VA @VA @VA @VA @VB @S @Vis same for A and B in eqm. Must be related to P. – Ideal gas: S = Nk lnV + function(N,V) ▯ @S▯ Nk P ▯ @S▯ ) = = ! P ▯ T @V N;U V T @V N;U 5 ▯ The "Thermodynamic Equation" – S = S(U;V;N) entropy is a function of state for a fluid in eqm. – Let N = constant (dN = 0). Then from basic calculus →for small changes in U,V, and V: ▯ ▯ ▯ ▯ ▯ ▯ @S @S @S dS = dU + dV + dN @U N;V @U U;N @U U;V = 1dU + P dV T T ) dU = TdS ▯ PdV ▯ Heat flow and Energy – Consider the first law applied to a small change in state carried out quasistati- cally: ▯U = Q + W ! dU = ▯Q + ▯W Note that the ▯’s are NOT differentials. They are just changes b/c Q and W are path-dependent. dU = ▯Q ▯ PdV when there is no other work done. ▯Q ) TdS = ▯Q or dS = T When quasistatic and only PV work. Quasistatic heat flow is directly connected to entropy. ▯ Heat Capacity and Entropy ▯ ▯ ▯ ▯Q ▯ @S CV= ▯ = T ▯T V @T V ▯ ▯ ▯ C = ▯Q ▯ = T @S P ▯T ▯ @T P P – By measuring heat capacities, you can determine entropy difference between 2 eqm. states of system: Z f Z f▯Q Z f CdT Sf▯ S i ▯S = dS = = i i T i T where the integral is along a quasistatic path (so S and T are well-defined) – Example: Find the entropy difference between 1kg of liquid water at room TP, and vapor at 100°C Obvious quasistatic path, isobaric: 6 Z ▯Q ▯S = Sv▯ S l T Z 377KC pT Qlatent = + 300K T ▯ T ▯oil 377 2260kJ ▯ 4:2kJK ▯1ln + = 7:0kJK ▯1 300 373K ▯ The Third Law of Thermodynamics – Try to calculate ▯S is the final state is at T = 0. Z 0CdT ▯S = Ti T ▯S diverges unless C ! 0 as T ! 0 – The heat capacity for ANY quasistatic process vanishes as T →0 – Corollary I: S→0 as T→0 where S = k ln ;0S = k l0 By quantum mechanics, there is no degenerate ground state (at ground state, = 1). is quantum ground state degeneracy. 0 – Corollary II: ▯ ▯ @S ! 0 as T ! 0 @X T Where X is a macroscopic variable (eg. V, B, ...). Ground state will remain at multiplicity of 1. – Corollary III: Impossible to achieve T = 0 by any conceivable process. ▯ Emperical Second Law of Thermodynamics (ESL) – No process is possible whose sole result is the transfer of heat from a colder to a hotter body – Consider 2 systems/bodies each at thermal eqm. in weak thermal contact Allow a little bit of heat Q to flow from A to B. Then by definition: Q Q ▯ 1 1 ▯ dStotal dSA+ dS B ▯ + = Q ▯ TA TB TB TA ESL→T ▯AT ) BS total0 So the ESL implies the Second Law, total entropy never decreases. 7 ▯ Chemical Potential – Consider 2 systems that can exchange particles of a certain kind (eg. electrons between metals) N A N B N is constant dNA= ▯dN B Now S = S + S will be maximal A B @S @SA @SB @SA @S B ) = 0 = + ) = @N A @NA @N A @NA @N B @S Therefore@N is the property that determines particle ("chemical") eqm. – Define chemical potential: ▯ ▯ @S ▯ = ▯T @N U;V – units: energy per particle, intensive property – If systems are already in thermal eqm, U is constant, T is Ahe same, then ▯ = ▯B is the condition that determines particle/chemicalA> ▯ B then particles will flow from A to B. ▯ Grand Thermodynamic Identity (GTI) – ▯ ▯ ▯ ▯ ▯ ▯ @S @S @S S(U;V;N) ! dS = dU + dV + dN @U V;N @V U;N @N V;U 1 P ▯ = dU + dV + dN T T T ) dU = TdS ▯ PdV + ▯dN – 3 conjugate pairs of state variables intensive extensive statistical T —– S [TS] = energy mechanical P —– V [PV] = energy particle ▯ —– N [▯N] = energy ▯ The many thermodynamic identitiesly 3 variables of state can be chosen idepen- dently, the other are determined (dependent). This leads to lots of equations. 