Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.

Successfully reported this slideshow.

Like this presentation? Why not share!

- The kinetic theory of gases by Ashwani Kumar 1645 views
- Lecture 13 ideal gas. kinetic mod... by Albania Energy As... 1909 views
- Lecture 05 mechanical waves. tran... by Albania Energy As... 2354 views
- Fisika Modern (10) statistical physics by jayamartha 1311 views
- Statistical mechanics by Kumar 2385 views
- Statistik Maxwell-Boltzmann & Inter... by Samantars17 359 views

3,164 views

Published on

Lecture 14 maxwell-boltzmann distribution. heat capacities

No Downloads

Total views

3,164

On SlideShare

0

From Embeds

0

Number of Embeds

4

Shares

0

Downloads

75

Comments

0

Likes

4

No embeds

No notes for slide

- 1. Lecture 14 Maxwell-Bolztmann distribution. Heat capacities. Phase diagrams.
- 2. Molecular speeds Not all molecules have the same speed. If we have N molecules, the number of molecules with speeds between v and v + dv is: dN = Nf (v )dv f (v ) = distribution function f (v )dv = probability of finding a molecule with speed between v and v + dv
- 3. Maxwell-Boltzmann distribution higher T higher speeds are more probable 3/2 2 m f (v ) = 4π v 2e −mv /(2kT ) ÷ 2π kT Maxwell-Boltzmann distribution
- 4. Distribution = probability density f (v )dv v2 ∫v 1 = probability of finding a molecule with speed between v and v + dv f (v )dv = probability of finding molecule with speeds between v1 and v2 = area under the curve Normalization: ∫ ∞ 0 f (v )dv = 1
- 5. Most probable speed, average speed, rms speed 2kT m v mp = Most probable speed (where f(v) is maximum) ∞ ∫ vf (v )dv = = ∫ ∫ f (v )dv v 0 ∞ vf (v )dv = ( ... ) = ∞ 0 0 v 2 3kT = ∫ v f (v )dv = ( ... ) = 0 m vrms = ∞ 2 v 2 3kT = m 8kT πm Average speed Average squared speed rms speed
- 6. Molar heat capacity How much heat is needed to change by ΔT the temperature of n moles of a certain substance? m = nM n = number of moles M = mass of one mole (molar mass) Q = mc ∆T = nMc ∆T Q = nC ∆T C = Mc C = molar heat capacity
- 7. Constant volume or constant pressure? The tables of data for specific heats (or molar capacities) come from some experiment. For gases, the system is usually kept at constant volume CV For liquids and solids, the system is usually kept at constant pressure (1 atm) CP You can define both for any system. You need to know what’s in the table!
- 8. Heat capacity for a monoatomic ideal gas Average total kinetic energy K total = 3 3 NkT = nRT 2 2 3 d K total = nRdT 2 From the macroscopic point of view, this is the heat entering or leaving the system: 3 nRdT = nCV dT 2 3 CV = R 2 dQ = nCV dT Molar heat capacity at constant volume for monoatomic ideal gas Point-like particles
- 9. Beyond the monoatomic ideal gas Until now, this microscopic model is only valid for monoatomic molecules. Monoatomic molecules (points) have 3 degrees of freedom (translational) Diatomic molecules (points) have 5 degrees of freedom: 3 translational + 2 rotational) Principle of equipartition of energy: each velocity component (radial or angular) has, on average, associated energy of ½kT The equipartition principle is very general.
- 10. Diatomic ideal gas K = K tr + Krot 3 2 = kT + kT 2 2 5 Average energy = kT per molecule 2 Average total energy K total = The same temperature involves more energy per molecule for a diatomic gas than for a monoatomic gas.. 5 5 NkT = nRT 2 2 dQ = nCV dT DEMO: Mono and diatomic “molecules” 5 CV = R 2 Including rotation Molar heat capacity at constant volume for diatomic ideal gas
- 11. Monoatomic solid Simple model of a solid crystal: atoms held together by springs. K1 = 3 kT 2 Vibrations in 3 directions But we also have potential energy! PE1 = K1 For N atoms: Molar heat capacity at constant volume for monoatomic solid UN = N KE1 + N PE1 = 3NkT = 3nRT (for any harmonic oscillator) d U = 3nRdT dQ = nCV dT CV = 3R
