Chapter 1. The Nature of Physical Chemistry and the Kinetic Theory of Gases

Chapter 1 The Nature of Physical Chemistry and the Kinetic Theory of Gases Physical chemistry is the application of the methods of physics to chemical problems Work Kinetic and Potential Energy Thermal Equilibrium Gas Laws Kinetic-Molecular Theory of Gases Maxwell Distribution of Molecular Speeds

Work Force = Mass x Acceleration, F = ma Units (kg m/s 2 ) dv dp F = ma = m =, where p is the linear momentum dt dt SI unit of Force is the Newton 1 N = 1 kg m/s 2 Work = Force x Distance, Units (kg m 2 /s 2 ) Work is a form of energy dw = F dl SI unit of Energy is the Joule 1 J = 1 kg m 2 /s 2

Hooke's Law & Harmonic Oscillator For an ideal (Hooke s Law) linear spring, the restoring force is proportional to the displacement: F restore = -k H x What is the work done by an external force, F ext, exactly equal and opposite to the restoring force, when it extends the spring from x=0 to x? F restore F = F = x ext restore kh

Hooke's Law & Harmonic Oscillator dw = F dl = F dl x dw ext x = k xdx = H restore k x 2 x= x 0 0 x= 0 H 2 2 kh x w =, work done to (slowly) extend the spring 2 from 0 to x A particle under a restoring force that obeys Hooke's law is called a harmonic oscillator.

Kinetic Energy The difference in kinetic energy between the initial & final states is the work performed in the process. Kinetic (motional) Energy is E k = 1/2 mu 2 with velocity u = d dt Can we show a relationship between Work and Kinetic Energy? Cast the previous expression as a function of time. t t d w = F ( ) = ( ) = ( ) restore d Frestore dt Frestore u dt dt 0 t0 t0 du Now Frestore = ma = m dt ( ) t 2 du mu w = m u dt = mu du = = Ek E dt 2 0 ut ( ) o ( ) ut t ut ut ( ) o k 1 0

Kinetic and Potential Energy for a conservative system Kinetic (motional) energy is E k = 1/2 mu 2 In the very important case in which the force on a particle depends only on its position, the forces can be described by a scalar function called the positional (or more commonly, potential) energy. Such a system is said to be conservative. When might a system be conservative? When will it most definitely not be conservative?

Potential Energy Function Consider the earlier work expression, w = F( ) d = EK E 1 K0 0 Since we know that the integrand is a function only of l, we can define a new function E p (l) [or V(l)], or ( ) = ( ) F d de F ( ) de p p = = d dv d More generally, we have the vector form, E p ( aka V) is the potential energy function F = V = E P

Potential Energy de dx F p = kx Consider the harmonic oscillator, x = kx x Integrating 2 x EP = k + cons ' t 2

Potential Energy Using the harmonic oscillator results, E p = 1/2 k h x 2 you can show the following: w = (E p1 E p0 ) = E k1 E k0 E k1 + E p1 = E k0 + E p0 The sum of the kinetic and potential energy remains constant in a transformation. True in general for a conservative system

Systems, Properties & Equilibrium System - Part of the world on which we focus attention Surroundings - Everything else System + Surroundings = Universe Properties - Examples: temperature, volume, pressure, mass, refractive index, density Two types of properties at equilibrium: Intensive - do not change with quantity of matter present (i.e., when system is subdivided) Extensive scales linearly with quantity of matter present Equilbrium - state when a system s macroscopic properties do not change with time

Systems, Properties & Equilibrium System - Part of the world on which we focus attention Surroundings - Everything else System + Surroundings = Universe

Temperature Zeroth Law of thermodynamics Two bodies in thermal equilibrium with a third body are in equilibrium with each other Thermal Energy k B T k B is the Boltzmann constant T is the Kelvin Temperature Scale (K) T( C) + 273.15 = T(K) Measure Temperature based on Thermal Expansion: l t = l 0 (1 + αt + βt 2 ) α is typically ~10-5 K -1

Zeroth Law of Thermodynamics

Empirical description of one-component gas What properties are required in order to characterize a one-component gas when it is in equilibrium? Amount of the Gas: mole, n. The number of atoms in 12 g of 12 C Temperature: T related to random motion Pressure: P, force per unit area A 70 kg person exerts a force (mg) of 686 N (Newton) upon the ground. The pressure depends upon the contact area. Volume that the gas occupies: V Molar volume V V V,an intensive property m n the ratio of two extensive properties is intensive These four properties are all that is needed to characterize a gas at equilibrium. The relationship between them is called an Equation of State. v

Boyle s Law (empirical) Describes relationship of pressure (P) and volume (V) when amount (n) and temperature (T) are constant SI unit for amount (n) of a substance is the mole Avogadro s constant, N A (or L) = 6.022 x 10 23 mol 1 n = N/N A V 1/P or PV = cons t, Boyle's Law Th l f fi d t f i i l ith The volume of a fixed amount of gas varies inversely with the pressure if the temperature is maintained constant.

Boyle slaw

Boyle s Law (experimental) PV = constant (hyperbolae) T 4 Is T 4 >T 1 or is T 4 <T 1?? Temp T 1

Charles s s Law Absolute T Scale 1 kg of each gas & P=1 atm Temperature

Avogadro slaw Describes relationship of volume (V) and amount (n) when pressure (P) and temperature (T) are constant V n, Avogadro's Law The volume of a gas varies directly with the amount if the pressure and temperature are constant.

