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# Lecture 25 - PowerPoint PPT Presentation

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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Goals:

• Chapters 18, micro-macro connection

• Third test on Thursday at 7:15 pm.

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

• 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

• Average distance particle moves between collisions:

N/V: particles per unit volume

• The mean free path at atmospheric pressure is:

λ=68 nm

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

• 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

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

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

PV = (1/3) Nmvrms2

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

• 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 at 10°C has the value K

• 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 theorem at 10°C has the value K

• 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 at 10°C has the value K

• 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 at 10°C has the value K

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

B)

A)

very unlikely at 10°C has the value K

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

probability=(1/2)N

Second Law of thermodynamics at 10°C has the value K

• 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