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Physics 1710 —Warm-up Quiz Answer Now ! 0 34% 48 of 140 0 What is the minimum number of additional dietary Calories (kcal) that a 100 kg mountain climber must burn to scale a 1000 m cliff? About 29 Cal About 230 Cal About 1000 Cal About 29,000 Cal About 230,000 Cal 0

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slide1

Physics 1710—Warm-up Quiz

Answer Now !

0

34% 48 of 140

0

What is the minimum number of additional dietary Calories (kcal) that a 100 kg mountain climber must burn to scale a 1000 m cliff?
  • About 29 Cal
  • About 230 Cal
  • About 1000 Cal
  • About 29,000 Cal
  • About 230,000 Cal
physics 1710 chapter 21 kinetic theory of gases

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Physics 1710Chapter 21 Kinetic theory of Gases

Solution:

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

So:

∆Q = W (in this case)

W = mgh

W = 100 kg (9.80 N/kg)(1000 m) = 9.80 x 10 5 J

∆Q = (9.80 x 10 2 kJ)/(4.183 kJ/kcal)

∆Q = 234. kcal

physics 1710 c hapter 20 heat 1 st law of thermo

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Physics 1710 Chapter 20 Heat & 1st Law of Thermo

  • Examples:
  • Touching a hot stove
  • Feeling the air rising from it
  • Feeling the glow

Conduction

Convection

Radiation

How does heat get from one place to another?

physics 1710 c hapter 20 heat 1 st law of thermo4

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Physics 1710 Chapter 20 Heat & 1st Law of Thermo

Conduction:

P = kA |dT/dx |

Examples:

Thermos bottles

Blankets

Double pane windows

Newton’s law of cooling P= h A(T 2 – T1)

Pans

R factor or R value

P= A(T 2 – T1)/∑i Ri

physics 1710 c hapter 20 heat 1 st law of thermo5

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Physics 1710 Chapter 20 Heat & 1st Law of Thermo

Convection:

Heat transfer by material transfer

Forced convection (fluids)

External force produces material transfer

Natural Convection

Buoyancy-driven flow

Newton’s law of cooling applied

P = h A(T 2 – T1)

h depends on flow conditions

physics 1710 c hapter 20 heat 1 st law of thermo6

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Physics 1710 Chapter 20 Heat & 1st Law of Thermo

Radiation:

Stefan-Boltzmann Law

P = εσAT4

Wien’s Law

P∝T4

σ = 5.6696 x 10-8 W/m2‧K4

Emissivity 0< ε <1; ε ~ ½

Reflectivity (albedo) R = (1- ε)

Energy balance

P in - εσA(Tave )4 = 0

physics 1710 c hapter 20 heat 1 st law of thermo7

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Physics 1710 Chapter 20 Heat & 1st Law of Thermo

Summary:

Heat is transferred by

Conduction—energy diffusion

Convection—mass transport

Radiation—electromagnetic waves

physics 1710 chapter 21 kinetic theory of gases8

No Talking!

Think!

Confer!

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How does heat get to the earth from the sun? What factors are important in the average temperature of the planet? (1) conduction; (2) convection; (3) radiation; (4) conduction and convection; (5) convection and radiation.

Physics 1710Chapter 21 Kinetic theory of Gases

Peer Instruction Time

slide9

Physics 1710 — e-Quiz

Answer Now !

10

35% 50 of 140

0

How does heat get to the earth from the sun? What factors are important in the average temperature of the planet?
  • Conduction
  • Convection
  • Radiation
  • Conduction and convection
  • Convection and Radiation
physics 1710 c hapter 20 heat 1 st law of thermo10

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Physics 1710 Chapter 20 Heat & 1st Law of Thermo

Global Warming?:

P in = ( 1- ελ) Psun

Tave = [( 1- ελ) Psun /(εGH σA)]1/4

Must understand every parameter

to be accurate.

physics 1710 chapter 21 kinetic theory of gases11

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Physics 1710Chapter 21 Kinetic theory of Gases

1′ Lecture:

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

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

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

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

nV = no e –E/kT

physics 1710 chapter 21 kinetic theory of gases12

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Physics 1710Chapter 21 Kinetic theory of Gases

Molecular Model of Ideal Gas

Key concept: gas is ensemble of non-interacting atoms or molecules.

