Physics 1710 —Warm-up Quiz

Physics 1710Physics 1710—Warm-up Quiz—Warm-up QuizWhat is the minimum number of additional dietary What is the minimum number of additional dietary Calories (kcal) that a 100 kg mountain climber must Calories (kcal) that a 100 kg mountain climber must burn to scale a 1000 m cliff?burn to scale a 1000 m cliff?

A B C D E

2%

50%

0%

27%

21%

A.A. About 29 CalAbout 29 CalB.B. About 230 CalAbout 230 CalC.C. About 1000 CalAbout 1000 CalD.D. About 29,000 CalAbout 29,000 CalE.E. About 230,000 CalAbout 230,000 Cal

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

Physics 1710Physics 1710 Chapter 21 Kinetic theory of Chapter 21 Kinetic theory of GasesGases

Solution:Solution:

∆∆E = ∆Q – W = 0 (no change in climbers internal E = ∆Q – W = 0 (no change in climbers internal energy)energy)

So:So:∆∆Q = W (in this case) Q = W (in this case)

W = mghW = mgh

W = 100 kg (9.80 N/kg)(1000 m) = 9.80 x 10 W = 100 kg (9.80 N/kg)(1000 m) = 9.80 x 10 55 J J

∆ ∆Q = (9.80 x 10 Q = (9.80 x 10 22 kJ)/(4.183 kJ/kcal) kJ)/(4.183 kJ/kcal) ∆ ∆Q = 234. kcalQ = 234. kcal

How does heat get from one place to How does heat get from one place to another?another?

Physics 1710Physics 1710 C Chapter 20 Heat & 1hapter 20 Heat & 1stst Law of Law of ThermoThermo

Conduction

Convection

Radiation

Examples:

• Touching a hot stove

• Feeling the air rising from it

• Feeling the glow

Conduction:Conduction:

• PP = kA = kA ||dT/dx dT/dx ||• Examples:Examples:

– Thermos bottlesThermos bottles– BlanketsBlankets– Double pane windowsDouble pane windows– Newton’s law of cooling Newton’s law of cooling PP = h A(T = h A(T 2 2 – T– T11))– PansPans– R factor or R value R factor or R value

PP = A(T = A(T 2 2 – T– T11)/)/∑∑i i RRii


Convection:Convection:• Heat transfer by material transferHeat transfer by material transfer

• Forced convection (fluids)Forced convection (fluids)– External force produces material transferExternal force produces material transfer

• Natural ConvectionNatural Convection– Buoyancy-driven flowBuoyancy-driven flow– Newton’s law of cooling appliedNewton’s law of cooling applied

PP = h A(T = h A(T 2 2 – T– T11))h depends on flow conditionsh depends on flow conditions


Radiation:Radiation:

• Stefan-Boltzmann LawStefan-Boltzmann LawPP = = εσεσ AT AT44

•Wien’s Law Wien’s Law

PP ∝∝TT4 4

• σσ = 5.6696 x 10 = 5.6696 x 10-8-8 W/m W/m22‧K‧K44

• Emissivity 0< Emissivity 0< ε ε <1; <1; εε ~ ½ ~ ½• Reflectivity (albedo) R = (1- Reflectivity (albedo) R = (1- εε))• Energy balanceEnergy balance

PP in in - - εσεσ A(T A(Tave ave )) 4 4 = 0= 0


Summary:Summary:

• Heat is transferred by Heat is transferred by o Conduction—energy diffusionConduction—energy diffusiono Convection—mass transportConvection—mass transporto Radiation—electromagnetic wavesRadiation—electromagnetic waves


No Talking!No Talking!Think!Think! Confer!Confer!

Peer Instruction Peer Instruction TimeTime

How does heat get to the earth from the How does heat get to the earth from the sun? What factors are important in the sun? What factors are important in the average temperature of the planet? (1) average temperature of the planet? (1)

conduction; (2) convection; (3) radiation; (4) conduction; (2) convection; (3) radiation; (4) conduction and convection; (5) convection conduction and convection; (5) convection

and radiation.and radiation.


How does heat get to the earth from the How does heat get to the earth from the sun? What factors are important in the sun? What factors are important in the average temperature of the planet? average temperature of the planet?

Physics 1710 — Physics 1710 — e-Quize-Quiz

Answer Now !

