Adiabatic Expansion/CompressionCalculate the cooling in a the reversible adiabatic expansion of an ideal gas.

P1, V1, T1First Law: Since the process is adiabatic, q = 0.

l dV (d f )

dU q w

P

Also w = -pex dV (definition)Since the process is reversible, pex = pintand

  nRT

P V T

A

The last term arises because the gas is ideal.

revnRTdU dw pdV dVV

P2, V2, T2

V

ideal.Note how we have used the given parameters.

nRTC d dll f l h d d b d f d h VnRTC dT dVV

C R

Recall from earlier that dU = CV 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

VC nRdT dVT V

So w s mp fy y ng y to o ta n

We can now integrate each side

Adiabatic Expansion/Compression

VC nRdT dVT V

P1, V1, T1

f f

i i

T VV

T V

C nRdT dVT V

PA

and

ln ln ln

i iT V

f fV i

i i f

T VC VnR T V V

P2, V2, T2

i i f

VCnR

f fV iT TC V

V

ln ln ln

f fV i

i i f

C VnR T T V Recall that ln ln , soaa b b

VC

Exponentiating both sides,

nRf i

i f

T VT V

Adiabatic Expansion/Compression

Repeating, ln ln lnf fV i

i i f

T VC VnR T V V

We could equally well have included the C/nR term with the volumes to obtain

nRCT V V VR R

VC

nR

ln ln ln lnf f i i

i V i V f f

T V V VnR nRT C V C V V

VC

and for a reversible adiabatic expansionif i

f

VT TV

2233For an ideal gas, , and so

2i

V f if

VC nR T TV

V V

nR nRC C Now i if fT V TV

We knew for our expansion that Vi < Vf, so we have proven that Tf < Ti in a reversible adiabatic expansion.

Adiabatic and Isothermal ProcessesWe 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

23PV PV V

2

32 2 1 1 1

22 2 5 5

PV PV VnR nR VPV PV

3i

f if

VT TV

, then

2 2 5 52 2 1 13 3 3 3

2 1 2 2 1 1 or PV PVV V PV PVnR nR

Adiabatic:

Now consider an isothermal expansion.In an isothermal expansion of an ideal gas, T1 = T2, and Boyle’s Law gives

1 1 o r PV PV PV PV This gives a way to unify things:1 1 2 2 1 1 2 2 o r PV PV PV PV

1 1 2 2PV PVwhere 1 for an isothermal expansion and

This gives a way to unify things:

where 1 for an isothermal expansion and5and for an adiabatic expansion.3

Adiabatic Expansion/Compression

 We now can look back at the early picture with deeper understanding

Ideal Gas Eq. of StateSurface

Adiabatic Compression - Example

5 53 3

f f i iPV PV23

if i

VT TV

If ideal, Adiabatic:fV

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 blowingnear treetop level. C ti t th t t i ?Can we estimate the temperature 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 Tdivide = 280 K?

35

5 5 53 0.63 3 gives and 0.625 0.754f fi i

f f i ii f i f

V VP PPV PVV P V P

20 666 So if T = 280K T =307K0.6663 1 1.207

0.754f i

i f

T VT V

So if Ti = 280 K, Tf =307 K,an increase of 49F

Exam 1 Review

Deduced from Combination of Gas Relationships:

Ideal Gas Law

Deduced from Combination of Gas Relationships:

V 1/P, Boyle's Law

V , Charles's Law

V n, Avogadro's Law

PV = nRT =Nk TTherefore, V nT/P or PV nT

PV = nRT =NkBTwhere R = gas constant (per mole)or kB = gas constant (per molecule)or kB = gas constant (per molecule)

The empirical Equation of State for an Ideal Gas

Ideal Gas Equation of StateIdeal Gas Equation of State

Ideal Gas LawIdeal Gas Law

PV = nRTwhere R = universal gas constant

R = PV/nT

R = 0 0821 atm L mol–1 K–1

R = 8 314 J mol–1 K–1 (SI unit)

R 0.0821 atm L mol KR = 0.0821 atm dm3 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

Real GasesCompressibility PVPVZ

nRT RTm

Z = 1 at all P, T Ideal Gas Behavior

Now look at real gases

Ideal gas

gat some temperature T

Look at a broader 0 – 800 atm region

van der Waals equation of stateW q f

Physically-motivated corrections to Ideal Gas EoS.y yFor a real gas, both attractive and repulsive intermolecularforces are present. Empirical terms were developed to help accountfor both.

2

2 2

nRT an RT aPV nb V V b V

V nb V V b V

General Principle!!General Principle!!

E i di t ib t d iblEnergy is distributed among accessibleconfigurations in a random process.

The ergodic hypothesis

Consider fixed total energy with multiple particles and various possible energies for p p gthe particles.

Determine the distribution that occupies h l i f h il bl “Ph the largest portion of the available “Phase

Space.” That is the observed distribution.

Energy Randomness is the basis of an exponential distribution of of an exponential distribution of

occupied energy levelsn(E) A exp[-E/<E>]

Average Energy <E> ~ kBTr g En rgy E B

n(E) A exp[-E/kBT]

This energy distribution is known as the Boltzmann Distribution.

