Chapter 18Th l ti f ttThermal properties of matter

18 1 Equations of state18.1 Equations of state18.2 Molecular properties of matter18 3 Ki ti M l l d l f id l18.3 Kinetic-Molecular model of an ideal gas18.4 Heat capacities18.5 Molecular speeds18 6 Phase of matter18.6 Phase of matter Relating the macroscopic properties of matterto the microscopic behavior of moleculesto the microscopic behavior of molecules

Ideal gasIdeal gas• An ideal gas is a theoretical gas composed of a set of g g p

– randomly-moving – point particles (volume of container >>volume the gas

occupies)– interact only through elastic collisions . Conservation of

momentum applies m v =m v for all collusionsmomentum applies m1v1=m2v2 for all collusions.• The ideal gas concept is useful because it obeys the ideal

gas law, a simplified equation of state, and is amenable to analysis under statistical mechanics.

Equation of state • Relates state variables, pressure p, volume V,

temperature T. p– For a simple relationship there is an equation.– When the relationship is complicated we use graphs and

tables but still call them equations of statetables but still call them equations of state

Simple equations of state IEquation of state for materialEquation of state for material

( ) without neglecting effect of pressure:[1 ( ) ( )]V V T T kβ

Equation of state for solid material

0 0 0

0

[1 ( - ) - ( - )]k the compressibility is the fractional volume change per unit pressure:

/ 1

V V T T k p p

V V V

β= +

Δ Δ0

0

/ 1p

V V VkV p

Δ Δ= − = −

Δ Δ

Q

p

Simple equations of state II Ideal gas lawIdeal gas lawThe relationship between its state variables

of an ideal gas is: pV nRT=of an ideal gas is:

Any gas that obeys the is an ideal gas.

pV nRT

pV nRT

=

=

R is the ideal gas constant.

In SI units, 8.314472 (15) / . Al 0 08206

R J mol KR L

=/ l KAlso 0.08206 R L=

-3 3 2

. / . 10 , /

Other forms of the ideal gas law

atm mol KL m atm N m= =Ot e o s o t e dea gas aw

totaltotal

mm nM pV RTM

= → =

A setup to study behavior of the gassesAnd for density: pM

RTρ =

Ideal gas problems IIdeal gas problems I• Using equations of state we can find an unknown state

variable or relate before and after states of a system that undergoes a change and find some unknown variablesundergoes a change and find some unknown variables.

• Identify• Set up: one state problems, two or many state

problems target ariables don’t forget other forms of theproblems, target variables, don’t forget other forms of the state law.

• Execute: use consistent set of units, T is always absolute temperature p is the absolute pressureabsolute temperature, p is the absolute pressure,

• Evaluate the answer if it makes sense. • STP is:

– standard pressure 1.03 x105 Pa – standard temperature 00C or 273.15 K – Pa=N/m2

Example 1:Example 1:

• The approximate number of air molecules in a 1The approximate number of air molecules in a 1 m3 volume at room temperature (300 K) and atmospheric pressure is:A. 41B. 450C. 2.5 × 1025

D. 2.7 × 1026

E. 5.4 × 1026

R = 8.2 × 10–5 m3.atm/mol.KR 8.2 10 m .atm/mol.K NA = 6.02 × 1023 mol–1.

Example 2Example 2

• An air bubble doubles in volume as it risesAn air bubble doubles in volume as it rises from the bottom of a lake with density of 1000 kg/m3. Ignoring any temperature g g g y pchanges, the depth of the lake is:

• A) 21 m)• B) 0.76 m• C) 4.9 mC) 4.9 m• D) 10 m• E) 0 99 m• E) 0.99 m

Van Der Vaals Equationl l h l d l l f h hMolecules have volume and molecules exert force on each other

No gas is ideal in reality. V D V l ti t k th ti i t t

2

2

Van Der Vaals equation takes these corrections into account

( - )anp V nb nRTV

⎛ ⎞+ =⎜ ⎟

⎝ ⎠

moral density is small fo

VnV

⎝ ⎠

r dilute gases. V

2

The effective volume decreases for higher density: ( - ). effV V nb

an

=

⎛ ⎞2The pressure increases for higher density:

and are empirical (observational/experimental) constan

effanp pV

a b

⎛ ⎞= +⎜ ⎟⎝ ⎠

ts. p ( p )3 2 -5 3

2For CO 0.364 . / and 4.27 10 /a j m mol b m mol= = ×

ExampleExample2 Calculate volume of one mole of CO gas at STP

using the ideal gas law and Van Der Waals euation and express the difference in % Which one youand express the difference in %. Which one you expect to be higher?