Ideal Gas Law Deduced from Combination of Gas Relationships: V 1/P, Boyle'sLaw V, Charles's Law V n, Avogadro's Law Therefore, V nt/p or PV nt PV = nrt where R = universal gas constant t The empirical Equation of State for an Ideal Gas

Ideal Gas Equation of State Isochores const V

Ideal Gas Law PV = nrt where R = universal gas constant R = PV/nT R = 0.08210821 atm L mol 1 K 1 R = 0.0821 atm dm 3 mol 1 K 1 R = 8.314 J mol 1 K 1 (SI unit) Standard molar volume = 22.4 L mol 1 at 0 C and 1 atm Real gases approach ideal gas behavior at low P & high T

ConcepTest #1 One mole of ice is similar in volume to A. The Arapahoe Glacier B. The iceberg that sank the Titanic C. A Toyota Prius D. An ice cube in a cold drink E. A snowflake

Measuring Pressure, P = gh is density (kg/m 3 ), g is acceleration of gravity (m/s 2 ), hi is height h (m) P(P (Pa) Barometer was early form of pressure measuring device

Atmospheric Pressure = 760 mm Hg = 10,300 mm H 2 O P = P 0 exp(-mgz/rt)

ConcepTest #2 A steel vessel contains 1 mole of gas at 100K. 2 moles of gas are added and the temperature is increased to 200K. How does the pressure change? A. P increases by a factor of 4 B. P decreases by a factor of 4 C. Pi increases by a factor of f6 D. P decreases by a factor of 6 E. P does not change

ConcepTest #3 An ideal gas is compressed at constant temperature. Which picture below best describes how the volume V of the gas behaves as the pressure P is increased? A V B V P P C V D V P P

Dalton slaw Definition: Partial pressure P i is the pressure exerted by one component of a gas mixture (at total pressure P t ) P i = x i P t where x i is the mole fraction (x i = n i /n total ) Dalton s Law: P t = P 1 +P 2 +P 3 + + P i The total pressure is equal to the sum of the partial pressures that each individual component gas would exert if it were alone

ConcepTest #4 A mixture of gases contains 4 g of He and 4 g of H 2. The total pressure is 300 Pa. What is the partial pressure of helium? A. 100 Pa B. 150 Pa C. 200 Pa D. 250 Pa

Sample problem A (non-rigid) diving bell has an air space of 3.0 m 3 when on the deck of a small boat. What is the volume of the air space when the bell has been lowered to a depth of 50 m? The density of sea water is 1.025 g cm -3. What do we need to assume? Constant T, so use Boyle: PV=cons t Ideal gas behavior for air Water temperature is unchanged at 50 m Why is rigid id important information? i p p gh 1.0 atm gh f i

Real Gases Compressibility, Z Z PV PV nrt RT Z = 1 Ideal Gas behavior Z < 1 PV less than expected Attractive forces Z > 1 PV greater than expected Repulsive forces Z Ar 1 Ideal Gas P

Real Gases data! Compressibility PV PVm Z nrt RT Z = 1 at all P, T Ideal Gas Behavior Ideal gas Now look at real gases at some temperature T Look at a broader 0 800 atm region We need new eq of state for each gas These data are many gases at one T. So next look at one gas at many T

Boyle Temperature, T B Z Z P, T T B is the temperature corresponding to the greatest extent of near-ideal behavior. We can determine T B analytically. Z T, P P df dt lim p0 f T 0 at T T, the Boyle temperature t B

van der Waals equation of state Physically-motivated y y corrections to Ideal Gas EoS. For a real gas, both attractive and repulsive intermolecular forces are present. Empirical terms were developed to help account for both. 1. Repulsive forces: make pressure higher than ideal gas Excluded volume concept (nb) P nrt V nb Volume of one molecule of radius r is V mol = (4/3) r 3 Closest approach of two molecules with radius r is 2r.

ConcepTest #5 nrt Excluded volume P V nb The volume of one molecule of radius r is V mol = 4/3 r 3 The closest approach of two molecules l with radius r is 2r. What is the excluded volume for the two molecules? A. 2V mol B. 4V mol C. 8V mol D. 16 V mol

van der Waals equation of state Physically-motivated corrections to Ideal Gas EoS. For a real gas, both attractive and repulsive intermolecular forces are present. Empirical terms were developed to help account for both. 1. Repulsive forces: make pressure higher than ideal gas (or, equivalently, make the volume smaller) Do the latter: Excluded volume P nrt V nb Volume of one molecule of radius r is V mol = 4/3 r 3 Closest approach of two molecules with radius r is 2r. The excluded volume V exc is 2 3 V mol = 8V mol for two molecules. So we might estimate that b 4V N This assumes binary collisions only. Always true? NO! mol A

van der Waals equation of state Physically-motivated y y corrections to Ideal Gas EoS. For a real gas, both attractive and repulsive intermolecular forces are present. Empirical terms were developed to help account for both. 2. Attractive forces: make pressure lower than ideal gas Pressure depends wall collisions, both on frequency and their force. Not easy to show, but we expect a pressure correction of the form a(n/v) 2, giving the van der Waals Equation of State 2 nrt an RT a P V nb V V b V 2 2

3D van der Waals eqn of state T= T/Tc

Real Gases CO 2 CO 2 Look at 50 C isotherm. Behavior is near ideal gas Look at 20 C isotherm. ABC Compression At C, liquid condensation begins D liquid- vapor mixture at P vap (20 C) E last vapor condenses F Steep rise in pressure A liquid or solid is much less compressible than a gas For T >T there is a single phase For T >T c, there is a single phase, with no liquid formed.

van der Waals Isotherms near T c v d W loops are not physical. Why? Patch up with Maxwell construction van der Waals Isotherms, T/T c

van der Waals Isotherms near T c Look at one of the van der Waals isotherms at a temperature re of 0.9 T c 1.5 G A D compress the gas at constant T, F G compress the liquid phase (steep and not very compressible) D F vapor condensing (gas and liquid coexist) These are stable states Reduced Pre essure, P r 1.0 0.5 F C B g, l D T r = 0.9 A F C supercooled liquid D B superheated gas 0.0 0.3 0.5 0.7 1 2 3 5 7 10 These are metastable t states tt Reduced d Volume, V r Metastable example: C B a non-physical artifact of vdw Use a very clean glass. Add water and heat (patched up with Maxwell construction) for a while with a microwave oven (superheat) eat) Add a drop of sand or perhaps touch with a spoon.