1 2 3 4 5

2% 2%

28%

8%

60%1.1. ConductionConduction

2.2. ConvectionConvection

3.3. RadiationRadiation

4.4. Conduction and convectionConduction and convection

5.5. Convection and RadiationConvection and Radiation

41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80

Global Warming?:Global Warming?:

PP in in = ( 1- = ( 1- εελλ) ) PPsunsun

TTave ave = [( 1- = [( 1- εελλ) ) PPsun sun /(/(εεGH GH σσA)]A)]1/41/4

Must understand every parameter Must understand every parameter

to be accurate.to be accurate.


11′ Lecture:′ Lecture:

• The Ideal Gas Law results from the cumulative The Ideal Gas Law results from the cumulative action of atoms or molecules.action of atoms or molecules.

• The average kinetic energy of the atoms or The average kinetic energy of the atoms or molecules of an ideal gas is equal to 3/2 kT.molecules of an ideal gas is equal to 3/2 kT.

• Energy average distributes equally (is Energy average distributes equally (is equipartitioned) into all available states.equipartitioned) into all available states.

• The distribution of particles among available The distribution of particles among available energy states obeys the Boltzmann distribution energy states obeys the Boltzmann distribution law. law.

nnVV = n = noo e e –E/kT–E/kT


Molecular Model of Ideal GasMolecular Model of Ideal Gas•Key concept: gas is ensemble of non-interacting Key concept: gas is ensemble of non-interacting atoms or molecules.atoms or molecules.

• Pressure due to a single molecule Pressure due to a single molecule at wall of vessel:at wall of vessel:

PP11 = -F = -F1 1 /A = -(∆p/A = -(∆px x / ∆t)/A/ ∆t)/AImpulse:Impulse:

∆∆ppxx = - mv = - mvxx – (mv – (mvxx ) = -2 mv ) = -2 mvxx

∆∆t = 2d /vt = 2d /vxx

Thus:Thus:

PP11 = - F = - F1 1 /A = mv/A = mvxx2 2 /(d‧A) /(d‧A)


PP11 = - F = - F1 1 /A = mv/A = mvxx2 2 /(d‧A) /(d‧A)

Total Pressure:Total Pressure:

P = N<PP = N<P11 >= Nm<v >= Nm<vxx2 2 >/(d‧A) >/(d‧A)

P = (N/V) m<vP = (N/V) m<vxx2 2 > >

Average vAverage vxx2 2 = <v= <vxx

2 2 >: >:

<v<v2 2 >=<v >=<vxx2 2 > + <v > + <vyy

2 2 > + <v > + <vzz2 2 > >

<v<v2 2 >=3<v >=3<vxx2 2 >; <v>; <vxx

2 2 >= 1/3 <v >= 1/3 <v 22>>

P = ⅔(N/V)(½ m<v P = ⅔(N/V)(½ m<v 22>)>)


P = ⅔(N/V)(½ m<vP = ⅔(N/V)(½ m<v2 2 >) >)ButBut

P = (N/V) kTP = (N/V) kTThusThus

T = 2/(3k)(½ m<v T = 2/(3k)(½ m<v 22>) >)

½ m<v½ m<v22> = 3/2 kT> = 3/2 kT

½ m<v½ m<vxx22> = 1/3 [½ m<v> = 1/3 [½ m<v22> ] = ½ kT> ] = ½ kT


Principle of Equipartition of Energy:Principle of Equipartition of Energy:

•Each degree of freedom contributes 1/2 Each degree of freedom contributes 1/2 kT to the energy of a system.kT to the energy of a system.


Each molecule in a gas contributes 3 Each molecule in a gas contributes 3 degrees of freedom to the system:degrees of freedom to the system:

N( 1/2 m<v N( 1/2 m<v 22>) = 3N(½ kT) = 3/2 nRT>) = 3N(½ kT) = 3/2 nRT

√√<v <v 22> = v> = vrmsrms = √(3 kT/m) = √(3RT/M) = √(3 kT/m) = √(3RT/M)


(Molar) Specific Heat of an Ideal Gas:(Molar) Specific Heat of an Ideal Gas:

∆∆Q = n CQ = n CV V ∆T (constant volume)∆T (constant volume)

∆∆Q = n CQ = n CP P ∆T (constant pressure)∆T (constant pressure)

W = ∫ P dV; at constant volume W = 0.W = ∫ P dV; at constant volume W = 0.

