Appendix II. Conversion Factors and Defi nitions
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554
Quantity
3
cm
CO2 concentration
coulomb (C)
coulomb V
curie (Ci)
dalton (Da)
day
degree (angle) ( )
dyn
dyn cm
dyn cmÀ1
dyn cmÀ2
einstein
erg
erg cmÀ2 sÀ1
eV
eV moleculeÀ1
farad (F)
Fahrenheit ( F)
foot (ft)
foot2
foot3
footcandle (fc)
foot-pound
foot sÀ1
g
g cmÀ2
Appendix II
Equals
À6
Quantity
3
10 m
1 ml
see Table 8.2
1 J VÀ1
1 ampere s
1J
3.7 Â 1010 Bq
3.7 Â 1010
disintegrations sÀ1
1.661 Â 10À24 g
(1/12 mass of 12C)
86,400 s
1440 minutes
0.01745 radian
1 erg cmÀ1
1 g cm sÀ2
10À5 N
1 erg
10À7 J
10À3 N mÀ1
0.1 N mÀ2
10À6 bar
1 mol (6.022 Â 1023)
photons
1 dyn cm
1 g cm2 sÀ2
10À7 J
6.242 Â 1011 eV
2.390 Â 10À8 cal
2.390 Â 10À11 kcal
10À3 J mÀ2 sÀ1
10À3 W mÀ2
10À7 W cmÀ2
1.433 Â 10À6 cal cmÀ2
minuteÀ1
1.602 Â 10À19 J
1.602 Â 10À12 erg
96.49 kJ molÀ1
23.06 kcal molÀ1
1 coulomb VÀ1
( F À 32)(5/9) C
30.48 cm
0.3048 m
9.290 Â 10À2 m2
2.832 Â 10À2 m3
28.32 liter
7.480 gallons
(U.S.)
1 lumen ftÀ2
10.76 lux
1.356 J
1.097 km hourÀ1
0.6818 mile hourÀ1
1 dyn s2 cmÀ1
10À3 N s2 mÀ1
10 kg mÀ2
À3
g cm
g dmÀ2 hourÀ1
gallon (British)
gallon (U.S.)
gallon (U.S.) acreÀ1
grain (based on wheat)
grain footÀ3
hectare (ha)
hertz (H)
horsepower (hp)
hour (h)
inch (in.)
inch2
inch3
joule (J)
J kgÀ1
J molÀ1
J sÀ1
kcal
kcal hourÀ1
kcal molÀ1
kDa
kelvin (K)
kg
kg hectareÀ1
kg mÀ2
kg mÀ3
kg H2O
kJ molÀ1
km
km2
km hourÀ1
Equals
1000 kg mÀ3
27.78 mg mÀ2 sÀ1
4.546 liter
4.546 Â 10À3 m3
3.785 liter
3.785 Â 10À3 m3
0.1337 ft3
9.354 liter hectareÀ1
0.06480 g
2.288 g mÀ3
104 m2
2.471 acres
1 cycle sÀ1
745.7 W
3600 s
25.40 mm
6.452 Â 10À4 m2
1.639 Â 10À5 m3
0.01639 liter
1Nm
1Ws
1 m3 Pa
1 kg m2 sÀ2
1 coulomb V
107 erg
0.2388 cal
10 cm3 bar
2.388 Â 10À4 cal gÀ1
0.2388 cal molÀ1
W
4.187 kJ
4.187 Â 1010 erg
1.163 W
4.187 kJ molÀ1
0.04339 eV moleculeÀ1
1.661 Â 10À21 g (1000 Â
1/12 mass of 12C)
(K À 273.15) C
2.2046 pounds (lb)a
0.001 tonne (metric ton)
0.893 pound (lb) acreÀ1
0.1 g cmÀ2
1.422 Â 10À3 pound (lb) in.À2 (psi)
9.807 Pa (9.807 N mÀ2) (at sea level,
45 latitude: see g, Appendix I)
10À3 g cmÀ3
1 liter H2O at 4 C
0.01036 eV moleculeÀ1
1000 m
0.6214 mile
100 hectares
247.1 acres
0.3861 mile2
0.2778 m sÀ1
27.78 cm sÀ1
0.6214 mile hourÀ1
(continued)
555
Conversion Factors and Definitions
Quantity
knot
kW
kW hour
lambert
langley
langley dayÀ1
langley minuteÀ1
lb
liter
liter atm
liter bar
liter hectareÀ1
liter H2O
ln
log
lumen (L) ftÀ2
lumen mÀ2
lux (lx)
m
m2
m3
M
mbar
mho
mg cmÀ3
mg dmÀ2 hourÀ1
Mg hectareÀ1
Equals
Quantity
À1
1.852 km hour
1.151 mile hourÀ1
1000 J sÀ1
3.600 Â 106 J hourÀ1
859.8 kcal hourÀ1
3.600 Â 106 J
8.598 Â 105 cal
0.3183 candela cmÀ2
3183 candela mÀ2
1 cal cmÀ2
4.187 Â 104 J mÀ2
0.4846 W mÀ2
697.8 W mÀ2
pound
0.001 m3
0.03531 ft3
0.2200 gallon (British)
0.2642 gallon (U.S.)
