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APPLICATIONS OF THE FIRST LAW
• Kinetic theory and internal energy: predicting the specific heats of dry air
• Applications of the first law
• Kirchoff’s theorem
• Poisson’s equation
• Potential temperature
KINETIC THEORY AND INTERNAL ENERGY
Without delving into the kinetic theory of gases, we can make use of one of its results in order
to obtain some insight into internal energy.
1
each molecule possesses an amount of energy per degree of freedom equal to kT
2
where k is Boltzmann’s constant=1.381023 JK-1 .
We can think of the number of degrees of freedom as the number of variables required to
fully describe the motion of the molecules. So a monatomic molecule has 3 degrees of
freedom (3 velocity components are required to specify its translational
 motion), a diatomic
molecule has 5 degrees of freedom (3 translational plus 2 angular velocities), while a
triatomic molecule has 6 degrees of freedom (3 translational, 3 angular).
Dry air consists primarily of molecular Nitrogen (N2) and Oxygen (O2) and so is
essentially diatomic with 5 degrees of freedom.
Hence the internal energy per mole is given by:
1
1
nkNA T  nR * T
2
2
(4.1)
where n is the number of degrees of freedom and NA is Avogadro’s number equal to
NA=6.021023 mol-1 . Note: Boltzmann’s constant may therefore be thought of as the universal gas constant per
molecule.

The internal energy per kilogram (i.e., the specific internal energy) is therefore given by:
1 R*T 1
n
 nRT
2 M
2
(4.2)
5
where M is the molecular weight of the gas. Thus, we would predict that for dry air, u  2 RT
and since c v  du then c  5 R . Moreover, since cp-cv=R, then we can also predict
dT
v
7
2
2
that for dry air c p  R . Using R=287.05 Jkg-1K-1 , our prediction becomes 
for dry air:

cp=1005
Jkg-1K-1 
. The table below shows that this prediction compares rather well with
observations.

30 kPa
100 kPa
-40oC
1004
1006
+40oC
1006
1007
Table 4.1: Measured specific heat capacity at constant pressure, cp, for dry air.
POISSON’S EQUATIONS
Despite the fact that energy in the atmosphere/ocean system ultimately comes from
radiative heating, adiabatic processes in the atmosphere are of interest for several reasons.
Often it is because real atmospheric processes occur quickly in comparison with the time
scale for heat transfer, and so may be considered to be approximately adiabatic.
Alternatively, we may wish to make the adiabatic assumption simply because we are ignorant
of the heat transfer and consequently must ignore it or give up. Poisson’s equations describe
relationships between the state variables T, p, and  for adiabatic processes.
When q=0 (adiabatic process), the first version of the first law may be written:
cvdT  pdv  0
(4.5)
Substituting for p from the ideal gas law, dividing by cvT, and using R=cp-cv we have, after
some manipulation:
(4.6)
d ln( Tv 1)  0
where =cp/cv (=7/5 for an ideal gas). This implies that, for an ideal gas undergoing an adiabatic
process:
(4.7)
Tv  1  const

Eq. (4.7) is one of three Poisson equations.
Eq. (4.7) can be used to explain a number of atmospheric phenomena. For example:
1. When you compress the air in a bicycle pump, v decreases and hence T increases and the
air warms. (Sometimes the bicycle pump is noticeably warmer after use.)
2. A similar explanation can be offered for the heating of subsiding (i.e., descending) air that
gives rise to a Chinook.
3. Conversely, in a rising, expanding thermal, v increases and so T falls. This is a dynamical
component that contributes to the decrease of temperature with height in the troposphere.
(The other component is the height dependence of absorption of longwave radiation emitted by
the Earth.)
Starting with the second version of the first law, Eq. (3.8) and following a similar manipulation,
one may arrive at the second Poisson equation for an adiabatic process in an ideal gas:
Tp  const
(4.8)
where =R/cp (=2/7 for an ideal gas). Finally, by combining Eqs. (4.7) and (4.8), we arrive at
the third Poisson equation:

