North, Gerald R., Erukhimova Tatiana L. Atmospheric Thermodynamics: Elementary Physics and Chemistry
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Problems 67
3.4 The coefficient of expansion is defined by
β =
1
V
∂V
∂T
p
. (3.61)
Compute β for an ideal gas.
3.5 Show that for any gas
∂p
∂T
V
=
β
κ
T
. (3.62)
(Hint: dV =
∂V
∂p
T
dp +
∂V
∂T
p
dT , see the Calculus refresher in this chapter.)
3.6 Find the internal energy of 1 kg of dry air at STP.
3.7 Suppose the atmosphere has its pressure given by p(z) = p
0
e
−z/H
with p
0
= 1 atm
and T (z = 0) = 273 K. Now suppose a 1 kg parcel is lifted adiabatically one scale
height H . How much work does the parcel do on the environment in the process?
What is the change in its specific internal energy, u?
3.8
Isobaric process A 1 kg parcel of dry air has temperature 285 K and pressure
1000 hPa. It is heated by contact with the dry ground to a temperature of 295 K.
(a) What is Q? (b) What is the change of the parcel’s specific internal energy? (c)
What is the change in the parcel’s specific enthalpy?
3.9
Isothermal process 1 kg of dry air at 300 K and 1000 hPa is expanded isothermally
(pretty unusual in the atmosphere) from a volume of 2 m
3
to twice that value. (a) What
is the work done by the gas in this expansion? (b) What is the heat absorbed? (c) What
is the change in enthalpy?
3.10
Isochoric process 1m
3
of dry air at 1000 hPa and 290 K is enclosed in a rigid box.
What is its density (kg m
−3
)? 3 g of liquid water are evaporated into the box. What
is the increase of temperature in this box whose volume is held fixed? What are the
changes in internal energy and enthalpy? Sketch a diagram of the change in the V –p
plane.
3.11
Adiabatic process A parcel of mass 1 kg is lifted adiabatically from 800 hPa, where
its temperature is 270 K, to 600 hPa. What is the new temperature? What are the
changes in internal energy and enthalpy? Sketch a diagram of the change in the V –p
plane.
3.12
Isobaric process A parcel of essentially dry air is held at a fixed altitude where the
pressure is 950 hPa. It is heated by infrared radiation being absorbed by some water
vapor in the parcel. The heating rate is 20 J kg
−1
s
−1
. What is the rate of change of the
temperature, specific enthalpy and specific internal energy? How does the potential
temperature change per unit time? Sketch a diagram of the change in the V –p plane.
3.13 An air column is composed of dry air and the density of the air is given by ρ(z) =
ρ
0
e
−z/H
, where ρ
0
= 1.25 kg m
−3
and H = 10 km.
(a) What is the mass of air (kg) lying above 1 m
2
?
(b) How many idealized “air” molecules are above the 1 m
2
?
68 The First Law of Thermodynamics
(c) Then approximately how many “air molecules” are there in the entire Earth’s
atmosphere?
3.14 The speed of sound in air can be computed from the formula v
sound
=
√
1/ρκ
X
, where
κ
X
is the compressibility holding the parameter X constant. Compare the sound speeds
(taking ρ = p/RT ) when X = κ
T
and X = κ
θ
. The latter fits the data. Do you recall
from physics why the adiabatic compressibility gives the correct answer instead of
the isothermal compressibility? See Problem 3.3 above.
3.15 Suppose the atmosphere satisfies p(z) = p
0
e
−z/H
and that it is isothermal (T (z) = T
0
).
What is the potential temperature θ as a function of z? Sketch a graph.