∆∆EEint int = ∆Q = n C= ∆Q = n CV V ∆T∆T

EEint int = n C= n CV V TT

CCV V = (1/n) d E= (1/n) d Eint int /dT /dT

CCV V = 3/2 R = 3/2 N= 3/2 R = 3/2 Noo kT = 12.5 J/mol‧K kT = 12.5 J/mol‧K


∆∆EEint int = ∆Q –W = nC= ∆Q –W = nCP P ∆T - P∆V∆T - P∆V

nCnCV V ∆T= nC∆T= nCP P ∆T – n R ∆T∆T – n R ∆T

CCP P - C- CV V = R = R

CCP P = 5/2 R= 5/2 R

γγ = C = CP P / C/ CV V = (5/2 R)/(3/2 R) = 5/3 = (5/2 R)/(3/2 R) = 5/3

γγ = 5/3 = 5/3


Adiabatic Expansion of an Ideal Gas:Adiabatic Expansion of an Ideal Gas:For adiabatic case:For adiabatic case:

dEdEint int = n C= n CV V dT = - PdVdT = - PdVSo thatSo that

dT = -P dV /(nCdT = -P dV /(nCV V ))

Also:Also:PV = nRTPV = nRT

PdV + VdP = nR dTPdV + VdP = nR dT

PdV + VdP = -RP /(nCPdV + VdP = -RP /(nCV V ) dV) dV


PdV + VdP = -R/(nCPdV + VdP = -R/(nCV V ) PdV) PdVRearranging:Rearranging:

dP/P = [1 – R/(nCV)] dV/VdP/P = [1 – R/(nCV)] dV/V

dP/P = - dP/P = - γγ dV/V dV/V

ln P = - ln P = - γγ lnV + ln K lnV + ln K

PV PV γγ = constant= constant


Bulk Modulus of an Ideal Gas:Bulk Modulus of an Ideal Gas:B = -∆P/ (∆V/V)B = -∆P/ (∆V/V)

P= K V P= K V - - γγ

dP = - dP = - γγ KV KV - - γγ - 1 - 1dVdV

B = -dP/(dV/V)B = -dP/(dV/V)

B= (B= (γγ KV KV 11dV)/(dV/V)dV)/(dV/V)

B= B= γγ KV KV - - γγ

B = B = γγ P P


Velocity of Sound in Ideal Gas:Velocity of Sound in Ideal Gas:

v = √(B/v = √(B/ρ)ρ)

v = v = √[ √[ γγP/(P/(ρρoo P/ P P/ Po o )])]

v = v = √[ √[ γγPPo o //ρρoo ] ]

N.B.: no pressure dependenceN.B.: no pressure dependenceFor air (ideal)For air (ideal)

v = v = √[ √[ γγPPo o //ρρoo ] ]

v = v = √[(√[(5/3)(101 kPa/ 1.26 kg/m5/3)(101 kPa/ 1.26 kg/m33 )] )]v = 365 m/s (Cf 343 m/s)v = 365 m/s (Cf 343 m/s)


Law of AtmospheresLaw of Atmospheres

dP = -mg ndP = -mg nV V dydy

P = nP = nV V kTkT

dP = kT dndP = kT dnVV

kT dnkT dnVV= -mg n= -mg nV V dydy

dndnVV/ n/ nV V = -(mg/kT) dy = -(mg/kT) dy

nnV V = n = noo e e –(mgy/kT)–(mgy/kT)


Boltzmann Distribution FunctionBoltzmann Distribution Function

nnV V = n = noo e e –(mgy/kT)–(mgy/kT)

nnV V = n = noo e e –U/kT–U/kT

nnV V (E) = n (E) = noo e e –E/kT–E/kT


SummarySummary::

• The Ideal Gas Law results from the cumulative The Ideal Gas Law results from the cumulative action of atoms or molecules.action of atoms or molecules.

• The average kinetic energy of the atoms or The average kinetic energy of the atoms or molecules of an ideal gas is equal to 3/2 kT.molecules of an ideal gas is equal to 3/2 kT.

½ m<v½ m<v22> = 3/2 kT> = 3/2 kT

• Energy average distributes equally (is Energy average distributes equally (is equipartitioned) into all available states. equipartitioned) into all available states.

•Each degree of freedom contributes 1/2 kT to the Each degree of freedom contributes 1/2 kT to the energy of a system.energy of a system.


Summary (cont’d.)Summary (cont’d.)

γγ = C = CP P / C/ CVV

PV PV γγ = constant= constantB = B = γγ P P

• The distribution of particles among available The distribution of particles among available energy states obeys the Boltzmann distribution energy states obeys the Boltzmann distribution law.law.

nnVV = n = noo e e –E/kT–E/kT


Physics 1710 —Warm-up Quiz

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