Lecture 1: State Variables
The syllabus, calendar, lecture notes, and
problem assignments are on the web, at
www.chem.uic.edu/rjgordon/. The web
address is in my email signature. If you did
not receive an email message from me,
contact me at rjgordon@uic.edu
immediately.
Office hours every day Monday through
Thursday. Monday and Thursday will be
covered by one of the graders, and Tuesday
and Wednesday by me. Occasionally I will
swap hours with them.
The second session on Friday is mandatory
and will usually deal with problem solving,
but new material may also be presented.
The Friday sessions will also be used to
make up three Wednesday lectures that have
been cancelled. (See the calendar.)
1
The course will cover the first nine and a
half chapters of Atkins. See the syllabus
and calendar for page assignments, etc.
2
Physical chemistry is difficult because it
draws from many areas of chemistry and
physics. It requires knowledge of classical
mechanics, thermodynamics, electricity and
magnetism, quantum mechanics, statistical
mechanics, as well as chemistry. It is also
very problem oriented. The course has a
dual agenda.
Tips on how to solve problems:
1. Do not use a calculator or computer
until the very end. Keep the solution as
analytical as possible.
2. Keep track of the dimensions of the
variables. Try to use dimensionless
variables as much as possible.
3. When you do get a numerical answer,
judge whether it has a physically
realistic size and proper units.
3
Scope of the Course
Chemistry 342: A central problem of 19th
century physics: understanding the
macroscopic states of matter and their
transformations (e.g., phase transitions,
chemical reactions, equilibrium, kinetics).
The link to the microscopic world is through
the kinetic theory of gases. The emphasis in
this course is on equilibrium
thermodynamics.
Chemistry 344: A central problem of 20th
century physics: understanding the
microscopic properties of matter and
deriving from them the macroscopic
observables; relating chemical structure,
spectroscopy, and kinetics to quantum
mechanics.
4
Q. How do we describe the macroscopic
state of matter?
A. By a list of state variables. By
definition, such variables do not depend on
the previous history of the system.
Compared with the ~1025 microscopic
variables needed to describe a macroscopic
system, only a few state variables are needed
to describe a macroscopic object. Our goal
is to identify these variables and the physical
laws that relate them.
There are two types of state variables:
extensive ones and intensive ones.
Example of a swimming pool: chemical
composition of the liquid, amount of mass,
temperature, pressure.
5
Examples of Extensive Variables
Number of moles of each chemical
component: n1, n2, …,nr. These are
dimensionless quantities. Knowing them
is equivalent to knowing the mass of the
system.
Energy of the system: U (kg m2 s-2; joules)
There are different sources of energy.
Kinetic energy can come from internal
motions of a molecule or crystal, such as
vibration, bending, and torsional motion.
Kinetic energy can also come from external
motion, such as translation and rotation.
Potential energy can come from electronic
energy, and contributes to the chemical
energy of a system. They all add up to the
total energy, U. Understanding the different
types of energy and the conversion frpm one
type to another is a major topic of this
course.
6
Volume: V (m3)
Magnetic dipole moment: I (joules/Tesla)
Electric dipole moment: m (C m)
Length (e.g., of a rubber band, polymer): L
(m)
Surface area (bubble, droplet): A (m2)
Entropy: S, a measure of disorder or chaos.
It is dimensionless.
7
Examples of Intensive Variables
Pressure: P. Defined as the force per unit
area: Newtons/m2 = pascal
1 bar = 100,000 Pascal
1 atmosphere = 1.01325 bar = 101,325
pascal = 760 Torr
Temperature: T. A quantity that is
proportional to the translational energy per
particle. (Hotter particles move faster.) It
has the dimensions of energy.
Chemical Potential: mi, chemical energy per
mole of substance i (joules)
Magnetic field, B (Tesla)
Electric field: E (Volts/m)
Tension, TL (joules/m)
8
Surface tension: g (Joules/m2)
Relation between extensive and intensive
variables:
P and V: Applying pressure to an object
reduces its volume.
-PV = (newtons/m2) x (m3) = newtons x
meters =force x distance = joules
Note the negative sign.
mi ni =(joules/mole) x (number of moles) =
joules
BI = (tesla) x (joules/tesla) = joules
Em = (V/m) x (Cm) = CV = joules
TL L = (J/m) x (m) = joules
gA = (Joules/m2) x (m2) = joules
9
The tricky case is temperature. The
extensive variable associated with
temperature is entropy. TS = joules
The Euler Relation
U = TS - PV + Simini + IB + mE + TLL + gA
+….
What we have done is replace the myriads of
variables needed to describe the microscopic
system with just a handful of macroscopic
state variables. For example, we do not care
about the charge on each particle, just the
overall dipole moment of the bulk sample.
If we need greater precision, we might
specify also the quadrupole moment, as well
as some higher multipole moments, but the
list is still short.
10
But not all macroscopic variables are state
variables. Example of a swimming pool that
can be filled with either rain water or water
from the faucet. There is no way of telling
from the present state of the pool how much
of each was used to fill the pool. All we can
say is that the sum of the two sources equals
the total water content of the pool:
Water in pool = Waterrain + Wfaucet
Similarly, the energy content of a
thermodynamic system is the sum of work
done on the system (ordered energy input)
and heat added to the system (chaotic
energy):
U=w+q
The size of w and q individually depends on
the path taken (history of the system) and
not on the state of the system.
11
Lecture 2: Ideal Gas Law
Assumptions of an ideal gas:
1. Particles are point masses
2. No forces between the particles
(elastic collisions)
Relation Between State Variables T, P, n, V
Figure 1.
Momentum
transfer by a gas
particle to a
surface
m = mass of one particle
n = number of moles
nNA = number of particles
(NA = Avogodro’s number)
V = volume of chamber
rn = number density = number of particles/V
= n NA/V
12
A = area of wall
v = velocity of particle in one dimension
pinitial = momentum of one particle striking
the wall = mv
pfinal = momentum of particle bouncing off
the wall = -mv
Dp = momentum change resulting from a
collision with the wall= -2mv
In time Dt, a particle moves a distance vDt.
Half of the particles in a volume AvDt strike