4
The Second Law of Thermodynamics
The Second Law of Thermodynamics deals with changes in the conditions of
a system and its surroundings under transitions from one thermodynamic state
to another. We will introduce a new function of state by considering a simple
quasi-static series of changes for an ideal gas. First recall that some quantities are
functions of the state only, while others depend on the path taken between two
states. According to the First Law, the internal energy U is a function only of
state with changes in going from state A to state B, denoted U , depending only
on the initial state coordinates (e.g., p
A
, V
A
, T
A
) and the final state coordinates
(p
B
, V
B
, T
B
). Recall that the enthalpy H is also a function of state. Changes in other
quantities such as the amount of work done by the system on the surroundings in
the (quasi-static) transition depend upon the path taken between the two states. In
this case the work done by the system on the surroundings may be written,
W
A→B
=
B
A
p(V )dV (4.1)
and the function p(V ) is a specific curve in the V –p plane defining the path taken
in going from A to B. The path must be quasi-static and reversible, otherwise the
path p(V ) may not even be defined, nor could we use pdV in computing d
W. The
change in internal energy depends only on the difference
U = U
B
− U
A
(4.2)
no matter what the path (even if the process is irreversible). For example, in the case
of the ideal gas the internal energy (and enthalpy) depend only on the temperature,
independent of the history of the system. For more complex systems U and H
might depend on other state variables as well, but not on the history of the system.
Note that the heat (induced by a temperature difference between the system and its
environment) taken into the system during the transition must also depend on the
69
70 The Second Law of Thermodynamics
V
p
A
B
Figure 4.1 Illustration of two paths joining points A and B in a V –p state diagram.
path since
Q
A→B
= W
A→B
+ U
B
− U
A
(4.3)
and the work done by the system surely depends on path as we have seen earlier (see
Figure 4.1). Next we explore quasi-static transitions of an ideal gas to see whether
there might be another function of the state of the system.
Consider the case of an ideal gas where the mass
M is fixed. We can use the
definition of the enthalpy to obtain:
dH =
Mc
p
dT = dQ + V dp. (4.4)
If we multiply through by the integrating factor 1/T we obtain
+
Mc
p
dT
T
=
dQ
T
+ V
dp
T
. (4.5)
Using the Ideal Gas Law, V /T =
MR/p and solving for dQ/T:
dQ
T
=−
MR
dp
p
+
Mc
p
dT
T
=−
MR dlnp +Mc
p
dlnT
=
Mc
p
dln
T
p
R/c
p
= d
Mc
p
ln(Tp
−κ
)
(4.6)
where as before κ = R/c
p
. The last expression says that dQ/T is a perfect
differential. That is, its change in going from A to B does not depend on the path
chosen for the sequence of quasi-static infinitesimal steps.Anecessary and sufficient
condition for the differential to be perfect is that integrals around arbitrary closed
loops result in no change in the function.
The Second Law of Thermodynamics 71
Calculus refresher: perfect differential Suppose F is defined by:
F =
path
A(x, y) dx +
path
B(x, y) dy (4.7)
where A(x, y) and B(x, y) are well-behaved functions. It might happen that there exists
a function f (x, y) such that
∂f
∂x
= A(x, y) and
∂f
∂y
= B(x, y), (4.8)
if so
df =
∂f
∂x
dx +
∂f
∂y
dy,or
∂A
∂y
=
∂B
∂x
(4.9)
is the exact differential of f and
F =
path
df = f (upper) − f (lower) (4.10)
which means that F is independent of the path along which the integral is taken
joining the upper and lower limits. It might be easier to see this if we introduce a
parameter t which defines the curve
x = x(t), y = y(t) (4.11)
(for example, a parabola y = x
2
could be written x(t) = t, y(t) = t
2
) this defines a
curve in the x–y plane (the path). The integral can be written:
F =
t
2
t
1
A
dx
dt
+ B
dy
dt
dt
=
t
2
t
1
∂f
∂x
dx
dt
+
∂f
∂y
dy
dt
dt
=
t
2
t
1
df
dt
dt =
f (x(t
2
),y(t
2
))
f (x(t
1
),y(t
1
))
df
= f (x(t
2
), y(t
2
)) − f (x(t
1
), y(t
1
