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Home , Thermodynamic equations

THERMODYNAMICS

A GENERALLY GLOOMY SUBJECT THAT TELLS US THAT THE UNIVERSE IS RUNNING DOWN, EVERYTHING IS GETTING MORE DISORDERED AND GENERALLY GOING TO HELL IN A HANDBASKET

James Trefil - NY THE LAWS OF

THE THIRD OF THEM,THE SECOND LAW, WAS RECOGNIZED FIRST

THE FIRST, THE ZEROTH LAW, WAS FORMULATED LAST

THE FIRST LAW WAS SECOND

THE THIRD LAW IS NOT REALLY A LAW

P. Atkins ~ The Second Law

Zero’th and third laws First law Conservation of

THE SECOND LAW QUESTIONS

DGm – Wh a t i s i t? Define T, S.

What is thermodynamics?

What are the ? THERMODYNAMICS

Expresses relationships between the macroscopic properties of a system without regard to the underlying physical (i.e., molecular) structure.

E.g., OF AN

PV = nRT •Where does this come from?

•What is the molecular machinery? ie. can we obtain this equation from a molecular theory ?

•Can we also obtain equations of state for liquids and solutions?

•How do we describe interactions? DEFINITIONS

1. AND

F Force F P = A

Cross section area = A Volume V BOYLE’S LAW 1/V

P V = constant EXPERIMENT

1 P ~ THEORY V (at constant T,n)

P DEFINITIONS 2. TEMPERATURE - -What is it? Normal scales — °C, °F arbitrary

Thermodynamic definition: - something that determines the direction will flow

CHARLES LAW

HEAT V ~ T (at constant n, P) V Is there an absolute If there is no heat flow, the scale of two bodies are at the same temperature ? temperature (zero’th law of thermodynamics)

T

O O -273 C 0 C

Note: third law. Can’t get to absolute zero in a finite number of steps DEFINITIONS

3. The amount of stuff; n.

• Don’t confuse and weight

• Numbers of molecules - - large !

• Use moles (mol) as a unit (Latin - - massive heap)

AVOGADRO’S PRINCIPLE 1 mol of particles = 12 V ~ n # of atoms in 12 gms of C 23 (constant T,P) 1 mol = 6.022 x 10 FINAL DEFINITIONS

ENERGY AND ENTROPY

Energy

• Started with the concepts of HEAT and • ~200 years ago these were considered to be different things and had different units

WORK NEWTON Force x distance moved

HEAT Caloric A weightless form of matter that flowed in and out of materials FINAL DEFINITIONS

UNITED BY CONCEPT OF ENERGY, defined as the capacity to do work

1Joule = 1 kg m2 s-2 Joule - a wierd Manchester Brewer interested in the mechanical equivalent of heat

Thermodynamics grew up around the question of the transformation of energy, particularly HEAT MECHANICAL WORK

NOTE: It’s easy to turn mechanical work into heat

Turning heat into mechanical work is much harder. Cannot turn 100% of the heat into work

THERE IS A MISSING QUANTITY THE FIRST LAW dE = dQ - PdV

Heat Q F dl = P dV (F dl = P Adl) dl

ENTHALPY: HEAT SUPPLIED AT CONSTANT PRESSURE

dQ p = d H

H = E + P V GETTING WORK OUT OF HEAT

WORK HEAT - easy

HEAT WORK - harder

CARNOT - Impossible to take heat at a certain temperature and convert it to work with no other changes in the system or the surroundings

NO USEABLE WORK

HOT USEABLE WORK COLD

THE CONCEPT OF ENTROPY AROSE FROM THIS ANALYSIS

__DQ DS = rev T ENTROPY - THEMODYNAMIC DEFINITION

DQrev DS = T and From Carnot’s analysis DQ S > irrev D T

W = – PdV Analogy Q = – TDS

(note sign change - direction of Q)

i.e., S is the quantity that defines the relationship between heat and temperature ENTROPY 16 TONS

PE = mgh ......

