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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

Chapter 16

Thermal Propertiesof Matter



State variables
State Variables

  • State variable = macroscopic property of thermodynamic system

  • Examples: pressure p

    volume V

    temperature T

    mass m


State variables1
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
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 state1
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 ‘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 n a
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 gas1
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
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 state1
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 state2
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’


Non ideal gases
Non-Ideal Gases?

  • Ideal gas equation:

  • Van der Waals equation:

Notes


Pv diagram for an ideal gas
pV–Diagram for an Ideal Gas

Notes


Pv diagram for a non ideal gas
pV–Diagram for a Non-Ideal Gas

Notes



Ideal gas equation
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 equation1
Ideal Gas Equation

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


Ideal gas equation2
Ideal Gas Equation

pV = nRT

pV = NkT

  • k = R/NA

  • ‘ RT per mol’ vs. ‘kT per molecule’



Assumptions
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
Kinetic Model

  • molecules collide with container walls

  • assume perfectly elastic collisions

  • walls are infinitely massive (no recoil)


Elastic collision
Elastic Collision

  • wall:

    infinitely massive, doesn’t recoil

  • molecule:

    vy: unchanged

    vx : reverses direction

    speed v : unchanged


Kinetic model1
Kinetic Model

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

  • Each molecule has a different speed

  • Consider averaging over all molecules


Kinetic model2
Kinetic Model

  • average over all molecules:

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

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

    = 3 (vx2)av


Kinetic model3
Kinetic Model

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

  • Derive result:


Kinetic model4
Kinetic Model

  • Compare to ideal gas law:

    pV = nRT

    pV = NkT


Kinetic energy
Kinetic Energy

  • average translational KE is directly proportional to gas temperature T


Kinetic energy1
Kinetic Energy

  • average translational KE per molecule:

  • average translational KE per mole:


Kinetic energy2
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 model5
Kinetic Model

  • ‘root-mean-square’ speed vrms:


Molecular speeds
Molecular Speeds

  • For a given T, lighter molecules move faster

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


Molecular speeds1
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 speeds2
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 speeds3
Molecular Speeds

  • Maxwell-Boltzmann distribution function


Molecular speeds4
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 speeds5
Molecular Speeds

  • average speed

  • rms speed


Molecular collisions
Molecular Collisions?

  • We assumed:

  • molecules = point particles, no collisions

  • Real gas molecules:

  • have finite size and collide

  • Find ‘mean free path’ between collisions



Molecular collisions2
Molecular Collisions

  • Mean free path between collisions:


Announcements
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 revisited1
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 revisited2
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 revisited3
Heat Capacity Revisited

  • important case:the volume V of material is held constant

  • CV = molar heat capacity at constant volume


C v for the ideal gas
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


C v for the ideal gas1
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


C v for the ideal gas2
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
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


A look ahead
A Look Ahead

(Etot)av = (f/2) nRT

CV= (f/2)R

Monatomic gas:f = 3

Diatomic gas:f = 3, 5, 7


C v for the ideal gas3
CV for the Ideal Gas

  • What about gases with other kinds of molecules?

  • diatomic, triatomic, etc.

  • These molecules are not pointlike


C v for the ideal gas4
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


C v for the ideal gas5
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
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 energy1
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
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


C v for the ideal gas6
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


A look ahead1
A Look Ahead

(Etot)av = (f/2) nRT

CV= (f/2)R

Monatomic gas:f = 3

Diatomic gas:f = 3, 5, 7


C v for the ideal gas7
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


C v for the ideal gas8
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


C v for the ideal gas9
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 c v for ideal gases
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)


C v for solids
CV for Solids

  • Each atom in a solid can vibrate about its equilibrium position

  • Atoms undergo simple harmonic motion in all 3 directions


C v for solids1
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


C v for solids2
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


C v for solids3
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
Phase Changes

  • ‘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

Do Exercise 16-39

pT Phase Diagram


Pv diagram for a non ideal gas1
pV–Diagram for a Non-Ideal Gas

Notes


Announcements1
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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