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Thermodynamics
Heat Capacity
Calorimetry
Enthalpy
Thermodynamic cycles
Adiabatic processes
NC State University
Motivation
The enthalpy change DH is the change in energy at constant
pressure. When a change takes place in a system that is
open to the atmosphere, the volume of the system changes,
but the pressure remains constant. In any chemical reactions
that involve the creation or consumption of molecules in the
vapor or gas phase there is a work term associated with the
creation or consumption of the gas.
Molar Enthalpy
Enthalpy can be expressed as a molar quantity:
Hm = H
n
We can also express the relationship between enthalpy and
internal energy in terms of molar quantities:
Hm = Um + PVm
For an ideal or perfect gas this becomes:
Hm = Um + RT
Usually when we write DH for a chemical or physical change
we refer to a molar quantity for which the units are kJ/mol.
Enthalpy for reactions
involving gases
If equivalents of gas are produced or consumed in a
chemical reaction, the result is a change in pressure-volume
work. This is reflected in the enthalpy as follows.
DH = DU + PDV
at const. T and P
which can be rewritten for an ideal gas:
DH = DU + DnRT
at const. T and P
The number of moles n is the number of moles created or
absorbed during the chemical reaction. For example,
CH2=CH2(g) + H2(g)
CH3CH3(g) Dn = -1
We arrive at this value from the formula
Dn = nproducts - nreactants = 1 - 2 = -1
The measurement of heat
We must carefully distinguish between heat and temperature.
When we add heat to the system its temperature increases.
We can use measurement of the temperature to determine
how much heat has been added. However, we need to know
the heat capacity of the system in order to do this.
Heat supplied
Heat capacity =
Temperature rise
The heat capacity is called C. If we perform a heat exchange
at constant volume then we designate the heat capacity as CV.
If the process occurs at constant pressure we call the heat
capacity CP.
qV,P
CV,P =
DT
Molar and Specific Heat
Capacities
We use molar heat capacities for pure substances. As the
name implies the units are J/mol-K for the molar heat
capacity. We write the molar heat capacity at constant
volume as CV,m.
For mixtures we cannot use a molar heat capacity and so
we use the specific heat capacity, which is the heat capacity
per gram of material with units of J/g-K.
Heat Capacity for
a Diatomic Molecule
For a diatomic molecule there is contribution from rotations
as well as translations. This means that as heat is added to
the system the rotational levels can be populated in addition
to an increase in molecular speed. The kinetic theory of
gases considers only the speed. An approximate rule is that
we obtain a contribution to the heat capacity, CV of 1/2nR for
each degree of freedom. We saw that for a monatomic gas
the heat capacity was CV = 3/2nR. A diatomic gas has two
rotational degrees of freedom and so the heat capacity is
approximately CV = 5/2nR. What does this say about CP?
Well, the relationship between CP and CV holds for all gases
so CP = 7/2nR for a diatomic “ideal” gas.
The temperature dependence
of the enthalpy change
Based on the discussion the heat capacity from the last
lecture we can write the temperature dependence of the
enthalpy change as:
DH = CPDT
Note that we can use tabulated values of enthalpy at 298 K
and calculate the value of the enthalpy at any temperature
of interest. We will see how to use this when we consider
the enthalpy change of chemical reactions (the standard
enthalpy change). The basic physics of all temperature
dependence is contained in the above equation or more
frequently in the equation below as molar quantity:
DH m = CP,mDT
Another view of
the heat capacity
At this point it is worth noting that the expressions for the
heat capacity at constant volume and constant pressure
can be related to the temperature dependence of U and H,
respectively.
DH = CPDT
CP = DH = H
DT
T
DU = CVDT
P
CV = DU = U
DT
T
V
The heat capacity is the rate of change of the energy with
temperature. The partial derivative is formal way of saying
this.
Variation in Reaction Enthalpy
with Temperature
Since standard enthalpies are tabulated at 298 K we need
to determine the value of the entropy at the temperature of
the reaction using heat capacity data. Although we have
seen this procedure in the general case the calculation for
chemical reactions is easier if you start by calculating the
heat capacity difference between reactants and products:
DrCP =  CP(products) –  CP(reactants)
and then substitute this into the expression:


D r H (T2) = D r H (T1) +
T2
D r CPdT
T1
If the heat capacities are all constant of the temperature
range then:
D r H (T2) = D r H (T1) + D r CPDT
Calorimetry
The science of heat measurement is called calorimetry.
