Page:Scientific Papers of Josiah Willard Gibbs.djvu/306

The increase of the volume of liquid will be

${\frac {\eta _{S(1)}}{\gamma '(\eta _{M}''-\eta _{M}')}},$

(589)

and the diminution of the volume of vapor

${\frac {\eta _{S(1)}}{\gamma ''(\eta _{M}''-\eta _{M}')}}\cdot$

(590)

Hence, for the work done (per unit of surface formed) by the external bodies which maintain the pressure, we shall have

$W={\frac {p\eta _{S(1)}}{\eta _{M}''-\eta _{M}'}}\left({\frac {1}{\gamma ''}}-{\frac {1}{\gamma '}}\right),$

(591)

and, by (514) and (131),

$W=-p{\frac {d\sigma }{dt}}{\frac {dt}{dp}}=-p{\frac {d\sigma }{dp}}=-{\frac {d\sigma }{d\log p}}\cdot$

(592)

The work expended directly in extending the film will of course be equal to $\sigma$.

Let us now consider the case in which there are two component substances, neither of which is confined to the surface. Since we cannot make the superficial density of both these substances vanish by any dividing surface, it will be best to regard the surface of tension as the dividing surface. We may, however, simplify the formula by choosing such substances for components that each homogeneous mass shall consist of a single component. Quantities relating to these components will be distinguished as on page 266. If the surface is extended until its area is increased by unity, while heat is added at the surface so as to keep the temperature constant, and the pressure of the homogeneous masses is also kept constant, the phase of these masses will necessarily remain unchanged, but the quantity of one will be diminished by $\Gamma _{'}$, and that of the other by $\Gamma _{''}$. Their entropies will therefore be diminished by ${\frac {\Gamma _{'}}{\gamma '}}\eta _{V}'$ and ${\frac {\Gamma _{''}}{\gamma ''}}\eta _{V}''$, respectively. Hence, since the surface receives the increment of entropy $\eta _{S}$, the total quantity of entropy will be increased by

$\eta _{S}-{\frac {\Gamma _{'}}{\gamma '}}\eta _{V}'-{\frac {\Gamma _{''}}{\gamma ''}}\eta _{V}'',$

which by equation (580) is equal to

$-\left({\frac {d\sigma }{dt}}\right)_{p}\cdot$

Therefore, for the quantity of heat $Q$ imparted to the surface, we shall have

$Q=-t\left({\frac {d\sigma }{dt}}\right)_{p}=-\left({\frac {d\sigma }{d\log t}}\right)_{p}\cdot$

(593)