- 12. pT diagram Melting curve (solid/liquid transition) solid Triple point Sublimation curve (gas/solid transition) liquid Critical point gas Vapor pressure curve (gas/liquid transition)
- 13. pT diagram for water DEMO: Boiling water with ice p solid Critical point liquid demo 1 atm gas 0°C 100°C T
- 14. In-class example: pT diagram for CO2 Which of the following states is NOT possible for CO 2 at 100 atm? A. Liquid B. Boiling liquid C. Melting solid D. Solid E. All of the above are possible.
- 15. Melting (at ~ -50°C) solid liquid Boiling (at ~ -5°C) For a boiling transition, pressure must be lower:.
- 16. Triple point for CO2 has a pressure > 1 atm. At normal atmospheric pressure (1 atm), CO 2 can only be solid or gas. Sublimation at T = -78.51°C
- 17. pT diagram for N2 At 1 atm, Tboiling = 77 K Tmelting = 63 K Triple point for N2: p = 0.011 atm, T = 63 K p 1 atm solid liquid Triple point DEMO: N2 snow 63 K demo gas 77 K T
- 18. pV diagrams Convenient tool to represent states and transitions from one state to another. Expansion at constant pressure p states A B process (isobaric process) VB VA V Example: helium in balloon expanding in the room and warming up If we treat the helium in the balloon is an ideal gas, we can predict T for each state: TA/B = pVA/B nR DEMO: Helium balloon
- 19. ACT: Constant volume This pV diagram can describe: p A. A tightly closed container cooling down. pB B. A pump slowly creates a vacuum inside a closed container. C. Either of the two processes. pA B A (isochoric process) In either case, volume is constant and pressure is decreasing. In case A, becauseT decreases. In case B, because n decreases. V
- 20. Isothermal curves For an ideal gas, p = nRT V T1 <T2 <T3 <T4 (For constant n, a hyperbola for each T ) Each point in a pV diagram is a possible state (p, V, T ) Isothermal curve = all states with the same T
- 21. ACT: Free expansion A container is divided in two by a thin wall. One side contains an ideal gas, the other has vacuum. The thin wall is punctured and disintegrates. Which of the following is the correct pV diagram for this process? A 1 Initial state B 2 Initial state Final state Final state 3 C Initial state Final state 4 D Initial state Final state
- 22. Final state has larger V, lower p During the rapid expansion, the gas does NOT uniformly fillV at a uniform p ⇒ hence it is not in a thermal state. ⇒ hence no “states” during process ⇒ hence this process is not represented by line A 1 Initial state B 2 Initial state Final state Final state 3 C Initial state Final state 4 D Initial state Final state
- 23. Beyond the ideal gas When a real gas is compressed, it eventually becomes a liquid… Decrease volume at constant temperatureT2: • At point “a”, vapor begins to condense into liquid. • Between a and b: Pressure and T remain constant as volume decreases, more of vapor converted into liquid. • At point “b”, all is liquid. A further decrease in volume will required large increase in p.
- 24. The critical temperature Critical point for water: 647K and 218 atm For T >> Tc, ideal gas. p Supercritical fluid solid Triple point liquid Critical point gas T critical temperature = highest temperature where a phase transition happens.
- 25. pVT diagram: Ideal gas States are points on this surface.
- 26. pVT diagram: Water Phase transitions appear as angles.

No public clipboards found for this slide

×
### Save the most important slides with Clipping

Clipping is a handy way to collect and organize the most important slides from a presentation. You can keep your great finds in clipboards organized around topics.

Be the first to comment