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Chemical Thermodynamics : Georg Duesberg Chemical Thermodynamics : Georg Duesberg The Properties of Gases Kinetic gas theory – Maxwell Boltzman distribution, Collisions Real (non-ideal) gases – fugacity, Joule Thomson effect Mixtures of gases – Entropy, Chemical Potential Liquid Solutions - Electrolyte activity, Henry’s and Raoult’s Law Thermodynamics of Mixtures - Colligative Properties

Chemical Thermodynamics : Georg Duesberg

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Page 1: Chemical Thermodynamics : Georg Duesberg

Chemical Thermodynamics : Georg Duesberg

Chemical Thermodynamics : Georg Duesberg

The Properties of Gases Kinetic gas theory – Maxwell Boltzman distribution, Collisions Real (non-ideal) gases – fugacity, Joule Thomson effect Mixtures of gases – Entropy, Chemical Potential Liquid Solutions - Electrolyte activity, Henry’s and Raoult’s Law Thermodynamics of Mixtures - Colligative Properties

Page 2: Chemical Thermodynamics : Georg Duesberg

Chemical Thermodynamics : Georg Duesberg

Readings

•  Atkins - PC •  Mortimer – PC •  Pitzer and Brewer – Thermodynamics •  Kittel Kroemer - Thermal Physics

http://www.tcd.ie/Chemistry/staff/people/duesberg/ASIN/20web/2027-10-09/teaching.html Or also via my chemistry staff page - link to ASIN page

Page 3: Chemical Thermodynamics : Georg Duesberg

Chemical Thermodynamics : Georg Duesberg

Chapter 1

The Properties of Gases

Page 4: Chemical Thermodynamics : Georg Duesberg

Chemical Thermodynamics : Georg Duesberg

Pressure

molecules/atoms of gas are constantly in motion

Ar

pressure = force area

p = F / A

Page 5: Chemical Thermodynamics : Georg Duesberg

Chemical Thermodynamics : Georg Duesberg

Kinetic theory and Gas Laws

•  When a gas is compressed at constant temperature,

•  the molecules have less volume to move and hit the wall of the container more frequently.

•  As a result, pressure will increases.

the pressure of a gas increases when it is compressed at constant temperature? – Boyles Law

Page 6: Chemical Thermodynamics : Georg Duesberg

Chemical Thermodynamics : Georg Duesberg

Boyle’s Law

pressure – volume

relationship

(temperature is constant) Boyle

(1627-1691) p ∞ 1/V

Kinetic theory and Gas Laws

Page 7: Chemical Thermodynamics : Georg Duesberg

Chemical Thermodynamics : Georg Duesberg

•  When a gas is heated, the gas molecules move faster and

•  hit the wall of the container violently.

•  The volume of gas must increase to keep the pressure constant.

•  So that the gas molecules hit the wall less frequently.

The volume of a gas increases when heated at constant pressure - Charles’ Law

Kinetic theory and Gas Laws

Page 8: Chemical Thermodynamics : Georg Duesberg

Chemical Thermodynamics : Georg Duesberg

Gay-Lussac’s Law (also Charles law) temperature – volume

relationship

(pressure is constant)

Gay-Lussac

(1778-1850)

V ∞ T

Page 9: Chemical Thermodynamics : Georg Duesberg

Chemical Thermodynamics : Georg Duesberg

p ∞ 1/V

p = const/V => p × V = const

p2 × V2 = const p1 × V1 = const

p1 × V1 = p2 × V2 The pressure-volume dependence of a fixed amount of perfect gas at different temperatures. Each curve is a hyperbola (pV = constant) and is called an isotherm.

ISOTHERMS

Kinetic theory and Gas Laws

Page 10: Chemical Thermodynamics : Georg Duesberg

Chemical Thermodynamics : Georg Duesberg

•  As temperature rises, the molecules move faster

•  The molecules will hit the walls of the container frequently and violently

•  Hence, the pressure increases

The pressure of a fixed volume of gas increases with temperature.