1.057 quarts
24.20 cal
101.3 J
100 J
23.88 cal
0.1069 gallon (U.S.) acreÀ1
1 kg H2O at 4 C
2.303 log
0.4343 ln
1 footcandle
10.76 lux
1 lux
0.09290 footcandle
1 lumen mÀ2
0.09290 footcandle
100 cm
3.280 ft
39.37 inch
10.76 ft2
10À4 hectare
35.32 ft3
1.308 yard3
220.0 gallons (British)
264.2 gallons (U.S.)
mol literÀ1
1000 mol mÀ3
0.1 kPa
100 Pa
103 dyn cmÀ2
0.9869 Â 10À3 atm
1 ohmÀ1
1 siemens
1 ampere VÀ1
1 kg mÀ3
2.778 Â 10À2 mg mÀ2 sÀ1
0.4461 ton (U.S.) hectareÀ1
À1
ms
micron
mile
mile2
mile hourÀ1
MJ
ml
mm Hg
mM
month (mean)
MPa
mol mÀ3
nautical mile
ng cmÀ2 sÀ1
newton (N)
Nm
N mÀ1
N mÀ2
ounce (avoirdupois)
ounce (troy)
pascal (Pa)
photosynthesis
poise (P)
pound (lb)a
pound acreÀ1
Equals
100 cm sÀ1
3.600 km hourÀ1
2.237 mile hourÀ1
1 mm
1609 m
1.609 km
5280 ft
640 acres
2.590 Â 106 m2
2.590 km2
0.4470 m sÀ1
44.70 cm sÀ1
1.609 km hourÀ1
0.2778 kW hour
1 cm3
10À6 m3
133.3 Pa
1 mol mÀ3
2.630 Â 106 s
106 N mÀ2
106 J mÀ3
10 bar
1.020 Â 105 kg mÀ2
(at sea level, 45 latitude; see g,
Appendix I)
9.872 atm
1 mM
1 mmol cmÀ3
1.151 mile
10 mg mÀ2 sÀ1
1 kg m sÀ2
105 dyn
1J
103 dyn cmÀ1
1 Pa
10À6 MPa
10 dyn cmÀ2
10À2 mbar
28.35 g
31.10 g
1 N mÀ2
1 J mÀ3
1 kg mÀ1 sÀ2
10À5 bar
9.869 Â 10À6 atm
see Table 8.2
1 dyn s cmÀ2
0.1 N s mÀ2
0.1 Pa s
0.4536 kg
1.121 kg hectareÀ1
703.1 kg mÀ2
0.07031 kg cmÀ2
(continued)
556
Appendix II
Quantity
À2
pound inch
(psi)
radian (rad)
revolution
minuteÀ1 (rpm)
s cmÀ1
s mÀ1
siemens (S)
therm
ton (U.S.)
ton (U.S.) acreÀ1
tonne (metric ton)
tonne hectareÀ1
tonne mÀ3
torr
Equals
Quantity
Equals
6.895 kPa
0.06895 bar
0.06805 atm
57.30
6 sÀ1
0.1047 radian sÀ1
100 s mÀ1
10À2 s cmÀ1
1 mho (ohmÀ1)
1 ampere VÀ1
1 Â 105 Btu
1.055 Â 108 J
2000 pounds (1b)
907.2 kg
1 ton (short)
0.8929 ton (long)
2.242 tonne hectareÀ1
2.242 Mg hectareÀ1
1000 kg
1 Mg
1.102 ton (U.S.)
