Chapter19:TheKine0cTheoryofGases

Thermodynamics = macroscopic picture

Gases micro -> macro picture

Avogadro’sNumberOne mole is the number of atoms in 12 g sample of carbon-12

C(12)—6 protrons, 6 neutrons and 6 electrons 12 atomic units of mass assuming mP=mn

NA=6.02 x 1023 mol-1

So the number of moles n is given by

n=N/NA

Another way to do this is to know the mass of one molecule: then

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N =Msample

mmole−massNA

Ideal Gases, Ideal Gas LawIt was found experimentally that if 1 mole of any gas is placed in containers thathave the same volume V and are kept at the same temperature T , approximately allhave the same pressure p. The small differences in pressure disappear if lower gasdensities are used. Further experiments showed that all low-density gases obey theequation pV = nRT . Here R = 8.31 K/mol ⋅K and is known as the "gas constant."The equation itself is known as the "ideal gas law." The constant R can be expressed as R = kNA. Here k is called the Boltzmann constant and is equal to 1.38 ×10-23 J/K.

If we substitute R as well as n = NNA

in the ideal gas law we get the equivalent form:

pV = NkT . Here N is the number of molecules in the gas.The behavior of all real gases approaches that of an ideal gas at low enough densities.Low densities means that the gas molecules are far enough apart that they do notinteract with one another, but only with the walls of the gas container.

IdealGasLaw

pV = nRT pV = NkBT

n = number of moles N = number of particles

gas constant

R = 8.315 J/(mol⋅K) = 0.0821 (L⋅atm)/(mol⋅K) = 1.99 calories/(mol⋅K)

Boltzmann Constant

kB = 1.38×10-23 J/K

R = kBNA

WorkDonebyIsothermal(ΔT=0)Expansion/CompressionofIdealGas

On p-V graph, the green lines are isotherms… … each green line corresponds to a system at a constant temperature.

Relates p and V

From ideal gas law, this means that for a given isotherm:

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pV = constant ⇒ p = nRT( ) 1V

The work done by the gas is then:

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Wby = pdVVi

V f

∫ =nRTV

dV = nRT

Vi

V f

∫ dVVVi

V f

∫

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⇒ Wby,isothermalΔT =0

= nRT lnVf

Vi

= nRT ln pip f

For expansion, we have: Vf > Vi Wby > 0

For compression, we have: Vf < Vi Wby < 0

Checkpoint 1: An ideal gas has an initial pressure of 3 pressure units and an initial volume of 4 volume units. The table gives the final pressure and volume of the gas in five processes. Which processes start and end on the same isotherm?

– a b c d e

•  p12 6 5 4 1

•  V 1 2 7 3 12

pV=nRT in this case T is a constant so pV=C=12

Sample problem #19-2

One mole of oxygen expands at a constant temperature T of 310 K from an initial volume Vi of 12 L to a final volume Vf of 19 L. How much work is done by the gas during the expansion.

We calculated W for isothermal W=nRT ln (Vf /VI)

W= (1 mole)(8.31J/mole K)(310K) ln(19/12)

W=1180 J

Work Done by Isobaric (Δp = 0) Expansion of an Ideal Gas

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⇒ Wby,isobaricΔP=0

= pΔV

= nRΔT€

Wby = pdVVi

V f

∫

Adiaba0cExpansionofanidealgas(Q=0)

Remember “Adiabatic means Q = 0” or, by 1st Law of Thermo ⇒ ΔEint = -Wby

In this case pVγ = constant where γ = cp/cV = (R+ cV )/cV

example: monatomic gas γ = 5/3

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p1V1γ = p2V2

γ

T1V1γ −1 =T2V2

γ −1

Compare with Isothermal Expansion (ΔT = 0)

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T1 =T2 ⇔ p1V1 = p2V2[ ] isothermal

Because a gas is thermally insulated, or expansion/compression happens suddenly ⇒ adiabatic

Pressure, Temperature, & RMS speed

The is F/A

p =FxA=1L2

mvxi2

Li=1

n

∑ =mL3

vxi2

i=1

n

∑

p =nmNA

L3(vx2 )avg =

nM (vx2 )avg

V

=nM (v2 )avg3V

=nMvrms

2

3V

v2 = vx2 + vy

2 + vz2

vx2 =

v2

3

Assume the collision of the gas molecule with the wall is elastic then the :

Δpx = (−mvx ) − (mvx ) = 2mvx-

The molecule travels to the back wall, collides and comes back. The time it takes is 2L/vx.