Review of Key Concepts from Zeb Physical basis comes later today

Maxwell distribution of molecular speeds Volume element (dv x dv y dv z ) transformation from cartesian to spherical polar coordinates: x = r sin cos, y = r sin sin z = r cos Volume element transform: dv x dv y dv z 4v 2 dv mv 3 2 2kT B 4 dp 4 B v e dv 2 For a normalized distribution, 0 P v dv 1 We use this constraint to set B, giving 3 2 3 mv 2 2 2 Mv 2 M 2 2RT m 2kT B Pvdv 4 ve dv 4 ve dv 2kT 2RT B

Characteristics of N(v), Maxwell speed distribution Mv M 2 2RT N v dv 4 v e dv 2 RT 3 2 nm A 4 2nkT A B m 4 2kT B v v vn v dv 2 3 2 n 2 Amv 2 2nAkBT ve 3 2 mv 2 2 2kT B ve 0 2 2 2 v v v N vdv 0 v v v rms 2 2 dv dv

Characteristics of N(v), Maxwell speed distribution 3 2 mv 2 mm 2 2kT B N v dv 4 v e dv 2kT B We (Zeb) showed how to calculate the average kinetic energy per molecule, and obtained a very important result: 1 1 KE mv m v 2 2 2 2 1 2 1 3kT B 3 mc m k T B 2 2 m 2 and is independent of mass! 2

ConcepTest Maxwell speed distributions are shown for three situations. (I) If the curves represent 3 different gases at the same T, which curve shows the gas of greatest molar mass? (II) If the curves represent the same gas at 3 different T, which curve shows the highest T? xxxxxxxxxxxxxx blue green xxxxxxxxxxxxxxxxxxxxxxxxxxxxx red 400 m/s 800 m/s 1200 m/s 1600 m/s A. blue; blue C. red; blue B. blue; red D. red; red

Let s look at some actual numbers to get a better intuition Molecule Mass (kg/mol) v MP (m/s) v Avg (m/s) v rms (m/s) KE, kj/mol H 2 2 x 10-3 1570 1770 1900 3.7 He 4 x 10-3 1110 1260 1360 3.7 N 2 28 x 10-3 420 475 515 3.7 Ar 40 x 10-3 350 400 430 37 3.7 With the microscopic distribution of molecular velocities now related With the microscopic distribution of molecular velocities now related to a macroscopic property T, let s see how this result gives physical, microscopic level, insight into the empirical Ideal Gas Equation of State.

Three Postulates of Kinetic Theory of Gases 1. Gas composed of hard spherical particles that are small relative to the mean distances between them. 2. Particles are in constant motion and have kinetic i energy. 3. Neither attractive nor repulsive forces exist between the particles except on contact (collision).

ConcepTest #2 Which of the following gases deviates most from the postulates of kinetic molecular theory? (i.e., deviates most from ideal gas behavior) A. He B. H 2 C. N 2 D. NH 3 E. CH 4

Kinetic-Molecular Theory Consistent with Ideal Gas Law, Dalton s Law, Graham s Law and many other observations W l l t th th t th t th ll We calculate the pressure that the gas exerts on the walls. Remember that pressure is the force per unit area, F/A.

Consider a molecule colliding elastically with a wall velocity, v x Wall l x-coordinate velocity, -v x

Rlti Relationship between bt Force and Change of Momentum Momentum,,p = mv F = ma = m dv/dt = d(mv)dt = dp/dt What is change of velocity on each collision? v = v final v initial i i = v x ( v x ) = 2v x What is change of momentum on each collision? p = 2v x m

Collision rate for one molecule with a wall Wall v x -v x Box with dimensions xbyybyz y x (1ength of 1 side) Number of collisions in time t = v x t/2x Number of collisions i per unit time = v x /2x

Force (momentum m m change) per Unit Time for One Particle F= ma=dp/dt = (Change of Momentum per Collision) x (Number of Collisions per Unit Time) Since the collision is elastic, v xafter = -v x before Thus for one collision, i Δp = p x before p xafter = mv x (-mv x ) = 2mv x dp/dt = F x = (2v x m) (v x /2x) = mv x2 /x This is the force exerted on the yz wall by one particle.

Pressure is Force per Unit Area y z x P x = F w /A = F x /yz F 2 x = (2v x m) (v x /2x) = mv x2 /x P x = mv x2 /xyz P x = mv x2 /V

What changes are required when N molecules l and a distribution ib ti of velocities are considered? Multiply l by N Replace v x2 with v 2 x P x = P y = P z = P implies and v v v 2 2 2 x y z v 2 3 P x = mv x2 /V 2 2 P m v /V P Nm v /3V x x y z x V= xyz

Fundamental Equation from Simple Kinetic Theory of Gases 2 PV Nm v /3 Parallels the ideal gas law: PV nrt Nm v 2 nrt 3 2 3RT 2 3RT 3kBT v v vrms M M m with M (molar mass) = mn /n

ConcepTest #3 At a given temperature, which of the following gases has the smallest v rms? A. He B. H 2 C. N 2 D. NH 3 E. CH 4

Connection between Kinetic Energy & Temperature Average Kinetic Energy per Molecule 1 2 2 Substitute this expression into the pressure equation to obtain the relationship between kinetic energy and T: m 3 E RT (per mole) or k 2 3 k k B T (per molecule) 2 v

ConcepTest #4 Flask A contains 8 g H 2, and Flask B contains 8 g He. The flasks have the same volume and temperature. t Compare the gas density (g/cm 3 ), the kinetic energy per mole, and the total kinetic energy of the gases in the two flasks. Density E k /n Total KE A. A>B A>B A>B B. A>B A=B A>B C. A=B A=B A=B D. A=B A=B A>B E. A=B A>B A>B

Fundamental Equation from Simple Kinetic Theory of Gases 2 PV Nm v /3 is identical to the ideal gas law: nrt 2 Nm v 3 PV nrt 2 3RT 2 3RT 3kBT v v v rms M M m with M (molar mass) = mn /n Next we must deal with collisions and finite molecular sizes.

Collisions Between Molecules A single molecule A travels with velocity v A in a container of a number of B molecules A collision occurs each time the distance between the centers of molecule A and one of the molecules B is d AB where d AB (d A + d B )/2 and d A is the diameter of molecule A d B is the diameter of molecule B d AB is the collision diameter for AB collisions

A(u A ) colliding with N B stationary B(0) molecules The density of B is N B /V The collision frequency Z A for one A molecule with N B B is Z A = d 2 AB v A N B /V, s -1

Going from one A molecule to N 3 A A molecules/cm When we have a density N A /V of A molecules, we express the total rate of A-B collisions as N d v N N s 2 1 A AB A B A Z Z AB A V 3 1 V V cm and Z AB is the Collision i Density We have still ignored the fact that the B molecules are moving. Thus we need to replace v A with the average relative speed between A and B, v rel. 11 2 2 2 rel A B v v v So (avg. orientation is )