pV-DiagramspV Diagrams• pV=nRT • we plot p vs V for a fixed• we plot p vs. V for a fixed

temperature.• We change the temperature and

plot p vs. V again.plot p vs. V again. • Each plot is an isotherm (fixed

temperature)• We can fit the data with the ideal

gas law for an ideal gas• Real gasses deviate from the

ideal gas curves and that is how l b t th f b t

pV-isotherms of an ideal gas

we learn about the forces between the gas molecules and their shape, volume, etc.

pV isotherms of an ideal gas

pV-diagram of a id lnot so ideal gas

• What is happeningWhat is happening between a and b?

• Why after b the plot y phas such a sharp slope?

• Can we create this plot if the system is i l d?isolated?

Molecular i fproperties of matter

• Matter composed of atoms• Matter composed of atoms and molecules ranging from 10-10 m to 10,000 times that in size

1( )U rr

∝

in size. • There is zero force between

molecules at r=r0 the ilib i di t

2

1( )

( )

rF rr

dUF r

∝

= −equilibrium distance• What happens when

temperature rises?

( )rF rdr

= −

p– Solids– Liquids– Gasses

Ideal gasIdeal gas• An ideal gas is a theoretical gas composed of a set of g g p

– randomly-moving – point particles (volume of container >>volume the gas

occupies)– interact only through elastic collisions . Conservation of

momentum applies m v =m v for all collusionsmomentum applies m1v1=m2v2 for all collusions.• The ideal gas concept is useful because it obeys the ideal

gas law, a simplified equation of state, and is amenable to analysis under statistical mechanics.

Atomic and molecular massAtomic and molecular mass

N=nNA

Kinetic-molecular model of an id lideal gas

• Applying Newton’s laws of motion and equation of• Applying Newton s laws of motion and equation of state of a large number of gas molecules in a container (ideal) to get the relationship between ( ) g pthe temperature and kinetic energy of the gas molecules A ti• Assumptions – Number of molecules is very large (~1023)– Molecules behave as point particles.Molecules behave as point particles. – Molecules are in constant motion (obey Newton’s laws)– Collisions are perfectly elastic (no momentum loss) – Container walls are perfectly rigid (no momentum loss in

collision with walls

Momentum change in molecular lli icollisions

Momentum change associated with a molecule's collision with

l l

the wall (yz plane) is:( ) 2P m v m v m vΔ = − − =one molecule ( ) 2x x xP m v m v m vΔ = =

Momentum change due to molecules’ collusion with area A during the time dtNumber of collisions during the time on the area :dt A

collusion with area A during the time dt

( )

time on the area : 1 | |2 x

dt AN A v dtV

⎛ ⎞⎜ ⎟⎝ ⎠

the volume which Only half of number of molecules atoms in it will the molecules per unit volume hit surfin direction

2

x

V

+

⎝ ⎠

ace during the time

T l h i d i h

Adt

Total momentum change associated with

collisions during the time on the area :dt AA d

2

122

xx x

NA v dtdp m v

VNA d

= =

2x

xNAmv dtdp

V=

From collisions to pressureThe force exerted on the wall due to collisions of the

2

molecules: (Newton's second Law)x xdP NAmvFdt V

= =

2

Pressure is then: xNmvFpA V

= =

For large number of molecules in a box, at any given moment one third of the molecules are moving in the direction,x

2 2 2 2( ) ( ) ( ) ( )

Since the directions x, y, and z are identical we have av x av y av z avv v v v= + +

2 2 2 2 21( ) ( ) ( ) and ( ) ( )3x av y av z av x av avv v v v v= = =

2xNmvp

V= →

2( )13

avNm vpV

=

Kinetic energy of the molecules and temperature of an ideal gastemperature of an ideal gas