ΔpΔt

=2mvx2l / vx

=mvx

2

LIf we calculated the average velocity and use the fact that the number in the sum is nNA then:

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vx2( )avg = vxi

2

i=1

n

∑ nNA

(v2 )avg = vrmsRMS = Root-Mean-Square

RMSSpeeds

vrms =3pVnM

1/2

=3nRTnM

1/2

vrms =3RTM

pV = nRT For ideal gas

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p =nMvrms

2

3V

We have pressure

The RMS velocity depends on:

Molar mass & Temperature

Problem #19-3: Here are five numbers: 5, 11, 32, 67, and 89. (a) What is the average value navg? (b) What is the rms value nrms of the numbers?

(n2 )avg =1n

ni2

i=1

n

∑ =52 +112 + 322 + 672 + 892

5= 2714.41

(n2 )avg = 52.1

(b)

navg =5 +11+ 32 + 67 + 89

5= 40.8

(a)

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navg ≠ n2( )avg = nrms

Problem #19-21: (a) Compute the rms speed of a nitrogen molecule at 20.0 0C (each N atom has 7 protons and 7 neutrons). At what temperatures will the rms speed be (b) half that value and (c) twice that value?

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vrms =3RTM

Remember to use: T = 20 °C + 273 = 293 K

Ifweknowthe“average”speedofpar0cles,wethenknowthekine0cenergy,

Whatistherela0onshipbetweentransla0onalkine0cenergy&internalenergy?

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KE =12m vrms( )2 =

12m

3RTM

Transla0onalKine0cEnergy&Internalenergy

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KE =32kBT

The KE of all ideal gas molecules depends only on the temperature (not mass!)

Monoatomic ideal gas : He, Ar, Ne, Kr… (no potential energies)

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Eint,monotonic = NA32kBT( ) = 3

2nRT The internal energy

of an ideal gas depends only on the temperature

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ΔE int,monotonic = 32 nR ΔT( )

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KE =12m vrms( )2 =

12m

3RTM

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M = mNA

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k = R NAB

Problem #19-26: What is the average translational kinetic energy of 1 mole nitrogen molecules at 1600 K?

MolarSpecificHeatatConstantVolume:MonatomicIdealGas

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ΔEint =Q = ncVΔT

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cV ,monatomic =32R=12.5 ⋅ J/mol ⋅K

Molar Specific Heat at Constant Volume (Wby=0)

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Q = ncVΔT & Wby = 0

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ΔE int = n 32 R( )ΔT = n CV( )ΔT =Q

MolarSpecificHeat:Monatomicidealgas

Molar Specific Heat at Constant Pressure (Wby=pΔV)

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ΔEint =Q−Wby = ncpΔT − pΔV

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ncVΔT = ncpΔT − nRΔT

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cV = cp − R or cp = cV + R

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cp,monatomic =52R= 20.8 ⋅ J/mol ⋅K

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ΔEint =Q = ncVΔT

Molar Specific Heat at Constant Volume (Wby=0)

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cV ,monatomic =32R=12.5 ⋅ J/mol ⋅K

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Wby = pdV∫= area under p −V graph

Rememberspecialcases…

3)Cyclical process (closed cycle) ΔEint,closed cycle =0 net area in p-V curve is Q

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ΔEint = 0⇒ Q =Wby

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ΔE int = −W[ ]adiabatic

Adiabatic expansion/contraction - NO TRANSFER OF ENERGY AS HEAT Q = 0

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ΔE int =Q

Constant-volume processes (isochoric)-

NO WORK IS DONE W = 0

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Wby = pdVVi

V f =Vi

∫ = 0

QΔV = 0 = nCVΔT

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Wby = pdVVi

V f =Vi

∫ = pΔV

QΔP= 0 = nCPΔT

Constant-pressure processes

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ΔE int =Q − pΔVCV = CP − R

Chapter 20: Entropy and the Second law of thermodynamics

0th law Thermal Equilibrium: A = B & B = C then A = C Q = ncΔT Q → 0 as ΔT → 0

1st law Conservation of energy: ΔEint = Q - Wby= Q + Won Change in Internal energy = heat added minus work done by