(4.9)
pv
 const

Where = cp / cv=7/5. Eq. (4.9) is the most familiar of Poisson’s equations, although they
are all equivalent.
POTENTIAL TEMPERATURE
Meteorologists use Eq. (4.8) to define a quantity known as potential temperature (a
thermodynamic concept that seems to be unique to meteorology). Suppose we start with a
parcel of air in some arbitrary initial state specified by T, p. Let us move the air parcel
adiabatically to a pressure of 100 kPa and call the temperature which it achieves the
potential temperature, . By using Eq. (4.8) in the initial and final states, it can be easily shown
that:
100 
(4.10)
  T 
 p 
Where p must be express in kPa (since we have used 100 kPa in the numerator). Because the
potential temperature of an air parcel is conserved under dry adiabatic processes, it may be used
as a tracer for air parcels.

Isentropic coordinates are ones in which potential temperature is used as the vertical coordinate
instead of height. In such a coordinate system, a parcel of dry air undergoing only adiabatic
processes will always remain on the same coordinate surface. [NOTES: Why they are called
isentropic coordinates will become clear when we discuss entropy and the Second Law of TD.
We will also take into account the presence of water vapour. A similar concept is
used in oceanography. These are isopycnal coordinates.]
The first law of thermodynamics may be expressed in yet a third version using potential
temperature. Let us begin with the definition of potential temperature, Eq. (4.10), take
logarithms and then total differentials:
dln   dln T  dln p
(4.11)
Multiplying by cp and using the ideal gas law and the second version of the first law:
c p d ln  

q
(4.12)
T
This third version of the first law has several useful features. It is somewhat simpler than the
previous two versions, it expresses the fact that the potential temperature is constant for
adiabatic processes and varies for diabatic processes, and it seems to imply that the quantity
on the right hand side isa function of state (since the potential temperature is a function of
state). We will encounter q/T again soon in conjunction with the Second Law of
Thermodynamics.
By considering Eq. (4.12) along with the first version of the first law for a cycle, and
remembering that a function of state does not change in a cyclic process (e.g.,  du  0 ), it is
easy to show that
c pTdln   pdv


The integral on the right hand side is the net work done during the cycle. Hence this equation
implies that on a graph with axes cpln and T, the area enclosed by acycle will equal the
work done
 during the cycle. We will make use of this fact later when we consider the
Tephigram.
THE SECOND LAW OF THERMODYNAMICS
• Thermodynamic surface for an ideal gas
• Second Law
• Entropy
THERMODYNAMIC SURFACE FOR AN IDEAL GAS
The ideal gas law is the equation of a surface in a three-dimensional space whose
coordinates are p, v, and T. This surface constitutes the ensemble of all states of an ideal gas
that are permitted by the ideal gas law. It is called the thermodynamic surface for that gas.
Any changes in the gas’ state variables simply reflects movement on this surface. The following
diagram illustrates this surface and on it are examples of isobaric, isothermal, isosteric, and
adiabatic processes. Such processes may all be represented in the form:
pv n  const
(5.1)
Where n=0 is an isobar, n=1 is an isotherm, n= is an adiabat, and n is an isostere.
[NOTE: an isosteric process is one for which the specific volume remains
 is also an isopycnic process because the density is constant. Processes
constant. Such a process
for which the total volume of the air parcel remains constant are called isochoric.]
SECOND LAW
Sometimes the use of the first law will give rise to a seemingly impossible result. How do we
know that it is impossible? Because it is entirely inconsistent with our collective experience. As
an example, consider the cooling of bathwater. We will assume this to be an isobaric process, so
that the first law is :
q  cw dTw
or
Q  mwcwdTw
(5.2)
Let us imagine that 100 litres of water cool from 50oC to 20oC. If that heat is given to 10 kg
of air in the bathroom (assuming the bathroom to be well insulated so that no heat is lost), what