)). (4.12)
Example 4.1 Consider the function
f (x, y) = sin xy. (4.13)
72 The Second Law of Thermodynamics
Then f
x
(x, y) = y cos xy and f
y
(x, y) = x cos xy. Then
df = y cos xy dx + x cos xy dy = d(sin xy). (4.14)
Example 4.2: integrating factor Consider the first-order linear differential equation
for the function y(t):
dy
dt
+ by = g(t) (4.15)
where b is a constant and g(t) is a given function. Now multiply through by the
integrating factor e
bt
. We can show that
d
dt
e
bt
y
= e
bt
g(t) (4.16)
and the left-hand side is rendered a perfect differential, by virtue of our knowing the
proper integrating factor. Now we are in position to solve the differential equation
by integrating each side:
e
bt
y(t)
t
t
0
=
t
t
0
e
bt
g(t
) dt
. (4.17)
Then
y(t) = y(t
0
)e
−b(t−t
0
)
+ e
−bt
t
t
0
e
bt
g(t
) dt
. (4.18)
Example 4.3 Consider the differential expression
dZ = 2xy
3
dx + 3x
2
y
2
dy. (4.19)
We want to consider integrals of this differential from the point (0,0) to the point
(1,1) along some different paths. First consider the straight line path, y = x, which
connects the two points. Then,
dZ = (2x
4
+ 3x
4
) dx = 5x
4
dx, (4.20)
Z
straight line
=
1
0
5x
4
dx = 1. (4.21)
Next consider the parabolic path, y = x
2
,dy = 2x dx, which also passes through
the two points:
dZ = 2x
7
dx + 3y
3
dy, (4.22)
4.1 Entropy 73
parabolic
dZ =
1
0
2x
7
dx +
1
0
3y
3
dy =
2
8
+
3
4
= 1. (4.23)
We obtain the same answer for the two different paths. In fact, it is possible to write
dZ = d(x
2
y
3
). (4.24)
In other words, dZ(x, y) is a perfect differential. Integrating from one point in the
x–y plane to another always yields the same answer.
On the other hand, suppose we had the function
dP = 2xy dx + 3x
2
dy. (4.25)
This differential is not perfect, but with the aid of the integrating factor, y
2
,wecan
create the perfect differential, dZ.
Example 4.4 In classical mechanics we encounter the same concept with
conservative forces, which implies the existence of a potential energy function.
If a force can be described by a potential energy function V (x, y, z) and if the force,
F(x, y, z) can be written as the gradient of the force:
F(x, y, z) · dr =−∇V (x, y, z) ·dr
=−
∂V
∂x
dx −
∂V
∂y
dy −
∂V
∂z
dz
=−dV . (4.26)
Or we can write
F(r) =−∇V (r). (4.27)
Hence the work done in going from A to B:
W
A→B
=
B
A
path
F · dr
= V (A) − V (B)
independent of path
. (4.28)
4.1 Entropy
In the case of the ideal gas mentioned above, we can use Poisson’s equation,
T = θ
p
p
0
κ
(4.29)
74 The Second Law of Thermodynamics
or
Tp
−κ
= θ(p
0
)
−κ
(4.30)
and inserting into (4.6) we obtain:
dQ
T
=
Mc
p
dln
θ(1000 hPa)
−κ
=
Mc
p
dlnθ
≡ dS (4.31)
where S is called the entropy. In chemistry and physics the entropy is usually
denoted S and the entropy per unit mass s; in some older meteorology contexts it
is denoted φ. The important thing is that entropy is a function only of state and
that its change in going from one state to another can be calculated by choosing a
reversible path joining initial and final states.
For an infinitesimal displacement along a reversible path (for a general
thermodynamic system, not just an ideal gas),
dS =
dQ
rev
T
[Second Law of Thermodynamics]. (4.32)
The extra subscript rev is added to remind the reader that the calculation of dQ/T
must be along a reversible path. Recall that a reversible path is one in which the
infinitesimal steps along the path are quasi-static and such that each can be reversed
to restore the system to its previous state. In performing the calculation, we should
find such an imaginary but realizable path joining the two states for the purposes of
calculation. Of course, in nature the transition might be (and often is) an irreversible
or spontaneous one from state A to state B, but there always exists a reversible path
joining the two states so that the change in entropy can always be calculated. As
indicated in the last equation, the very existence of such a function of state as
defined above constitutes one statement of the Second Law of Thermodynamics.