0.4461 ton (U.S.) acreÀ1
1 g cmÀ3
1 mm Hg
133.3 Pa (at sea level, 45
latitude; see g,
Appendix I)
transpiration
V
see Table 8.2
1 ampere ohm
1 J coulombÀ1
1 J sÀ1
1 kg m2 sÀ3
107 erg sÀ1
14.33 cal minuteÀ1
104 W mÀ2
1 J mÀ2 sÀ1
103 erg cmÀ2 sÀ1
1.433 Â 10À3 cal cmÀ2 minuteÀ1
1.433 Â 10À3 langley minuteÀ1
0.2388 cal mÀ2 sÀ1
2.388 Â 10À5 cal cmÀ2 sÀ1
2.388 Â 10À3 cal cmÀ1 CÀ1 sÀ1
0.1433 cal cmÀ1 CÀ1 minuteÀ1
1J
6.048 Â 105 s
0.9144 m
0.8361 m2
0.7646 m3
3.154 Â 107 s
5.256 Â 105 minutes
8760 hours
365 days
1.333 mbar (at sea level,
45 latitude; see g,
Appendix I)
1333 dyn cmÀ2 (at sea level,
45 latitude; see g,
Appendix I)
watt (W)
W cmÀ2
W mÀ2
W mÀ1 CÀ1
Ws
week
yard
yard2
yard3
year (normal
calendar)
year (sidereal)
year (solar)
mg cmÀ2 sÀ1
365.256 days
3.156 Â 107 s
365.242 days
3.156 Â 107 s
10 mg mÀ2 sÀ1
a
Sometimes it proves convenient to express force in units of mass. To see why this is possible, consider the force F exerted by
gravity on a body of mass m. This force is equal to mg, g being the gravitational acceleration (see Appendix I). Thus F/g can be
used to represent a force but the units are those of mass. One atmosphere is quite often defined as 760 mm Hg (or
1.033 kg cmÀ2), but the elevational and latitudinal effects on g should also be considered (see Appendix I).
Appendix III
Mathematical Relations
III.A. Prefixes (for units of measure)
a
f
p
n
m
m
ca
da
a
10À18
10À15
10À12
10À9
10À6
10À3
10À2
10À1
atto
femto
pico
nano
micro
milli
centi
deci
daa
ha
k
M
G
T
P
E
deka
hecto
kilo
mega
giga
tera
peta
exa
10
102
103
106
109
1012
1015
1018
Not recommended by SI (Systeme International, the internationally accepted
system for units).
III.B. Areas and Volumes
The following relations pertain to a cube of length s on a side, a cylinder of
radius r and length along the axis l, and a sphere of radius r.
Shape
Cube
Cylinder
Sphere
Area
2
6s
2prl + 2pr2
4pr2
Volume
3
s
pr2l
4
pr 3
3
V/A
s/6
rl/(2l + 2r)
r/3
III.C. Logarithms
The following relations are presented to facilitate the use of natural and
common logarithms, their antilogarithms, and exponential functions. For
those readers who are completely unfamiliar with such quantities, a textbook or handbook should be consulted.
ln (xy) = ln x + ln y
ln (x/y) = ln x À ln y
ln (1/x) = Àln x
ln x = 2.303 log x
ln 1 = 0
ln e = 1
ln 10 = 2.303
ln ex = x
eln y = y
ln xa = a ln x
ln (1/xa) = a ln x
ln (x/y) = Àln (y/x)
log x = (1/2.303) ln x = 0.434 ln x
log 1 = 0
log e = 1/(2.303) = 0.434
log 10 = 1
log 10x = x
10log x = x
557
558
Appendix III
x2 x3 x4
ỵ ỵ
2
3
4
2
3
x
x
ex ẳ 1 ỵ x ỵ ỵ ỵ
2! 3!