will be the final temperature of the air?
If the warming of the air also occurs isobarically then the first law for the air is:
Q  m a c p dTa
(5.3)
Equating the Q’s in Eqs. (5.2) and (5.3), we can solve for dTa using cp=1.0103 Jkg-1K-1
and cw=4.2103 Jkg-1K-1. The temperature change of the air is 1.3103 K!!!
We know this result
is impossible, but it follows from the first law, so what is wrong? Clearly,
in the final stages of the process as we have envisaged it, the air will be warmer than the
bathwater. And yet we have implicitly assumed that heat will continue to flow from the water
into the air to warm it. Experience tells us that heat simply does not flow spontaneously from
a colder to a warmer body. This is called the Clausius formulation of the second law.
Rudolf Clausius (1822-1888) was a German physicist who established the foundations of
modern thermodynamics in a seminal paper of 1850. He introduced the concepts of internal
energy and entropy (from the Greek for transformation). His great legacy to physics was the
concept of the irreversible increase of entropy (“Die Energie der Welt ist constant; die
Entropie strebt einen Maximum zu.” (1865)).
A more precise expression of the Clausius formulation is:
“It is impossible to construct a cyclic
engine that will produce the sole effect of transferring
heat from a colder to a hotter reservoir.”
The term “sole effect” means that no work can be done. Of course heat pumps exist, so it is
possible to transfer heat from a colder to a warmer reservoir, but only by doing work.
Oxford physicist P.W. Atkins in his book entitled “The Second Law” states it in yet
another way (p. 9):
“…although the total quantity of energy must be conserved in
any process….the distribution of that energy changes in an
irreversible manner.”
He also talks about a “fundamental tax,” stating that “Nature accepts the equivalence of heat
and work, but demands a contribution whenever heat is converted into work.” (p. 21). Note,
however, that there is no tax when work is converted into heat (for example, by friction).
Since we may think of heat as disordered motion, and work as ordered motion, it would appear
that the second law also has something to say about a fundamental disymmetry between order
and disorder.
In order to formulate the second law mathematically, we will take another look at the bathtub
problem, invoking a cyclic engine to transfer the heat between the room air and the water in the
tub, as in the following diagram.
We will focus our attention this time on the thermodynamics of the cyclic engine. Let us
suppose that TwTa . Let qa be the heat added to the engine by the room air, and let qw
be the heat added to the engine by the bath water. Our second law experience says that the
former will be positive and the latter negative (that is the air, being warmer, will give up heat
to the engine and the bathtub water will gain heat from the engine). Moreover, the first law
requires that the magnitudes of these heat transfers must be the same. Hence we may write
qa =q= -qw where we insist that the quantityq must be positive.
q
Let us now consider  T . Why we do this is not immediately obvious, although you should
recall from last lecture that q/T is a function of state. In this case:


Tw  Ta 


 
 q
 0
T
Tw
Ta
Tw Ta
T
T
 w a 
q
qw
qa
q q
The inequality in Eq. (5.4) holds in general for real (irreversible) processes. The equality
in Eq. (5.4) will only hold as the two temperatures become equal. Heat transfer under such
near-equilibrium conditions will be very slow and essentially reversible. If we define
entropy, s, as follows:
qrev
 ds
(5.5)
T
then, generally (dropping the line integral):