Getting the correct relative collision speeds Using this information we go back to correct the expression for the number of collisions per sec of one molecule of A with B molecules with density N B /V. Z A 1 2 2 2 2 AB A B B B d v v N V and the A-B collision density for N A /V molecules with N B /V is Z AB 1 2 2 2 2 AB A B A B d v v N N V 2

Consider collisions only between A molecules We found Z A d v N 2 AB A B V for a single A molecule with B molecules. If the B molecules become A s and are allowed to move, 1 then 2 2 2 v v v 2 v and d 2 AB becomes d 2 A A A A A So for a one component gas, Collision frequency Z A and the collision density s 1 2 2 d v N A A A V s 1 2 2 2d A v A N A 3 1 2 Z m s AA 2V

Mean Free Path Mean Free Path () = Distance Traveled in unit time Number of Collisions in unit time u A 2 d 2 AuAN A / V V 2 da 2 N A

ConcepTest #1 u A 2 d A 2 uan A / V In which of the following systems stems will the mean free path () be the shortest? A. Pure CCl 4, 1 Pa pressure B. Pure CCl 4, 10 Pa pressure C. Pure He, 1 Pa pressure D. Pure He, 10 Pa pressure

Mixture of A and B Using average speed of A, Using average relative speed of A and B, ua 1/2 2 2 u AB u A u B Number of collisions of one molecule of A with B molecules (per unit time); SI unit: s 1 Z A d 2 ABuAN B d 2 2 2 AB ua ub V Z A 1 2 N B V Total number of A-B collisions (per unit volume per unit time); SI unit: m 3 s 1 Z AB d 2 ABuAN A N B V 2 Z AB Using average speed of A, ua Pure A 2 d AB 2 2 ua ub V 2 Using average relative speed, u AA 2 u A 1 1 2 N A N B 1 Number of collisions i of one molecule l of A with A molecules l (per unit time); SI unit: s Z A d 2 AuAN A V Z A 2 d 2 AuAN A V Total number of A-A A collisions (per unit volume per unit time); SI unit: m 3 s 1 Z AA d 2 2 AuAN A Z 2V 2 AA 2 d A 2V 2 2 uan A 2

More on a gas of A molecules Consider a gas of A molecules with average speed v A and number density n A = N A /V. One can show (but we will not) that the flux, F, of molecules l passing in one direction through a plane of unit area is given by F n A v 4 A molecules cm s 2 1 This is a very useful relation to remember, but here we wish to focus on the average velocity, and see if we can understand the basis for the molecular speed distribution that seems ubiquitous. There is a key, universal, fundamental concept: Mother Nature does not play favorites, but rather selects in an unbiased manner from the allowed outcomes.

Mother Nature s choices One example of this is that a gas will uniformly fill a box that is free of any external forces. If there are external forces* (such as gravity), the gas assumes an exponential distribution of the form mgz mgz o N ( z ) e where mgz o is a sort of average gravitational potential and z o is a scale height of the gas. We see this characteristic form of an energy relation virtually everywhere: rates of reaction, energy distributions of gas particles... How does it arise? *This would give rise to be a positional (potential) energy V(z) that Zeb *This would give rise to be a positional (potential) energy V(z) that Zeb set to zero when he found the Maxwell speed distribution. The result here will have the same functional form, and the same underlying physics.

Possible arrangements of particles with fixed total energy While the real systems of interest typically have ~N A particles, we illustrate the concept on an embarrassingly small, but simple, system. Consider a system comprised of 4 particles. Any particle can have only discrete energies: E = 0, 1, 2, 3, 4... (quantized). Let the total energy of the system be 3 gizmos. What is the average energy per particle, E? A. 12 gismos B. 4 gismos C. 0.75 gismos D. 15 1.5 gismos E. Not enough information given

Possible arrangements of particles with fixed total energy Consider a system comprised of 4 particles. Any particle can have only discrete energies: E = 0, 1, 2, 3, 4... (quantized). Let the total energy of the system be three. What are the possible arrangements of energy? Energy 0 1 2 3 4 I. 3 0 0 1 II. 2 1 1 0 III. 1 3 0 0 IV. 0 4??? NO energy not conserved So there are exactly three available energy configurations.

Possible arrangements of particles with fixed total energy Consider a system comprised of 4 particles. Any particle can have only discrete energies: E = 0, 1, 2, 3, 4... (quantized). Let the total energy of the system be three. With all possible arrangements enumerated, We make the ergodic assumption, namely that the system visits all of the available configurations (3) with equal probability. What is the average occupancy of each energy level? Energy 0 1 2 3 4 I. 3 0 0 1 0 II. 2 1 1 0 0 III. 1 3 0 0 0 Totals/3 20 2.0 133 1.33 033 0.33 033 0.33 0 Plot n(e) Plot vs. n(e)vs. E E

Possible arrangements of particles with fixed total energy Energy 0 1 2 3 4 I. 3 0 0 1 0 II. 2 1 1 0 0 III. 1 3 0 0 0 Totals/3 2.0 1.33 0.33 0.33 0 3.0 2.5 Plot n(e) vs. E 2.0 n(e) 1.5 f( E ) 4 E 0.75 e 1.0 0.5 0.0 0 1 2 3 4 5 Energy Energy randomization leads to an exponential distribution of occupied energy levels at constant total energy

General Principle!! Energy is distributed ib t d among accessible configurations in a random process. The ergodic hypothesis Consider fixed total energy with multiple particles and various possible energies for the particles Determine the distribution that occupies the largest portion of the available Phase Space. That is the observed distribution.

PhET Simulation Related to Ideal Gases and Kinetic i Theory

PhET Kinetic Theory Simulation i http://phet.colorado.edu/get_phet/simlauncher.php Select Gas Property simulation

Thermodynamics The study of energy and its transformations The Zeroth Law of Thermodynamics Nature of Thermal Equilibrium Two bodies in contact via a heat conducting (diathermic) connection are in thermal equilibrium when there is no net heat flow between them. If A and B and A and C are in thermal equilibrium with each other, then B and C must also be in thermal equilibrium.