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0 ≤ ΔStotal

2nd law HEAT FLOWS NATURALLY FROM HOT OBJECT TO A COLD OBJECT Heat will NOT flow spontaneously from cold to hot

Today:

Halloffame

Boltzmann constant kB =1.38×10-23 J/K

Ludwig Boltzmann (1844-1906)

W = number of states

IrreversibleProcesses

Howtounderstandthis:Entropy

How to describe a system: P, T, V, Eint, and n

Entropy, S, like T,V, P, Eint, and n is state variable

What is a process? expansion, compression, temperature rise, add mass

Reversible process {processes can be done infinitely slowly to ensure thermal equilibrium}

[ Note: since T > 0, if Q is positive (negative) the ΔS is positive (negative) ]

How to define entropy? Easier to define Change of entropy during a process.

where Q is energy transferred to or from a system during a process

Temp in Kelvin Units: [J/K]

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ΔSpart = S f − Si =dQTi

f

∫

The entropy of a closed system (no energy and no mass comes in and out) never decreases.

It either stays constant (reversible process) or increases (irreversible process).

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0 ≤ ΔStotal

The Second Law of Thermodynamics

Entropy:IdealGasProcesses

1) For reversible process:

2) For isothermal process:

3) In general for gas, using 1st law:

4) For adiabatic (reversible) adiabatic compression/expansion (Q=0):

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ΔScycle,rev = 0 =dQT

∫

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ΔSisothermal =QT

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ΔE int = 0 ⇒ Q =W

W = nRT lnVf

Vi

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⇒ ΔSisothermal = nR lnVf

Vi

Now integrate:

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ΔSgeneral,gas = S f − Si = nR lnV f

Vi

+ nCV ln

T fTi

⇐ pV = nRT[ ]€

dE int = dQ − dWnCV dTT

=dQT

−pdVT

& p =nRTV

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ΔSrev,adiabatic = 0

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⇒dQT

∫ =nRTV

dVT

∫ +nCV dTT

∫

Entropy is a State Function

Checkpoint:AnidealgashasatemperatureT1attheiniDalstateishowninthep‐Vdiagram.ThegashasahighertemperatureT2atthefinalstatesaandb,whichcanreachalongthepathsshown.Istheentropychangealongthepathtostatealargerthan,smallerthan,orthesameasthatalongpathtostateb?

From i to a: ΔS = nCV lnT2T1

From i to b: ΔS = nCV lnT2T1+ nR lnVb

Va

Entropy:Liquid/solidprocesses

1)   For phase changes: Temperature = constant

2) For temperature changes:

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ΔSphase change =dQT

∫

=Qphase change

T

=mLT

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ΔSliquid / solid = S f − Si =dQT

∫

=mcdTT

∫

= mc lnTf

Ti

SampleProblem#20‐2:TwoidenDcalcopperblocksofmassm=1.5kg:BlockLisatTiL=600CandblockRisatTiR=200C.TheblocksareinathermallyinsulatedboxandareseparatedbyaninsulaDngshuTer.WhenweliVtheshuTer,theblockscometoequilibriumwithTf=400C.Whatistheentropyofthisirreversibleprocess?SpecificheatofCuis386J/kg‐K.

The left block is initially at 600C, and comes to equilibrium with final temperature 400C. Heat is transferred from the left block to the right.

dQ = mcdT ΔSL =dQTi

f

∫ =mcdTTTil

Tf

∫ = mc lnTf

TiL= −35.86J / K

Now set reservoir at 200C and put in contact with R. Raise the temperature slowly to 400C.

ΔSR = 38.23J / K ΔSrev = ΔSR + ΔSL = 2.4J / K

Sampleproblem20‐1:Suppose1.0molofnitrogengasisconfinedtotheleVsideofthecontainerinthefigure.Youopenthestop‐cockandthevolumeofthegasdoubles.Whatistheentropychangeofthegasforthisirreversibleprocess?

Free Expansion so ΔT = 0

ΔS =QT=nRT ln Vf /Vi )( )

T= nR ln

Vf

Vi

Put in numbers

ΔS = +5.76J / K

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Q = nRT lnVf

Vi