Download 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
pressure =
force
area
p=F/A
Ar
molecules/atoms of gas are constantly in motion
Chemical Thermodynamics : Georg Duesberg
Kinetic theory and Gas Laws
the pressure of a gas increases when it is compressed at
constant temperature? – Boyles Law
•  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.
Chemical Thermodynamics : Georg Duesberg
Kinetic theory and Gas Laws
Boyle’s Law
pressure – volume
relationship
(temperature is constant)
Boyle
(1627-1691)
Chemical Thermodynamics : Georg Duesberg
p ∞ 1/V
Kinetic theory and Gas Laws
The volume of a gas increases when heated at constant
pressure - Charles’ Law
•  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.
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
Kinetic theory and Gas Laws
ISOTHERMS
p ∞ 1/V
p = const/V => p × V = const
p1 × V1 = const p2 × V2 = const
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.
Chemical Thermodynamics : Georg Duesberg
p 1 × V1 = p 2 × V 2
Kinetic theory and Gas Laws
The pressure of a fixed volume of gas increases with
temperature.
•  As temperature rises, the
molecules move faster
•  The molecules will hit the
walls of the container
frequently and violently
•  Hence, the pressure increases
Chemical Thermodynamics : Georg Duesberg
Isobare
V∞T
V = const’ ×T
V/T = const’
V1 / T1 = const’
V2 / T2 = const
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.
Chemical Thermodynamics : Georg Duesberg
V1 / T1 = V2 / T2
Surface of states Isobare and Isotherm
Chapter 1 : Slide 12
Chemical Thermodynamics : Georg Duesberg
Avogadro’s Law
Avogadro
(1776-1856)
2 H2(g) + O2(g) → 2 H2O(l)
R = 8.314 J / mol / K
k=1.38X10-23J/K
NA = 6.022×1023 – Avogadro number
Chemical Thermodynamics : Georg Duesberg
n∞V
n 1 / V1 = n 2 / V 2
N Ak = R
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)
3
Vm/(dm
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
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
mol− )
Gas
useful :
p1V1 p 2 V2
=
n1T1 n 2 T2
IDEAL GAS EQUATION
(1) Boyle Law p ∞ 1/V
V ∞ 1/p
(2) Gay-Lussac’s Law V ∞ T
V∞T
(3) Avogadro’s Law n ∞ V
V ∞T × n / p
V∞n
N = nNA
p × V = const × n × T
p × V = k × nNA × T
p ×V = R × n ×T
k=1.38X10-23J/K
R = 8.314 J / mol / K
N Ak = R
Chemical Thermodynamics : Georg Duesberg
NA = 6.022×1023 – Avogadro number
Application: Barometric formula: p as a function of height
Consider a column of gas with unit
cross sectional area.
Chemical Thermodynamics : Georg Duesberg
Variation of pressure with altitude
Barometric formula: p as a function of height
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".
Chapter 1 : Slide 17
Chemical Thermodynamics : Georg Duesberg
Barometric Formula
As elevation increases, the height
of the atmosphere decreases and
its pressure decreases.
F = mg = ρVg = ρhSg
F ρghS
P= =
= ρgh
S
S
m N
= 2
2
s
m
mass
moles (M W )
=
volume
V
Rewrite PV = nRT as
Therefore,
Chemical Thermodynamics : Georg Duesberg
kg m
kg
x
x
m
=
m3 s 2
m2
density = ρ =
Write in differential form.
dP = − ρ gdh
Check units.
PM W
ρ=
RT
n
P
=
V RT
Continue Derivation of Barometric Formula
Substitute the expression for density into the differential eqn.
PM W g
dP = −
dh
RT
Divide both sides of the above equation by P and integrate.
⌠ dP = −⌠ M W g dh
⎮
⎮
⌡ RT
⌡ P
Integration of the left side and moving the constants outside
the integral on the right side of the differential equation gives,
MW g
MW g
ln P = −
dh = −
h + ln C
∫
RT
RT
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
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.
− M W gh
ln Ph =
+ ln P0
RT
Ph = P0 e
Chemical Thermodynamics : Georg Duesberg
M W gh
−
RT
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
21
Chemical Thermodynamics : Georg Duesberg
IG-09
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:
P ( x, x + dx; x, y + dy; x, z + dz ) ∝
e
−
mgy
kT dxdydz
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
22
Probability density function
e
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
dy
P ( y) ∝
mgy
−
kT dxdydz
y
dx
x
Chemical Thermodynamics : Georg Duesberg
Height distribution in a gas
P ( y) ∝
e
mgy
−
kT dxdydz
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) :
n( y)
=
n ( y = 0)
Chemical Thermodynamics : Georg Duesberg
e
mgy
−
kT
DALTON’S LAW
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
Dalton
(1801)
Chemical Thermodynamics : Georg Duesberg
constant temperature and volume
p = pA + pB + pC + ….
p×V=n×R×T
pA = n A × R × T / V
pB = n B × R × T / V
p = (nA + nB) × R × T / V
mole fraction x < 1
pA / p = nA /(nA + nB) = xA
p A = xA × p
n
p = Σ pi
i=1
Chemical Thermodynamics : Georg Duesberg
p = pA + p B
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
macroscopic
(gas cylinder)
microscopic
Maxwell
(1831-1879)
Chemical Thermodynamics : Georg Duesberg
(atoms/molecules)
Boltzmann
(1844-1906)
Kinetic Molecular Theory of Gases
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)
Chemical Thermodynamics : Georg Duesberg
Kinetic Molecular Theory of Gases: Assumptions
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:
-­‐v vafter = -vbefore
v Δvelocity
2v
Force = mass ×
=m
Δtime
Δt
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
Collisions of the gas molecules with a wall
L
Freaction
Chemical Thermodynamics : Georg Duesberg
Small volume, v=LA, adjacent to
wall where L is less than the mean
free path
As a result of a collision with the wall
the momentum of a molecule changes by
Kinetic Molecular Theory of Gases
Pressure = Forcetotal/Area
P=F/A
•  Ftotal = F1 collision x number of collisions in a particular time interval
Only molecules within a distance νxΔt
with νx > 0 can reach the wall on the
right in an interval Δt.
L = vx Δt
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.
Chemical Thermodynamics : Georg Duesberg
Collisions of the gas molecules with a wall
This means that Δt is given by:
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.
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
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)
v=LA
N/ V = density
Remembering Pascal’s law
dividing by A yields the pressure
everywhere.
P=F
L
Chemical Thermodynamics : Georg Duesberg
A
Kinetic theory: go from 1 to 3 dimensions
L
Velocity squared of a molecule:
2
2
v = vx + v y + vz
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
Equipartition principle
and isotropic!
Chemical Thermodynamics : Georg Duesberg
2
2
Kinetic theory
Combing these results yields
From the ideal gas law
And with c = <v>
v
2
3kT 2
3kT
=c=
=
2m
m
Relation between the absolute temperature and average kinetic
energy of a molecule.
Chemical Thermodynamics : Georg Duesberg
Kinetic theory
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?
Chemical Thermodynamics : Georg Duesberg