2( )13

avNm vpV

=

2 2

31 2 1 2( ) ( ) where3 3 2 3av av tr

V

pV Nm v N m v K⎡ ⎤= = =⎢ ⎥⎣ ⎦

2

3 3 2 31 = ( ) . 2tr avK N m v

⎢ ⎥⎣ ⎦⎡ ⎤⎢ ⎥⎣ ⎦2

2 3 Temperature of an ideal gas3 trpV KK nRT

⎢ ⎥⎣ ⎦⎫= ⎪→ =⎬ Temperature of an ideal gas 3

2 related to kinetic energy of the gas molecules

trK nRTpV nRT

→ =⎬⎪= ⎭

related to kinetic energy of the gas molecules.Average translational kinetic energy of n moles of ideal gas.

Average kinetic energy of a gas molecule and its temperaturemolecule and its temperature

23 1For n mole: ( ) 2 2tr avK nRT N m v= =

23 1For one molecule: ( ) 2 2

trav

K nRT m vN N

= =

⎛ ⎞ 23 1with we have ( ) 2 2

trA av

A

KN RN T m vn N N

⎛ ⎞= = =⎜ ⎟

⎝ ⎠

23

23

8.314 J/mole.K 6.033 10 molecules/molA

RkN

= =×

Boltzman constant :

-231.381 10 J/k = × molecule.K. Now we see the depends on temperature:

average translational kinetic energy of a gas molecule only

2 21 3 1 3( ) = and for one mole of gas: ( ) =2 2 2 2av A avm v kT N m v RT

Molecular speedsIf your were free to go how fast would you get out of this class?out of this class?

( )( ) n vf vN

=

N is total number of students (gas molecules)n( ) is n mber of st dentsn(v) is number of students that have speeds between v and v+dv

Molecular speeds and temperaturetemperature

If your were free to go and there was a monster in the class how fast would you get out of this class?If f t d th t i th lIf your were free to go and there was monster in the class and pizza outside, how fast would you get out of this class?

Most probable speed

Molecular speedsMolecules in a gas do not all have the same speed.g pFor total of molecules of them have speeds between & where ( )

N dNv v dv dN Nf v dv+ = ( )

( ) : robability of molecule having speed between , or mp

ff v dv p v v dvv

+most is the speed for the peak of the probable speedmp

t

p p

probability function.For higher temperatures K increases so is the .v

p p

( )

tr

0

For higher temperatures K increases so is the .

Avarage speed:

mp

av

v

v vf v dv∞

= ∫The root-mean-square speed of a gas molecule is d

( ) ( )efined:

∞

∫( ) ( )( ) ( )2 2

0 But what is ? rms av

v v v f v dv f v∞

= = ∫

The Maxwell Boltzmann Distribution( )v is a probability function for distributionf ( )v is a probability function for distribution

of the molecular speeds. F id l hi i ll d

f

M ll B lt di t ib tiFor an ideal gas this is called . It can be derived using statistical mechanics but we won't derive

Maxwell - Boltzman distribution

i h

( ) 23/ 2

2 - / 2

it here.

4 mv kTmf v v eπ ⎛ ⎞⎜ ⎟( )

( )3/ 2

/

4 2

8 kT

f v v enkT

mf ε

π

π

= ⎜ ⎟⎝ ⎠

⎛ ⎞⎜ ⎟( ) - /

2

8 2

1

kTmf v em nkT

επ ε⎛ ⎞= ⎜ ⎟⎝ ⎠

21 is kinetic energy of each molecule.2

mvε =

The Maxwell Boltzmann Distribution3/ 28 mπ ⎛ ⎞( )

( )

- /8Maxwell-Boltzman distribution: 2

2

kTmf v em nkT

df vkT

επ ε⎛ ⎞= ⎜ ⎟⎝ ⎠

( )mp

2v can be found by solving 0 . df vkT

m dv= =

( )0

8 can be found from av avKTv v vf v dvmπ

∞= = ∫

3 3 can be found using rmskT RTvm M

= =

( ) ( )2 2

0= rms av

v v v f v dv∞

= ∫

( )2

N ( ) ( )( ) ( ) ( )2

22 2av0 0

ote: or vav

v f v dv vf v dv v∞ ∞

≠ ≠∫ ∫

Kinetic-molecular theory problems • Using kinetic-molecular theory we can relate the macroscopic• Using kinetic-molecular theory we can relate the macroscopic

state variables (p, V, T) to the microscopic values such as molecular speeds.S t id tif th i bl d k h th• Set up: identify the variables and unknowns, choose the proper equations