Pressure due to a single molecule

at wall of vessel:

P1 = -F1 /A = -(∆px / ∆t)/A

Impulse:

∆px = - mvx – (mvx ) = -2 mvx

∆t = 2d /vx

Thus:

P1 = - F1 /A = mvx2 /(d‧A)

physics 1710 chapter 21 kinetic theory of gases13

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Physics 1710Chapter 21 Kinetic theory of Gases

P1 = - F1 /A = mvx2 /(d‧A)

Total Pressure:

P = N<P1 >= Nm<vx2 >/(d‧A)

P = (N/V) m<vx2 >

Average vx2 = <vx2 >:

<v2 >=<vx2 > + <vy2 > + <vz2 >

<v2 >=3<vx2 >; <vx2 >= 1/3 <v 2>

P = ⅔(N/V)(½ m<v 2>)

physics 1710 chapter 21 kinetic theory of gases14

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Physics 1710Chapter 21 Kinetic theory of Gases

P = ⅔(N/V)(½ m<v2 >)

But

P = (N/V) kT

Thus

T = 2/(3k)(½ m<v 2>)

½ m<v2> = 3/2 kT

½ m<vx2> = 1/3 [½ m<v2> ] = ½ kT

physics 1710 chapter 21 kinetic theory of gases15

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Physics 1710Chapter 21 Kinetic theory of Gases

Principle of Equipartition of Energy:

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

physics 1710 chapter 21 kinetic theory of gases16

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Physics 1710Chapter 21 Kinetic theory of Gases

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

N( 1/2 m<v 2>) = 3N(½ kT) = 3/2 nRT

√<v 2> = vrms = √(3 kT/m) = √(3RT/M)

physics 1710 chapter 21 kinetic theory of gases17

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Physics 1710Chapter 21 Kinetic theory of Gases

(Molar) Specific Heat of an Ideal Gas:

∆Q = n CV ∆T (constant volume)

∆Q = n CP ∆T (constant pressure)

W = ∫ P dV; at constant volume W = 0.

∆Eint = ∆Q = n CV ∆T

Eint = n CV T

CV = (1/n) d Eint /dT

CV = 3/2 R = 3/2 No kT = 12.5 J/mol‧K

physics 1710 chapter 21 kinetic theory of gases18

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Physics 1710Chapter 21 Kinetic theory of Gases

∆Eint = ∆Q –W = nCP ∆T - P∆V

nCV ∆T= nCP ∆T – n R ∆T

CP - CV = R

CP = 5/2 R

γ = CP / CV = (5/2 R)/(3/2 R) = 5/3

γ = 5/3

physics 1710 chapter 21 kinetic theory of gases19

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Physics 1710Chapter 21 Kinetic theory of Gases

Adiabatic Expansion of an Ideal Gas:

For adiabatic case:

dEint = n CV dT = - PdV

So that

dT = -P dV /(nCV )

Also:

PV = nRT

PdV + VdP = nR dT

PdV + VdP = -RP /(nCV ) dV

physics 1710 chapter 21 kinetic theory of gases20

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Physics 1710Chapter 21 Kinetic theory of Gases

PdV + VdP = -R/(nCV ) PdV

Rearranging:

dP/P = [1 – R/(nCV)] dV/V

dP/P = - γ dV/V

ln P = - γ lnV + ln K

PV γ= constant

physics 1710 chapter 21 kinetic theory of gases21

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Physics 1710Chapter 21 Kinetic theory of Gases

Bulk Modulus of an Ideal Gas:

B = -∆P/ (∆V/V)

P= K V - γ

dP = - γ KV - γ - 1dV

B = -dP/(dV/V)

B= (γ KV 1dV)/(dV/V)

B= γ KV - γ

B = γ P

physics 1710 chapter 21 kinetic theory of gases22

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Physics 1710Chapter 21 Kinetic theory of Gases

Velocity of Sound in Ideal Gas:

v = √(B/ρ)

v = √[ γP/(ρo P/ Po )]

v = √[ γPo /ρo ]N.B.: no pressure dependence

For air (ideal)

v = √[ γPo /ρo ]v = √[(5/3)(101 kPa/ 1.26 kg/m3 )]v = 365 m/s (Cf 343 m/s)

physics 1710 chapter 21 kinetic theory of gases23

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Physics 1710Chapter 21 Kinetic theory of Gases

Law of Atmospheres

dP = -mg nV dy

P = nV kT

dP = kT dnV

kT dnV= -mg nV dy

dnV/ nV = -(mg/kT) dy

nV = no e –(mgy/kT)

physics 1710 chapter 21 kinetic theory of gases24

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Physics 1710Chapter 21 Kinetic theory of Gases

Boltzmann Distribution Function

nV = no e –(mgy/kT)

nV = no e –U/kT

nV (E) = no e –E/kT

physics 1710 chapter 21 kinetic theory of gases25

0

Physics 1710Chapter 21 Kinetic theory of Gases

Summary:

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

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

½ m<v2> = 3/2 kT

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

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

physics 1710 chapter 21 kinetic theory of gases26

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Physics 1710Chapter 21 Kinetic theory of Gases

Summary (cont’d.)

γ = CP / CV

PV γ= constant

B = γ P

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

nV = no e –E/kT