q
T
 ds
(5.6)
where the inequality holds for irreversible processes and the equality for reversible
processes. Clearly for adiabatic irreversible processes, the entropy must always increase,
leading to the previouslymentioned statement by Clausius. [NOTE: See also Stephen
Hawking’s “A Brief History of Time” for a discussion of entropy and the arrow of time
(Ch. 9).] For adiabatic reversible processes, entropy remains constant.
Formally, we can think of entropy as a function of state whose increase gives a measure of
the energy of a system which has ceased to be available for work during a given process.
Note also from last lecture that ds=cpdln for reversible processes. Therefore, processes in
which potential temperature is conserved are also isentropic processes.
Although it is important to be aware of the second law of thermodynamics, the field of
atmospheric thermodynamics makes little use of it. We tend to assume that atmospheric
processes occur reversibly.
However, since our atmosphere is in actuality an enormous heat engine that transports heat from
the tropics to the poles, the concepts and ideas that arise from the second law are of some use.
SUMMARY OF REVERSIBLE AND IRREVERSIBLE PROCESSES:
From Zemansky and Dittman, “Macroscopic Physics”: “A reversible process is one that is
performed in such a way that, at the conclusion of the process, both the system and the local
surroundings may be restored to their initial states, without producing any changes in the
rest of the universe.” For example, imagine a weight on a pulley. Suppose that the weight is
lowered by a small amount so that work is done and a transfer of heat takes place from the
system to its surroundings. If the system can be restored to its original state, I.e., the weight is
lifted back up and the surroundings forced to part with the heat that they had gained during
the lowering, then the original process is reversible.
A little thought should convince you that no real process is reversible. The abstraction of
reversible processes, however, provides a clean theoretical foundation for the description of
real world, irreversible processes.
What real world processes makes things irreversible?
Dissipative effects, such as viscosity, friction, inelasticity, electric resistance, and magnetic
hysteresis, etc.
Processes for which the conditions for mechanical, thermal, or chemical equilibrium, i.e.,
thermodynamic equilibrium are not satisfied.
In atmospheric and oceanic sciences, it is common to assume that there are no dissipative
effects (particularly for flows at large spatial scales) and that motion of the atmosphere and
oceans is isentropic. Viscous and frictional dissipation do, of course, occur (particularly in
the atmospheric and oceanic boundary layers) but a good understanding of the dynamics can
be obtained from inviscid theory.
APPLICATIONS OF THE SECOND LAW
• Carnot cycle
• Applications of the second law
• Combined first and second law
• Helmholtz and Gibbs free energies
CARNOT CYCLE
The Carnot cycle is an ideal heat engine cycle that offers insights into other heat engines,
including the atmosphere. It is a reversible cycle, performed by an ideal gas, that consists of
two isothermal processes linked by two adiabatic processes, as illustrated below:
A
4
1
B
D
2
3
C
Nicolas Leonard Sadi Carnot (1796-1832) led a short but interesting life. He was able to
pursue his scientific research as a result of his appointment to the army general staff. He
died of scarlet fever followed by cholera, and all his papers were burned after his death.
Only three major scientific works survive. His work on heat engines led to the ideal heat
engine and cycle that now bear his name. Interestingly, he was able to make significant
scientific advances while still believing in the erroneous caloric theory of heat, in which
heat is taken to be a function of state.
For each of the four processes in the Carnot cycle, we will consider the change in internal
energy, the heat added to the working gas and the work done by the gas (i.e., the components
of the first law).
Isothermal expansion
at temperature T1
du1=0
Adiabatic expansion
between T1 and T3
du2=cv(T3-T1)
Isothermal
compression at T3
du3=0
Adiabatic
compression between
T3 and T1
du4=cv(T1-T3)
Entire Carnot cycle
du3=0
q1 
B
 pdv  RT ln v
vB
q2=0
A
w2=-u2
q1
T1
ds2=0

q3 
D

C

ds1 
1
A

w1=q1
vC w3=q3
pdv  RT3 ln
vD
q4=0
ds3 
w4=-u4
q3
T3
ds4=0

q=q1+q3
w= q1+ q3
ds=0
Table 1: Thermodynamics of the Carnot cycle
Notes: 1) q3 is negative because we consider q to be the heat added to the system. 2) The net work done in a Carnot cycle must equal the
difference between the heat input and the heat exhausted. This work is also equal to the area contained within the Carnot cycle on a p-v diagram.
3) In order to show that ds=0, we must show that vB/vA=vC/vD. This can be done by comparing the Poisson equations for the two adiabatic
processes.
The efficiency of the Carnot cycle is defined to be the ratio of the mechanical work done to the
heat absorbed from the hot reservoir. That is:
E
w T1  T3
T