A calorimeter consists of a container in a heat bath.
A physical or chemical process occurs in the container and
heat is added or removed from the heat bath. The
temperature increases or decreases as result. By knowing
the heat capacity of the bath we can measure the amount
of heat that has been added or removed from the system.
Energy in the
form of heat
flows into the bath.
Calorimetry
In the studies of biological systems there are two important
types of calorimetry.
1. Differential scanning calorimetry (DSC)
2. Isothermal titration calorimetry (ITC)
In DSC the temperature is increased at a constant heating
rate and the heat capacity is measured. DSC is used for
determining the parameters associated with phase transitions
e.g. protein unfolding, denaturation, DNA hybridization etc.
In ITC the temperature is held constant while one component
is added to another. The heat of interaction (e.g. binding) is
measured using this method. ITC is widely used to determine
the enthalpy of binding, e.g. for protein-protein and
protein-drug interactions among other types of biological
applications.
A note on using current flow in
calorimetry measurements
Energy is measured in Joules. Electrical energy is used
to delivery heat in calorimetry applications. In electrostatics
the units of energy are:
Joules = Coulombs * Volts
J = CV
Coulomb is a unit of charge and volt is a unit of potential.
A charge moving through a potential is a little like a waterfall.
The problem here is that we need a dynamic description
since “moving charge” is not “static”.
C
V
O
A note on using current flow in
calorimetry measurements
As charge moves through the potential energy (heat) is
Released.
Power = Amperes * Volts
W = IV
Power has units of energy per unit time. So as the current
flows through a wire at a certain rate, heat energy is added
to the system at that rate.
C
V
O
A note on using current flow in
calorimetry measurements
As charge moves through the potential energy (heat) is
Released.
Power = Amperes * Volts
W = IV
Power has units of energy per unit time. So as the current
flows through a wire at a certain rate, heat energy is added
to the system at that rate.
C
V
O
A note on using current flow in
calorimetry measurements
As charge moves through the potential energy (heat) is
Released.
Power = Amperes * Volts
W = IV
Power has units of energy per unit time. So as the current
flows through a wire at a certain rate, heat energy is added
to the system at that rate.
C
V
O
A note on using current flow in
calorimetry measurements
As charge moves through the potential energy (heat) is
Released.
Power = Amperes * Volts
W = IV
Power has units of energy per unit time. So as the current
flows through a wire at a certain rate, heat energy is added
to the system at that rate.
C
V
O
A note on using current flow in
calorimetry measurements
As charge moves through the potential energy (heat) is
Released.
Power = Amperes * Volts
W = IV
Power has units of energy per unit time. So as the current
flows through a wire at a certain rate, heat energy is added
to the system at that rate.
C
V
O
A note on using current flow in
calorimetry measurements
Over a period of time, t, the total energy added to the system
is:
Energy = Amperes * Volts * time
E = IVt
This equation is used in the text to describe a number of
calorimetry applications.
C
V
O
The standard state
The standard state of a substance is its pure form a 1 bar of
pressure or 1 molar concentration. The standard state can
have any temperature, but the temperature should be specified.
For example, the standard enthalpy of vaporization of water
Is the enthalpy of vaporization at 373 K (the boiling point) and
1 bar of pressure.
If we lower the temperature below 373 K (for example to 372 K)
then water no longer boils. We can say that the reaction is
defined at the standard state of 1 bar (or 1 atm in practice) of
pressure. A reaction can take place at concentrations other
than the standard state, but there will be a dependence of
the enthalpy on pressure. Vaporization is a special case since
the atmosphere is always at 1 atm at sea level.
Enthalpy of physical change
A physical change is when one state of matter changes into
another state of matter of the same substance. The difference
between physical and chemical changes is not always clear,
however, phase transitions are obviously physical changes.
Vaporization
DH fus
DH vap
Solid  Liquid  Gas
DH freeze
DH cond Condensation
Freezing
Fusion
DH sub
Solid  Gas
Vapor Deposition
DH vap dep
Sublimation
Properties of Enthalpy as a
State Function
The fact that enthalpy is a state function is useful for the
additivity of enthalpies. Clearly the enthalpy of forward
and reverse processes must be related by:
D forw ardH = – D reverseH
so that the phase changes are related by:
D fusH = – D freezeH
D vapH = – D condH
D subH = – D vap depH
Moreover, it should not matter how the system is transformed
from the solid phase to the gas phase. The two processes of
fusion (melting) and vaporization have the same net enthalpy
as sublimation.
