4 Molar, specific and “total” quantities
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4.4 Molar, specific and “total” quantities
61
If we divide both sides by the molar mass of the gas, we get
pv = RT /M
Example 4.2 Converting between molar and specific quantities The molar heat capacity
(Cp,m ) of lead is 26.4 J K−1 mol−1 . To calculate the specific heat capacity, we use the molar
mass:
cp =
26.4 J K−1 mol−1
= 0.127 J K−1 g−1
207.2 g mol−1
Key ideas and equations
r Anything that goes into a system is counted as positive, anything coming out is counted
as negative.
Suggested reading
Thermometry is a surprisingly complicated business. For instance, you may never have
thought about the fact that a mercury-in-glass thermometer will give different readings
depending on the temperature of the stem, and not just the bulb where we think the
measurement occurs, because the glass also expands with temperature. The channel into
which the mercury expands therefore has a different diameter depending on the temperature
of the stem. Then there’s the whole issue of calibration. The best starting point for reading
about temperature and thermometers, if this topic catches your fancy, is the web site of the
International Temperature Scale of 1990: http://www.its-90.com.
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5
The First Law of Thermodynamics
The First Law is about the inter-relationships between a few key quantities: energy, work and
heat. These concepts are central to such disparate fields of study as chemistry, engineering
and physiology, to name just a few.
In a nutshell, the first law asserts that energy is conserved. It therefore allows us to do
some simple bookkeeping on energy transfers and thus to determine how much energy a
system has used or stored. Although this may not sound very exciting, it is remarkably
useful.
5.1 Differentials
Thermodynamics can’t be studied without learning some new mathematics. Central among
the mathematical tools of thermodynamics are differentials. Differentials are tiny (infinitesimal) increments. They are useful for describing how a change in one quantity is related to
a change in another.
There are two kinds of differentials. The first, which are somewhat more straightforward
to understand, are exact differentials. Exact differentials represent small changes in a
property of a system. In thermodynamics, a measurable property is called a state variable.
A function that only depends on state variables is called a state function. Since state
functions only depend on the state variables, they also are properties of a system and thus
have exact differentials. You have seen exact differentials before, in your calculus course.
The differential appearing in an ordinary integral such as
x2
f (x) dx
x1
is always an exact differential. In particular, if dx is an exact differential,
x2
dx = x2 − x1 =
x
x1
Only the endpoints matter, and not the way in which x was changed. It doesn’t matter
if x was changed slowly or quickly, or even if it increased and decreased again before
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5.1 Differentials
63
stopping at its final value, only where it started and where it ended up. We say that x is
path independent. The following three-way correspondence follows:
x is a state variable
or function
⇐⇒
dx is an exact
differential
⇐⇒
x is path
independent
Examples of state variables include pressure, volume and temperature, to name just a
few.
The other case, that of inexact differentials, is best understood by example. We will
often have occasion to talk about dw, the amount of work done when some state variable
(for instance, the volume) is changed by a small amount. However, dw is not like dV ; since
work is not a state variable (no system is characterized by the amount of work it “has”),
it doesn’t make sense to talk about the “change in work.” Rather, when we integrate dw,
we will be adding up the small amounts of work done along some path (P) to compute the
total work done w. Symbolically,
P
dw = w
In this case, it may matter how we do the work. Some forces, like gravitation, are conservative. Work done by or against a conservative force is path independent. However, many
forces are non-conservative. For non-conservative forces, it is easy to devise paths that
make the same changes in a system (i.e. have the same initial and final states) but for which
completely different amounts of work are done. We will see an example of this in the next
section. We say that w is a path function, and that dw is an inexact differential. For inexact
differentials, we therefore have
x describes
a process
⇐⇒
dx is an inexact
differential
⇐⇒
x is path dependent
Exercise group 5.1
(1) Suppose that F is a state function, and consider the following process:
F1 =−1.41 u
(p1 , V1 ) −−−−−−−→ (p1 , V2 )
⏐
⏐
⏐ F =2.25 u
F4 ⏐
2
F3 =−3.09 u
(p3 , V1 ) ←−−−−−−− (p3 , V2 )
What is F4 ? Note that “u” represents the units of F , whatever they are.
Hint: Imagine starting at (p1 , V1 ) and following the arrows around. How has the state
changed? What does this imply about the net change in F ?
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The First Law of Thermodynamics
Fext
xe
x
Figure 5.1 Geometry of a piston system used to derive the equation for pressure–volume work. If the
cylinder has cross-sectional area A, the volume is V = A(xe − x), where xe is the position of the end
of the cylinder. Accordingly, dV = −A dx (area multiplied by change in height) or dx = −dV /A.