Thermodynamics The study of energy and its transformations ons Chapter 2 The First Law of Thermodynamics Conservation of Energy The total energy of the universe (or any isolated system) is constant

Basic Definitions System - volume of interest Open (reaction system vessel, - matter test tube, can pass between biological system cell, atmosphere, & surroundings etc.) Surroundings- volume outside system Matter system surroundings Energy Open System Closed system - matter cannot pass between system & surroundings Isolated system - Neither matter nor energy can pass between system & surroundings Thermal energy only: possible: diathermic not possible: adiabatic surroundings system Energy Closed System surroundings system Isolated System

Energy, Work and Heat Energy can be considered to be the capacity to do work For an closed system, doing work reduces its energy, and having work done on it increases its energy Work - generalized force over a generalized displacement Mechanical work = force x distance; -fdx, or in general dw F dr Expansion work = pressure x volume change; -pdv Electrical work = emf x charge displacement; EdQ Sign convention Work done by a system is negative Work done on a system is positive Work is the result of organized motion of molecules

Energy, Work and Heat (2) Heat is the change in energy in a system that is produced by a change in its temperature Heat is the result of disordered (thermal) motion of molecules

First Law of Thermodynamics U (internal energy) is the total energy of a system (kinetic + potential) Change in energy from initial state, i, to final state, f, U = U f U i Internal energy is a state function and an extensive property Value of a state function depends only on current state of the system, not how you get there - path independent (an exact differential; more later) As we will see, both heat transfer and work done on a system definitely depend on the path followed and are not state functions

ConcepTest #1 Which of the following is not a state function? A. Altitude B. Pressure C. Work D. Mass

First Law of Thermodynamics m U changes only by doing work or transferring heat to/from system If work is done on the system (heat in), ΔU > 0 If system does work (heat out), ΔU < 0 Thi i li th t f i l t d t (th This implies that for an isolated system (the universe), U is constant.

First Law of Thermodynamics m Mathematical statement of first law: U = q + w q = heat transferred to system w = work done on system The first law is simply a statement of the conservation of total energy for a system with defined d energy inputs and outputs t

ConcepTest #2 A system receives 575 J of heat from and delivers 325 J of work to its surroundings. What is the change in internal energy of the system? A. +900 J B. +250 J C. -250 J D. -900 J

Distinguish between System & Surroundings

Distinguish between System & Surroundings We also distinguish between thermal We also distinguish between thermal and non-thermal energy sources.

PV Work The gas in the cylinder is the system How much work is performed on the gas in a cylinder (system) when compressing the gas? Generically: w=-f dist Expansion: dw=-p ex dv V 1 2 w P V dv V ex

Expansion when P=0, w=-p dv ex V 2 work P dv P (at constant pressure) V V ex ex V 1 Compress Gas ΔV = V final -V initial is negative Work is positive, work is done on system System gains energy Expand Gas V = V final -V initial is positive Work is negative, work is done by system System loses energy

Reversible Processes A process effected by infinitesimal changes in a variable. Proceeds through h a sequence of equilibrium i states t One always remains on the surface of an equation of state Idealized process takes infinitely long to carry out The system remains in equilibrium throughout the process and can be reversed by an infinitesimal change in the variable. The work done by the system in a reversible expansion from A to B is the maximum work that the system can perform in changing from A to B

Reversible and Irreversible Work The path is a portion on of the eq. of state surface 1 The path is not Completely on the eq. of state surface 2 Reversible: system and surroundings in equilibrium P P P int ex Irreversible: system and surroundings not in equilibrium w P P P V V P V V 0 2 1 2 2 1 2 2 1 P P ex

Reversible compression work with an ideal gas Reversible, so P ext = P sys = P For constant P we have already shown that w = -P ext ΔV V2 V nrt V 2 w P V dv dv V1 V1 V Isothermal: T= cons t PV = nrt V 1 V 1 V 1 V nrt ln V ln V 2 1 V 1 nrt ln V 2 2 w nrt dv V 2 A ibl di b ti (i l t d i t ) i A reversible adiabatic (insulated piston) compression requires is a bit more complex and will not be needed here.

Reversible adiabatic compression work with an ideal gas First law: du = dq + dw = dq - PdV (insulated gas conta ainer) P ext = P PV = nrt V 1 V 2 Adiabatic: dq = 0 So du + PdV = 0 V 2 V nrt V 2 w P V dv dv V1 V1 V We might return to this case later, but for an ideal gas, we can derive an adiabatic equation of state: PV PV 1 1 2 2 CP 5 where for an ideal gas, C 3 and V V w C T T 2 1

Heat Transactions du = dq + dw exp + dw e 0 0 At constant volume, dw exp =0 0, and for no additional work, such as electrical work, we have du=dq V or U=q V

Math for Heat Transactions U is a state function It depends only on state, t not on path to get there U = U final -U initial This means mathematically that du is an exact differential and U i f du For now, consider a system of constant composition. U can then be regarded d as a function of V, T and P. Because there is an equation of state relating V, T, and P, any two are sufficient to characterize U. So we could have U(P,V), U(P,T) or U(V,T).

Math for Heat Transactions So we could choose U(p,V), U(pT) U(p,T) or U(V,T). Exact differential review: F(x,y) F F df dy dx x y y Let us choose U = U(V,T) When V V + dv at cons t T, U changes to x U U' U dv V T

Math for Heat Transactions Or in general, U U U' U dv dt V T T For infinitesimal changes, U U du dv dt V T T V V Some terms are familiar: U Heat capacity at constant volume C U V T TT V VV T Internal pressure at constant temp

Heat Capacity U U du dv dt V T T (V) T C V (T) If dv = 0, then du = C V (T)dT However, C V (T) is approximately constant over small temperature changes and above room temperature so integrate both sides: U = C V T Since constant volume: q V =C V T V

Internal Energy (5) Many useful, general relationships are derived from manipulations of partial al derivatives, ves, but I will (mercifully) spare you more. Suffice it to say that U is best used for processes taking place at constant volume, with only PV work: Then du = dq V and U= U 2 U 1 = q V The increase in internal energy of a system in a rigid container is thus equal to the heat q V supplied to it. We would prefer a different state function for constant pressure processes, one that naturally accounts for PV work on the surroundings: enthalpy.