Addivity of Enthalpies
Because the enthalpy is a state function the same magnitude
must be obtained for direct conversion from solid to gas as
for the indirect conversion solid to liquid and then liquid to gas.
DsubH = DfusH + DvapH
Of course, these enthalpies must be measured at the same
temperature. Otherwise an appropriate correction would need
to be applied as described in the section on the temperature
dependence of the enthalpy.
Chemical Change
In a chemical change the identity of substances is altered
during the course of a reaction. One example is the
hydrogenation of ethene:
CH2=CH2(g) + H2(g)
CH3CH3(g) DH = -137 kJ/mol
The negative value of DH signifies that the enthalpy of the
system decreases by 137 kJ/mol and, if the reaction takes
place at constant pressure, 137 kJ/mol of heat is released
into the surroundings, when 1 mol of CH2=CH2 combines
with 1 mol of H2 at 25 oC.
Standard Enthalpies of Formation
The standard enthalpy of formation DfHo is the enthalpy for
formation of a substance from its elements in their standard
states. The reference state of an element is its most
stable form at the temperature of interest. The enthalpy of
formation of the elements is zero.
For example, let’s examine the formation of water.
H2(g) + 1/2 O2(g)
H2O(l)
DHo = -286 kJ
Therefore, we say that DfHo (H2O, l) = -286 kJ/mol.
Although DfHo for elements in their reference states is zero,
DfHo is not zero for formation of an element in a different
phase:
C(s, graphite)
C(s, diamond)
DfHo = + 1.895 kJ/mol
Standard Enthalpy Changes
The reaction enthalpy depends on conditions (e.g. T and P).
It is convenient to report and tabulate information under a
standard set of conditions.
Corrections can be made using heat capacity for variations
in the temperature. Corrections can also be made for
variations in the pressure.
When we write DH in a thermochemical equation, we always
mean the change in enthalpy that occurs when the reactants
change into the products in their respective standard states.
Standard Reaction Enthalpy
The standard reaction enthalpy, DrHo, is the difference
between the standard molar enthalpies of the reactants
and products, with each term weighted by the
stoichiometric coefficient, ,
The standard state is for reactants and products at 1 bar
of pressure or 1 molar concentration. The unit of energy
used is kJ/mol.
The temperature is not part of the standard state and it is
possible to speak of the standard state of oxygen gas at
100 K, 200 K etc. It is conventional to report values at
298 K and unless otherwise specified all data will be
reported at that temperature.
Hess’s Law
We often need a value of DH that is not in the thermochemical
tables. We can use the fact that DH is a state function to
advantage by using sums and differences of known quantities
to obtain the unknown. We have already seen a simple
example of this using the sum of DH of fusion and DH of
vaporization to obtain DH of sublimation.
Hess’s law is a formal statement of this property.
The standard enthalpy of a reaction is the sum of the standard
enthalpies of the reactions into which the overall reaction
may be divided.
Application of Hess’s Law
We can use the property known as Hess’s law to obtain
a standard enthalpy of combustion for propene from the
two reactions:
C3H6(g) + H2(g)
C3H8(g)
DH = -124 kJ
C3H8(g) + 5O2(g)
3CO2(g) + 4H2O(l) DH = -2220 kJ
If we add these two reactions we get:
C3H6(g) + H2(g) + 5O2(g) 3CO2(g) + 4H2O(l) DH = -2344 kJ
and now we can subtract:
H2(g) + 1/2O2(g)
H2O(l)
DH = -286 kJ
to obtain:
C3H6(g) + 9/2O2(g)
3CO2(g) + 3H2O(l) DH = -2058 kJ
Understanding the
additivity of enthalpies
Calculate DrxnHo for the rxn :
2 SO3(g)  2 SO2(g) + O2(g)
given the following :
a) S(s) + O2(g)  SO2(g)
DfHo = -297 kJ
b) S(s) + 3/2 O2(g)  SO3(g) DfHo = -396 kJ
Enthalpy of solution
Note the trends in the heats of solution for these
common compounds.