the wall in time Dt. (The other half are
moving in the wrong direction.)
Dpwall = Total momentum imparted to the
wall in time Dt = (+2mv) (rn/2) (AvDt)
= mv2nNAADt/V
(Why is there a factor of ½ ?)
13
Force imparted to the wall = Dpwall/Dt =
mv2nNAA/V
Pressure on wall = P = Force/Area =
mv2nNA/V
PV = nNAmv2
½ mv2 = kinetic energy of one particle
½ NA mv2 = kinetic energy of one mole of
particles = constant x T = RT
PV = nRT
This is the equation of state of an ideal gas.
We define Boltzmann’s constant as k=R/NA.
The kinetic energy of one molecule moving
in one dimension is ½ kT. (In 3D it is 3/2
kT. This is an example of the equipartition
theorem.)
14
The problem is more complicated because
the particles don’t all have the same speed
and they don’t all move in one dimension,
but the proportionality PV ~ nT still holds.
R is called the gas constant. Its units are:
L atm mol-1 K-1 from PV=nRT
J mol-1 K-1 from U ∂ RT
15
How good is assumption 1?
Suppose the radius of a molecule is 3x10-8
cm. For 1 mole at 1 atm at 298 K, the
volume of all the molecules is
V0 = (4/3) pr3NA=68 cm3 =0.068 l.
If the gas pressure is 1 atm, its volume is
V=nRT/P = 1 x 0.08206x298/1 = 24.4 l.
The fractional error introduced by assuming
point masses = V0/V = 0.0028.
Clearly, we can make the error much smaller
by reducing P.
16
One way of defining the temperature is by
the equation of state of a very dilute gas.
Suppose a fixed volume of gas, V, is in
thermal equilibrium with some standard
system, such as water at its triple point.1
Let the pressure of the gas be P3.
We define the absolute temperature of the
triple point to be some number, q3. (In this
case, 273.16 K)
Under some other condition (for example,
the gas thermometer might be in thermal
equilibrium with boiling water), the same
mass and volume of gas has a pressure P.
The absolute temperature, q(P), of the gas in
that case is then given by the ideal gas law:
q(P) = (P/P3)q3.
1
Triple point of H2O: 6.11 mbar, 273.16K
17
To guarantee that the gas has ideal behavior,
we take a limit,
θ ( p ) = lim ( P / P3 )θ 3
P , P3 →0
We can also define the absolute temperature
in terms of two fixed points, such as the
melting and boiling points of water. We
arbitrarily divide the temperature interval
into 100 parts. Again assume a fixed mass
and volume of gas, and define the two
absolute temperatures to be q0 and q100.
The absolute temperature, q(P), at some
condition of interest is given by:
(q(P)-q0)/100 = (P-P0)/(P100-P0).
Again, take the limit of P0Ø0.
We define the Celsius temperature, T, by
T=q-q0.
18
Note: In the water example, it is an arbitrary choice
to define q0 = 273.15 K and q100 = 373.15 K. These
choices are unique to the Kelvin scale. An
alternative is given in problem 1.3. What all absolute
temperature scales have in common is that qØ0 in the
limit that PVØ0.
Note: An absolute temperature scale could also be
devised by using a fixed pressure and allowing the
volume to change when the thermometer comes into
contact with the various heat baths. In this case we
would use the notation q(V). In all of these
examples, we use one or two large heat baths as
references to define the temperature scale, and we use
the gas sample as a thermometer.
19
How is pressure measured?
Principle of the manometer: Compare the
pressure of the sample with the force of gravity.
Figure 2.
Manometer
P = pressure of the unknown gas.
r = mass density of the fluid in the
manometer.
h = height difference of the fluid in the two
arms of the manometer.
20
A = cross sectional area of the arms.
The force exerted by the gas on the fluid in
one arm of the manometer is PA.
Additional force exerted by the fluid in the
evacuated arm of the manometer is Dmg,
where Dm is the mass difference of the fluid
in the two arms.
But Dm=rAh, giving a force of rAhg.
Therefore PA=rAgh fl P = rgh
More generally,
dP
= ρg
dy
21
Buoyancy
Why does a cork float?
Let h be the height of the cork, and A its
cross sectional area. Its volume is V=hA.
The pressure difference from the top to the
bottom of the cork is: DR=rmediumgh.
The force difference is:
DF = rmediumghA = rmediumgV.
Gravitational force on the cork
= -mcorkg = -rcorkgV.
Net upward force on the cork
F = (rmedium - rcork)gV.
22
How does a balloon work?
Figure 3. Forces on a balloon.
The radius of the balloon is r.
23
Consider an infinitesmal ribbon. Its radius
is r cosq, its width is r dq, and its height
above the “equator” is h = r cosq.
The pressure difference on the ribbon above
and below the equator is
Pbottom - Ptop = rairg(2h)
The vertical component of the force is F cosq,
where F=PA and A = {(2pr sinq) r dq.
The difference between the vertical
components of the force on the ribbon is
DF = rairg(2h)cosq dA
= rairg(2rcosq)cosq {(2pr sinq) r dq}
Integrating over the entire surface of the
balloon:
∆F =
π /2
2
2
2
ρ
gr
cos
θ
(
2
π
r
sin θ ) dθ
∫ air
0
24
∆F = 4πρ air gr 3
π /2
2
cos
∫ θ sin θdθ
0
1
4 3
∆F = 4πρ air gr ∫ x dx = πr ρ air g = Vρ air g = mair g
3
0
3
2
Subtracting the force of the gas inside the
balloon pushing outward,
DF = (rair -rgas) Vg
25
Gas Mixtures: Dalton’s Law
Suppose there are n1 moles of substance 1
and n2 moles of substance 2 present in a
sample.
V and T are the same for both substances.
P1V= n1RT
P2V= n2RT
PV = nRT
P1= n1RT/V = (n1/(n1+n2)) nRT /V = x1P
Similarly, for any component i, with mole
fraction, xi,
Pi = x i P
Question: A certain gas has a pressure 300 Torr
and consists of 1 gram of H2 and 1 gram of He.
What are the partial pressures of each component?
26
Question: (Exercise 1.10). The density of air at 740
Torr and 27oC is 1.146g/l. Assuming that air consists
only of N2 and O2, what are their mole fractions and
partial pressures?
27
Lecture 3. Kinetic Theory of Gases, Part I
Discrete probability distributions
Example of size distribution of words in a text
Number of
letters, Li
Frequency
Probability
1
2
3
4
5
6
5
10
17
8
2
0
5/43 10/43 17/43 8/43 2/43 0
7
1
1/43
pi = ni/N
N = Sni
<L> = SLiPi = 1x5/43 + 2x10/43+…+7x1/43 = 2.9
<L2> = SLi2 Pi = 1x5/43 + 4x10/43+…+49x7/43 = 9.9
Lrms = (9.9)1/2 = 3.1
Most probable length = 3
28
Continuous distributions
f(v)dv is the probability of observing a
speed between v and v+dv.
f(v) is the probability density. What are its
dimensions?
Normalization:
∞
∫
f (v)dv = 1
−∞
Warning: Be careful to use the correct
limits of integration. They are not
necessarily ≤¶.
Probability of observing a speed between
v1 and v2 : P(v , v ) = f (v)dv
v2
1
2
∫
v1
Most probable speed: df/dv=0
Mean speed:
∞
v = v = ∫ vf (v )dv
−∞
RMS speed:
1/ 2
v rms
∞

