The Second Law: The Machinery
The Second Law: The Machinery
Sections 5.1 - 5.6
The laws of thermodynamics we have learned thus far now enable us to
begin finding relationships between properties that may have not been
thought to be related with one another - there are many interesting
“hidden” relationships that can be extracted - lets now combine the laws:
Combining First and Second Laws
The First Law of Thermodynamics:
Chapter 5 of Atkins: The Second Law: The Concepts
dU = dq + dw
Properties of the Internal Energy
Properties of the Gibbs Energy
The Chemical Potential of a Pure Substance
For reversible change in constant composition system with no nonexpansion work
dw rev = − pdV
Real Gases: The Fugacity
Therefore
Definition of Fugacity
Standard States of Gases
Relation Between Fugacity and Pressure
Prof. Mueller
Chemistry 451 - Fall 2003
(closed system, no non- expansion work)
Lecture 13 - 1
Prof. Mueller
When S and V are changed, dU ∝ dS and dU ∝ dV, from the fundamental equation,
suggesting that dU should be written as a function of S and V
As in chapter three, the equation above means that the change in U is proportional
to changes in S and V, with the coefficients being slopes of plots of U against S at
constant V, and U against V at constant S. For systems of constant composition
The state functions are exact differentials, meaning that they must pass
the test that indicate their independence of path taken:
⎛ ∂g ⎞
⎛ ∂h ⎞
df = gdx + hdy is exact if ⎜ ⎟ = ⎜ ⎟
⎝ ∂y ⎠ x ⎝ ∂x ⎠ y
We know that dU = T dS - p dV is exact, so it follows that:
A relationship between quantities that would not seem to be related on
first sight, has been generated. In fact, the four Maxwell relations can
be derived from the four state functions U, H, G and A:
⎛ ∂T ⎞
⎛ ∂p ⎞
⎜ ⎟ = −⎜ ⎟
⎝ ∂V ⎠S
⎝ ∂S ⎠V
⎛ ∂S ⎞
⎛ ∂V ⎞
⎜ ⎟ = −⎜ ⎟
⎝ ∂T ⎠ p
⎝ ∂p ⎠T
⎛ ∂U ⎞
⎜ ⎟ = −p
⎝ ∂V ⎠ S
Chemistry 451 - Fall 2003
Lecture 13 - 2
⎛ ∂T ⎞
⎛ ∂p ⎞
⎜ ⎟ = −⎜ ⎟
⎝ ∂V ⎠ S
⎝ ∂S ⎠V
⎛ ∂U ⎞
⎛ ∂U ⎞
dU = ⎜ ⎟ dS + ⎜ ⎟ dV
⎝ ∂S ⎠V
⎝ ∂V ⎠ S
Prof. Mueller
Chemistry 451 - Fall 2003
Maxwell Relations
Reversible change : T dS same as dq, and -p dV same as dw
Irreversible change : T dS > dq (Clausius inequality), and -p dV > dw
Sum of dw + dq = Sum of T dS + -p dV, system of constant composition
⎛ ∂U ⎞
⎜ ⎟ =T
⎝ ∂S ⎠V
Fundamental Equation
dU is an exact differential, independent of path, so the same values for
dU are obtained regardless of change being reversible or irreversible
Properties of Internal Energy
Thermodynamic
definition of
temperature
dU = TdS − pdV
dqrev = TdS
Lecture 13 - 3
Prof. Mueller
⎛ ∂T ⎞ ⎛ ∂V ⎞
⎜ ⎟ =⎜ ⎟
⎝ ∂p ⎠ S ⎝ ∂ S ⎠ p
⎛ ∂p ⎞ ⎛ ∂S ⎞
⎜ ⎟ =⎜ ⎟
⎝ ∂T ⎠V ⎝ ∂V ⎠T
Chemistry 451 - Fall 2003
Lecture 13 - 4
1
Variation of Internal Energy with Volume
Same arguments that apply for internal energy, U, are applied to G.
