Part1B(Advanced Physics) Statistical Physics

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1 PartB(Advanced Physics) Statistical Physics Course Overview: 6 Lectures: uesday, hursday only 2 problem sheets, Lecture overheads + handouts. Lent erm (mainly): Brief review of Classical hermodynamics: Statistical Mechanics: Microscopic picture: Boltzmann distribution, Entropy and disorder. Easter erm (mainly): Applications, including: Condensed matter systems Collective phenomena, phase transitions Books/Website: Lecturer: Dr. W. Allison wa4@cam.ac.uk el: x3746 Books: see Course Handbook February 05 Lecture

2 Lecture : Revision, Basic Concepts Aims: Revision: Zeroth Law - concept of temperature First law conservation of energy Functions of state Second law increase of entropy in irreversible changes hermodynamic potentials specifically the Helmholtz free energy, F, (V constant) owards a microscopic picture of Entropy Entropy and configuration Entropy of mixing (Gibbs paradox) First First Law Law d U = dw + d Q Second Law Law S = d Q rev 0 Free Free Energy F = U S F 0 February 05 Lecture 2

3 Definitions A thermodynamic system Co-ordinates: ordinates: Intensive, independent of size, eg: Pressure, p; Magnetic field, B; emperature, etc. Extensive, proportional to size, eg: Volume, V; magnetic dipole moment, m; Entropy, S etc. Functions of State: Any function that assumes a unique value for every state of the system. Examples: emperature,. (Zeroth law) Internal energy, U. (First law) Entropy, S. (Second law) there are many others... February 05 Lecture 3

4 Equilibrium and Reversibility Equilibrium State Common knowledge suggests that an isolated system will evolve (irreversibly) to a time invariant state of EQUILIBRIUM. Most things we encounter are metastable: eg: Nuclei could transform to Fe (or make a black hole). hermodynamics still works. Reversible changes Reversible changes must be quasi-static (infinitely slow). Others are irreversible and dissipative (eg. stirring a viscous liquid, I 2 R heating in resistor.) February 05 Lecture 4

5 emperature - Zeroth Law Zeroth Law (An empirical observation) If two systems are separately in thermal equilibrium with a third then they must be in equilibrium with each other. Every system may be labelled with a parameter, which describes the direction of energy flow when any two systems are brought into contact. Leads directly to the concept of temperature (the third system may be a thermometer). EMPERAURE is a FUNCION OF SAE. In the next lecture we will investigate the concept of temperature from a statistical point of view. February 05 Lecture 5

6 Heat and Energy First law of thermodynamics: Energy is conserved if heat is taken into account. he principle is expressed in the equation: Change in in internal energy d U = dw + d Q Work done on the system For a REVERSIBLE change: dw In a fluid: dw so, = i X = p dv Note, W and Q are not Functions of of State x d Q = du + i d i Force p dv Heat given to the system ( e.g. p dv ) Displacement Note the sign, if if dv>0, work is is done by the system and dw<0) February 05 Lecture 6

7 Second Law Not all processes allowed by the First Law can occur; for example, Consider an isolated system consisting of two equal masses starting at temperatures 00K and 200K: Initial state Final state he first law allows both processes. Experience (and experiment) show that only process B can happen spontaneously. Second Law: defines those processes that can occur. Related to the nature of irreversible events. February 05 Lecture 7

8 Irreversible changes Irreversible cycles: Recall Carnot s theorem Carnot engine : Less efficient engine Q Q = 2 2 Q ' Q < ' 2 2 Carnot heorem Q 2 > Q 2 Generalising to any cycle: S = d Q rev = 0 i.e. in any cycle, d Q 0 d Q < 0 February 05 Lecture 8

9 he Second Law What happens to S in an irreversible change? Going round the partly reversible loop: 2 d Q irrev d Qrev 2 d Q d Q ds irrev d Q rev For an isolated system dq=0, ds 0. he entropy of an isolated system cannot decrease Gives us the arrow of time. 0 = S he second law February 05 Lecture 9

10 Free Energy Free energy is another function of state useful in bridging to statistical mechanics (see later lectures). F = U - S (evidently a fn. of state since U, & S are fns. of state). For a system that can exchange heat with its surroundings but do no work (pdv=0, constant volume) d Qsurroundings = dusystem d Q Q d system system = du system U system he free energy decreases in any spontaneous, irreversible change, as the system moves towards equilibrium. he Free Energy is a minimum at equilibrium. A hermodynamic Potential. = February 05 Lecture 0 S system ( U S ) = F 0 system system system

11 Entropy and configuration Joule expansion: Irreversible expansion of a perfect gas into a vacuum. Insulating walls We can work with the initial and final states (common strategy in problem solving) when dealing with functions of state. What is the final state, f, p f? Use the first law: U = Q + W Consider the system as a whole W=0, no work done on (or by) the universe Q=0, isolated adiabatic change. First Law gives U=0. For a perfect gas: U=3R/2. f = i. p follows from gas laws p V = p 2 V 2. February 05 Lecture

12 Joule expansion: Entropy change Entropy change: o calculate the entropy change, we need a reversible route from initial to final state. Initially: p o V o o Finally: p V or, if volume doubles, p o /2,2V o, o. Choose an isothermal expansion = 0 U = 0 du 0 d S = d S S = d Q + dw = d S = R V R = R ln dv dv V ( V V ) P dv In the case above V o 2V o S = R ln 2 > 0 Note: entropy increases without heat input. Leads to the notion that entropy is related to configuration (disorder). 0 For mole February 05 Lecture 2

13 Entropy of mixing Change in entropy on mixing A and B are two components separated by a partition. Mixing is irreversible so expect entropy to increase when the partition is removed. For f moles, S f S, when only V changes: Gas A: S A = f R ln(v/fv) S S mix mix Gas B: S B = (-f) R ln(v/(-f)v) = S = R A + S B = R f ln + f ( f ) ( f ln( f ) + ( f ) ln( f )) > 0 S > 0, an irreversible process Note: final state is more disordered ln f February 05 Lecture 3

14 Gibbs Paradox What if the 2 gases are identical? here can be no increase in entropy since initial and final states are identical. S mix = 0. Particles are indistinguishable (impossible to separate them even in principle). If the particles are distinguishable, no matter how small the difference between A and B there is a finite entropy change. o understand the paradox we need a microscopic view of entropy. We start on such an approach next lecture. February 05 Lecture 4

Problem: Calculate the entropy change that results from mixing 54.0 g of water at 280 K with 27.0 g of water at 360 K in a vessel whose walls are

Problem: Calculate the entropy change that results from mixing 54.0 g of water at 280 K with 27.0 g of water at 360 K in a vessel whose walls are perfectly insulated from the surroundings. Is this a spontaneous