Page:Scientific Papers of Josiah Willard Gibbs.djvu/58

But we may also estimate the work and heat of any path by means of an integration extending over the elements of an area, viz: by the formulæ of page 7,

$W^{C}=H^{C}=$ ∑C ${\frac {1}{\gamma }}\delta A,$

(8)

$W^{S}=$ ∑[S, v'', p0, v'] ${\frac {1}{\gamma }}\delta A,$

(9)

$H^{S}=$ ∑[S, η'', t0, η'] ${\frac {1}{\gamma }}\delta A.$

(10)

In regard to the limits of integation in these formulæ, we see that for the work of any path which is not a circuit, the bounding line is composed of the path, the line of no pressure and two vertical lines, and for the heat of the path, the bounding line is composed of the path, the line of absolute cold and two horizontal lines.

As the sign of $\gamma$, as well as that of $\delta A$, will be indeterminate until we decide in which direction an area must be circumscribed in order to be considered positive, we will call an area positive which is circumscribed in the direction in which the hands of a watch move. This choice, with the positions of the axes of volume and entropy which we have supposed, will make the value of $\gamma$ in most cases positive, as we shall see hereafter.

The value of $\gamma$, in a diagram drawn according to this method, will depend upon the properties of the body for which the diagram is drawn. In this respect, this method differs from all the others which have been discussed in detail in this article. It is easy to find an expression for $\gamma$ depending simply upon the variations of the energy, by comparing the area and the work or heat of an infinitely small circuit in the form of a rectangle having its sides parallel to the two axes.

Let $N_{1}N_{2}N_{3}N_{4}$ (fig. 8) be such a circuit, and let it be described in the order of the numerals, so that the area is positive. Also let $\epsilon _{1},\,\epsilon _{2},\,\epsilon _{3},\,\epsilon _{4}$ represent the energy at the four corners. The work done in the four sides in order commencing at $N_{1}$, will be $\epsilon _{1}-\epsilon _{2},\,0,\,\epsilon _{e}-\epsilon _{4},\,0.$ The total work, therefore, for the rectangular circuit is

$\epsilon _{1}-\epsilon _{2}+\epsilon _{3}-\epsilon _{4}.$

Now as the rectangle is infinitely small, if we call its sides $dv$ and $d\eta$, the above expression will be equivalent to

$-{\frac {d^{2}\epsilon }{dvd\eta }}dvd\eta .$