The external force Fext results in an external pressure on the piston of pext = Fext /A or Fext = pext A.
Therefore, the differential of work dw = Fext dx = −pext dV .
5.2 Pressure–volume work
The essence of the First Law is the equivalence of heat and work. The type of work that will
most concern us in this book is pressure–volume work. This is just ordinary mechanical
work, but we will write the equations in a form particularly suited to thermodynamics.
Suppose that we have a gas in a piston either being compressed by or expanding against an
external force Fext directed perpendicular to the surface of the piston (Figure 5.1). Recall
that mechanical work is defined as a force times the distance over which that force is
applied, so the work done on or by the system (the gas) during a small displacement dx is
dw = Fext dx
We’ll worry about adding up the dws (integrating) later.
Now suppose that the force is distributed over an area A so that we actually measure
not force, but pressure p. Recall that F = pA. Furthermore, a small change dV in the
volume V is related to a small change dx in the position of the piston by dx = −dV /A
(see Figure 5.1). The negative sign accounts for the fact that if the force (which pushes
in) and displacement are in the same direction, the volume decreases. We arrive at the
fundamental relationship
dw = −pext dV
(5.1)
If we want the total work, we just integrate this equation:
w=
P
dw = −
P
pext dV
(5.2)
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5.2 Pressure–volume work
65
along whatever path P is followed during the process, i.e. how pext changes during the
process.
The negative sign in Equation (5.1) is important: since pressures are always positive,
the work will always have the opposite sign to the change in volume. Thus, during a compression, the volume decreases and the work is positive. According to our sign convention
and in accord with common sense, this means that during a compression work is done on
the system. Conversely, an expansion represents a volume increase and negative work, i.e.
work done by the system.
Example 5.1 Work at constant pressure Suppose that the external pressure is constant
during a process. Then
w = −pext
V2
dV = −pext (V2 − V1 ) = −pext
V
V1
One special case of this formula deserves mention. Suppose that a gas expands against a
vacuum. This means that the external pressure is zero. Then w = 0.
There are some special types of processes (i.e. special paths) that we use all the time in
thermodynamics, either because they are theoretically significant, or because they describe
(or approximate) some conditions we encounter in real processes. An isothermal process
is one in which the temperature is constant. A reversible process is one in which the
system stays in equilibrium with its surroundings at all times. During the expansion of
a gas, reversibility would imply that the external and internal pressures are equal during
the entire path. This is not something we can really do; if the pressure is the same on
both sides of the piston, there is no net force, and the piston doesn’t move. In a real
(irreversible) expansion, we would have to make the external pressure lower than the internal
pressure.
If we can’t really do a reversible expansion, why do we care about this kind of process?
Consider Figure 5.2, which shows three different paths for an isothermal expansion from
V1 to V2 . The solid curve is the pV isotherm, i.e. the line along which T is a constant
at equilibrium. For an ideal gas, this would be the curve pV = constant. The reversible
expansion corresponds to following this curve. According to Equation (5.2), the work done
during this expansion is just the area under this curve between V1 and V2 (because pext = p
in a reversible process), give or take the sign, which will be negative since the system is
doing useful work during an expansion. In a real process, we would have to make pext
smaller than p. We could just go directly to the final pressure (the long-dashed line in the
figure). The work done would then be (again, ignoring the sign) the area of the rectangle
under the long-dashed line. Of course, we could reduce the pressure in steps. The shortdashed line in the figure shows what would happen if we reduced the pressure in three equal
steps. The work done in this case would be the area under the “staircase”. You can see that
taking more steps gives us an amount of work between the reversible work and the result
for a single pressure reduction. You should also be able to see that, no matter how many
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October 29, 2011
The First Law of Thermodynamics
p
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V1
V2
V
Figure 5.2 Expansion of a gas. The solid curve is the pV isotherm.
pressure-reduction steps we use, we can’t produce more work than the reversible process
would, we can only approach this limit. This is why reversible processes are important:
A reversible process maximizes the work done by a system.
We can approach reversibility by taking lots of small steps, and we know that the reversible
work done by a system is a maximum (i.e. w is maximally negative along a reversible path),
so we now have an upper bound on what a machine can accomplish. Note that we haven’t
for the moment proven this to be a general principle, only observed that it is true for the
expansion of a gas. We can do better, and will return to this important principle later in the
course.
Example 5.2 Reversible, isothermal expansion of an ideal gas Let’s calculate the work
done during the isothermal, reversible expansion of an ideal gas. Because the process is
reversible, pext = p, the pressure of the gas in the piston. Therefore,
w=−
V2
p dV
V1
To evaluate the integral, we need to write p as a function of V . Because this is an ideal gas,
p = nRT /V . Furthermore, this is an isothermal expansion so T is constant:
w = −nRT
V2
V1
V2
V1
dV
= −nRT ln
= nRT ln
V
V1
V2
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5.2 Pressure–volume work
67
In the exercises that follow, you may need to do some slightly harder integrals. If you
need a little review, have a look at Appendix I.
