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
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
Chemical Thermodynamics : Georg Duesberg
Chapter 1
The Properties of Gases
Chemical Thermodynamics : Georg Duesberg
Pressure
molecules/atoms of gas are constantly in motion
Ar
pressure = force area
p = F / A
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
Chemical Thermodynamics : Georg Duesberg
Boyle’s Law
pressure – volume
relationship
(temperature is constant) Boyle
(1627-1691) p ∞ 1/V
Kinetic theory and Gas Laws
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
Chemical Thermodynamics : Georg Duesberg
Gay-Lussac’s Law (also Charles law) temperature – volume
relationship
(pressure is constant)
Gay-Lussac
(1778-1850)
V ∞ T
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
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
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
Chemical Thermodynamics : Georg Duesberg
Chapter 1 : Slide 12
Surface of states Isobare and Isotherm
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
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
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
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.
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
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===
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 +−=−= ∫
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.
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
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
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
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−
==
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 + ….
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
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
Chemical Thermodynamics : Georg Duesberg
Chapter 2
Kinetic gas theory
Chemical Thermodynamics : Georg Duesberg
Kinetic Molecular Theory of Gases
Maxwell
(1831-1879)
Boltzmann
(1844-1906)
macroscopic
(gas cylinder)
microscopic
(atoms/molecules)
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
-‐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
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
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
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=
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
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 =
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
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.
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