Kinetic theoryKinetic theory

class 15:(ThT Q)(ThT Q)

Did you complete at least 70% of Ch 21 1 3 ?Chapter 21:1-3 ?

A YA.Yes We will use the CSReviewB.No ReviewKT: using online demo.

& formulae& formulaeDegrees of freedom H t it tiHeat capacity ratio:

γ = CP/CV & other γ P VCP-CV relations

Today: y1. Kinetic theory of gases. 2. Degrees of Freedom 3 γ the adiabatic exponent

Wed noon

►3. γ, the adiabatic exponent

Remember: the online schedule has precedence.

Consider a mole of He at 300K. If we compress the gas to halfIf we compress the gas to half its volume, which process , prequires the most work? pp.A.isothermalB adiabatic fB.adiabaticC.both same f

fa

Note: lines go backward in compression; but from

iin compression; but from where do you start?

Microscopic view of how atoms hi i llhitting wall creates pressure.

From CS: Chapter 21:Kinetic Theory:The ideal gas law works for all atoms and

molecules at low pressure. It is rather amazing gthat it does. Kinetic theory explains why. The properties of an ideal gas can be understood by thi ki f it N idl i ti l fthinking of it as N rapidly moving particles of mass m. As these particles collide with the container walls momentum is imparted to thecontainer walls, momentum is imparted to the walls, which we call the force of gas pressure. In this picture the pressure is related to the p paverage of the square of the particle velocity by

View adiabatic online.22 1( )NP

2( )

3 2P mv

V

Average translational kinetic energyAverage translational kinetic energy21 3

2 2bmv K T

2 2avg

RKBA

RKN

3RTVBoltzmann’s Constant

rmsVM

Amazing Results!Amazing Results!• <KE> depends only on T! Now, we can call the p y ,

total kinetic energy of the gas the thermal energy of the gas. For an ideal gas, this is the internal energy E So E is proportional tointernal energy, Eint. So Eint is proportional to T. 3

rmsRTV rmsVM

• Air at 20°C ►Vrms = 500 m/s = 1100 mphrms p

Which molecule has the highest l i ?average velocity?

In our classroom there are molecules of N2In our classroom there are molecules of N2and CO2. N2 has less mass than CO2(M=28 vs 44) Which molecule has the(M 28 vs. 44). Which molecule has the highest average velocity?

A NitrogenA. NitrogenB. Carbon DioxideC. You can’t trick us, Prof. Allred, they are

the same.

Formulae22 1( )NP mv

Formulae

1 ( )3 2

P mvV

1. .2. Using the ideal gas law, we obtain the

average translational kinetic energy per21 3

2 2 Bmv k T,

average translational kinetic energy per molecule:

2 2 B

3. The rms speed is then given by2

rms3 3Bk T RTv v

m M

Degrees of Freedom:gRoughly speaking, a degree of freedom is a way in

which a molecule can store energy. For instance since there are three differentinstance, since there are three different directions in space along which a molecule can move,

1 th th d f f d f th1. there are three degrees of freedom for the translational kinetic energy.

2. There are also three different axes of rotation2. There are also three different axes of rotation about which a polyatomic molecule can spin, so we say there are three degrees of freedom for the rotational kinetic energythe rotational kinetic energy.

3. There are even degrees of freedom associated with the various ways in which a molecule can ib t d ith th diff t l l ivibrate, and with the different energy levels in

which the electrons of the molecule can exist.

It takes more Q to raise temperature at constant P than constant V, because someconstant P than constant V, because some Q goes to work. Higher

temperaturestemperatures are the upperblue lines.blue lines. For ideal gases the blue lines also mark the li flines of constant Eint.

Molar Specific Heat of an Ideal Gas C V lat Constant Volume:

VQ nC T 5 polyatomic2VC

3 monatomic2VC R 5 diatomic

2VC R

2

Real gases deviate from these formulas because in addition to the translational and rotational degrees of freedom they also have

.

rotational degrees of freedom, they also have vibrational and electronic degrees of freedom. These are unimportant at low temperatures due to quantum mechanical effects but becomequantum mechanical effects, but become increasingly important at higher temperatures. The rough rule is: No of degrees of freedomNo. of degrees of freedom

2VC R

Consider a mole of He in a sealed jar d l f N i th W h tand a mole of N2 in another. We heat

each gas to raise its temperature from 20o C to 100o C. For which gas will the ∆Eint be greatest?int g

A.HeB.N2C both sameC.both same

Molar Specific Heat of an Ideal Gas C Pat Constant Pressure:

The internal energy of an ideal gasThe internal energy of an ideal gas depends only on the temperature:

C C RPQ nC T P VC C R .