. . . . . 16 . . . THUD !! . TONS . . .

doesn't levitate ! ...... 16 . . . . heat TONS . .

doesn't happen doesn't happen spontaneously spontaneously OTHER MANIFESTATIONS OF ENTROPY

HOT COLD

Happens spontaneously

Also happens spontaneously

ENERGY AND MATTER TEND TO DISPERSE CHAOTICALLY - THE SECOND LAW THE SECOND LAW OF THERMODYNAMICS AND FREE ENERGY

For a process to occur spontaneously,S must increase; eg mixing

DStot = DS sys + DSsurr But,if heat is released by the system

S = - DQ sys = - DH sys D surr T T Define free energy

D G = -TDS tot =DH -TDS FREE ENERGY

DG ® advantage of using the free energy is that it contrives a way to relate overall changes to changes in the system alone

WHY IS IT CALLED THE FREE ENERGY?

– Because it is the maximum amount of non-expansion work that can be obtained from a system (T,P const)

DG – DH ® heat tax! SUMMARY

WE HAVE DISCUSSED THE ORIGIN OF THE THERMODYNAMIC EQUATIONS

PV = nRT

DGm = DHm - TDSm

NOW WHAT?

NEXT STEP INTRODUCE THE MOLECULES

1. From F = ma derive the and provide a molecular machinery

2. Show that by looking at averages and distributions of “mechanical” properties we can obtain S, G etc. FEYNMAN - LECTURES ON PHYSICS

FEYNMAN - “If in a cataclysm all human knowledge was destroyed except one sentence that could be passed on to future generations – – – what would contain the most information in the fewest words ?”

“ALL THINGS ARE MADE OF ATOMS – LITTLE PARTICLES THAT MOVE AROUND IN PERPETUAL MOTION, ATTRACTING EACH OTHER WHEN THEY ARE A LITTLE DISTANCE APART, BUT REPELLING UPON BEING SQUEEZED INTO ONE ANOTHER .”

BASIC LAWS OF PARTICLES BASIC LAWS OF STUFF

Mechanics (various forms) Thermodynamics Electricity + magnetism

THE PROBLEM IS TO LINK THEM PRESSURE

If there is no opposing force on the piston, each collision moves it a little bit.

How much force, F, do we need to put on the piston to prevent movement?

First: Remember P = F/A

Second: How much force is imparted to the piston by the collisions ? .. d . (mx ) F = mx = dt i.e., WE NEED TO CALCULATE HOW MUCH MOMENTUM PER SEC IS DELIVERED BY THE COLLISIONS PRESSURE

THIS IS EASY! a) Calculate momentum delivered by one collision b) Count # of collisions/sec

v x

Assume perfectly elastic conditions

Then the particles have

+ mvx momentum before - mvx momentum after

Change = mvx – (-mvx) = 2mvx PRESSURE

Now calculate the # of collisions in t FIRST: Define the number n N = (# of particles/unit vol) V v x In a time t, only those particles that are close Enough and have sufficient v x velocity will hit the piston this doesn’t get x there in t v = x t

Vol occupied by the molecules that will make it is v x t A

The number hitting the piston = N vx t A

# per sec = N vx A PRESSURE

HENCE

F = NvxA.2mvx

F 2 P = = 2 Nm v x A a) All molecules don’t have the same v BUT x b) Some are moving away from the piston

1 2 S o r e p l a c e v 2 w i th v x 2 x 2 P = N m v x

2 2 2 N o w v = v = v x y z

2 1 2 2 2 2 v = v + v + v v x 3 x y z = 3 THE - KINETIC THEORY

2 m v 2 HENCE P = N 3 2

2 n m v2 = 3 V 2

2 OR P V = n KE 3

COMPARED TO P V = n'R T

IS n’ THE # OF MOLES ? IS KE ~ T ?

Recall the thermodynamic definition

TWO BODIES ARE AT THE SAME T IF THERE IS NO HEAT FLOW BETWEEN THEM. TEMPERATURE - KINETIC THEORY

COLD HOT

High density Low density Low v High v PISTON

WHAT HAPPENS AT EQUILIBRIUM?