Maxwell Speed Distribution LawMaxwell Speed Distribution Law

2B

3 2mu 2k T21 dN m4

Bmu 2k T2

B

4 u eN du 2 k T

1 dN is the fraction of molecules per unit speed intervalN duN du

Maxwell Speed Distribution Law 2u

0

1

N u du

1

1 2

fraction of particles with u between u and u

u

N u du

Most probable speed, ump20 for u = u mp mp

dN kTudu m

0 1 2

Average speed, <u> or ū 8kTu u N u dum

0 m

M n s d sp d 2> 2 2 3kTu u N u du

Mean squared speed, <u2> 0

u u N u dum

3kTRoot mean square speed 2 3

rmskTu um

Distinguish betweenDistinguish betweenSystem & Surroundings

Internal EnergyInternal Energy

Internal Energy (U) is the sum of all potential and kinetic energy for all

U is a state function

potential and kinetic energy for all particles in a system

U is a state functionDepends only on current state, not on path

U = Ufinal - Uinitial

Internal EnergyInternal 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

i n b :

U = q (heat absorbed by the system)

given by:

U = q (heat absorbed by the system)+ w (work done on the system)

Reversible and Irreversible WorkW

Work done on

P1(V2-V1)

Work done onthe gas =

also the min work required tocompress the gas

P1(V2 V1)

ConcepTest #3Two 10 kg weights sit on a piston, compressing the air

underneath. One of the weights is removed, and

p

gthe 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 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 L k i d i i h i h i i 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 or if both are removed at once.

Internal Energy (2)Internal Energy (2)U is a state functionU is a state function

It depends only on state, not on path to get thereU = Ufi l - Ui iti lU = Ufinal Uinitial

This means mathematically that dU is anf

exact differential: i

U dU

For now consider a system of constant compositionFor 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,q gany two are sufficient to characterize U.So we could have U(P,V), U(P,T) or U(V,T).

Enthalpy DefinedEnthalpy DefinedEnthalpy, H U + PV

At Constant P,

H = U + PV

U = q + w

H = U + PV

q= qP = U - w, w = -PV

q = U + PV= HqP = U + PV= H

At constant V, q = U = H

Comparing H and UComparing H and Uat constant PH = U + PV

1. Reactions that do not involve gasesV 0 and H U

2. Reactions in which ngas = 0V 0 d H UV 0 and H U

3 Reactions in which n 0 3. Reactions in which ngas 0 V 0 and H U

Heat Capacity atHeat Capacity atConstant Volume or Pressure

CV = dqV/dT = (U/T)V

P ti l d i ti f i t l Partial derivative of internal energy with respect to T at constant V

CP = dqP/dT = (H/T)P

Partial derivative of enthalpy Partial derivative of enthalpy with respect to T at constant P

Ideal Gas: CP = CV + nR

Th h i l E tiThermochemical Equations

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l)

(a combustion reaction) cH = – 890 kJ

R ti t b b l d

Phases must be specified

Reaction must be balanced

H is an extensive property

Sign of H changes when reaction is reversed

Standard StateStandard State

The Standard State of an element is defined to The Standard State of an element is defined to be the form in which it is most stable at 25 °C and 1 bar pressureand 1 bar pressureSome Standard States of elements:

Hg (l) O2 (g) Cl2 (g) Ag (s) C (graphite)

The standard enthalpy of formation (fH°)of an element in its standard state is of an element in its standard state is defined to be zero.

Enthalpies of FormationEnthalpies of Formation

The standard enthalpy of formation ( H°)The standard enthalpy of formation ( f H )of a compound is the enthalpy change for the formation of one mole of compound from the f f f p felements in their standard states.

Designated by superscript o: H°Designated by superscript o: H

For example, CO2:

C (graphite) + O2 (g) CO2 (g)

H° = 393 5 kJ/mol Appendix DrxnH = -393.5 kJ/mol Appendix D

f H° CO2 (g) = -393.5 kJ/mol

E h l i f R iEnthalpies of Reaction

The enthalpy of reaction can be calculated f th th l i f f tifrom the enthalpies of formationof the reactants and products

rxnH° = fH°(Products) rxn f ( )

- fH°(Reactants)

Example: Find rxn H°(usin Standard Enthalpies f F rmati n)(using Standard Enthalpies of Formation)

CH (g) + 2 O (g) CO (g) + 2 H O (l)CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l)

f H° (from Appendix D, text):CH4 (g) -74.6 kJ/molO2 (g) 02 (g)CO2 (g) -393.5H2O (l) -285 8H2O (l) 285.8

rxnH° = -393.5 -2 (285.8) – 0 – (-74.6) kJ/mol

Therefore, rxnH° = -890.5 kJ/mol

Exam 1Exam

~ 5/6 problems (weights given) – budget your time Closed book Don’t memorize formulas/constants Don t memorize formulas/constants

You will be given things you needExam will not be heavily numeric, but will emphasize concepts

If a question seems lengthy do another problem & come back laterIf a question seems lengthy, do another problem & come back later Understanding homework will be useful