Enthalpy Defined Enthalpy, H U + PV dh du d PV du PdV VdP At Constant P, H = U + PV U = q + w q = q P = U - w, and w = -PV So q P = U + PV H If, in addition, V is constant, then U = H = q

Comparing H and U at constant P H = U + PV A B C D 1. Reactions that do not involve gases V 0 and H U 2. Reactions in which n gas = 0 V V 0 and H H UU 3 Reactions in which n 0 3. Reactions in which n gas 0 V 0 and H U

ConcepTest #1 Which of the following reactions has the largest difference between H and U? A. NH 3 (g) + HCl (g) NH 4 Cl (s) B. CO (g) + Cl 2 (g) COCl 2 (g) C. ZnS (s) + 3/2 O 2 (g) ZnO (s) + SO 2 (g) D. ZnO (s) + CO (g) Zn (s) + CO 2 (g)

Help Session Thursday 7-9 PM JILA Auditorium Enter through door on the east side of the underpass between JILA and LASP

ConcepTest #1 Which of the following reactions has the largest difference between H and U? A. NH 3 (g) + HCl (g) NH 4 Cl (s) n gas = -2 B. CO (g) + Cl 2 (g) COCl 2 (g) n gas = -1 C. ZnS (s) + 3/2 O 2 (g) ZnO (s) + SO 2 (g) n gas = 0 D. ZnO (s) + CO (g) Zn (s) + CO 2 (g) n gas = 0

Comparing H and U at constant P (2) H = U + PV 1. Reactions that do not involve gases V 0 and H U 2. Reactions in which n gas = 0 V V 0 and H H UU 3. Reactions for which ngas 0 H U + n gas RT with Ideal Gases assumed

Heat Capacity at Constant Volume or Pressure Constant Volume dq V U C V dt T V Constant Pressure dq P H U PV C P dt T T T P P P Ideal Gas: PV = nrt and hence C P = C V + nr

Heat Capacity, C Heat Capacity (J K -1 ) Heat needed to raise T of system by 1 K q = CT Specific Heat Capacity (J K -1 kg -1 ) Heat needed to raise T of 1 kg by 1 K q = C S mt Molar Heat Capacity (J K -1 mol -1 ) Heat needed to raise T of 1 mole by 1 K q = C m nt

Calorimetry Measure H and U for Reactions Isolate sample, bomb, and water bath from surroundings Initiate reaction Heat released causes temperature rise in system

Calorimetry Problem For the complete combustion of 1 mole ethanol in a bomb calorimeter, 1364.5 kj is released at 25 C. Determine c Uand c H. C 2 H 5 OH (l) + 3 O 2 (g) c 2 CO 2 (g) + 3 H 2 O (l) c U= q V = 1364.5 kj Now for c H n = -1 At 25 C, RT =8.314 J/ K 298.15 K = 2.5 kj c H= c U + nrt = 1364.5 2.5 = 1367.0 kj/mol

Endothermic & Exothermic Processes H = H final -H initial H H Positive Positive amount of heat absorbed by the system Endothermic Process H Negative Negative amount of heat absorbed (i.e. heat released by the system) Exothermic Process

ConcepTest #2 For which of the following reactions is the indicated sign of H incorrect? A. H 2 O (s) H 2 O (l) H = + B. CH 4 +2O 2 CO 2 +2HO 2 H= = C. H 2 O (g) H 2 O (l) H = + D. 2 H (g) H 2 (g) H =

Thermochemical h Equations CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (l) (a combustion reaction) c H= 890 kj Reaction must be balanced Phases must be specified H is an extensive property Sign of H changes when reaction is reversed

Calculation l of rxn H I. Hess s Law H of an overall process is the sum of the Hs for the individual steps II Use of Standard Enthalpies of II. Use of Standard Enthalpies of Formation

I. Hess s Law Calculate H for the reaction: H 2 O (solid at 0 C) H 2 O (gas at 100 C) H 2 O (s, 0 C) H 2 O (l, 0 C) fus H = 6.0 kj/mol H 2 O(l (l, 0 C) H 2 O(l (l, 100 C) CT = 7.5 5kJ/mol H 2 O (l, 100 C) H 2 O (g, 100 C) vaph = 40.6 kj/mol H 2 O (s, 0 C) H 2 O (g, 100 C) rxn H = 54.1 kj/mol

II. Standard State The Standard State of an element is defined to be the form in which it is most stable at 25 C and 1 bar pressure Some Standard States of elements: Hg (l) O 2 (g) Cl 2 (g) Ag (s) C (graphite) The standard enthalpy of formation ( f H ) of an element in its standard state is defined to be zero.

II. Standard Enthalpy of Formation The standard enthalpy of formation ( f H ) of a compound is the enthalpy change for the formation of one mole of compound from the elements in their standard state. Designated by superscript o: H For example, CO 2 : C (graphite) + O 2 (g) CO 2 (g) rxn H = -393.5 kj/mol Appendix D f H CO 2 (g) = -393.5 kj/mol

ConcepTest #3 For which of the following reactions is rxn H = f H of the product? A. 2Ag(s) +Cl 2 (g) 2AgCl (s) B. Ca (s) + C (diamond) + 3/2 O 2 (g) CaCO 3 (s) C. H 2 (g) +1/2O 2 (g) H 2 O (l) D. H 2 O (l) + 1/2 O 2 (g) H 2 O 2 (l)

Enthalpies of Reaction The enthalpy of reaction can be calculated from the enthalpies of formation of the reactants and products rxnh = f H (Products) - f H (Reactants)

Example: Find rxn H (using Standard Enthalpies of Formation) CH 4 (g) + 2 O 2 (g) CO 2 (g) +2HO 2 (l) f H (from Appendix D, text): CH 4 (g) -74.6 kj/mol O 2 (g) 0 CO 2 (g) -393.5 H 2 O (l) -285.88 rxn H = -393.5-2 (285.8) 0 (-74.6) kj/mol So rxn H = -890.5 kj/mol

Bond Dissociation Enthalpy BDE is enthalpy for breaking a specific bond Water has two O-H BDEs: HOH H + OH OH H + O H = 499 kj/mol H = 428 kj/mol HOH 2H + O H = 927 kj/mol Bond Enthalpy is the average value: O H bond enthalpy is 963/2 = 463 kj/mol

Sample Problem The enthalpy of formation of CH 4 (g) is -74.6 kj/mol The enthalpy of sublimation of graphite is 716.7 7 kj/mol The bond dissociation enthalpy of H 2 (g) is 436.0 kj/mol Determine the C-H bond enthalpy in CH 4 CH 4 (g) C (graphite) + 2H 2 (g) C (g) + 4H (g) -74.6 0 0 716.7 872 Answer: (1663.3/4) 3/4) kj/mol = 415.8 kj/mol This result is the average CH bond strength in methane, NOT D(H 3 C H) (g)

