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Chapter 16 - PowerPoint PPT Presentation

Chapter 16. Thermal Properties of Matter. Macroscopic Description of Matter. State Variables. State variable = macroscopic property of thermodynamic system Examples: pressure p volume V temperature T mass m. State Variables.

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Chapter 16

Thermal Propertiesof Matter

Macroscopic Descriptionof Matter

• State variable = macroscopic property of thermodynamic system

• Examples: pressure p

volume V

temperature T

mass m

• 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

• State variables: p, V, T, m

• The relationship among these:‘equation of state’

• sometimes: an algebraic equation exists

• often: just numerical data

• 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 state variables of a gas are easy to study:

• p, V, T, mgas

• often use: n = number of ‘moles’ instead of mgas

• 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

• 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!

• 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

• 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

• State variables: p, V, T, and mgas= nM

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

Derive ‘Law of Atmospheres’

• Ideal gas equation:

• Van der Waals equation:

Notes

pV–Diagram for an Ideal Gas

Notes

pV–Diagram for a Non-Ideal Gas

Notes

Microscopic Descriptionof Matter

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

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

pV = nRT

pV = NkT

• k = R/NA

• ‘ RT per mol’ vs. ‘kT per molecule’

Kinetic-Molecular Theory of an Ideal Gas

• gas = large number N of identical molecules

• molecule = point particle, mass m

• molecules collide with container walls= origin of macroscopic pressure of gas

• molecules collide with container walls

• assume perfectly elastic collisions

• walls are infinitely massive (no recoil)

• wall:

infinitely massive, doesn’t recoil

• molecule:

vy: unchanged

vx : reverses direction

speed v : unchanged

• For one molecule: v2 = vx2 + vy2 + vz2

• Each molecule has a different speed

• Consider averaging over all molecules

• average over all molecules:

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

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

= 3 (vx2)av

• (Ktr)av= total kinetic energy of gas due to translation

• Derive result:

• Compare to ideal gas law:

pV = nRT

pV = NkT

• average translational KE is directly proportional to gas temperature T

• average translational KE per molecule:

• average translational KE per mole:

• 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

• ‘root-mean-square’ speed vrms:

• For a given T, lighter molecules move faster

• Explains why Earth’s atmosphere contains alomost no hydrogen, only heavier gases

• 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)

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

• Maxwell-Boltzmann distribution function

• 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

• average speed

• rms speed

• We assumed:

• molecules = point particles, no collisions

• Real gas molecules:

• have finite size and collide

• Find ‘mean free path’ between collisions

• Mean free path between collisions:

• 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)

• Now introduce ‘molar heat capacity’ C

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

• 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

• 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

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

(Etot)av = (f/2) nRT

CV= (f/2)R

Monatomic gas:f = 3

Diatomic gas:f = 3, 5, 7

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?

• 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

• 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

• 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

(Etot)av = (f/2) nRT

CV= (f/2)R

Monatomic gas:f = 3

Diatomic gas:f = 3, 5, 7

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

CV for Solids

• f = 3 + 3 = 6 degrees of freedom

(Etot)av = (f/2) nRT

= 3 nRT

CV= (f/2)R = 3 R

Phase Changes Revisited

• ‘phase’ = state of matter = solid, liquid, vapor

• during a phase transition : 2 phases coexist

• at the triple point : all 3 phases coexist

pT Phase Diagram

pV–Diagram for a Non-Ideal Gas

Notes

• 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