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4 Molar, specific and “total” quantities

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



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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



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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