SOLUTE
hydrochloric acid
ammonium nitrate
ammonia
potassium hydroxide
caesium hydroxide
sodium chloride
potassium chlorate
acetic acid
sodium hydroxide
ΔHo in kJ/mol
-74.84
+25.69
-30.50
-57.61
-71.55
+3.88
+41.38
-1.51
-44.51
Tabulated bond enthalpies
Bond
H—H
H—C
C—C
H—N
N—N
H—O
O—O
H—F
F—F
H—Cl
Cl-Cl
H—Br
Br-Br
H—I
I—I
pm
74
109
154
101
145
96
148
92
142
127
199
141
228
161
267
kJ/mol
436
413
348
391
170
366
145
568
158
432
243
366
193
298
151
Bond
C—C
C=C
C—N
CºC
C—O
C—S
O—O
C—F
O=O
C—Cl
C—Br
N—N
C—I
NºN
pm
154
134
147
120
143
182
148
135
121
177
194
145
214
110
kJ/mol
348
614
308
839
360
272
145
488
498
330
288
170
216
945
Thermodynamic cycles
The energies and enthalpies of ionic solids are dominated
by Coulombic interactions. The lattice enthalpy can be
calculated from Coulombic interactions. Using the
calculated lattice enthalpy and experimental data one can
obtain the enthalpy of formation of an ionic solid by means
of the Born-Haber cycle. This is illustrated on the next slide
for a generic salt of a monovalent ion M+X-.
The reactions needed to obtain M+X- in the solid phase
to M+ + X- in the vapor phase are given below. The overall
process of separating the ions in an ionic solid into the
constituent ions is written as
M+X- (s)  M+ (g) + X- (g)
Lattice enthalpy
Enthalpies of Ionization
The molar enthalpy of ionization is the enthalpy that
accompanies the removal of an electron from a gas phase
atom or ion:
H(g)
H+(g) + e-(g)
DHo = +1312 kJ/mol
For ions that are in higher charge states we must consider
successive ionizations to reach that charge state. For
example, for Mg we have:
Mg(g)
Mg+(g)
Mg+(g) + e-(g)
Mg2+(g) + e-(g)
DHo = +738 kJ/mol
DHo = +1451 kJ/mol
We shall show that these are additive so that the overall
enthlalpy change is 2189 kJ/mol for the reaction:
Mg(g)
Mg2+(g) + 2e-(g)
DHo = +2189 kJ/mol
Electron Gain Enthalpy
The reverse of ionization is electron gain. The corresponding
enthalpy is called the electron gain enthalpy. For example:
Cl(g) + e-(g)
Cl-(g)
DH = -349 kJ/mol
The sign can vary for electron gain. Sometimes, electron
gain is endothermic.
The combination of ionization and electron gain enthalpy
can be used to determine the enthalpy of formation of salts.
Other types of processes that are related include molecular
dissociation reactions.
The Born-Haber cycle
The thermodynamic cycle is illustrated. The scheme shows
that if one knows all of the energies except one (difficult to
measure) quantity, one can calculate it using Hess’ law.
Lattice Enthalpy
M+ + X-
M+XEnthalpy
of Formation
Ionization
Potential
Electron
Affinity
Enthalpy
of Vaporization
1/2 X2 (g) + M (s)
Bond Dissociation
M (s)
X (g)
Using enthalpies of formation
Calculate the enthalpy change for the following rxn:
__ NH3(g) + __ O2(g)  __ NO(g) + __ H2O(g)
Using enthalpies of formation
Calculate the enthalpy change for the following rxn:
4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(g)
Solution: Look up the enthalpies of formation of each
molecule in the thermodynamic tables. Remember that
the enthalpy of formation of O2 is zero.
Using enthalpies of formation
Calculate the enthalpy change for the following rxn:
4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(g)
Solution: Look up the enthalpies of formation of each
molecule in the thermodynamic tables. Remember that
the enthalpy of formation of O2 is zero.
Enthalpies of Combustion
Standard enthalpies of combustion refer to the complete
combination with oxygen to carbon dioxide and water.