=  ∫ v 2 f (v)dv 
− ∞

29
One-dimensional Maxwell-Boltzmann
distribution
f(v) = C exp{-1/2 mv2/kT} = C exp{-v2/a2}
where ½ ma2 = kT and m is the atomic mass
a º speed sound º 400 m/s
Define the dimensionless variable: x ª v/a
f(x) = C exp(-x2)
Normalization:
∞
C ∫ e − x dx = C π = 1
2
−∞
f(v) = (1/ap1/2) exp(-v2/a2)
Note the limits of integration.
Most probable velocity is zero.
Average velocity is also zero because
∞
∫
xe − x dx = 0
2
−∞
30
Root mean square velocity:
v
2
=
1
α π
∞
2
∫v e
−∞
−v 2 / α 2
α2
dv =
π
∞
2 −x
∫ x e dx =
2
−∞
α2
2
vrms = a/21/2
½ m <v2> = ¼ m a2 = ½ kT = <E>
31
Lecture 4: Kinetic Theory of Gases, Part II
Three-dimensional distribution
f(vx,vy,vz)dvxdvydvz = f(vx)f(vy)f(vz) dvxdvydvz
= C3 exp{-(vx2 + vy2 + vz2)/a2} dvxdvydvz
v2 = vx2 + vy2 + vz2; x2 = v2/a2
dvxdvydvz = v2 dv sinq dq df
Take the average over all angles. What remains is
f(x) = C x2 exp(-x2), where 0 § v § ¶
∞
Normalization:
∫x
2 −x2
e
dx = π / 4
0
Note the limits of integration.
f(v) = (4/a3p1/2) v2 exp {-v2/a2}
Note: x2dx = v2dv/a3
32
Most probable speed:
df/dx = 0 fl xmp=1 fl vmp = a
Average speed:
x =
4
π
∞
∫x e
3 − x2
dx =
0
2
π
<v> = (2/p1/2) a
Show that ½ m<v2> = 3/2 kT = <E>. This a
further example of the Equipartition
Theorem.
Question: What fraction of the molecules
have speeds greater than some value v0?
∞
Answer:
∫ f (v)dv =
v0
4
π
∞
∫x e
2 − x2
dx
x0
Integrate by parts: u=x, dv = x exp(-x2)dx,
du=dx, v = -(1/2) exp(-x2)
33
Result:
2 x0
π
e
− x 02
+
2
π
∞
∫e
−x2
dx =
x0
2 x0
π
e
− x02
+ 1 − erf ( x0 )
The error function is defined by
erf ( z ) =
2
π
z
∫e
− x2
dx
0
Question: In two dimensions, the same
analysis shows the speed distribution is
proportional to v exp(-v2/a2). What is the
most probable speed, and what is the
probability of finding a speed greater than
this value?
34
Kinetic Theory of Collisions
Collision frequency:
Let molecules A and B have radii ra and rb..
Figure 4.
Collision cross
section.
Suppose molecule B is stationary, and A moves with
an average speed <v>.
Area swept out by A is p(ra+rb)2 = sab.
sab is the collision cross section. In general, it is the
effective target area, and may be energy dependent.
Volume swept out by A in one second is sab<v>.
Number of collisions experienced by one A molecule
in one second is
zab = sab <v> (nb/V)NA = sab <v> b
N
where
N
b
is the number density of B.
35
Equivalently,
sab <v>(Pb/RT)NA = sab <v>(Pb/kBT).
Dimensions are sec-1.
For sab = 100 Þ2 = 1 nm2, Pb = 1 Torr = 133 pa,
<v> = 400 m/s, T=300K fl zab ~ 4x106 s-1,
or 4 collisions /ms/Torr.
10 −18 ⋅ 400 ⋅ 133
6
=
4
⋅
10
1.38 ⋅ 10 − 23 ⋅ 300
For constant volume, za ~ T1/2
Average speed =
v =
4
α = 145.5 (T/Mamu)1/2 m/s
π
The total number of collisions between species A and
species B per second per unit volume is
ZAB = zab (na/V)NA = sab <v> NA2 (na/V) (nb/V)
= sab <v> NA2 ca cb
The number of moles of A that react per m3 per
second is sab <v> NA ca cb
36
Assuming sab = 1 nm2 and <v> = 400 m/s, the rate
constant is
k(T) = sab <v> NA
= 2.4x108 m3/mole/sec
= 2.4x1014cm3/mole/sec
k(T)/NA= 4x10-10 cm3/molecule/sec.
This is the rate constant for elastic collisions.
Chemical reactions are generally much slower.
Proper average over scattering angles introduces a
factor of 2 .
Proper average over speeds gives the rate constant
k (T ) = ∫ σ (v )vf (v )dv .
37
Mean Free Path
The time between collisions is 1/zab. The distance a
molecule travels between collisions is
l=<v>/zab = (21/2 sab Nb)-1
In liquid water, assume sº0.5x10-18 m2
c = 1/18 moles/cm3 = 3.35x1028 molecules/m3
l=0.42x10-10 m = 0.42 C
For 1 Torr N2, 298 K, s=0.42 nm2,
