Page:Scientific Papers of Josiah Willard Gibbs.djvu/53

the distances of the isometrics being proportional to the differences of the logarithms of the volumes, the distances of the isopiestics being proportional to the differences of the logarithms of the pressures, and so with the isothermals and the isodynamics,—the distances of the isentropics, however, being proportional to the differences of entropy simply.

The scale of work and heat in such a diagram will vary inversely as the temperature. For if we imagine systems of isentropics and isothermals drawn throughout the diagram for equal small differences of entropy and temperature, the isentropics will be equidistant, but the distances of the isothermals will vary inversely as the temperature, and the small quadrilaterals into which the diagram is divided will vary in the same ratio: $\therefore \lambda \sim 1\div t.$ (See p. 5.)

So far, however, the form of the diagram has not been completely defined. This may be done in various ways: e.g., if $x$ and $y$ be the rectangular co-ordinates, we may make

${\begin{cases}x=\log v,\\y=\log p;\end{cases}}\,$ or $\,{\begin{cases}x=\eta ,\\y=\log t;\end{cases}}$ or ${\begin{cases}x=\log v,\\y=\eta ;\end{cases}}\,$ etc.

Or we may set the condition that the logarithms of volume, pressure and of temperature, shall be represented in the diagram on the same scale. (The logarithms of energy are necessarily represented on the same scale as those of temperature.) This will require that the isometrics, isopiestics and isothermals cut one another at angles of 60°.

The general character of all these diagrams, which may be derived from one another by projection by parallel lines, may be illustrated by the case in which $x=\log v,$ and $y=\log p.$

Through any point $A$ (fig. 6) of such a diagram let there be drawn the isometric $vv'$, the isopiestic $pp'$, the isothermal $tt'$ and the isentropic $\eta \eta ^{\prime }$. The lines $pp'$ and $vv'$ are of course parallel to the axes. Also by equation (h)

$\tan tAp=\left({\frac {dx}{dy}}\right)_{t}=\left({\frac {d\log p}{d\log v}}\right)_{t}=-1$

and by (g) $\tan \eta Ap=\left({\frac {dy}{dx}}\right)_{\eta }=\left({\frac {d\log p}{d\log v}}\right)_{\eta }=-{\frac {c+a}{c}}.$

Therefore, if we draw another isometric, cutting $\eta \eta ^{\prime }$, $tt'$, and $pp'$ in $B$, $C$ and $D$,

${\frac {BD}{CD}}={\frac {c+a}{c}},{\frac {BC}{CD}}={\frac {a}{c}},{\frac {CD}{BC}}={\frac {c}{a}}\cdot$