When a system has a change of state resulting in ∆G, this results from
changes in H, S, and T. Again, for infinitesimal changes:
In Chapter 3, we defined a coefficient called the internal pressure
⎛ ∂U ⎞
⎟
⎝ ∂V ⎠T
πT = ⎜
which can be written as a thermodynamic equation of state , in terms of p and
T, and can therefore be applied to any substance. We said that we would prove
that:
⎛ ∂p ⎞
⎟ −p
⎝ ∂T ⎠V
π T = T⎜
We can obtain πT by dividing the equation for dU by dV and substituting in:
⎛ ∂U ⎞
⎛ ∂U ⎞
dU = ⎜ ⎟ dS − ⎜ ⎟ dV
⎝ ∂S ⎠V
⎝ ∂V ⎠ S
⎛ ∂U ⎞
⎜ ⎟ =T
⎝ ∂S ⎠V
⎛ ∂U ⎞
⎜ ⎟ = −p
⎝ ∂V ⎠ S
⎛ ∂U ⎞ ⎛ ∂ U ⎞ ⎛ ∂S ⎞ ⎛ ∂U ⎞
⎜ ⎟ =⎜ ⎟ ⎜ ⎟ +⎜ ⎟
⎝ ∂V ⎠T ⎝ ∂S ⎠V ⎝ ∂V ⎠T ⎝ ∂V ⎠S
Prof. Mueller
⎛ ∂S ⎞
= T⎜ ⎟ − p
⎝ ∂V ⎠T
dG = dH - TdS - SdT
Since H = U + pV:
⎛ ∂p ⎞ ⎛ ∂S ⎞
⎜ ⎟ =⎜ ⎟
⎝ ∂T ⎠V ⎝ ∂V ⎠T
Lecture 13 - 5
From the relationship below, we can derive two new partial derivatives
⎛ ∂G ⎞
⎜ ⎟ = −S
⎝ ∂T ⎠ p
G decreases when
the T is increased at
constant p, because
S is positive
⎛ ∂G ⎞
⎜ ⎟ =V
⎝ ∂p ⎠T
Because we control p and T, G is a very important quantity in chemistry.
G carries the combined consequences of the 1st and 2nd laws!
Prof. Mueller
Lecture 13 - 6
Temperature Dependence of G
In many chemical and physical transformations, the equilibrium
composition of a system depends on the Gibbs energy - so we must know
the response of the Gibbs energy to temperature changes
Since
⎛ ∂G ⎞
⎜ ⎟ = −S
⎝ ∂T ⎠ p
we write
⎛ ∂G ⎞
G−H
⎜ ⎟ =
⎝ ∂T ⎠ p
T
⎛ ∂G ⎞ G
H
⎜ ⎟ − =−
⎝ ∂T ⎠ p T
T
And now rewrite the LHS using:
⎛ ∂ ⎛ G ⎞⎞
1 ⎛ ∂G ⎞
G
∂ ⎛ 1 ⎞ 1 ⎛ ∂G ⎞
⎜ ⎜ ⎟⎟ = ⎜ ⎟ + G ⎜ ⎟ = ⎜ ⎟ − 2
∂T ⎝ T ⎠ T ⎝ ∂T ⎠ p T
⎝ ∂T ⎝ T ⎠⎠ p T ⎝ ∂T ⎠ p
G increases when p
is increased at
constant T (since V is
positive)
Chemistry 451 - Fall 2003
Chemistry 451 - Fall 2003
and we can rearrange to
G decreases most
sharply when S is
large (G of gaseous
phase is more
sensitive to T then
liquids or solids)
Prof. Mueller
dG = Vdp - SdT
thus,
Gibbs Energy, Pressure & Temperature
dG = Vdp − SdT
dH = dU + pdV + Vdp
For a closed system doing no non- expansion work, dU can be replaced
by the fundamental equation, dU = T dS - p dV
dG = ( TdS - pdV ) + pdV + Vdp - TdS - SdT
Change in G is proportional to changes in pressure and temperature
(use a Maxwell relation)
Chemistry 451 - Fall 2003
Properties of the Gibbs Energy
=
Lecture 13 - 7
Prof. Mueller
1 ⎡⎛ ∂G ⎞ G ⎤
⎢⎜ ⎟ − ⎥
T ⎣⎝ ∂T ⎠ p T ⎦
Chemistry 451 - Fall 2003
Lecture 13 - 8
2
Temperature Dependence of G
Pressure Dependence of G
This results in the Gibbs- Helmholz equation (which relates G
and H):
Integrate dG
=
Vdp - SdT
with respect to p at constant T (drop dT term)
Gm (p f )− Gm (pi ) =
⎛ ∂ ⎛ G ⎞⎞
H
⎜ ⎜ ⎟⎟ = − 2
T
⎝ ∂T ⎝ T ⎠⎠ p
:
Calculate the ∆Gm of (a) H2O (l) treated as an incompressible fluid , and (b) H2O
treated as a perfect gas, when pressure is increased isothermally from 1.0 bar
to 2. 0 bar at 298 K
(a) incompressible liquid, V
= constant,
Gm (p f )− Gm (pi ) =
(a)
(We shall see that we want the GH equation in this form, since the