Sample Problem Find ΔH for H 2 O (s) 0 C H 2 O (g) 100 C sublimation: Happens all around: snow, ice in fridge, dry ice This is clearly not a constant T process. Since H is a state function, define any path to get to product. Make it simple to use. Calculate ΔH for that path. Here is one: H 2 O (s) H 2 O( (g) fusion H fusion + 6 kj/mol H vap +40.6 kj/mol vaporization H 2 O () 0 ºC C p,l =33.6 J/mol K heating H 2 O () 100 ºC 100C H H C, dt H 0C fusion p vap CpT for H O 2

ConcepTest #0 Find H for the reaction: CH 4 (g) + H 2 O(l) H 2 CO(g) +2H(g) 2 using the following standard heats of formation: CH 4 (g) H 2 CO(g) H 2 O(g) -74.8 kj/mol -108.6 kj/mol -241.8 kj/mol a. Less than 208.0 0 kj/mol b. Equal to 208.0 kj/mol c. Greater than 208.0 0 kj/mol d. Not possible to tell

ConcepTest #0 Solution Find H for the reaction: CH 4 (g) + H 2 O(l) H 2 CO(g) + 2H 2 (g) using the following standard heats of formation: on: CH 4 (g) H 2 CO(g) H 2 O(g) -74.8 kj/mol -108.6 kj/mol -241.8 kj/mol ΔH = [ΔH f (H 2 CO(g))+ ΔH f (2H 2 (g)) ] [ΔH f (CH 4 (g)) + ΔH f (H 2 O (l)) ] We need this H o (H 2 O(l) ) = H o (H 2 O(g) ) + H fus (H 2 O(g) ) = -241.8 + F (a negative number) ΔH =[-108.6 +2x0] [-74.8 +(-241.8 +F)] = 208 F Since F is negative, ΔH must be > 208 kj/mol

Thermochemistry Revisited Consider some chemical reaction is taking place at constant P ex, and the temperature changes from T i to T f as the reaction proceeds. We can write rxn rxn i p H H T C products T where H rxn is the enthalpy change for the stated conditions, and ΔT = (T f T i ). STANDARD ENTHALPY CHANGE Refers to enthalpy change under set of standard d conditions. The standard d enthalpy change ΔH rxn for a process A B is the change in enthalpy H B H A when A and B are in their standard states. The components of A and B are also separated from each other. That s double talk! What is a standard state?? STANDARD STATE The standard state of a substance at a specified temperature is its pure form at that temperature. Moreover, for the very special case that T=298.14 K and P = 1 atm, we call H rxn the standard reaction enthalpy and designate this condition with a superscript o. 298 H o at P = 1atm (or sometimes or even ) H process process However, the standard enthalpy change of a Rxn can be defined at any temperature

Rx Enthalpy as f(t) Consider the reaction 2A + 3B C + 2D at two temperatures T 1 and T 2. What is the difference in the standard enthalpies of formation at these temperatures? For constant C p, T o o 2 o o H H C C r r T r r 1 dt T T 2 1 P P 2 1 C C C r o o o P P,m P,m products reactants The are the coefficients in the reaction above. m is molar.

Ammonia Heat of Formation Thus, suppose we want to determine the heat of formation of NH 3 at 400 K. ½ N 2 + 3/2 H 2 NH 3 H 400 H C T f f P kj? 400 298 46.11 mol How to obtain ΔC P? Cp(NH 3 ) = 35.06 J/mol K Cp(N 2 ) = 29.1 J/mol K Cp(H 2 ) = 28.8 J/mol K all from tabulated values 1 3 J J So C 35.06 29.1 28.8 22.7 P 2 2 mol K mol K kj 22.7 kj kj and Hf 400 46.11 400 298 48.43 mol 1000 mol mol get J to kj This approach assumes that the C P are all constant over this temperature range. This assumption is certainly valid for ideal gases; it is certainly false if there is a phase change, and it is not exact for real gases!

Propene Hydrogenation/Combustion What s propene? H 2 C=CH-CH 3 Does it matter where the double bond goes? What s the hydrogenation Rxn? H H C C H H C H H + H 2 H H H H C C C We know (from table) that ΔH for this reaction is -124 kj/mol at standard conditions. All components are gases. Now we use Hess Law to find the standard heat of reaction for the combustion of propene, CH 9 3 6 O2 3CO2 3HO 2 2 g g g We need to find a series of reactions that have known ΔH and add up to this one. H H H H

Propene Hydrogenation/Combustion 9 CH O 3CO 3HO 3 6 2 2 2 2 3 6 2 3 8 C H H g C H -124 C H 5O 3CO 4H O -2220 3(-393.5)+ 4 285.88 103.8 3 8 2 2 2 1 H O H 2 2g O 2g 286 - H H O 2 2 f 9 kj C H O 3CO 3H O 3 6 2 2 2-2058 2 mol

ConcepTest #1 Pressure 4 3 1 2 An ideal gas goes through the four step process shown. In which stages is work done? Vl Volume a. In 1 and 2 only d. In none of the stages b. In 1 and 3 only e. In all of the stages c. In 2 and 4 only

ConcepTest #2 Pressur re The curve shown at left represents the expansion n of an ideal gas. If this process is isothermal then we can also say: Volume a. There is no heat involved in this process. b. There is no work involved in this process. c. There is a transfer of thermal energy, or heat, into the gas. d. There is a transfer of thermal energy, or heat, out of the gas.

ConcepTest #3 Two 10 kg weights sit on a piston, compressing the air underneath. One of the weights is removed, and the air underneath expands from 18.3 to 20.0L. Then the second weight is removed and the air expands from 20.0 to 22.0L. How does the amount of work done compare if instead both weights were removed at once? Assume the same total change in volume. a. More work is done removing one weight at a time than removing both weights at once. b. Less work is done removing one weight at a time than in removing both weights at once. c. The same amount of work is done removing one weight at a time, or if both are removed at once.