1 < x
ln 1 ỵ xị ẳ x
1
III.D. Quadratic Equation
The quadratic equation has the following form:
ax2 ỵ bx ỵ c ẳ 0
a 0ị
Its two solutions (two roots) are
xẳ
b ặ
p
b2 4ac
2a
III.E. Trignometric Functions
Consider a right triangle of hypotenuse r :
r
α in radians or º
(2π radians = 360º)
y
α
x
y
r
x
cos a ẳ
r
y
tan a ẳ
x
sin a ẳ
1
sin a
1
1
cos a
1
Ơ
tan a
Ơ
a3 a 5 a7
ỵ ỵ
3! 5! 7!
a2 a4 a6
cos a ẳ 1 ỵ ỵ
2! 4! 6!
sin a ¼ a À
a in radians
a in radians
III.F. Differential Equations
As a final topic in this appendix, we will consider the application of the
integral calculus to the solution of differential equations. A differential
equation expresses the relationship between derivatives (first order as well
as higher order) and various variables or functions. The procedure in solving
559
Mathematical Relations
differential equations is first to put the relation into a form that can be
integrated and then to carry out suitable integrations so that the derivatives
are eliminated. To complete the solution of a differential equation, we must
incorporate the known values of the functions at particular values of the
variables, the so-called boundary conditions. We will illustrate the handling
of differential equations by a simple but extremely useful example.
One of the most important differential equations in biology has the
following general form:
dy
¼ Àky
dt
ðIII:1Þ
where k is a positive constant and t represents time. We encountered this
equation in Chapter 4 (Eq. 4.10) in discussing the various competing pathways for the deexcitation of an excited singlet state. Equation III.1 also
describes the process of radioactive decay, where y is the amount of radioisotope present at any time t. The equation indicates that the rate of change
in time of the amount of radioactive substance (dy/dt) is linearly proportional to the amount present at that time (y). Because the radioisotope
decays in time, dy/dt is negative, and hence there is a minus sign in Equation
III.1. Any process that can be described by Equation III.1, such as a chemical
reaction, is called a first-order rate process, and k is known as the first-order
rate constant.
To put Equation III.1 into a form suitable for integration, we must
separate the variables (y and t) so that each one appears on only one side
of the equation:
dy
¼ Àk dt
y
ðIII:2Þ
which follows from Equation III.2 upon multiplying each side by dt/y. (Note
that the same initial process of separation of variables applies to the integration of a more complicated example in Chapter 3, namely, Eq. 3.12.)
When the variables are separated so that a possible integrand, e.g., Àkdt,
is expressed in terms of only one variable, we can integrate that integrand.
On the other hand, the integrand
Àkydt cannot be integrated as it stands —
R
i.e., we cannot perform Àkydt — because we do not know how y depends
on t. In fact, the very purpose of solving Equation III.1 is to determine the
functional relationship between y and t.
Next, we will insert integral signs into Equation III.2, replace y with y(t)
to emphasize that y depends on the independent variable t, and perform the
integration from t = 0 to t = Ơ:
?
Z t
Z ytị
?ytị
dy
?
III:3ị
ẳ ln y?
ẳ ln ytị ln y0ị ẳ k dt ẳ kt
?y0ị
y0ị y
0
When we take exponentials of quantities in Equation III.3, we obtain the
following solution to the differential equation represented by Equation
III.1:
yðtÞ ¼ yð0Þ eÀkt
ðIII:4Þ
560
Appendix III
Because of the factor eÀkt, Equation III.4 indicates that y decays exponentially with time for a first-order rate process (e.g., Fig. 4-11). Moreover, y(t)
decreases to 1/e of its initial value [y(0)] when t satisfies the following
relation:
1
yðt Þ ¼ yð0Þ ¼ yð0Þ eÀkt
e
ðIII:5Þ
where the value of time, t, that satisfies Equation III.5 is known as the
lifetime of the process whose decay or disappearance is being considered.
Equation III.5 indicates that eÀ1 equals eÀkt, so the first-order rate constant
k is equal to the reciprocal of the lifetime t (see Eq. 4.14). Therefore, the
solution (Eq. III.4) of the partial differential equation (Eq. III.1) describing
a first-order rate process becomes
ytị ẳ y0ị et=t
III:6ị
Appendix IV
Gibbs Free Energy and
Chemical Potential
The concept of chemical potential is introduced in Chapter 2 (Section 2.2)
and used throughout the rest of the book. In order not to overburden the text
with mathematical details, certain points are stated without proof. Here we
will derive an expression for the chemical potential, justify the form of the
pressure term in the chemical potential, and also provide insight into how
the expression for the Gibbs free energy arises.
