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### Chapter 16

Thermal Propertiesof Matter

State Variables

- State variable = macroscopic property of thermodynamic system
- Examples: pressure p

volume V

temperature T

mass m

State Variables

- State variables: p, V, T, m
- I general, we cannot change one variable without affecting a change in the others
- Recall: For a gas, we defined temperature T (in kelvins) using the gas pressure p

Equation of State

- State variables: p, V, T, m
- The relationship among these:‘equation of state’
- sometimes: an algebraic equation exists
- often: just numerical data

Equation of State

- Warm-up example:
- Approximate equation of state for a solid
- Based on concepts we already developed
- Here: state variables are p, V, T

Derive the equation of state

The ‘Ideal’ Gas

- The state variables of a gas are easy to study:
- p, V, T, mgas
- often use: n = number of ‘moles’ instead of mgas

Moles and Avogadro’s Number NA

- 1 mole = 1 mol = 6.02×1023 molecules = NA molecules
- n = number of moles of gas
- M = mass of 1 mole of gas
- mgas = n M

Do Exercise 16-53

The ‘Ideal’ Gas

- We measure:

the state variables (p, V, T, n) for many different gases

- We find:

at low density, all gases obey the same equation of state!

Ideal Gas Equation of State

- State variables: p, V, T, n

pV = nRT

- p = absolute pressure (not gauge pressure!)
- T = absolute temperature (in kelvins!)
- n = number of moles of gas

Ideal Gas Equation of State

- State variables: p, V, T, n

pV = nRT

- R = 8.3145 J/(mol·K)
- same value of R for all (low density) gases
- same (simple, ‘ideal’) equation

Do Exercises 16-9, 16-12

Ideal Gas Equation of State

- State variables: p, V, T, and mgas= nM
- State variables: p, V, T, and r = mgas/V

Derive ‘Law of Atmospheres’

Ideal Gas Equation

pV = nRT

- n = number of moles of gas = N/NA
- R = 8.3145 J/(mol·K)
- N = number of molecules of gas
- NA = 6.02×1023 molecules/mol

Ideal Gas Equation

- k = Boltzmann constant = R/NA = 1.381×10-23 J/(molecule·K)

Assumptions

- gas = large number N of identical molecules
- molecule = point particle, mass m
- molecules collide with container walls= origin of macroscopic pressure of gas

Kinetic Model

- molecules collide with container walls
- assume perfectly elastic collisions
- walls are infinitely massive (no recoil)

Elastic Collision

- wall:

infinitely massive, doesn’t recoil

- molecule:

vy: unchanged

vx : reverses direction

speed v : unchanged

Kinetic Model

- For one molecule: v2 = vx2 + vy2 + vz2
- Each molecule has a different speed
- Consider averaging over all molecules

Kinetic Model

- average over all molecules:

(v2)av= (vx2 + vy2 + vz2)av

= (vx2)av+(vy2)av+(vz2)av

= 3 (vx2)av

Kinetic Model

- (Ktr)av= total kinetic energy of gas due to translation
- Derive result:

Kinetic Energy

- average translational KE is directly proportional to gas temperature T

Kinetic Energy

- average translational KE per molecule:
- average translational KE per mole:

Kinetic Energy

- average translational KE per molecule:
- independent of p, V, and kind of molecule
- for same T, all molecules (any m) have the same average translational KE

Kinetic Model

- ‘root-mean-square’ speed vrms:

Molecular Speeds

- For a given T, lighter molecules move faster
- Explains why Earth’s atmosphere contains alomost no hydrogen, only heavier gases

Molecular Speeds

- Each molecule has a different speed, v
- We averaged over all molecules
- Can calculate the speed distribution, f(v)(but we’ll just quote the result)

Molecular Speeds

f(v) = distribution function

f(v) dv = probability a molecule has speed between v and v+dv

dN = number of molecules with speed between v and v+dv

= N f(v) dv

Molecular Speeds

- Maxwell-Boltzmann distribution function

Molecular Speeds

- At higher T:more molecules have higher speeds
- Area under f(v) = fraction of molecules with speeds in range: v1 < v < v1 or v > vA

Molecular Speeds

- average speed
- rms speed

Molecular Collisions?