1 3
q
T1
T1
(6.1)
It can be shown that this is the maximum efficiency of any cyclic engine working between the
same two temperatures. This is known as Carnot’s theorem. Note that the efficiency of the
Carnot cycle can only be unity (i.e., 100%) if the cold reservoir (T3) has a temperature of
order to increase the efficiency of an engine, one would like to minimize
absolute zero. In
T3/T1.
EQUIVALENCE OF HEAT AND WORK, EFFICIENCY, AND THE FIRST LAW
Recall the first law: du=q- w. If we consider an engine cycle in a p-v diagram then the path
taken (clockwise for a heat engine, counterclockwise for a refrigerator) may be expressed as
the line integral of the first law around the cycle:
 du   q   w
Furthermore, since du is an exact differential its line integral vanishes, leaving us with:

 q   w
(6.2)
i.e. there is an equivalence of net heat gained/lost and net work done by/on the system.
Efficiency, E, may also be considered to be the ratio of Work Out to Heat In.
E
Work
Heat
Out
In
Consider the following idealized engine cycle:

The net heat input is QH=Q12+Q23. The net heat output is QC=Q34+Q41. The net heat in is
therefore Q=QH-QC and we now know that this must be the net work done, i.e., W= QH-QC .
The efficiency may then be expressed in terms of the heat as:
E
QH  QC
Q
1 C
QH
QH
Important Note: QH and QC here are MAGNITUDES of heat, and are therefore
always positive.
Example: Suppose that the working fluid is an ideal gas.
From the first law we have that dU=Q- W. The incremental work done is related to pressure
and volume by W=pdV. Also, the change in internal energy dU along any branch of the cycle is
equal to dU=CvdT=(Cp-nR)dT. NOTE: we are not using molar values here, so Cp-Cv=nR.
Branch 12: Volume constant
2
2
 Q   dU  C (T
v
1
2
 T1 )  0
So heat is absorbed.
1
Branch 23: Pressure constant
Since we are dealing with an ideal gas for which pV  nRT we can express the change in
volume in terms
of the change in temperature as:
nR
dT
 p
dV 
So the first law along this branch gives us: q=du+pdV=(cp-nR)dT+nRdT=cpdT from which

we get that:
3
3
Q  C p dT  C p (T3  T2 )  0 So, again, heat is absorbed.


2
2
Branch 34: Volume constant
4
4
 Q   dU  C (T
v
3
4
 T3 )  0
So heat is rejected.
3
Branch 41: Pressure constant

1
1
 Q  C  dT  C
p
4
p
(T4  T1 )  0
So, again, heat is rejected.
4
The efficiency of this engine, E, is therefore:

Cv (T3  T4 )  C p (T4  T1 )
QC
E  1
 1
QH
C p (T3  T2 )  Cv (T2  T1 )
Note that since we want the MAGNITUDE of QC, the signs in the numerator are such that it
is a positive quantity.

What would an efficiency of 100% mean in this case? Well, we can substitute in Ti=piVi/nR
for i=14 and show that in order for E to equal 1 we must have that:
Cv (T3  T4 )  C p (T4  T1 )  Cv ( p3V3  p4V4 )  C p ( p4V4  p1V1 )  0
where we’ve cancelled out the nR’s. Now note that V3=V4, V1=V2, p1=p4, and p2=p3 to show

V3Cv ( p3  p4 )  C p p4 (V4  V1 )  0
which can only be true if p3p4 and V4V1. In other words, 100% efficiency is only possible
if the cycle collapses to a single point on the p-V diagram, and no work is done!