Exercise group 5.2
(1) Calculate the work done when a gas expands from a volume of 1 L to 10 L against
a constant external pressure of 1 atm. (Hint and caution: the answer should be in SI
units.)
(2) Calculate the work done on or by the gas when 8 mol of nitrogen is compressed
isothermally and reversibly at 300 ◦ C from an initial volume of 1 L to a final volume of
30 mL. Assume that nitrogen behaves as an ideal gas. Explain the meaning of the sign
of your answer.
(3) 18 mol of nitrogen are held at a pressure of 1.08 atm and a temperature of 20 ◦ C. The
external pressure is changed very suddenly to 0.35 atm and is held constant during the
expansion. The gas is in good thermal contact with its surroundings, whose temperature
is a constant 20 ◦ C so that by the end of the expansion, the temperature of the gas has
returned to equilibrium with the surroundings.
(a) Treat nitrogen as an ideal gas and calculate the work done during the expansion.
Explain the meaning of the sign of the work.
(b) Suppose that the work done on or by the gas results from the raising or lowering
of a 2.3 kg mass. The work done when a mass is raised is w = mgh where h is the
height to which the mass is raised (negative if lowered). How far does the mass
have to move, and in what direction?
(4) 0.10 mol of an ideal gas is expanded isothermally at 50 ◦ C from an initial volume of
1.4 L to a final volume of 3.2 L. Calculate the work done along each of the following
paths: (a) reversible, (b) pressure dropped suddenly to the final pressure, (c) a two-step
process with two equal pressure drops.
(5) The ideal gas law is only valid for most gases at very low temperatures. Real gases are
more accurately described by the van der Waals equation:
p+
n2 a
V2
(V − nb) = nRT
where a and b are constants specific to a particular gas, R is the ideal gas constant, p
is the pressure, V the volume, and n the number of moles.
(a) Derive an equation for the work done during a reversible, isothermal expansion of
a van der Waals gas.
(b) For nitrogen, a = 0.141 Pa m6 mol−2 and b = 3.91 × 10−5 m3 mol−1 . Calculate
the work done when 1 kmol of nitrogen is expanded reversibly and isothermally at
a temperature of 300 K from 1 to 3 m3 .
(c) Compare the answer obtained in question 5b to what you would have calculated
from the ideal gas equation. Was the extra work worth the effort?
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The First Law of Thermodynamics
(6) Solids and liquids are often said to be incompressible, which means that changing the
pressure does not change their volumes. Because of this, we tend to ignore the effect of
pressure on solids and liquids when solving thermodynamic problems. However, solids
and liquids are not really incompressible. For copper, for instance, the relationship
between pressure and molar volume at constant temperature is
p=
Vm◦
1
ln
κT
Vm
+ p◦
where κT , Vm◦ and p◦ are constants with the values
κT = 7.25 × 10−12 Pa−1
Vm◦ = 7.093 × 10−6 m3 mol−1
p◦ = 101 325 Pa
Calculate the work done per mole when copper is compressed reversibly and isothermally from an initial pressure of 1 atm to a final pressure of 1000 atm. This is roughly
equivalent to the pressure change that a piece of copper would experience if it was
dropped from the ocean’s surface to the bottom of the Marianas trench, the deepest
ocean trench on Earth. Based on your calculation, do you think that ignoring pressure–
volume terms (such as the work) is justifiable for solids?
Hint: You may need an integral from the table in Appendix I.1.
5.3 The First Law of Thermodynamics
The internal energy U of a system is the total energy, excluding kinetic energy, associated
with the motion of the center of mass or with rotation around the center of mass, as well
as the potential energy due to the position of the body in space (e.g. gravitational potential
energy). The internal energy is therefore the energy stored in the atoms and molecules of a
piece of matter, including the energy associated with the intermolecular forces.
The first law of thermodynamics can be viewed as an extension of classical mechanical
results on the conservation of energy. It states that there are two ways to change the internal
energy of a system and that these two ways are equivalent: work (w) can be done on the
system or heat (q) can flow into it. Thus the change in internal energy can be written
U =q +w
(5.3)
A perpetual motion machine of the first kind is a machine that indefinitely sustains some
external energy-using process without external input of energy. The first law forbids such
machines; if energy goes out of the system without being replenished from an external
source, the internal energy decreases. This process cannot go on indefinitely.
We now come to an important fact:
The internal energy is a state function.