Kinetic theory and Gas Laws

Page 11: Chemical Thermodynamics : Georg Duesberg

Chemical Thermodynamics : Georg Duesberg

V ∞ T

V = const’ ×T

V/T = const’

V2 / T2 = const

V1 / T1 = const’

V1 / T1 = V2 / T2 The variation of the volume of a fixed amount of gas with the temperature constant. Note that in each case they extrapolate to zero volume at -273.15° C.

Isobare

Page 12: Chemical Thermodynamics : Georg Duesberg

Chemical Thermodynamics : Georg Duesberg

Chapter 1 : Slide 12

Surface of states Isobare and Isotherm

Page 13: Chemical Thermodynamics : Georg Duesberg

Chemical Thermodynamics : Georg Duesberg

2 H2(g) + O2(g) → 2 H2O(l) n ∞ V n1 / V1 = n2 / V2

Avogadro’s Law

Avogadro

(1776-1856)

R = 8.314 J / mol / K

k=1.38X10-23J/K RkNA =NA = 6.022×1023 – Avogadro number

Page 14: Chemical Thermodynamics : Georg Duesberg

Chemical Thermodynamics : Georg Duesberg

Avogadro principle: Volume of real gases At a given T and p, equal volumes of gases contain the same number of molecules,

Vm = V/n. Table below presents the molar volumes of selected gases at standard conditions (SATP 25°C and 100kPa)

Gas Vm/(dm3 mol−1)

Perfect gas 24.7896*

Ammonia 24.8

Argon 24.4

Carbon dioxide 24.6

Nitrogen 24.8

Oxygen 24.8

Hydrogen 24.8

Helium 24.8

22

22

11

11

TnVp

TnVp:useful =

At STP Vm = 22.414 m3/kmol at 0 °C and 101.325 kPa dm3 mol-1. At IUPAC = 22.711 m3/kmol at 0 °C and 100 kPa m3/kmol

Page 15: Chemical Thermodynamics : Georg Duesberg

Chemical Thermodynamics : Georg Duesberg

(1) Boyle Law p ∞ 1/V

p × V = const × n × T

(2) Gay-Lussac’s Law V ∞ T

IDEAL GAS EQUATION

(3) Avogadro’s Law n ∞ V

V ∞ 1/p

V ∞ T

V ∞ n

V ∞ T × n / p

p × V = R × n × T

R = 8.314 J / mol / K

p × V = k × nNA × T

AnNN =

k=1.38X10-23J/K

RkNA = NA = 6.022×1023 – Avogadro number

Page 16: Chemical Thermodynamics : Georg Duesberg

Chemical Thermodynamics : Georg Duesberg

Application: Barometric formula: p as a function of height

Variation of pressure with altitude

Consider a column of gas with unit cross sectional area.

Page 17: Chemical Thermodynamics : Georg Duesberg

Chemical Thermodynamics : Georg Duesberg

Chapter 1 : Slide 17

Boundary condition: ground level pressure is p0 so that p = p0 exp(-Mgh/RT)

An exponential decrease of p with height. Equal Δh's always give the same proportional change in p. Note the assumptions: 1) Ideal gas behavior 2) Constant g 3) Isothermal atmosphere Mgh is the gravitational potential energy. We will often see properties varying in proportion to exp(-E/RT) = exp(-ε/kBT)

where E is a form of molar energy (ε is a molecular energy) because these are examples of "Boltzmann distributions".

Barometric formula: p as a function of height

Page 18: Chemical Thermodynamics : Georg Duesberg

Chemical Thermodynamics : Georg Duesberg

Barometric Formula As elevation increases, the height of the atmosphere decreases and its pressure decreases.

Write in differential form.

gdhdP ρ−=

( )VMmoles

volumemass W=== ρdensity

Rewrite PV = nRT as RTP

Vn=

Therefore, RTPMW=ρ

Check units.

22223 mN

sm

mkgmx

smx

mkg

==hSgVgmgF ρρ ===

ghSghS

SFP ρ

ρ===

Page 19: Chemical Thermodynamics : Georg Duesberg

Chemical Thermodynamics : Georg Duesberg

Continue Derivation of Barometric Formula Substitute the expression for density into the differential eqn.

dhRTgPMdP W−=

Divide both sides of the above equation by P and integrate.