IV.A. Entropy and Equilibrium
A suitable point of departure is to reconsider the condition for equilibrium.
The most general statement we can make concerning the attainment of
equilibrium by a system is that it occurs when the entropy of the system plus
its surroundings is at a maximum. Unfortunately, entropy has proved to be
an elusive concept to master and a difficult quantity to measure. Moreover,
reference to the surroundings ---- the “rest of the universe” in the somewhat
grandiloquent language of physics ---- is a nuisance. Consequently, thermodynamicists sought a function that would help describe equilibrium but
would depend only on readily measurable parameters of the system under
consideration. As we will see, the Gibbs free energy is such a function for
most applications in biology.
The concept of entropy (S) is really part of our day-to-day observations.
We know that an isolated system will spontaneously change in certain
ways----a system proceeds toward a state that is more random, or less ordered,
than the initial one. For instance, neutral solutes will diffuse toward regions
where they are less concentrated (Fig. 1-5). In so doing, the system lowers its
capacity for further spontaneous change. For all such processes DS is positive, whereas DS becomes zero and S achieves a maximum at equilibrium.
Equilibrium means that no more spontaneous changes will take place; entropy is therefore an index for the capacity for spontaneous change. It would
be more convenient in some ways if entropy had been originally defined with
the opposite sign. In fact, some authors introduce the quantity negentropy,
which equals ÀS and reaches a minimum at equilibrium. In any case, we
must ultimately use a precise mathematical definition for entropy, such as
dS = dQ/T, where dQ refers to the heat gain or loss in some reversible
reaction taking place at temperature T.
561
562
Appendix IV
We can represent the total entropy of the universe, Su, as the entropy of
the system under consideration, Ss, plus the entropy of the rest of the
universe, Sr. We can express this in symbols as follows:
S u ẳ Ss ỵ Sr
or
dSu ẳ dSs ỵ dSr
IV:1ị
An increase in Su accompanies all real processes----this is the most succinct
way of stating the second law of thermodynamics. Su is maximum at equilibrium.
The heat absorbed by a system during some process is equal to the heat
given up by the rest of the universe. Let us represent the infinitesimal heat
exchange of the system by dQs. For an isothermal reaction or change, dQs is
simply ÀdQr because the heat must come from the rest of the universe. From
the definition of entropy,1 dS = dQ/T, we can obtain the following relationship:
dSr ẳ
dQr
dQ
dU s ỵ PdV s
ẳ sẳ
T
T
T
IV:2ị
The last step in Equation IV.2 derives from the principle of the conservation
of energy for the case when the only form of work involved is mechanical----a
common assumption in stating the first law of thermodynamics. It is thus
possible to express dQs as the sum of the change in internal energy (dUs) plus
a work term (PdVs). The internal energy (Us) is a function of the state of a
system, i.e., its magnitude depends on the characteristics of the system but is
independent of how the system got to that state. PVs is also a well-defined
variable. However, heat (Qs) is not a function of the state of a system.
As we indicated previously, equilibrium occurs when the entropy of the
universe is maximum. This means that dSu then equals zero. By substituting
Equation IV.2 into the differential form of Equation IV.1, we can express this
equilibrium condition solely in terms of system parameters:
dU s ỵ PdV s
0 ẳ dSs ỵ
T
IV:3ị
or
TdSS ỵ dU s ỵ PdV s ẳ 0
Equation IV.3 suggests that there is some function of the system that has an
extremum at equilibrium. In other words, we might be able to find some
expression determined by the parameters describing the system whose derivative is zero at equilibrium. If so, the abstract statement that the entropy
of the universe is a maximum at equilibrium could then be replaced by a
statement referring only to measurable attributes of the system----easily
measurable ones, we hope.
1. This definition really applies only to reversible reactions, which we can in principle use to
approximate a given change; otherwise, dQ is not uniquely related to dS.