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5.4 Calculus of differentials
69
We know this because the internal energy has been defined to exclude external influences
so that its value can only depend on the state of the system. On the other hand, as we have
seen in Section 5.2, the work is a path function. Since U = q + w, the influence of the
path on w must somehow cancel out when we add it to q to give us the change in the state
function U . It follows that the heat must also be a path function.
Work and heat are path functions.
In fact, these two quantities are the only path functions we will routinely see in this course.
It is often useful to write the first law in differential form:
dU = dq + dw
(5.4)
The only hiccup is that we must be clear about what the differentials mean: dU , being the
differential of a state function, is an exact differential and we can integrate it to get the
change in internal energy:
final state
dU = Ufinal − Uinitial =
U
initial state
On the other hand, dq and dw, being differentials of path functions, are inexact differentials.
They can only be integrated for a given path to give the total heat and work:
q=
dq
and
P
w=
dw
P
Exercise group 5.3
(1) An induction coil is a device that, when placed in a changing magnetic field, generates
electricity. Reasonably large time-varying magnetic fields are produced by high-voltage
power lines. Some years ago, a farmer whose property was crossed by high-voltage
lines placed a large induction coil under the lines and supplied his farm with electricity
from it. He was found guilty of theft. What was he stealing and from whom? Explain
(briefly) the scientific basis of the case.
5.4 Calculus of differentials
Equations written in terms of differentials are just shorthand versions of differential equations (equations relating derivatives). For instance, dx = α dy is short for
dx
dy
=α
dz
dz
where z is some progress variable (time, amount of reactant used, etc.). We use the differential notation when the identity of z doesn’t matter. The differential relationship dx = α dy
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The First Law of Thermodynamics
can then be thought of as a statement about how a small change in x is related to a small
change in y.
It follows from the relationship between derivatives and differentials that the usual rules
of differentiation apply also to the latter. In particular, if x and y are variables and a is a
constant:
d(x + y) = dx + dy
d(ax) = a dx
d(xy) = y dx + x dy
(sum rule)
(product rule)
5.5 Heat and enthalpy
We focus our attention on a system that is only capable of doing pressure–volume work.
We do this because that type of system comes up most often in chemistry. Suppose that
we heat such a system at constant volume. Then dw = −pext dV = 0 because the volume
doesn’t change. Therefore, the first law gives us
dU = dqV
(5.5)
The subscript in the above equation indicates that the volume is held constant, i.e. it specifies
the path. Therefore,
U = qV
At constant volume, the change in internal energy equals the heat
transferred to the system.
It would be nice if there were a state function whose changes corresponded to heat at
constant pressure because a number of experiments are performed under those conditions.
In thermodynamics, whenever we start a sentence with “it would be nice if there were a state
function. . . ,” we just make one up that has the desired property. At constant pressure, pext =
p (otherwise the pressure wouldn’t stay constant) so, from the First Law and considering
only pressure–volume work,
dU = dqp − p dV .
Let’s rearrange this equation slightly:
dqp = dU + p dV
(5.6)
Note that the last term on the right-hand side is −dwp for a system that can only perform
pressure–volume work. Inspired by this equation, we define the enthalpy H by
H = U + pV
(5.7)
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5.6 Heat capacity
71
H is a state function because it is the sum of U , which is a state function, and of the product
of p with V , each of which is a state variable. Differentiating this expression, we get
dH = dU + p dV + V dp
At constant pressure, the last term is zero. Considering also Equation (5.6), we conclude
that
dH = dqp =⇒
H = qp
(5.8)
This is exactly what we were looking for:
At constant pressure, the enthalpy change equals the heat transferred to the system.
When a system is composed only of solids and liquids (called condensed phases),
an important simplification is possible: since solids and liquids are incompressible and
since they are relatively dense, (pV ) is always very small for processes involving only
condensed phases provided p isn’t huge. It then follows that H ≈ U .
5.6 Heat capacity
You might guess that the larger the temperature change, the larger the amount of heat
transferred to (or from) a body. That is essentially correct. Under most conditions, the heat
transferred is almost exactly proportional to the temperature change. In other words,
q∝
T
Heat is a path function so we need to specify the path carefully. If we do that, the amount
of heat required to cause a given temperature change is very reproducible. At constant
volume,
q V = CV T
and at constant pressure,
qp = Cp T
Again note that the notation qX indicates that X is held constant during the process. CV and
Cp are the heat capacities of the system, respectively at constant volume and at constant
pressure. In general, these two heat capacities are different.
The above equations are valid provided the relevant heat capacity is approximately
constant, which it is for most substances when the change in temperature is not too large.
When the heat capacity is not constant, we must rewrite these equations in differential form:
dqX = CX dT
(5.9)
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