For example, for methane we have:
CH4(g) + 2O2(g)
CO2(g) + 2H2O(l) DcombHo = -890 kJ/mol
Enthalpies of combustion are commonly measured in a
bomb calorimeter (a constant volume device). Thus,
DUm is measured. To convert from DUm to DHm we need
to use the relationship:
DHm= DUm + DgasRT
The quantity Dgas is the change in the stoichiometric
coefficients of the gas phase species. We see in the
above express that Dgas= -2. Note that H2O is a liquid.
Enthalpy of formation from combustion
The enthalpy of combustion DcombHo for propane,
C3H8, is -2220 kJ. What is DfHo of C3H8?
Enthalpy of formation from combustion
The enthalpy of combustion DcombHo for propane,
C3H8, is -2220 kJ. What is DfHo of C3H8?
Solution: Write down the balanced equation
Look up the enthalpies of formation of the CO2 and H2O.
Remember that the enthalpy of formation of O2 is zero.
The tabulated values of enthalpy are given at 298 K,
Keep in mind that H2O is in the liquid phase.
Enthalpy of formation from combustion
The enthalpy of combustion, DcombHo for propane,
C3H8, is -2220 kJ. What is DHof of C3H8?
Solution: Write down the balanced equation
Enthalpy of formation
What is the enthalpy of combustion for propane,
C3H8 assuming that water is in the vapor phase?
Solution: Write down the balanced equation
Look up the enthalpies of formation of the CO2 and H2O.
Remember that the enthalpy of formation of O2 is zero.
Expansion following combustion
What volume of gas is produced if 50 grams of
C3H8 is combusted at 373 K at 1 atm? You may
assume all molecules are in the vapor phase.
Expansion following combustion
What volume of gas is produced if 50 grams of
C3H8 is combusted? You may assume all molecules
are in the vapor phase (T = 373 K).
Solution: Write down the balanced equation
Calculate the change in number of moles of gas, Dngas.
Then calculate DV.
Expansion following combustion
What volume of gas is produced if 50 grams of
C3H8 is combusted? You may assume all molecules
are in the vapor phase.
Solution: Write down the balanced equation
Calculate the change in number of moles of gas, Dngas.
Then calculate DV.
Dngas = S nproducts – S nreactants
Dngas = 3 + 4 – 1 – 5 = 1
Work done following combustion
How much work can be extracted for an expansion
against a constant pressure of 10 atm if 50 grams of
C3H8 is combusted?
You may assume all molecules are in the vapor
phase at 373 K.A
Work done following combustion
How much work can be extracted for an expansion
against a constant pressure of 10 atm if 50 grams of
C3H8 is combusted?
You may assume all molecules are in the vapor
phase.
Comparing fuels on a mass basis
We can consider the heat released upon combustion
as the energy available to do work in an engine. If
we consider hydrocarbon fuels (methane, ethane,
propane, butane, and octane) as well as ethanol, we
can compare them both on a per mole and per mass
basis.
The mole basis is obtained using the enthalpy of
combustion.
Comparing fuels on a mass basis
Fuel
DcombHo
CH4
C2H6
C3H8
C4H10
C8H18
C2H6O
-882
-1561
-2219
-2878
-5430
-1370
Comparing fuels on a mass basis
Fuel
DcombHo Mm
CH4
C2H6
C3H8
C4H10
C8H18
C2H6O
-882
-1561
-2219
-2878
-5430
-1370
16
30
44
58
114
46
Comparing fuels on a mass basis
Fuel
DcombHo Mm q (kJ/gram)
CH4
C2H6
C3H8
C4H10
C8H18
C2H6O
-882
-1561
-2219
-2878
-5430
-1370
16
30
44
58
114
46
55.1
52.0
50.4
49.6
47.6
29.8
Methane vs. Methanol as a fuel
On a per mole basis methane and methanol have
similar enthalpies of combustion. On a mass basis
methanol provides 22 kJ/g
while methane produces 55 kJ/g.
Methanol is convenient because it is a liquid.
Atmospheric methane: a greenhouse gas
Methane Capture:
An important ecological problem
On a per mole basis methane is converted to methanol with a
heat of reaction of -164.5 kJ/ mol.
The direct conversion shown above does not work
well because it is slow unless the temperature is so high that
one risks complete oxidation to CO2.
Methane Conversion via Syngas
The conversion of CH4 to syngas is the most common
industrial method for producing CH3OH from CH4.