equilibrium constants of reactions are related to G/T rather than
just G)
m
∫
pf
pi
∫
pf
pi
Vm dp
(b) perfect gas, V
m
= RT/ p
Vm dp = Vm (p f − pi )
= (18.0 ×10−6 m3 mol-1 )× (1.0 ×10 5 Pa)= +1.8 J mol-1
Gm (p f )− Gm (pi ) =
(b)
∫
pf
pi
⎛p ⎞
RT
dp = RT ln⎜ f ⎟
p
⎝ pi ⎠
= (2.48 kJ mol-1 )× ln(2.0) = +1.7 kJ mol-1
∆G = +ve, though the increase for gas is 1000 times that for liquid
Prof. Mueller
Chemistry 451 - Fall 2003
Lecture 13 - 9
Pressure Dependence of G, Solid & Liquid
For liquids and solids, ∆V is very small, and in a lot of cases may be
neglected (under lab conditions, Vm∆p is small!)
Prof. Mueller
Chemistry 451 - Fall 2003
Pressure Dependence of G, Gases
Molar volumes of gases are very large, so Gibbs energy may be strongly
dependent upon the pressure
Hence, we can normally assume that Gibbs
energies for solids and liquids are
independent of pressure - however, for
geophysical problems which involve very
high temperatures and pressures, the
volume effect on Gibbs energies cannot be
ignored!
If pressure is increased by 10x at room
temperature, then molar Gibbs energy
increases by about a factor of 6 kJ mol-1.
Suppose that for a solid phase transition that
∆trsV = +1.0 cm3 mol-1 , then at a high
pressure of 3.0 Mbar (i. e., 3.0 x 106 bar),
Gibbs energy of transition from ∆trsG (1 bar)
is calculated as:
As well, if we set p = p o (standard pressure
of 1.0 bar), the molar Gibbs energy of
perfect gas at some pressure p is related to
its standard value by
∆trsG (3 Mbar)
Prof. Mueller
Gm (p f )− Gm (pi ) =
∫
pf
pi
Vm dp =
∫
pf
pi
RT
dp
p
⎛ p⎞
Gm (p f )= Gmo + RT ln⎜ o ⎟
⎝p ⎠
= ∆trs G (1 bar) + (1.0 x 10 -6 m 3 mol -1 )
x (3.0 x10 11 Pa - 1.0 x 10 5 Pa)
= ∆trs G (1 bar) + (3.0 x 10 2 kJ mol -1 )
Chemistry 451 - Fall 2003
Lecture 13 - 10
Lecture 13 - 11
Prof. Mueller
Chemistry 451 - Fall 2003
Lecture 13 - 12
3
Chemical Potential
Chemical Potential
Chemical potential, µ, of a pure substance is defined as
⎛ p⎞
⊕⎟
⎝p ⎠
⎛ ∂G ⎞
⎟
⎝ ∂n ⎠T ,p
µ = µ ⊕ + RT ln⎜
µ =⎜
Chemical potential is useful for demonstrating how the Gibbs energy of a
system changes as a substance is added to it. For a pure substance,
µ ∝ ln p , standard state at po
Gibbs energy is G = n x G m , so
⎛ ∂G ⎞
⎟ = Gm
⎝ ∂n ⎠T ,p
µ =⎜
Here, the chemical potential is the same as the
molar Gibbs energy. For example, chemical
potential of a perfect gas can be rewritten as
This equation you will use many times in the next month, so make sure
you understand it. It simply says that we can define a molar free energy
for any substance (in this case an ideal gas, but we will generalize as time
goes on and the equation will always look the same just with appropriate
fudge factors built in) as just the molar free energy at the standard
condition plus some term that depends on the temperature and the log of
the relative amount of the substance (expressed as a pressure in the case
of a gas or later a concentration in the case of a solute in a liquid).