Whoever said that doing no work is inefficient???????
In class example: Efficiency of the gasoline engine.
The atmosphere as a heat engine:
Our atmosphere is heated in the tropics (say Ttropics=313K) and cools at the poles (say
Tpoles=233K). If we were to imagine our atmosphere as an enormous heat engine that converts
net incoming heat into work (kinetic energy) then the efficiency of this engine is only 26%.
In fact, the actual efficiency is much lower than this.
Sandström’s theorem (see Houghton, “The Physics of Atmospheres”) states that a steady
circulation in the atmosphere can be maintained only if the heat source is situated at a higher
pressure than the heat sink. Some reflection on the Carnot cycle will reveal that the theorem
must be valid. Because the atmosphere of the Earth is transparent to solar radiation, the
requirements of the theorem are satisfied for Earth. What about Venus?
APPLICATIONS OF THE SECOND LAW
• Impossible processes
• Kelvin formulation of the Second Law
• Combined First and Second Laws
• Helmholtz and Gibbs Free Energies
1.
Impossible Processes
Impossible processes are ones that violate the second law of thermodynamics, i.e., ones for
which:
q
ds 
T
(Recall that for adiabatic reversible processes ds=0. These processes are therefore called
isentropic and have many applications to the atmosphere and oceans. Adiabatic irreversible
processes have ds>0, since adiabatic processes have q=0.)

2. Kelvin Formulation of the Second Law
This may be postulated as follows: “It is impossible to construct a cyclic device that will produce
work and no other effect than the extraction of heat from a single heat source.” This result can be
readily deduced by considering an isothermal cyclic engine in contact with a single reservoir,
and examining the consequences of the second law (which will demonstrate that, in fact, heat
must be lost from the engine to the reservoir, and hence that work must be done on the engine).
If it were not for the second law, as expressed by Kelvin, our energy worries would vanish.
We could simply extract as much work as we needed from a very large heat source such as the
oceans or the interior of the Earth.
3. Combined First and Second Laws
If we combine the first law in the form:
du  q  pdv
with the second law in the form:
or dh  q  vdp
Tds  q
(6.2)
(6.3)

then the combined laws take the form:
du  Tds  pdv

or dh  Tds  vdp
(6.4)
4. Helmholtz and Gibbs Free Energies
The Helmholtz and Gibbs free energies are functions of state. i.e., they are perfect differentials.
 defined, respectively, by:
They are
f  u  Ts
g  h  Ts
(6.5)
Using them, the combined first and second laws (Eq. 6.4) may be written as:
df  sdT  pdv
dg  sdT  vdp
(6.6)
where, again, the equality hold for reversible processes and the inequality holds for
irreversible processes.

The Helmholtz free energy is particularly useful when considering chemical reactions that
occur isothermally. You’ll note that the change of the Helmholtz energy during a reversible,
isothermal process equals the work done on the system, and for a reversible, isothermal, and
isochoric process the Helmholtz energy is conserved.
The Gibbs free energy is particularly useful when considering phase changes, which are
isothermal and isobaric (and can be conceived of as occuring reversibly). Hence during such
phase changes the Gibbs free energy remains constant. The Gibbs energy can be used to
uniquely determine the triple point temperature and pressure of a substance, since g must be
the same for the solid, liquid, and vapour phases (2 equations, 2 unknowns (T,p)).
For irreversible processes, the Gibbs free energy decreases as the substance approaches its
new equilibrium. It is therefore clear that an equilibrium state is one at which the Gibbs
energy is at a minimum.
Hermann von Helmholtz (1821-1894) had become the patriarch of
German science by 1885. He began his career as an army doctor. As an
MD and professor of physiology, he published a massive volume of work
on physiological optics and acoustics. He also invented the
ophthalmoscope. At the same time he began to publish work unrelated
to physiology--a paper on the hydrodynamics of vortex motion in 1858,
and an analysis of the motion of violin strings in 1859, for example.
In 1871, Helmholtz turned his attention full-time to physics, and made
significant contributions in areas ranging from electrodynamics to
thermodynamics and meteorology.
Josiah Willard Gibbs (1839-1903) was a Yale graduate and professor,
who turned his attention to thermodynamics in the early 1870’s. He
contributed to the use of geometrical methods and thermodynamic
diagrams. Perhaps his most significant contribution was to the
understanding of thermodynamic equilibrium, which he viewed as a
natural generalization of mechanical equilibrium, both being
characterized by minimum energy. He also made a substantial
contribution to statistical mechanics.