•Forces balance FOR COLLISIONS OF •Atoms gain or lose energy depending on PAIRS OF MOLECULES wether the piston is moving towards them 2 2 or away from them during the collisions 1 m v = 1 m v 2 1 1 2 2 2

THEN – IF TWO GASES ARE AT THE SAME T, THE MEAN KE OF THE CENTER OF MASS MOTIONS ARE EQUAL. KINETIC THEORY - THE CONSTANTS k and R

K E ~ T

1 m v 2 = 3 k T 2 2 2 P V = n 1 m v 2 = n k T 3 2

HENCE P V = n'R T At same T, P, V, n is a constant!

Absolute scale of temp V z ero i s w h en

1 m v 2 = 0! T 2 IMPORTANT POINTS

•A simple consideration of the motion of particles gives a fundamental understanding of P, T, the ideal gas law, absolute T and absolute zero, etc.

•Can we stretch this approach and ultimately get a molecular interpretation of S , DG , etc.? THE DISTRIBUTION OF ENERGY AND MATTER

All of the thermodynamic quantities we have dealt with so far deal with how much material is present and,on average,how fast the molecules are moving

WHAT ABOUT

a) The distribution of velocities?

b) The distribution of molecules in space?

i.e., THIS WILL LEAD TO A DESCRIPTION OF ENTROPY, DG etc. IN TERMS OF STATISTICAL ARGUMENTS EXAMPLE The distribution of molecules in the atmosphere

ASSUMPTIONS Constant T ! No wind ! n P V = n k T o r P = Nk T ( N = V ) i.e., if we know P, we know n, if P is a constant BUT, in the atmosphere it varies

h + dh P at (h + dh) must be less than P at h by an h Area A amount that is proportional to the weight of gas in Adh

# of moles in Adh = NAdh

Force Nadh = Area A EXAMPLE (cont.)

h + dh h Area A Force Nadh = Area A

Ph+dh - Ph = dP = - mgNdh

P = NkT, or dP = kTdN

dN mgN HENCE = - d h kT

-m g h/k T -P E /k T N = No e = No e THIS IS A GENERAL RESULT BOLTZMANN’S LAW N ~ e -P E /k T -K E /k T SIMILARLY n > u ~ e

PROPERTIES OF PROPERTIES OF MOLECULES STUFF (Bulk Materials)

Thermodynamics How do we (P, T, V, G, S etc get from the + relationships molecules to between them) bulk properties ?

Can bridge the “gap”by considering the average properties of all the particles of the system and the distribution of matter and energy.

2 2 m v o b ta i n e d P V = n { 3 2 }

ENTROPY A measure of the distribution of energy and matter in a system BOLTZMANN’S TOMB

We have seen that Entropy is associated with the distribution of energy and matter in a system.This can be expressed formally in terms of the equation carved on Boltzmann’s tomb

S = k lnW

Which,today,is normally written

S = k ln W

Where W is the number of arrangements Available to the system[At a given V, E, N] ENTROPY

S = klnW Why does the relationship have this form ?

Consider an ordered pack of cards - 1 arrangement

Now shuffle; what is the number of Possible arrangements ?

66 W = 52! (~4.45 X 10 )

Now consider two packs shuffled separately = W W1W2 S = kln W = kln W1W2 BUT S IS A THERMODYNAMIC PROPERTY THAT MUST BE = klnW1 + klnW2 ADDITIVE = S 1 + S 2 FREE ENERGY REVISITED

For this to occur S must increase ie. DS > 0

DStot = DS sys + DSsurr

BUT,IF HEAT IS RELEASED BY THE SYSTEM

S = - DQ sys = - DH sys D surr T T

D G = -T D Stot =DH -TDS WHAT YOU SHOULD KNOW

•Describes relationships between macroscopic properties THERMODYNAMICS •2nd law: In a spontaneous process S always increases

•Free energy: What is it?

Relates the “mechanical” properties of STATISTICAL MECHANICS atoms and molecules to macroscopic or Thermodynamic quantities

Pressure EXAMPLES: Temperature (Heat is motion!)

WHAT IS ENTROPY?