c=5.4x10-5 moles/liter, = 3.24x1022 molecules/m3,
l =5.2x10-5 m = 5.2x105C
N
Rough rule of thumb:
l = 1000 Angstroms at 1000 Torr.
(The previous result is within a factor of two of the rule.)
How low must the pressure of N2 be to have l = 1 m?
P = 5x10-5 Torr.
38
Effusion
The leak rate through a pinhole is given by the flux of
molecules through the hole. The volume swept out in
one second is A<v>. The number of molecules in
that volume is <v> A. The number of molecules
hitting the area A is approximately
N
G = N <v>/2 m-2sec-1
More rigorous treatment:
r
r
Γ = ρ N ∫ (v cosθ ) f (v )d 3 v = ρ N ∫∫∫ (v cosθ ) f (v)v 2 sin θdθdϕdv
N
Note: rN has the same meaning as .
Note: volume of the paralellepiped of molecules
striking the surface per second is v cosq A.
Figure 5.
Effusion
39
Γ = ρ Nα π
−3
−3 / 2
π /2
∞
2π
∫ v dvθ ∫ sin θdθ ϕ∫ dϕv cosθ e
2
v =0
=0
−v 2 / α 2
=0
Note: Integral over angles is p.
Γ = ρ Nαπ
−1 / 2
∞
ρ Nα 1
∫0 x e dx = 2 π = 4 ρ N v
3 −x2
Result:
flux = rate of effusion per unit area = rN<v>/4
Number of molecules passing through a hole of area
A in one second is rN <v>A/4.
Suppose the pressure is 10-5 Torr. How long does it
take to form a monolayer coating of N2 at 298 K,
assuming s=0.4 nm2?
Number of molecules per m2 is 1/s = 2.5x1018
rN =3.24x1017 m3, <v>=475m/s
3.24x1017 t x475/4 = 2.5x1018
t = 0.065 sec
40
A vacuum chamber has a volume of 1 m3 and initially
has a perfect vacuum inside. Suddenly a leak
develops. The leak turns out to be a pinhole of radius
0.1 mm. The chamber is sitting in a lab at one atm
external pressure and 298 K. Assume the lab air is
pure nitrogen. How long will it take for the pressure
to rise to 1 mTorr inside of the chamber?
Answer: The number of molecules, N, hitting area A
in time t is given by N = rN vAt
¼
rN = nNA/V = PNA/RT = 6.022x1023/(0.08206x298)
= 2.461018 /liter = 2.46x1025 m-3
v=475 m/s
A=3.1416x10-8 m2
N = 2.46x1025 /760/104 = 3.23x1018
Solving for t gives t=0.035 seconds.
41
Lecture 5. Real Gases
The ideal gas law,
Z = PV/nRT = PVm/RT = 1
is based on the assumption of noninteracting particles.
The quantity Vm=V/n is the molar volume.
Real atoms and molecules have an
intermolecular potential that is responsible
for chemical bonds and van der Waals
clusters. The long-range potential between
neutral molecules varies as -R-6. Short
range interaction is always repulsive.
Example of a chemical bond is Na2:
Re=0.308 nm, De = 16.6 kcal/mol, D0=16.4 kcal/mol.
Extreme case of a weak van der Waals
cluster is He2,
Re=0.75 nm, De = 0.1 kcal/mol, D0=2.6 x 10-6 kcal/mol.
42
Figure 6.
Potential
Energy
Curve.
43
Result is a non-ideal equation of state,
Z = 1 + B£ P + C£ P2 + …
Z= 1 + B/Vm + C/Vm2 +…
The “virial coefficients, B£, C£, … B, C,…
can be related to the intermolecular
potential.
It is also useful to work with empirical
equations of state, just as it is useful to work
with empirical potential energy functions.
The most famous one is the van der Waals
equation of state:
P = nRT/(V - nb) - a(n/V)2
P = RT/(Vm - b) - a/Vm2
The first term contains a correction for
repulsive forces. The second term is caused
by attractive forces.
44
Isotherms are plots of P vs Vm for constant
T. See Fig. 1.23.
Figure 8. Van
der Waals
Isotherms.
Ideal gas isotherms are hyperbolas, with
∂P
<0
.
∂Vm
The non-ideal isotherm has regions of
∂P
>0
which correspond to a phase
∂Vm
transition.
45
At low temperatures the isotherm has a
minimum and a maximum. As T increases,
the min and max get closer and finally
merge at a point of inflection, called the
critical point.
This point is defined by the conditions
 ∂P

