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Lecture 25. Goals:. Chapters 18, micro-macro connection. Third test on Thursday at 7:15 pm. Nitrogen molecules near room temperature. Percentage of molecules. 15. 10. 5. (m/s). 0-100. 1000-1100. 900-1000. 700-800. 800-900. 300-400. 500-600. 200-300. 400-500. 500-600. 100-200.

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lecture 25
Lecture 25

Goals:

  • Chapters 18, micro-macro connection
  • Third test on Thursday at 7:15 pm.
slide2

Nitrogen molecules

near room temperature

Percentage of

molecules

15

10

5

(m/s)

0-100

1000-1100

900-1000

700-800

800-900

300-400

500-600

200-300

400-500

500-600

100-200

600-700

atomic scale
Atomic scale
  • What is the typical size of an atom or a small molecule?

A) 10-6 m

B) 10-10 m

C) 10-15 m

r

r ≈1 angstrom=10-10 m

mean free path
Mean free path
  • Average distance particle moves between collisions:

N/V: particles per unit volume

  • The mean free path at atmospheric pressure is:

λ=68 nm

slide6

Consider a gas with all molecules traveling at a speed vx hitting a wall.

  • If (N/V) increases by a factor of 2, the pressure would:

A) decrease

B) increase x2

C) increase x4

  • If m increases by a factor of 2, the pressure would:

A) decrease

B) increase x2

C) increase x4

  • If vx increases by a factor of 2, the pressure would:

A) decrease

B) increase x2

C) increase x4

slide7

P=(N/V)mvx2

  • Because we have a distribution of speeds:

P=(N/V)m(vx2)avg

  • For a uniform, isotropic system:

(vx2)avg= (vy2)avg= (vz2)avg

  • Root-mean-square speed:

(v2)avg=(vx2)avg+(vy2)avg+(vz2)avg=Vrms2

microscopic calculation of pressure
Microscopic calculation of pressure

P=(N/V)m(vx2)avg

=(1/3) (N/V)mvrms2

PV = (1/3) Nmvrms2

micro macro connection
Micro-macro connection

PV = (1/3) Nmvrms2

PV = NkBT (ideal gas law)

kBT =(1/3) mvrms2

  • The average translational kinetic energy is:

εavg=(1/2) mvrms2

εavg=(3/2) kBT

slide10

The average kinetic energy of the molecules of an ideal gas at 10°C has the value K1. At what temperature T1 (in degrees Celsius) will the average kinetic energy of the same gas be twice this value, 2K1?

(A) T1 = 20°C

(B) T1 = 293°C

(C) T1 = 100°C

  • Suppose that at some temperature we have oxygen molecules moving around at an average speed of 500 m/s. What would be the average speed of hydrogen molecules at the same temperature?

(A) 100 m/s

(B) 250 m/s

(C) 500 m/s

(D) 1000 m/s

(E) 2000 m/s

equipartition theorem
Equipartition theorem
  • Things are more complicated when energy can be stored in other degrees of freedom of the system.

monatomic gas: translation

solids: translation+potential energy

diatomic molecules: translation+vibrations+rotations

equipartition theorem1
Equipartition theorem
  • The thermal energy is equally divided among all possible energy modes (degrees of freedom). The average thermal energy is (1/2)kBT for each degree of freedom.

εavg=(3/2) kBT (monatomic gas)

εavg=(6/2) kBT (solids)

εavg=(5/2) kBT (diatomic molecules)

  • Note that if we have N particles:

Eth=(3/2)N kBT =(3/2)nRT (monatomic gas)

Eth=(6/2)N kBT =(6/2)nRT (solids)

Eth=(5/2)N kBT =(5/2)nRT (diatomic molecules)

specific heat
Specific heat
  • Molar specific heats can be directly inferred from the thermal energy.

Eth=(6/2)N kBT =(6/2)nRT (solid)

ΔEth=(6/2)nRΔT=nCΔT

C=3R (solid)

  • The specific heat for a diatomic gas will be larger than the specific heat of a monatomic gas:

Cdiatomic=Cmonatomic+R

entropy
Entropy
  • A perfume bottle breaks in the corner of a room. After some time, what would you expect?

B)

A)

slide15

very unlikely

  • The probability for each particle to be on the left half is ½.

probability=(1/2)N

second law of thermodynamics
Second Law of thermodynamics
  • The entropy of an isolated system never decreases. It can only increase, or in equilibrium, remain constant.
  • The laws of probability dictate that a system will evolve towards the most probable and most random macroscopic state
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