⎮⌡⌠

⎮⌡⌠−= dhRTgM

PdP W

Integration of the left side and moving the constants outside the integral on the right side of the differential equation gives,

ChRTgMdh

RTgMP WW lnln +−=−= ∫

Page 20: Chemical Thermodynamics : Georg Duesberg

Chemical Thermodynamics : Georg Duesberg

Continue Derivation of Barometric Formula Evaluating the integral between the limits of P0 at zero height and Ph at height h, gives

0lnln PRTghMP W

h +−

=

RTghM

h

W

ePP−

= 0

The constant of integration C can be determined from the initial condition P(h = 0) = P0, where P0 is the average sea level atmospheric pressure.

Page 21: Chemical Thermodynamics : Georg Duesberg

Chemical Thermodynamics : Georg Duesberg

21

Sample calculation •  Calculate the pressure on Mount Carrauntoohil (1,038 m)

under normal conditions? h = 1082 m Temperature as 25°C T = 298 K P0 = 101.3 kPa = 1 bar (760 Torr)

m(air) = 29 g/mol (N2 = 28 amu, O2 =32 amu) g = 9.81 ms-2 Standard gravity R = 8.314 J / mol / K p = p0 exp(-mgh/RT) = 89.5 kPa = 671 Torr

Page 22: Chemical Thermodynamics : Georg Duesberg

Chemical Thermodynamics : Georg Duesberg

IG-09

22

Height distribution in a gas

• Energy (E = Mgy) being considered is significantly higher than a quanta of energy. E is nearly continuous.

• Easier to think of probability density functions:

( ), ; , ; ,mgykTP x x dx x y dy x z dz dxdydze−

+ + + ∝

P is the probability of finding a molecule between x & x + dx, y & y + dy and z & z + dz

NB: dx, dy and dz are large compared to a molecule but small compared to the size of the system

Page 23: Chemical Thermodynamics : Georg Duesberg

Chemical Thermodynamics : Georg Duesberg

Probability density function

( )mgykTP y dxdydze−

dx

dy

y

x

The directions parallel to the ground (x & z) do not contribute to the probability density function; only the height (y) above ground has an influence

Page 24: Chemical Thermodynamics : Georg Duesberg

Chemical Thermodynamics : Georg Duesberg

Height distribution in a gas

( )mgykTP y dxdydze−

For an ideal gas at constant temperature T, the probability density P(y) is related to the number density (# of molecules N per unit volume V ) n(y) :

( )( )0

mgykT

n yn y e−

==

Page 25: Chemical Thermodynamics : Georg Duesberg

Chemical Thermodynamics : Georg Duesberg

DALTON’S LAW

Dalton (1801)

pure gases

gas mixtures

(atmospheres) the total pressure of a gas mixture, p,

is the sum of the

pressures of the individual gases (partial pressures) at a

constant temperature and volume

p = pA + pB + pC + ….

Page 26: Chemical Thermodynamics : Georg Duesberg

Chemical Thermodynamics : Georg Duesberg

pA = nA × R × T / V

p × V = n × R × T

pB = nB × R × T / V

p = pA + pB

p = (nA + nB) × R × T / V

pA / p = nA /(nA + nB) = xA

mole fraction x < 1

pA = xA × p

p = Σ pi i=1

n

Page 27: Chemical Thermodynamics : Georg Duesberg

Chemical Thermodynamics : Georg Duesberg

Dalton’s Law

Suppose we have two gases in a container: nA moles of gas A and nB moles of gas B.

We can define individual partial pressures

pA = nART/V and pB = nBRT/V .

Dalton’s Law is that the measured total pressure p is the sum of the partial pressures of all the components:

p = pA+pB+… = (nA+nB+…)RT/V.

Mole fractions: define xJ for species J as nJ/n

where n = (nA +nB+…).