 ∂V
m




T
= 0,
 ∂2P

 ∂V 2
m




T
= 0.
Solving these equations for the van der
Waals equation gives
Vm,c = 3b
RTc = 8a/27b
Pc = a/27b2
Define the dimensionless (reduced)
variables:
Pr = P/Pc
Vr = Vm/Vm,c
Tr = T/Tc
46
Substituting back into the van der Waals
equation gives a universal curve:
Pr = 8Tr/(3Vr - 1) - 3/Vr2.
The fact that it is universal is called the Law
of Corresponding States. This implies only
that the real equation if state can be well
described by just two parameters.
Figure 9.
Law of
Corresponding
States
47
We also find that
Zc = PcVc/RTc = 3/8.
Expanding the van der Waals equation of
state as a virial series:
P = RT/(Vm - b) - a/Vm2
Z = PVm/RT = Vm/(Vm - b) - a/(RTVm)
= 1/(1- b/Vm) - a/(RTVm) + …
@ 1 + b/Vm - a/(RTVm) + b2/Vm2
b − a / RT b 2
+ 2 + ...
Z = 1+
Vm
Vm
48
Lecture 6: The First Law
1. The internal energy of an isolated system
is constant. This is equivalent to the
statement that U is a state variable.
Because U is a state variable, the change in U caused
by a change of state does not depend on the path.
We define w>0 as the work done on the system. It
follows that w<0 corresponds to work by the system
on its surroundings. Similarly, q>0 is for an
endothermic process and q<0 is for an exothermic
process. If we consider w and q as the only sources
of energy change, then
DU = Uf - Ui = w + q
regardless of the path-dependent values of w and q.
49
Types of paths:
∏
Insulating walls fl q=0
This is called an adiabatic path, and the work
done is adiabatic work, wad.
w=wad
DU=wad
∏No work done: w=0
DU=q
∏ General case: DU= w + q
Important concept: We are free to choose any path
we wish connecting a given initial and final state in
order to calculate DU. In particular, we may choose
an adiabatic path in order to define energy and heat.
Mechanical definition of energy change: DU=wad
Mechanical definition of heat: q = wad - w
Mechanical equivalent of heat: w=Jq
What is the value of J? In 1849, Joule came up with
a value of 1 cal at 15±C = 4.15 J.
The value today is 4.1840.
50
Example of the swimming pool: A pool may be
filled either by rain water or by faucet water.
Suppose we wish to measure the amount of rain
water, but have no direct way of doing so. (It is hard
to put a rain gauge on the clouds!) Instead, we use
the following strategy. First, we cover the pool with
a piece of plastic, and fill the pool with faucet water.
We carefully measure the amount of water with a
gauge attached to the tap. Next, we drain the pool,
remove the sheet, and allow rain water to enter the
pool for a period of time. Finally, cover the pool with
plastic and fill the rest of the pool with faucet water,
again using the gauge on the tap. The difference
between two measurements equals to the amount of
rain water that fell into the pool.
In this analogy, the plastic sheet is equivalent to the
insulating walls, the first gauge measurement is wad,
the second measurement is w, and the rain water is q.
The definition of heat is therefore wad - w.
51
2. The work done on an adiabatic system to
change its state from a specified initial state
to a specified final state is independent of
the type of work done.
Other types of paths:
∏ Isothermal path: Tf = Ti
∏ Isochoric path: Vf = Vi
∏ Isobaric path: Pf = Pi
∏ Cyclic path:
Initial and final states are the same,
DU=0
52
3. A system that undergoes cyclic behavior
cannot deliver any work without the
expenditure of an equivalent amount
somewhere in the system. In other words, a
perpetual motion machine “of the first kind”
cannot be built. Example: water running
down a hill, turning a turbine, and then
running back up hill.
One of the most useful concepts is that of a
∏ reversible path: For such a path, the
system is in thermal and mechanical
equilibrium throughout.
53
Lecture 7: Applications of the First Law
Expansion of a gas:
Figure 10. Work done by
an expanding gas.
dw = Fdz = - Fexdz = - PexAdz = - PexdV
54
Consider the special case of a free expansion
(or sudden compression): Pex is constant.
V fin
w = − ∫ Pex dV = − Pex (V f − Vi ) = − Pex ∆V
Vin
We cannot say anything about DU for this process
unless additional information is given that would
allow us to calculate q.
Suppose V=Vf is fixed (expansion to a definite final
volume). Then PfVf= nRTf, but Pf and Tf are not
determined. We have insufficient information to
calculate q and DU.
Suppose Pf=Pex is fixed (expansion against a constant
external force). Then PfVf= nRTf, but Vf and Tf are
not determined.
What type of additional information do we need?
1. If the process is isothermal, Tf = Ti. Then
knowing either Pf or Vf is sufficient to determine the
final state.
2. If the process is adiabatic, q=0, DU = w,
and Uf = Ui + w.
55
Isothermal Processes
Ti, Vi, Pi fl Tf, Vf, Pf
Ti = Tf
For an ideal gas, DU=0.
1) Sudden isothermal expansion: Pex = Pf < Pi
Vf
w = − ∫ Pex dV = − Pf
Vi
Vf
∫
dV = − Pf (V f − Vi )
Vi
But PiVi = PfVf or Vf = Pi Vi /Pf
w = -Vi (Pi - Pf)
q = Vi (Pi - Pf)
Numerical example:
Pi = 2 atm, Pf = 1 atm, Vi = 0.1 liter
w = -0.1 liter atm = -0.1[10-3 m3 x 101,325 pa]
=-10.13 J
56
2) Reversible isothermal expansion
Vf
Vf
Vf
P
dV
= − nRT ln
= − PiVi ln i
V
Vi
Pf
Vi
w = − ∫ PdV = − nRT ∫
Vi
= -0.2 ln 2 liter atm =-14.04 J
Figure 11.
Indicator diagram for
isothermal expansion.
Blue is reversible, red
is irreversible.
Conclusion: More work is extracted in the reversible
process.
57
3) Sudden compression: Pex = Pf > Pi
Numerical example:
Pi = 1 atm, Pf = 2 atm, Vi = 0.2 liter
w = 0.2 liter atm = 20.27 J
4) Reversible compression:
w = 0.2 ln 2 liter atm = 14.04 J
Figure 12.
Indicator diagram for
isothermal
compression.
Blue is reversible, red is
irreversible.
Conclusion: More work is required in the
irreversible process.
58
Claim: Reversible isothermal expansions always
deliver more work than irreversible ones.
That is, -wrev > - wirr
where wrev = -nRT ln(Vf/Vi)
and
wirr = -Pf (Vf - Vi)
Proof: In both cases, Vf = nRT/Pf
Claim: Vf - Vi < Vf ln (Vf/Vi) = - Vf ln (Vi/Vf)
That is, 1 - Vi/Vf < - ln (Vi/Vf)
Define: x = 1 - Vi/Vf
Claim: x < ln(1-x) for 0<x<1
This is equivalent to x < x + x2/2 + x3/3 + …
QED
59
Lecture 8. Energy and Heat Capacity
How does the energy vary with temperature?
U = U(T,V)
 ∂U 

