1.5. Other key ideas from physics and chemistry[[Student version, December 8, 2002]] 23
liberate a lot more energy than others (burning a match)? No, the dynamite just liberates its energy
muchfaster;the energy liberatedperchemical bondis roughly comparable to any other reaction.
Example One important chemical reaction is the one happening inside the batteries in your
channel-changer. Estimate the chemical energy released in this reaction.
Solution: Printed on the battery we find that its terminals differ in potential by
∆V=1. 5 volt.This means that the battery imparts an energy of roughlye∆V =
1. 6 · 10 −^19 coul× 1. 5 volt=2. 4 · 10 −^19 Jto each electron passing through it. (The
value of the fundamental chargeeused above is listed in Appendix B.) If we suppose
that each electron passing across the battery enables the chemical reaction inside
to take one step, then the energy just calculated is the change in chemical bond
energies (minus any thermal energy given off).In contrast to chemical reactions, the radioactive decay of plutonium liberates about a million
times more energy per atom than a typical chemical reaction. Historically this was the first solid
clue that something very different from chemistry was going on in radioactive decay.
1.5.4 Low-density gases obey a universal law
The founders of chemistry arrived at the idea that atoms combine in definite proportions by noticing
that gases combine in simple, fixed ratios of volume. Eventually it became clear that this obser-
vation reflects the fact that thenumberof gas molecules in a box at atmospheric pressure is just
proportional to its volume. More precisely, one finds experimentally that the pressurep,volume
V,number of moleculesN,and temperatureTof any gas (at low enough density) are related in a
simple way called theideal gas law:
pV=NkBT. (1.11)
Here the temperatureTis understood to be measured starting from a special point calledabsolute
zero;other equations in this book, such as Equation 1.4, also useT measured from this point.
In contrast, the Celsius scale assigns zero to the freezing point of water, which turns out to be
273 ◦Cabove absolute zero. Thus room temperatureTrcorresponds to about 295 degrees above
absolute zero (we will define temperature more carefully in Chapter 3). The quantitykBappearing
in Equation 1.11 is called theBoltzmann constant;itturns out to be about 1. 38 · 10 −^23 joules per
degree Celsius. Thus the numerical value ofkBTat room temperature iskBTr=4. 1 · 10 −^21 J.A
less cumbersome way of quoting this value, and an easier way to memorize it, is to express it in
units relevant to cellular physics (piconewtons and nanometers):
kBTr≈ 4. 1 pN·nm. most important formula in this book (1.12)Takeaminute to think about the reasonableness of Equation 1.11: If we pump in more gas (N
increases), the pressure goes up. Similarly if we squeeze the box (V decreases), or heat it up (T
increases),pagain increases. It’s all quite reasonable, but the formula makes it very precise.
The form of Equation 1.11 may look unfamiliar. Chemistry texts generally write it aspV=nRT,
wherenis the “amount of substance” (number of moles) andRTis about 2500 joules per mole at
room temperature. Dividing 2500JbyNmoleindeed gives the quantitykBTrin Equation 1.12.