We give it a special name because later we will apply it to mixtures and talk about the chemical
For an ideal gas, the chemical potential at some pressure is just the
chemical potential at the standard pressure plus the change in entropy
associated with changing to the new pressure. In general, we split things
into terms that are properties of the molecules in question and terms that
simply have to do with changing the amount of room or total number of
states that are available to the molecules.
Prof. Mueller
Prof. Mueller
⎛ p⎞
⊕⎟
⎝p ⎠
µ = µ ⊕ + RT ln⎜
potentials of individual components. However, remember that the chemical potential of some
compound is just its molar Gibbs free energy.
Chemistry 451 - Fall 2003
Lecture 13 - 13
Real Gases & Fugacity
Chemistry 451 - Fall 2003
Lecture 13 - 14
Standard States of Real Gases
Consider the pressure dependence of the chemical potential of a real gas
To adapt to the case of a real gas, we
replace the true pressure p by an effective
pressure f, called the fugacity
⎛ f ⎞
⊕⎟
⎝p ⎠
µ = µ ⊕ + RT ln⎜
Fugacity, from Latin for “fleetness”; refers to
the “tendency to escape”. Fugacity has the
same units as pressure, and is a bit of a
“fudge factor” for treating real gases
In physical chemistry, since many properties
of materials are derived from chemical
potentials, fugacities are used to describe pressures
A perfect gas is in standard state when pressure is p o (1 bar): pressure
arises from kinetic energy, with no interactions taken into account
The standard state of a real gas is a hypothetical state in which the gas is
at p o and behaving perfectly
We choose a hypothetical standard state in order to standardize the
interactions between gas particles (for different types of gases) by setting
all interactions to zero. We do not choose the standard state as a gas at
pressure approaching zero, since the chemical potential of a gas will
approach negative infinity at this point: (µ → – ∞ as p → 0)
Standard chemical potential of gases, µ o , arises solely from internal
molecular structure and properties of gases, and not from the interactions
between particles
As p → 0, µ coincides with a perfect gas. At intermediate p, attractive
forces dominate (f < p), and at high pressures, repulsion gives f > p
Prof. Mueller
Chemistry 451 - Fall 2003
Lecture 13 - 15
Prof. Mueller
Chemistry 451 - Fall 2003
Lecture 13 - 16
4
Fugacity and Compression Factor, Z
Fugacity vs. Pressure
f = φp
Fugacity is written as:
For all gases
where is the dimensionless fugacity coefficient , which depends on the
nature of the gas, the temperature and the pressure. Thus,
⎛ p⎞
+ RT ln φ
⊕⎟
⎝p ⎠
∫
where µo refers to the standard state gas influenced only by kinetic energy
The second term describes the perfect gas, and RT ln φ expresses the
effect of the molecular forces of a real gas: all gases become perfect as p
approaches 0, so f → p as p → 0, and φ → 1 as p → 0
where Z = pV/nRT = pV
do this on the next slide:
Prof. Mueller
0
m
⎛ Z −1⎞
⎟ dp
p ⎠
p
p′
p
which is rearranged to:
ln φ = B′p + C ′p + L
2
The full VdW equations
of state are shown in
terms of fugacity
coefficients as
functions of reduced
pressure (and labeled
as curves at reduced
temperatures, T/ T c ),
providing estimates of
fugacities for a wide
range of gases
Prof. Mueller
Chemistry 451 - Fall 2003
Lecture 13 - 19
m
= RTZ/ p, and substituting in gives us proof, q.e.d.