T
∂
V

dU = 
 ∂U 
 dV
 ∂V  T
dT + 
For an ideal gas:
 ∂U 

T
∂

V
dU = 
dT = CV (T ) dT
The change in energy along any path is
Tf
∆U = ∫ CV (T )dT
Ti
CV is the heat capacity at constant volume.
U = Utrans + Urot + Uvib + Uelec
For any atom or molecule: Utrans = (3/2)nRT
(3 degrees of freedom)
60
For a linear molecule: Urot = nRT (2 d. f.)
For a non-linear molecule: Urot = 3/2 nRT (3 d. f.)
nN Ahν
U
=
For a diatomic molecule: vib exp(hν / kT ) − 1
This result is derived by setting Ev = vhn,
P(Ev) = Cexp(-Ev/kT), and taking the sum over v,
∞
U vib = ∑ Pv Ev
0
For a polyatomic molecule, take the sum over all
vibrational modes.
In most cases, Uelec is nearly zero.
Cv,trans = 3/2 nR
Cv,rot = nR or 3/2 nR
Cv,vib = nRf2 where
hν exp( − hν / 2kT )
f =
kT (1 − exp( −hν / kT ))
61
Derive the last result. Show that at low T, f=0, and at
high T, f=1.)
This can be done by expanding the demominator:
1 - exp(-hn/kT) º 1 - (1 - hn/kT) = hn/kT
These results are all examples of the equipartition
theorem.
Cv is an extensive quantity, with units J K-1. It is
practical to define the intensive equivalents:
Molar heat capacity: Cv,m = Cv/n (J K-1 mol-1)
Specific heat capacity: Cv,m/M (J K-1 g-1)
62
Physical meaning of the heat capacity:
∏The difference in energy between any two states is
given by
Tf
∆U = ∫ CV dT
Ti
.
This result assumes ideal behavior.
It is also true that
dU = dw + dq = -PdV +dq
But at constant volume, dw=0.
Therefore, after integration,
DU = qV
where qv is the heat absorbed at constant volume.
63
Figure 13.
Example of a
thermodynamic
cycle
State 1: n=1 mole, P1= 1 atm, T1=300 K, V1=24.62 l
State 2: P2=P1,
T2=2T1,
V2=2V1
State 3: P3=(1/2)P1, T3=T1,
V3=2V1
Path A: 1 Ø 2
Path B: 2 Ø 3
Path C: 3 Ø 1
at constant pressure
at constant volume
at constant temperature
64
Calculate w, q, and DU along each reversible path
and for the entire cycle. Assume the gas consists of
atoms.
DU = (3/2)nRDT = 12.47 DT Joules
DUA = 3.74 kJ
DUB = -3.74 kJ
DUC = 0
wA = -P1 DV = -24.62 liter atm = -2.495 kJ
wB = 0
wc =-nRT ln(V2/V1) = 1.729 kJ
wcycle= -0.766 kJ
q = DU - w for each step.
65
Lecture 9: Enthalpy and Heat Capacity
Suppose the heat is not supplied at constant volume.
Typically experiments are performed in the open air,
i.e., at constant pressure. But in that case, the system
must also undergo PV work, and more heat is
required to get the same increase in T. This heat is
called the enthalpy, H.
Definition:
H = U + PV
Clearly, H is a state variable. What other properties
does it have?
dH = dU + PdV + VdP
= dq + dw + PdV + VdP
= dq + VdP
Note: The statement that dw = -PdV assumes that
the system is in mechanical equilibrium.
At constant pressure, dP = 0 and
dH = dqp
DH = qp
66
For an ideal gas, PV = nRT fl H = U + nRT
For constant n, DH = DU + nRDT
For constant T, DH = DU + RTDn
For constant n and T, DH = DU
Note: For an ideal gas with T, DU=0.
Note: DH is not equal to DU + PDV +VDP.
Heat capacity at constant pressure:
CP
 ∂H 
=

