Concise Physical Chemistry

c03 JWBS043-Rogers September 13, 2010 11:24 Printer Name: Yet to Come

LINE INTEGRALS IN GENERAL 39

The work of the leftmost and rightmost steps cancel− 219 +219, but the topmost

horizontal produces−6126 J of work and the bottom horizontal takes up only 5552 J

of work to return the machine to its original state. The cycle produces− 6126 +

5552 =−574 J of work (negative because work goes from the machine into the

outside world) and has been returned to its original state, ready to produce work

−574 J,−574 J,−574 J,...over infinitely many work cycles, running forever.

Five or six generations of garage scientists have made the sad discovery that such

machinesnever work. However sad it may be for would-be millionaire inventors, the

impossibility of a perpetual motion machine leads the physical chemist to a treasure

trove of thermodynamic theory through thesecond law of thermodynamicsto the

inspired work of the American thermodynamicists J. W. Gibbs and G. N. Lewis.

3.3 LINE INTEGRALS IN GENERAL

If you have a straight brass rod—say, 70 cm in length and 1.0 cm^2 in cross section—it

is a simple matter to determine its weight even though the rod may be inaccessible

to you as part of a machine, so that you cannot simply weigh it. Multiply the density

of the brass used in its manufacture,ρin g/cm^3 ,by70×1.0=70 cm^3 = 70 ρand

you have the answer in grams. We have essentially set up anxaxis and performed

the integration

M=ρ

∫ 70

0

dx=ρ(70−0)

Most machinists (and lots of other people) would laugh at how we have made a simple

problem complicated.

But suppose the rod is bent (Fig. 3.2). Now setting up anxaxis for the bar is not

the best way of finding its mass. Onexincrement is not the same as another because

the rod is not collinear with the axis.

The corresponding integral is

M=ρ

∫b

a

ds=ρ(b−a)

y

x

FIGURE 3.2 Different segments of a curved rod.