• Execute: use consistent set of units – M usually is given in g/mol in the tables. The SI unit is

kg/mol.– Either work on per mol base or per molecule base Don’t mix– Either work on per mol base or per molecule base. Don t mix

up.– R is the gas constant per mole pV=nRT– k is the gas constant per molecule pV=NkT– T is always absolute temperature, p is the absolute pressure– For unit check assume N has units of moleculesFor unit check assume N has units of molecules

• Evaluate the answer if it makes sense.

Molecular dspeeds

Calculating molecular kinetic energyof an ideal gasof an ideal gas

a) What is the approximate average translational kinetic energy of a molecule of oxygen gas at a

f 270C? (6 21 10 21 J)temperature of 270C? (6.21 x10-21 J)b) What is the total random translational kinetic energy

of the molecules in one mole of oxygen gas? (3740of the molecules in one mole of oxygen gas? (3740 J)

c) What is the root-mean-square speed of oxygen molecules at this temperature? (v (O )=484 m/s)molecules at this temperature? (vrms(O2)=484 m/s)

d) What is the rms speed of the hydrogen molecules at this temperature? (vrms(H2)=1934 m/s) p ( rms( 2) )

e) Why there is almost no H2 molecules in earth’s atmosphere? (Earth’s escape speed is 11.2 km/s).

f) What is the approximate average translationalf) What is the approximate average translational kinetic energy of a molecule of hydrogen gas at 270C? How come hydrogen molecules are faster?

In real life gas molecules collideCollisions between moleculesIn real life gas molecules collide. How often? Ho far the tra el bet een t o collisions?How far they travel between two collisions? The number of collisions a molecule ( ith di ) h d i th(with radius ) can have during the time if it travels with speed of

rdt v is :dN

22 4(2 ) ( ) and per unit time is:

hi i if l l l i i f id

N dN r vNdN r vdtV dt V

ππ= =

This is true if only one molecule is moving. If we consider all of the molecules are moving, more collisions happen:

24 2 (the analdN r vNdt V

π= ysis is complicated)

Mean-free path of a molecule24 2dN N24 2 Number of collisions per unit time

t is the average time ellepsed before the next collision

dN r vNdt V

π=

mean

2

t is the average time ellepsed before the next collision 1 here 1. Why?/ 4 2mean

Vt dNdN dt r vNπ

= = =

the distance a molecule travels btween two collisions is called: Mean free path

mean free 22

1 1 4 2 4 2mean

V VvtN rr N

λ λππ

= = → =

How mean free path of a molecule is affected by:a) concentration of the gas molecules b) size of the molecules?c) temperature?

Calculating the mean free pathCalculating the mean free patha) Calculate mean free path of a molecule of air at

270C d 1 t C id th l l h270C and 1 atm. Consider the molecules as spheres of radius 2.0 angstrom. How large is this distance compared to size of a molecule? (5 8x10-8m)compared to size of a molecule? (5.8x10 8m)

b) Estimate mean free time of an oxygen molecule. (1.2 x10-10s)x10 s)

andVvt pV NkTλ = = =2

and 4 2

/

meanvt pV NkTr N

NkT p kT

λπ

λ = =2 24 2 4 2

r N r p

λπ π

= =

Constant volume heat capacity of an ideal gasof an ideal gas

Heat is energy in transit. When a box full of gas is h t d th t i th t t l t d theated the system gains energy that translated to increase of kinetic energy of each molecule. To avoid calculating work done during the volume change we assume volume is constant .