In the case of dry parcels of ideal gas (nearly always satisfied in the dry
atmosphere) we can use the formula
dS = Mc
p
dlnθ (4.33)
or
S = Mc
p
ln θ (4.34)
and in terms of specific quantities (kg
−1
):
ds = c
p
dlnθ, s = c
p
ln θ, (4.35)
4.1 Entropy 75
where an arbitrary integration constant has been set to zero, since it is never actually
needed. In these last formulas s refers to the entropy per unit mass (units J K
−1
kg
−1
)
or the specific entropy. We will use
s (units J K
−1
mol
−1
) to indicate entropy per
mole. For a parcel of dry air, knowing the potential temperature is equivalent
to knowing its entropy. Hence, the change in entropy for a parcel undergoing a
reversible transition (through a series of quasi-static changes) can be computed
simply by computing the change in its potential temperature
S = S
B
− S
A
= Mc
p
ln
θ
B
θ
A
. (4.36)
It is obvious that for a reversible adiabatic process, the change in entropy vanishes,
since
¯
dQ = 0 along the reversible path. For so-called diabatic processes (ones
in which some heat is exchanged between the system and its surroundings), the
formula defining the change in entropy can be used (dS = dQ
rev
/T ).
Example 4.5 Show that dQ
rev
/T = Mc
p
dlnθ really works for an ideal gas. First
write
dQ = dH − V dp. (4.37)
Substituting for dH and V :
dQ =
Mc
p
dT − MRT
dp
p
. (4.38)
After dividing through by T we simply repeat the steps leading to (4.6).
Example 4.6 Compute the change in entropy for a parcel of mass M at pressure
level p
0
being heated diabatically (e.g., by radiation heating) from temperature
T
0
to T .
Answer: Since the heating is done at a fixed level, the pressure p remains constant
at a value p
0
. We can proceed to compute the change in entropy by using
dS =
dQ
T
(4.39)
S =
T
T
0
c
p
M
dT
T
= c
p
M ln
T
T
0
. (4.40)
Example 4.7 How much does the potential temperature of the parcel change in this
heating if it takes place at 500 hPa?
76 The Second Law of Thermodynamics
Answer: Since θ = T
(
p/p
0
)
−κ
, p = constant = 500 hPa, and p
0
= 1000 hPa, we
have θ = T
(
p/p
0
)
−κ
= (T − T
0
)
(
500/1000
)
−0.287
, for an isobaric heating
process.
Example 4.8 A parcel of ideal gas is taken from (V
0
, p
0
) to (2V
0
,
1
2
p
0
) by first (path
a) expanding isobarically to 2V
0
, then the pressure is reduced at constant volume
to
1
2
p
0
(path b). What is the change in entropy?
Answer: Along path a we compute S
a
=
dQ/T = Mc
p
2T
0
T
0
dT /T =
Mc
p
ln 2. Along path b, S
b
=
dQ/T =
dU /T = Mc
v
T
0
2T
0
dT /T =
−
Mc
v
ln 2. Now S
a
+ S
b
= M(c
p
− c
v
) ln 2 =
MR ln 2.
Example 4.9 Using the same initial and final states as in the previous example,
compute the change in entropy but along the isothermal path joining the two states
(path c).
Answer: Along c we have
dQ/T =
dW /T = (1/T
0
)
2V
0
V
0
p dV =
(
MRT
0
/T
0
)
2V
0
V
0
dV /V = MR ln 2, which is the same as in the alternative
reversible two-step path chosen in the previous example.
Example 4.10: change in phase Compute the change in entropy for 1 kg of water
being evaporated to gas at 373 K at constant pressure.
Answer: First note that 1 kg of water is 55.56 mol. The change in entropy is
Q/(373 K). Since the pressure is held constant, Q = ν
vap
H
◦
= 55.56 mol ×
40.66 kJ mol
−1
=2259 kJ. The change in entropy is then S = 2259 kJ/(373 K)=
6.06 kJ K
−1
.
4.2 The Second Law of Thermodynamics
There are several equivalent statements of the Second Law of Thermodynamics.
For our present purposes it suffices to state the law as follows.
There exists a function S of the extensive parameters of the system that is a
function of state and whose changes from one state to another can be calculated by
dS =
dQ
rev
T
(4.41)
and such that
dS ≥ 0 (4.42)
for an isolated system.
The subscript is again a reminder that the calculation must be conducted along
a reversible path.