The internal energy of an ideal gas depends only on the temperature:depends only on the temperature:

E nC Tint VE nC T

Consider a mole of N2. We heat the 2gas to raise its temperature from 20o C to 100o C For which case20o C to 100o C. For which case will the ∆Eint be greatest?A.Constant

volumevolumeB.Constant

PPressureC both sameC.both same

Adiabatic Processes in an Ideal G C /CGas: γ = CP/CV.

An adiabatic process is one in which no heatAn adiabatic process is one in which no heat is exchanged between the system and the environment. When an ideal gas expands or contracts adiabatically, not only does its pressure change, as expected from the ideal

l b t it t t h ll.

gas law, but its temperature changes as well. Under these conditions the final pressure, Pf, can be computed from the initial pressure Pcan be computed from the initial pressure, Pi, and from the final and initial volumes, Vf and Vi, by t tP V PV PV y or constantf f i iP V PV PV

γ = CP/CV.

The quantity γ is called the di b iadiabatic exponent.

For monatomic gases: the nobel gases: He, Ar, Ne g g , ,etc. γ=Cp/Cv =(5/2R)/(3R/2) =5/3 =1.67diatomics like N2 and O2; γ=7/2R/(5R/2) =1.4

Note that this doesn’t mean that the ideal gas law no longer holds; it does, and in fact it can be combined with the adiabatic law for pressure

.

.

combined with the adiabatic law for pressure given above to obtain the adiabatic law for temperatures: This is useful for computing T.

1 constantTV

Consider a mole of He at 300K. If we compress the gas to halfIf we compress the gas to half its volume, which process , prequires the most work? pp.1. isothermal2 adiabatic f2. adiabatic3. both same f

fa

Note: lines go backward in compression; but from

iin compression; but from where do you start?

• 1 mol He at 300 K– Compress to volume isothermally1

2

fV• W = ln 1730f

i

VnRT JV

• 1 mol He at 300K– Compress to ½ volume adiabatically

1 1• Eq1 (21,20) in textbook: 1 1i i f fT V T V

1

476if

VT T K

Here is why the

intE Q W W

476f if

T T KV

adiabatic process

takes more work. The k i i iQ

int32

vW E nC T nR T

work going in raises the temperature, this increases the2

2194J

increases the pressure which makes it more difficult to compress the gas.

Quiz 14-4 (participation point)Quiz 14 4 (participation point)

Consider a mole of He gas and a mole ofConsider a mole of He gas and a mole of N2 gas, both at the same initial pressure and temperature. Which p prequires the most work to adiabatically compress to half its

l ?volume?A. HeB. N2C. Both the same

• Work → Internal energy (kinetic energy)

• He: all W → translational K• He: all W → translational K• N2: W ► translational K

+ rotational K► energy from work spreads out.► energy from work spreads out.► molecules don’t go as fast.► temperature doesn’t rise as far► temperature doesn t rise as far.►pressure doesn’t rise as much.►less pressure ► less work►less pressure ► less work.

Compressions (& rarefactions) in sound waves are adiabatic because they happen too rapidly for any appreciable pp p y y ppamount of heat to flow. This is why the adiabatic exponent γ appears in theadiabatic exponent, γ, appears in the formula for the speed of sound in an ideal gas:ideal gas:

RTvM

Sound WavesC i di b i

Compressions are adiabatic

VR

V

Air: γ =1.40; M=29 g/mol v = 343 m/s at 20 ̊CHe: Larger γ ► Larger v & Smaller M ► Larger vSo v in He is very large!Standing wave in a tube: f= v/2L; if v↑ then f ↑

(Squeaky voice and flute note frequency ↑ )(Squeaky voice and flute note frequency ↑.)Computational Hint: when computing v using M, remember to convert M from grams to kgremember to convert M from grams to kg.

2VFL

13-1 -3 are based on this material #2 have13 1 3 are based on this material. #2 have the heat capacity of Cu (review) and heating of gassesheating of gasses

13.4 has work involved. We will do a sample numerical problemsample numerical problem.

13-413 4

We heat up 2.43 mol of some gas from 21.7C toWe heat up 2.43 mol of some gas from 21.7C to [05] C, holding the pressure constant. The molar specific heat at constant volume for this gas is CV = [06] J/mol · K.

(a) Assuming that this gas behaves like an ideal gas, find the value of the molar specific heat at constant pressure CP .

(b) Fi d h f h h fl i h(b) Find the amount of heat that flows into the gas.(c) Find the amount of work done on the gas. (d) Find the change of internal energy of the gas.