3 and and 2tr v trdK nRdT dQ nC dT dQ dK= = =2

3 / 2

3vnC dT nRdT=

3 molar heat capacity of an ideal gas. 2vC R=

Molar heat capacity of real gassesMolar heat capacity of real gasses3C R=23 (8.314 J/mol.K)

vC R

C = (8.314 J/mol.K)212.47 J/mol.K.

v

v

C

C =

Check with real dataThe ideal gas model is good for monoatomic gasses but it is way far off for polyatomic p ygasses. We need a better model

Corrections to heat capacity3 Is a good approximation for constant-volume molar heatC R= Is a good approximation for constant-volume molar heat 2

capacity of monoatomic gasses. It fails for multiatomic gasses. Why?P l t i l l h dditi l ki ti d t

vC R=

Polyatomioc molecules may have additional kinetic energy due to rotation of the molecules or vibration of the atoms around an eq ilibri m state When a pol atomic gas is heeted the thermal enrgequilibrium state. When a polyatomic gas is heeted the thermal enrgy leads to rise in all kinds of kinetic enrgy.Temperature change due to heat does not follow a simple pathTemperature change due to heat does not follow a simple path.

Equipartition of energyPrinciple of equipartition of energy:

1Average kinetic enrgy per molecule per velocity component: kTAverage kinetic enrgy per molecule per velocity component:2

Number of degrees of freedom of a mlecule is the df

kT

N

number of velocity components needed to describe motion of the molecule.

1Average kinetic enrgy per molecule per degree of freedom:2

kT

1/molecule 2

avtotal dfK N kT= ×

Equipartition of energy

1/moleculeavK N kT= ×/molecule 2total dfK N kT= ×

Example using equi-partition of energy principle answer the f ll i tifollowing questions:

a) What is the average kinetic energy per molecule of a monoatomic gas?(3/2)kT g ( )

b) What is the average kinetic energy per molecule of a diatomic gas indicating rotation? (5/2)kT

Q(J)Nonlinear

nonlinear

T (K)Cv Experimental values for hydrogen gas H2.

You add heat to a constant volume of hydrogen gas at a constant rate. Does the temperature increase at a constant rate (between 25 and 5000 K)?

Heat capacities of solids, Dulong Petit RuleDulong Petit Rule

Each atom is bound to its position by interatomic forces. Each atom is like a simple harmonic oscillator in its place.If i di l d b A i ill h h i l f

2

If an atom is displaced by A it will have mechanical energy of 1( ) kA2mech atomE U= = we can prove av avK U K+ =2

An atom in a solid is a harmonic oscillator with three possible modes of oscillation so three degrees of freedom. For each degree of freedom there is two components of energy (kinetic and potential)Acording to equipartition of energy

1 1total

1 1(E ) 3 3 32 2

For a mole of solid:

atom kT kT kT= × + × =

totE 3 33 3 (8.314 / . ) 24.9 / .

A

v

N kT RTC R J mol K J mol K

= =

= = × =

Experimental values of Cv for solidsa) Why some solids have good agreement with Dulong Petit rule and some don’t?b) Why the rule fails for all of them at low temperatures?b) Why the rule fails for all of them at low temperatures?

SummarySummary

Phases of matter• What causes a phaseWhat causes a phase change in matter?

• Can an ideal gas go under phase change? Phase eq ilibri m a set of A typical PT diagram• Phase equilibrium: a set of condition under which phase change happens.

• Phase diagram represents

yp g

the phase change conditions on a graph.

• How may phases are present at each point on apresent at each point on a phase diagram?

• How many phases exist at a triple point on a phase diagram?diagram?

• Study phase change along the lines a, b, s, d).

• What happens at critical i ?point?

PVT surface diagrams

Three important lines in pstudy of phase diagrams: 1. Constant pressure heating (a-f)2 Isotheral compression (g m)2. Isotheral compression (g-m)3. Isothermal solidification directly

from vapor phase (n-q)

Ideal gas phase diagram

Three important lines in study of phase diagrams. Which color represents• Constant pressure heating?• Isothermal compression?• Isothermal solidification?

ExerciseExercise1. At an absolute pressure of 2.00x10-13 atm, a

partial vacuum easily obtained in laboratories.partial vacuum easily obtained in laboratories. Calculate the mass of nitrogen present in a volume of 300 cm3 if the temperature of the gas is 22.0oC. The molar mass of nitrogen is 28.0 g/mol.

2. The size of an oxygen molecule is about 2.0x10-10 m. Make a rough estimate of the

t hi h th fi it l f thpressure at which the finite volume of the molecules should cause noticeable deviations from ideal gas behavior at room temperaturefrom ideal-gas behavior at room temperature (T=300K).