The first step or production of syngas is called cracking. The
second step requires a Cu or Pd catalyst. The overall process
is only about 10% efficient.
Photochemical Methane Conversion
An alternative approach is to use light to activate water in
order to make a hydroxyl radical, which then reacts with
methane.
This process requires sufficient light energy to cleave the
HO-H bond in H2O.
Methane Monooxygenase
Heating of a fuel in an engine
The combustion of a fuel leads to heating of the gas
produced in the reaction. The heating can be calculated
using the heat capacity (or specific heat).
For octane, cp = 255.7 J/mol/K.
What is the s (the specific heat)?
Heating of a fuel in an engine
Consider the fact that after combustion the octane fuel has
been converted into CO2 and H2O in the vapor phase. For
H2O vapor, cp = 33 J/mol/K. What is the final temperature if
12 microliters of octane are combusted?
Focus on energy
The work done in the internal combustion
engine is called pressure volume work.
For a simple irreversible
stroke the work is:
work = P DV
In a 3.0 L 6 cylinder
engine DV = 0.5 L.
Assuming the initial
volume is 80 microliters, what is P?
We can estimate the
pressure from the
amount of octane is injected and combusted
From combustion to useful work
A typical fuel injector will inject 12 microliters of
octane fuel per stroke. What is the pressure of
the gas created by combustion in a volume of 80
microliters?
In our first examination of this problem we will
ignore the heating and simply calculate the
conversion from the volume of liquid fuel to the
volume of H2O and CO2 according to the reaction
stoichiometry.
Calculating pressure
First convert from volume to moles:
The change in the number of moles is:
Therefore the pressure is:
The pressure is ~100 atm.
Solar Heating
Equilibration
In an equilibration two objects that are initially at different
temperatures come to equilibrium at a final temperature.
For example, in a solar water heater we can imagine that a
slab of black stone acts as the absorber. If the stone has a
mass of 20 kg and a temperature of 80 oC.
What is the final temperature of 1 L of water that is initially at
10 oC? (for H2O s1 = 4.18 J/g-oC and stone s2 = 0.5 J/g-oC )
Equilibration
In an equilibration two objects that are initially at different
temperatures come to equilibrium at a final temperature.
For example, in a solar water heater we can imagine that a
slab of black stone acts as the absorber. If the stone has a
mass of 20 kg and a temperature of 80 oC.
What is the final temperature of 1 L of water that is initially at
10 oC? (for H2O s1 = 4.18 J/g-oC and stone s2 = 0.5 J/g-oC )
Biological implications
Obviously, the food we eat release heat in our bodies. This
heat is used both to maintain body temperature and for
processes that build up our bodies (anabolic processes).
We often talk about calories in the food we eat. The calorie
on a cereal box is equal to 1000 calories in the chemical
nomenclature. 1 calorie = 4.184 Joules. Therefore, the
Intake required for an average man of 12 MJ/day is about
3000 calories per day (in the sense of diet).
Differential relationships
for enthalpy
We have defined a relationship between the enthalpy and
Internal energy
H = U + PV
The infinitesimal change in the state function H results in
H + dH = U + dU + (P + dP)(V + dV)
Therefore
dH = dU + PdV + VdP
Now we substitute dU = dq + dw into this expression
dH = dq + dw + PdV + VdP
Since dw = - PdV
dH = dq + VdP
At constant pressure, dP = 0, and we have
dH = dqP
Path Functions
We have seen that work and heat are path functions. The
magnitude of the work and heat depends not just on the
final values of the T and P, but also on the path taken.
We can summarize the paths and their implications in the
table below.
Path
Isothermal
Constant V
Constant P
Adiabatic
Condition
DT = 0
DV = 0
DP = 0
q=0
Result
w = -q
w = 0, DU = CvDT
w = -PDV, qP = CpDT
DU = w
State Functions
At present we have introduced two state functions:
Internal Energy
DU
Enthalpy
DH
State functions do not depend on the path, only on the
value of the variables.
We can make the analogy with elevation. The potential
energy at an elevation h, which we call V(h) does not
depend on how we got to that elevation. If we compare
V(h1) in Raleigh to V(h2) on Mt. Mitchell the difference
V(h2) - V(h1) is the same regardless of whether we drive
to Mt. Mitchell through Statesville or Asheville. The work
we do to get (i.e. how much gas we use in a car!) is a path
function.