8 – For example: Take S, V, N to be independent. Therefore U = U(S, V, N) ▯ ▯ ▯ ▯ ▯ ▯ @U @U @U ) dU = dS + dV + dN compare with GTI @S V;N @V S;N @N V;S ▯ ▯ ▯ ▯ @U @U P = ▯ @V @N S;N S;V – -P= rate of increase of energy with volume (done adiabatically so S is constant) – ▯ = energy increase when one particle is added at constant volume (adiabati- cally) – If you start with V(S, V, N) instead we obtain another equation for ▯: ▯ ▯ @V ▯ = P @N U;S ▯ Mixtures – For example air is a mixture of mos2ly2N ;O2;C2 ;H O and Ar. – Now we have another independent variable for each component →U(S;V;N ;N ;N ;:::) 1 2 3 – Each component has its own chemical potential ▯ ▯ @U X ▯1= dU = TdS ▯ PdV + ▯ iN i @N 1 S;V;2 3N ;::: i – If 2 mixtures can exchange particles of type X, then in eqm. ▯ is the same for x both. ▯ Partial Pressure: Mixture of gases – For a mixture of gases, the total amount of particles is equal to the sum of the particles of each gas: X N = N i i – Applying this to the ideal gas law: NkT X N kT X N kT P = = i = P P = i V V i i V i i th we get i , the coPtribution to pressure fromgas. i – For real gases, P 6iPi(but it’s close) – However, ▯ = @U is still well-defined and determines eqm. i @Ni 9 ▯ Chemical Potential for ideal Gas – Combining the Sackur-Tetrode eq. with the chemical potential and substituting in the ideal gas law: ▯ ▯ " ▯ ▯2=3 # @S 5 4▯mU 5 ▯ = ▯T S = Nk ▯ lnN + ln 2 V + @N U;V 2 3h 2 ▯ ▯ 5 IGL ! kT lnP ▯ 2ln(kT) + const: ▯ ▯ @▯ kT V = = @P T P N – uneven P →uneven ▯ →flow from high P to low P →eqm. w/ uniform P ▯ 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 hQ ;l are per cycle – First Law for one cycle: ▯U = 0 = Q h Q ▯lW ) W = Q ▯ h l i.e. a fractionof heat from hot reservoir is dumped in the cold one Qh 10 ▯ Efficiency – Define efficiency e as: W Q h Q l Q l e = = = 1 ▯ 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 ) ▯ Qh + Q l▯ 0 ! Ql ▯ Tl ! e ▯ 1 ▯ Tl 0 < e < 1 Th Tl Q h Th Th Q ▯ Minimum change in entropy happens when quasistatiT, solower 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 P 1W ) X = ▯ ▯ 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 higheh T (eg. hot steam)


Buy Material

Are you sure you want to buy this material for

50 Karma

Buy Material

BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.


You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

Why people love StudySoup

Steve Martinelli UC Los Angeles

"There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

Jennifer McGill UCSF Med School

"Selling my MCAT study guides and notes has been a great source of side revenue while I'm in school. Some months I'm making over $500! Plus, it makes me happy knowing that I'm helping future med students with their MCAT."

Jim McGreen Ohio University

"Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

Parker Thompson 500 Startups

"It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

Become an Elite Notetaker and start selling your notes online!

Refund Policy


All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email


StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here:

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.