- We assumed:
- molecules = point particles, no collisions
- Real gas molecules:
- have finite size and collide
- Find ‘mean free path’ between collisions

Molecular Collisions

- Mean free path between collisions:

Announcements

- Midterms:
- Returned at end of class
- Scores will be entered on classweb soon
- Solutions available online at E-Res soon
- Homework 7 (Ch. 16): on webpage
- Homework 8 (Ch. 17): to appear soon

Heat Capacity Revisited

DQ = energy required to change temperature of mass m by DT

c = ‘specific heat capacity’

= energy required per (unit mass × unit DT)

Heat Capacity Revisited

- Now introduce ‘molar heat capacity’ C

C = energy per (mol × unit DT) required to change temperature of n moles by DT

Heat Capacity Revisited

- important case:the volume V of material is held constant
- CV = molar heat capacity at constant volume

CV for the Ideal Gas

- Monatomic gas:
- molecules = pointlike(studied last lecture)
- recall: translational KE of gas averaged over all molecules

(Ktr)av = (3/2) nRT

CV for the Ideal Gas

- Monatomic gas:

(Ktr)av = (3/2) nRT

- note: your text just writesKtr instead of (Ktr)av
- Consider changing T by dT

CV for the Ideal Gas

- Monatomic gas:

(Ktr)av = (3/2) nRT

d(Ktr)av = n (3/2)R dT

- recall: dQ = n CV dT
- so identify: CV= (3/2)R

In General:

If (Etot)av = (f/2) nRT

Then d(Etot)av = n (f/2)R dT

But recall: dQ = n CV dT

So we identify: CV= (f/2)R

CV for the Ideal Gas

- What about gases with other kinds of molecules?
- diatomic, triatomic, etc.
- These molecules are not pointlike

CV for the Ideal Gas

- Diatomic gas:
- molecules = ‘dumbell’ shape
- its energy takes several forms:

(a) translational KE (3 directions)

(b) rotational KE (2 rotation axes)

(c) vibrational KE and PE

Demonstration

CV for the Ideal Gas

- Diatomic gas:

Etot = Ktr + Krot + Evib

(Etot)av = (Ktr)av + (Krot)av + (Evib)av

- we know: (Ktr)av = (3/2) nRT
- what about the other terms?

Equipartition of Energy

- Can be proved, but we’ll just use the result
- Define:

f = number of degrees of freedom

= number of independent ways that a molecule can store energy

Equipartition of Energy

- It can be shown:
- The average amount of energy in each degree of freedom is:

(1/2) kT per molecule

i.e.

(1/2) RT per mole

Check a known case

- Monatomic gas:
- only has translational KEin 3 directions: vx, vy, vz
- f = 3 degrees of freedom

(Ktr)av = (f/2) nRT = (3/2) nRT

CV for the Ideal Gas

- Diatomic gas:
- more forms of energy are available to the gas as you increase its T:

(a) translational KE (3 directions)

(b) rotational KE (2 rotation axes)

(c) vibrational KE and PE

CV for the Ideal Gas

- Diatomic gas:

low temperature

- only translational KEin 3 directions: vx, vy, vz
- f = 3 degrees of freedom

(Etot)av = (f/2) nRT = (3/2) nRT

CV for the Ideal Gas

- Diatomic gas:

higher temperature

- translational KE (in 3 directions)
- rotational KE (about 2 axes)
- f = 3+2 = 5 degrees of freedom

(Etot)av = (f/2) nRT = (5/2) nRT

CV for the Ideal Gas

- Diatomic gas:

even higher temperature

- translational KE (in 3 directions)
- rotational KE (about 2 axes)
- vibrational KE and PE
- f = 3+2+2 =7 degrees of freedom

(Etot)av = (f/2) nRT = (7/2) nRT

Summary of CV for Ideal Gases

(Etot)av = (f/2) nRT

CV= (f/2)R

Monatomic:f = 3 (only)

Diatomic:f = 3, 5, 7 (with increasing T)

CV for Solids

- Each atom in a solid can vibrate about its equilibrium position
- Atoms undergo simple harmonic motion in all 3 directions

CV for Solids

- Kinetic energy :3 degrees of freedom
- K = Kx+ Ky + Kz
- Kx = (1/2) mvx2
- Ky = (1/2) mvy2
- Kz = (1/2) mvz2

CV for Solids

- Potential energy:3 degrees of freedom
- U = Ux+ Uy + Uz
- Ux = (1/2) kx x2
- Uy = (1/2) ky y2
- Uz = (1/2) kz z2

Phase Changes

- ‘phase’ = state of matter = solid, liquid, vapor
- during a phase transition : 2 phases coexist
- at the triple point : all 3 phases coexist

Announcements

- Midterms:
- Returned at end of class
- Scores will be entered on classweb soon
- Solutions available online at E-Res soon
- Homework 7 (Ch. 16): on webpage
- Homework 8 (Ch. 17): to appear soon

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