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Appendix II. Conversion Factors and Defi nitions

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.



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