P Adiabatic Expansion/Compression P 1, V 1, T 1 A B P 3, V 3, T 3 C P 2, V 2, T 2 V This is why we call U or H state functions. Consider the three paths, A, B and C UA UB UC and HA HB HC but qa qb qc and w w w A B C Suppose that path A in the 2 D slice in PV above characterizes the expansion, with P 1, V 1 and T 1 being the initial condition. Further, let path A be a reversible adiabatic path. What can we say experimentally? P 1, V 1, T 1 P 2, V 2, T 2 Is V 1 < V 2 or is V 1 > V 2? Is T 1 < T 2, is T 1 =T 2, Is P 1 < P 2 or is P 1 > P 2? or is T 1 > T 2? Now we calculate the cooling in a the reversible adiabatic expansion of an ideal gas.

Adiabatic Expansion/Compression Calculate the cooling in a the reversible adiabatic expansion of an ideal gas. P P 1, V 1, T 1 A du q w First Law: Since the process is adiabatic, q = 0. Also w = -p ex dv (definition) Since the process is reversible, p ex = p int and nrt du dwrev pdv dv V V P 2, V 2, T 2 Recall from earlier that du = C V dt by definition, and so we have We cannot integrate directly because T is changing with V. So we simplify by dividing by T to obtain We can now integrate each side C T V The last term arises because the gas is ideal. Note how we have used the given parameters. dt nr dv V CdT V nrt V dv

Adiabatic Expansion/Compression P 1, V 1, T 1 CV T dt nr V dv P A V a Recall that aln b ln b, so T f T i C T V dt V f V i nr dv V and T ln ln ln f V P CV f Vi 2, V 2, T 2 nr T i V i V f V nr C T T V f f V i ln ln ln nr Ti Ti V f C Exponentiating both sides, C V T nr f V i T V i f

Adiabatic Expansion/Compression C T V V f f V i Repeating, ln ln ln nr T V V i i f We could equally well have included the C/nR term with the volumes to obtain nr C T V f nr V f nr V i V i ln ln ln ln T C V C V V i V i V f f nr C V V i and T T for a reversible adiabatic expansion f i V f nr C f f nr C 3 V V V Now TV TV i i i For an ideal gas, C nr, and so T T V f i 2 V f We knew for our expansion that V i < V f, so we have proven that T f < T i in a reversible adiabatic expansion. 2 3

Adiabatic and Isothermal Processes We again make use of the ideal gas behavior to obtain another relationship. If we use the ideal gas equation of state to replace the temperatures in 2 3 V i T T f i V f, then Adiabatic: 2 3 PV PV V 2 2 1 1 1 nr nr V 2 PV PV 2 2 1 1 V V or PV PV 2 1 2 2 1 1 nr nr 2 2 5 5 3 3 3 3 Now consider an isothermal expansion. In an isothermal expansion of an ideal gas, T 1 = T 2, and Boyle s Law gives PV P V PV P V 1 1 or This gives a way to unify things: 1 1 2 2 1 1 2 2 PV PV 1 1 2 2 where 1 for an isothermal expansion and 5 and for an adiabatic expansion. 3

Adiabatic Expansion/Compression We now can look back at the early picture with deeper understanding Ideal Gas Eq. of State Surface

Adiabatic Compression - Example Adiabatic: 2 3 V i f i V f T T V If ideal, 5 5 3 3 PV f f i i PV Boulder can have windstorms, in which the temperature can rise substantially ( 50 F) in 30 minutes on a strong west wind. It is a bit like the jet stream blowing near treetop level. Can we estimate t the temperature t increase? Consider an adiabatic compression from the continental divide (P=0.5 Bar) to Boulder (P=0.8 Bar). What do we estimate for ΔT if T divide = 280 K? 5 3 5 5 3 5 V f P V i f P i PV f f i i V P V P i f i f 3 3 PV gives and 0.625 0.754 0.6 T f T i 2 3 V i 1 V 0.754 f 0666 0.666 So if T i = 280K K, T f =307K K, 1.207 an increase of 49F

Exam 1 Review Ideal Gas Law Deduced from Combination of Gas Relationships: V 1/P, Boyle's Law V, Charles's Law V n, Avogadro's Law Therefore, V nt/p or PV nt PV = nrt =Nk B T where R = gas constant (per mole) or k B = gas constant (per molecule) The empirical Equation of State for an Ideal Gas

Ideal Gas Equation of State

Real Gases Compressibility PV PVm Z nrt RT Z = 1 at all P, T Ideal Gas Behavior Ideal gas Now look at real gases at some temperature T Look at a broader 0 800 atm region

General Principle!! Energy is distributed ib t d among accessible configurations in a random process. The ergodic hypothesis Consider fixed total energy with multiple particles and various possible energies for the particles. Determine the distribution that occupies the largest portion of the available Phase Space. That is the observed distribution.

Energy Randomness is the basis of an exponential distribution of occupied energy levels n(e) A exp[-e/<e>] Average Energy <E> ~ k BT n(e) A exp[-e/k B T] This energy distribution is known as the Boltzmann Distribution.

Maxwell Speed Distribution Law 1dN Ndu 32 m 2 2 mu B B 4 ue 2 kt 2k T 1dN is the fraction of molecules per unit speed interval Ndu

Maxwell Speed Distribution Law N u du 1 u 2 u 1 N u du fraction of particles with u between u and u 0 1 2 Most probable speed, u dn 2 mp du 0 for u = u mp u mp kt m Average speed, <u> or ū u un u du 8kT m 0 m u 3kT 2 u 2 N u du m Mean squared speed, <u 2 > 0 Root mean square speed u rms u 2 3 3kT m

Distinguish between System & Surroundings

Internal Energy Internal Energy (U) is the sum of all potential and kinetic energy for all particles in a system U is a state function Depends only on current state, not on path U = U final -U initial

Internal Energy, Heat, and Work If heat (q) is absorbed by the system, and work (w) is done on the system, the increase in internal energy (U) is given by: U = q (heat absorbed by the system) + w (work done on the system)

Reversible and Irreversible Work Work done on the gas = also the min work required to compress the gas P 1 (V 2 -V 1 )

Internal Energy (2) U is a state function It depends only on state, not on path to get there U = U final - U initialiti This means mathematically that du is an f exact differential: U du For now, consider a system of constant composition. U can then be regarded as a function of V, T and P. Because there is an equation of state relating them, any two are sufficient to characterize U. So we could have U(P,V), U(P,T) or U(V,T). i