 ∂T  P
T2
∆H = ∫ C P dT
T1
67
How are CP and CV related?
For an ideal gas: H = U + nRT
 ∂H 

∂
T

P
CP = 
 ∂U 

∂
T

P
=
CP = CV + nR,
+ nR
CP,m = CV,m + R
Empirical result: CP,m = a + bT + c/T2
DH = a(T2 -T1) +
½ b(T
2
2
- T12) - c(T2-1 - T1-1)
Example: Heating of I2 at constant pressure
Phase
solid
liquid
vapor
A
40.12
80.33
37.40
b
0.04979
0
0.00059
Melting point = 386.8 K
Boiling point = 458.4 K
Enthalpy of fusion = 15.52 kJ/mol
Enthalpy of vaporization = 41.80 kJ/mol
68
c
0
0
-0.71e+5
Calculate the heat necessary to raise one mole of I2
from T1 to T2 at 1 atm.
a) T1 = 100 K, T2 = 200 K
qP = 40.12 (200-100) +
= 4.76 kJ
½ x 0.04979 (200 -100 )
2
2
b) T1 = 100 K, T2 = 400 K
½
x 0.04979 (386.82
qP = 40.12 (386.8-100) +
-1002) + 15,520 + 80.33(400-386.8) = 31.56 kJ
c) T1 = 100 K, T2 = 500 K
qP = 40.12 (386.8-100) +
x 0.04979 (386.82
-1002) + 15,520 + 80.33(458.4-386.8) + 41,800
+ 37.40(500-458.4) +
x 0.00059(5002-458.42)
-0.71x105(500-1-458.4-1) = 79.63 kJ
½
69
½
Lecture 10. Adiabatic Processes
General case: qad=0, DUad=wad
Irreversible adiabatic expansion: wad,irr = -Pf(Vf - Vi)
Reversible adiabatic expansion:
What shape does a reversible adiabat have in a P-V
diagram? (Recall that an isotherm is a hyperbola.)
We have two equally valid descriptions of the energy
change:
dU = CVdT (ideal gas)
dU = -PdV (for a reversible adiabat)
\ CVdT = -PdV = -nRT dV/V (ideal gas)
\ CVdT/T = -nR dV/V
If we assume that CV is constant, then integration
gives
ln(Tf/Ti) = -(nR/CV) ln (Vf/Vi) = (nR/CV) ln (Vi/Vf)
70
 Vi
=
=
V
Ti
PiVi
 f
Tf
Pf V f
Pf
Pi
=
V
 i
V
 f








nR
Cv
1+ nR / CV
1 + nR/CV = (Cv + nR)/CV = Cp/Cv ª g
\ PiVig = PfVfg
 Pf
= 
Ti  Pi
Tf
Equivalent result:



γ −1
γ
(Prove this by replacing V with nRT/P on both sides.)
Value of g for an ideal gas:
Atoms:
g = 5/3
Rigid rotor:
g = 7/5
Diatomic molecule with vibration fully active:
g = 9/7
71
Comparison of reversible and irreversible adiabatic
expansion of an ideal gas against a constant external
pressure:
Irrversible case: DUad,irr = wad,irr = -Pf(Vf - Vi)
(same form as the irreversible isothermal expansion)
wad,irr (= wad,rev) = CV(Tf - Ti)
Equating these results gives
PfVf - PfVi = CvTi - CvTf
nRTf - PfVi = CvTi - CvTf
(nR + CV)Tf = CvTi + PfVi
Tf = (CvTi + PfVi)/CP = Ti/g + PfVi/CP
= Ti/g + (Pf/Pi)PiVi/CP= Ti/g + (Pf/Pi)nRTi/CP
= Ti/g + (Pf/Pi)[(CP - CV)/CP]Ti
= Ti/g + (Pf/Pi)[1 - 1/g]Ti
= Ti [1/g + (g-1)/g (Pf/Pi)]
72
γ −1
 γ
 Pf
= Ti  
 Pi 
Tf
Reversible case:
Claim that more work is done reversibly. That is,
 Tf