Then, xA + xB + … = 1 and pJ = p xJ

Page 28: Chemical Thermodynamics : Georg Duesberg

Chemical Thermodynamics : Georg Duesberg

Chapter 2

Kinetic gas theory

Page 29: Chemical Thermodynamics : Georg Duesberg

Chemical Thermodynamics : Georg Duesberg

Kinetic Molecular Theory of Gases

Maxwell

(1831-1879)

Boltzmann

(1844-1906)

macroscopic

(gas cylinder)

microscopic

(atoms/molecules)

Page 30: Chemical Thermodynamics : Georg Duesberg

Chemical Thermodynamics : Georg Duesberg

Physical properties of gases can be described by motion of individual gas atoms/molecules

Assumptions:

1)  each macroscopic and microscopic particle in motion holds an kinetic energy according to Newton’s law

2)  They undergo elastic collisions

3)  They are large in number and are randomly distributed

4)  They can be treated as points of mass (diameter<< mean free path)

Kinetic Molecular Theory of Gases

Page 31: Chemical Thermodynamics : Georg Duesberg

       -­‐v   v  

Δt2vm

ΔtimeΔvelocitymassForce =×=

1)  According to Newton's law of action–reaction, the force on the wall is equal in magnitude to this value, but oppositely directed.

2.) Elastic collision with wall: vafter = -vbefore

Kinetic Molecular Theory of Gases: Assumptions

Page 32: Chemical Thermodynamics : Georg Duesberg

3. Avogardo Number – Brownian motion 4. Gases are composed of atoms/molecules which are separated from each other by a distance l much more than their own diameter d

d = 10-10 m

L = 10-3 m….. few m

molecules are mass points with negligible volume

Kinetic Molecular Theory of Gases: Assumptions

Page 33: Chemical Thermodynamics : Georg Duesberg

Chemical Thermodynamics : Georg Duesberg

L

reactionF

Collisions of the gas molecules with a wall

As a result of a collision with the wall the momentum of a molecule changes by

Small volume, v=LA, adjacent to wall where L is less than the mean free path

Page 34: Chemical Thermodynamics : Georg Duesberg

Chemical Thermodynamics : Georg Duesberg

Pressure = Forcetotal/Area P=F/A •  Ftotal = F1 collision x number of collisions in a particular time interval

Assume that in a time Δt every molecule (atom) in the original volume, v=LA, within the range of velocities

will collide with the wall.

Kinetic Molecular Theory of Gases

Only molecules within a distance νxΔt with νx > 0 can reach the wall on the right in an interval Δt.

tvL xΔ=

Page 35: Chemical Thermodynamics : Georg Duesberg

Chemical Thermodynamics : Georg Duesberg

Collisions of the gas molecules with a wall

The “reaction force” of a molecule on the wall is the negative of the average rate of change in the momentum of gas molecules in the volume v that collide with the wall in the time Δt.

This means that Δt is given by:

The total force on the wall is the sum of the average rate of momentum change for all molecules in the volume v=LA that collide with the wall

Here we have divided by 2 since only ½ of the molecules in our volume have a positive velocity toward the wall

Page 36: Chemical Thermodynamics : Georg Duesberg

Chemical Thermodynamics : Georg Duesberg

L

Collisions of the gas molecules with a wall (cont.)

We do the sum by noting that the total number of molecules in the volume v is (N/ V)

Remembering Pascal’s law dividing by A yields the pressure everywhere.

v=LA

N/ V = density

AFP =

Page 37: Chemical Thermodynamics : Georg Duesberg

Chemical Thermodynamics : Georg Duesberg

L

Kinetic theory: go from 1 to 3 dimensions

Velocity squared of a molecule: 2222zyx vvvv ++=

The average of a sum is equal to the sum of averages… All the directions of motion (x, y, z) are equally probable. Remember homogeneous and isotropic! Equipartition principle

Page 38: Chemical Thermodynamics : Georg Duesberg

Chemical Thermodynamics : Georg Duesberg

Kinetic theory

Combing these results yields

From the ideal gas law

And with c = <v>

mkT

mkTcv 32232===

Relation between the absolute temperature and average kinetic energy of a molecule.

Page 39: Chemical Thermodynamics : Georg Duesberg

Chemical Thermodynamics : Georg Duesberg

vrms of a molecule is “thermal speed”: The absolute temperature is a measure of the average kinetic energy of a molecule.

Example: What is the thermal speed of hydrogen molecules at 800K?

Kinetic theory