Prof. Mueller
Fugacity and van der Waals Gases
To evaluate φ , experimental data on Z is needed from low pressures up
to the pressure of interest - sometimes, the fugacity can be obtained from
1
the virial coefficients of a gas:
2
⎡ ⎛ f⎞
⎛ p ⎞⎤
− V perfect,m )dp = RT⎢ln⎜ ⎟ − ln⎜ ⎟⎥
⎝ p′⎠⎦
⎣ ⎝ f ′⎠
⎛ f p′ ⎞ 1 p
ln⎜ × ⎟ =
∫ (Vm − Vperfect,m )dp
⎝ p f ′ ⎠ RT p ′
0, the gas behaves perfectly, and f’ → p’
⎛f⎞
1 p
ln⎜ ⎟ = ln φ =
∫ (Vm − Vperfect,m )dp
RT p ′
⎝ p⎠
V perfect, m = RT/ p, V
Lecture 13 - 17
m
p′
/RT is the compression factor of the gas. Let’s
Chemistry 451 - Fall 2003
⎛ f ⎞
Vm dp = µ − µ′ = RT ln⎜ ⎟
⎝ f ′⎠
⎛ p⎞
V perfect ,m dp = µ perfect − µ′perfect = RT ln⎜ ⎟
⎝ p′ ⎠
∫ (V
If p’ →
∫ ⎜⎝
p
p′
the difference of the equations is
We can prove that the fugacity coefficient of a gas can be calculated from
ln φ =
∫
f is fugacity at pressure p and f’ is fugacity at pressure p’ . Perfect gas:
µ = µ ⊕ + RT ln⎜
p
Gm (p f )− Gm (pi ) =
Chemistry 451 - Fall 2003
Lecture 13 - 18
Fugacity Summary
Most gases:
Z < 1 at moderate pressures, attractive forces
Z > 1 at higher pressures, repulsive forces
Z < 1, moderate p
ln φ =
⎛ Z −1⎞
⎟ dp < 0
p ⎠
0
p
∫ ⎜⎝
Z > 1, high p
ln φ =
p
⎛ Z −1⎞
⎟ dp > 0
p ⎠
∫ ⎜⎝
0
f <p
f >p
µ < µ⊕
µ > µ⊕
particles stick together
particles driven apart
Prof. Mueller
Chemistry 451 - Fall 2003
Lecture 13 - 20
5
Fugacity Calculation
Suppose attractive interactions for ammonia gas can be neglected at
10.00 atm and 298.15 K. Find an expression for fugacity of a van der
Waals gas in terms of the pressure.
We neglect a from the VdW equation (for attractive forces) and write
p=
RT
,
Vm − b
Z = 1+
Evaluating the integral for ln φ :
ln φ =
⎛ Z −1⎞
⎟ dp =
p ⎠
0
p
∫ ⎜⎝
Then solving for f we have
with b = 3.707 x 10
-2
L mol
φ=
-1
p
⎛ b ⎞
bp
RT
∫ ⎜⎝ RT ⎟⎠dp = RT
bp
0
f
= e bp RT
p
f = pe bp RT
, then bp/ RT = 1.515 x 10 -2 , giving
f = (10.00 atm) x exp( 0.01515) = 10.2 atm
The repulsive term, b, increases the fugacity above the pressure, so the
“escaping tendency” is greater than if it were perfect
Prof. Mueller
Chemistry 451 - Fall 2003
Lecture 13 - 21
6