T
 i




Pf
Pi



 rev
γ −1
 γ



<
<
1
γ
 Tf

T
 i



 irrev
+ (1 − γ )
1
Pf
Pi
Define: x = Pf/Pi < 1 and a = (g-1)/g <1.
Want to show that xa - ax < 1 - a
Define f(x) = xa - ax.
Clearly, f(0) = 0 and f(1) = 1-a
\ All we need to show is that f(x) increases
monotonically with x. This is equivalent to showing
that df/dx > 0 for 0<x<1.
73
df/dx = axa-1 - a = a(x-(1-a) - 1) > 0
because x<1 and a<1.
QED
Application to the Carnot Cycle
Figure 15.
The Carnot Cycle.
This cycle consists of two isotherms and two
adiabats.
We will use the notation of figure 4.5.
State A: Thot, VA
State B: Thot, VB
State C: Tcold, VC
State D: Tcold, VD
74
Path 1:
Path 2:
Path 3:
Path 4:
Isotherm, T=Thot
Adiabat, Thot Ø Tcold
Isotherm, T=Tcold
Adiabat, Tcold Ø Thot
w1 = -nRThot ln (VB/VA) q1 = nRThot ln (VB/VA)
w2 = CV (Tcold - Thot)
q2 = 0
w3 = -nRTcold ln (VD/VC) q3 = nRTcold ln (VD/VC)
w4 = -CV (Tcold - Thot)
q4 = 0
Show that VA/VB = VC/VD
wtot = -nR(Thot - Tcold) ln (VB/VA) = -qtot
75
Lecture 11: Thermochemistry
Central concept: DHcycle = 0
Example 1: Enthalpy of combustion of CO at 298 K
CO + ½ O2 = CO2
Set up a cycle to illustrate the concept:
At 298 K and 1 atm,
DHf(CO) = - 110.53 kJ/mol
DHf(CO2) = - 393.51 kJ/mol
DHf(O2) = 0
76
DH(path 1) + DH(path 2) = DH(path 3)
or
DH(path 2) = DH(path 3) - DH(path 1)
or
DHreaction = DHproducts - DHreactants
DHcombustion = -393.51 -(-110.53) = -282.98 kJ/mol
77
Example 2: Enthalpy of combustion of CO at 2000 K
1) qP,reactants
CO + ½ O2 at 298 K --ö CO + ½ O2 at 2000 K
3) DHr, 298
∞
2) DHr, 2000 ∞
4) qP, products
--ö CO2 at 2000 K
CO2 at 298 K
DH(path 1) + DH(path 2) = DH(path 3) + DH(path 4)
qP,reactants + DHr, 2000 = qP,products + DHr, 298
DHr, 2000 = DHr, 298 + (qP,products - qP,reactants)
T2
qP ,react . = ∫ C P ,react .dT
Ti
T2
qP , prod . = ∫ C P , prod .dT
Ti
78
Q. What is the enthalpy of formation of Cr at 583 K?
A. Zero, because DHf is defined with respect to the
standard state at each temperature.
Q. What is the DHf of C6H6 at 583 K?
A.
353
∆H
583
f
= ∆H
298
f
(C6 H 6 ) + ∫ C P (C6 H 6 (liq))dT + ∆H vap (C6 H 6 )
298
583
583
353
298
+ ∫ C P (C6 H 6 (vap))dT − ∫ (6C P (C ( gr )) + 3C P ( H 2 ( g ))dT
79
Hess’ Law: The standard enthalpy of a reaction
equals the sum of the standard enthalpies of
individual reactions into which the overall reaction
can be divided.
Example 3: Hydrogenation and Combustion
C2H2 + 2H2 Ø C2H6
(1)
C2H6 + 7/2 O2 Ø 2CO2 + 3H2O (liq)
(2)
2H2O (liq) Ø 2H2 + O2
(3)
C2H2 + 5/2 O2 Ø 2CO2 + H2O (liq)
(4)
Let’s first add up the enthalpies of the first three
reactions:
DH1 = DHf(C3H6) - DHf(C2H2) = 52.26 - 226.73 =
-174.47 kJ/mol
DH2 = 2DHf(CO2) + 3DHf(H2O, liq) - DHf(C2H6) =
2(-393.51) + 3(-187.78) - 52.26 = -1402.62
kJ/mol
DH3 = -2DHf(H2O, liq) = 375.56 kJ/mol
DH1 + DH2 + DH3 = -1201.53 kJ/mol
80
Now let’s calculate the enthalpy of the last reaction:
DH4 = 2DHf(CO2) + DHf(H2O, liq) - DHf(C2H2)
= 2(-393.51) -187.78 26 - 226.73 = -1201.83
kJ/mol
Conclusion:
DH1 + DH2 + DH3 = DH4
We can demonstrate this result by formally
construction a thermodynamic cycle; however, I
don’t recommend doing this as a problem solving
method.
4
HCªCH+7/2 O2 + 2H2 Ø2CO2 + H2O(liq) + 2H2 +O2
1∞
C2H6 + 7/2 O2
Æ3
Ø
2CO2 + 3H2O(liq)
2
81
Example 4. Bond Enthalpies
CH4
4DH(C-H)
------------------Ø
C+H+H+H+H
∞ 431.8
CH3 + H
Æ 338.8
ö CH2 + H + H ö CH + H + H + H
471.1
421.7
4DH(C-H) = 431.8+471.1+421.7+338.8 = 1663.4
DH(C-H) = 415.9 kJ/mol
Subtle point: Bond energies and bond enthalpies are
equal only at 0K.
82