Page:Scientific Papers of Josiah Willard Gibbs.djvu/60

and in part vapor. The properties of such a mixture are very simple and clearly exhibited in the volume-entropy diagram.

Let the temperature and the pressure of the mixture, which are independent of the proportions of vapor, solid and liquid, be denoted by $t'$ and $p'$. Also let $V,L$ and $S$ (fig. 9) be points of the diagram which indicate the volume and entropy of the body in three perfectly defined states, viz: that of a vapor of temperature $t'$ and pressure $p'$, that of a liquid of the same temperature and pressure, and that of a solid of the same temperature and pressure. And let $v_{V},\eta _{V},v_{L},\eta _{L},v_{S},\eta _{S}$ denote the volume and entropy of these states. The position of the point which represents the body, when part is vapor, part liquid, and part solid, these parts being as $\mu ,\nu$, and $1-\mu -\nu$, is determined by the equations

$v=\mu v_{V}+\nu v_{L}+(1-\mu -\nu )v_{S},$

$\eta =\mu \eta _{V}+\nu \eta _{L}+(1-\mu -\nu )\eta _{S},$

where $v$ and $\eta$ are the volume and entropy of the mixture. The truth of the first equation is evident. The second may be written

$\eta -\eta _{S}=\mu (\eta _{V}-\eta _{S})+\nu (\eta _{L}-\eta _{S}),$

or multiplying by $t'$

$t'(\eta -\eta _{S})=\mu t'(\eta _{V}-\eta _{S})+\nu t'(\eta _{L}-\eta _{S}).$

The first member of this equation denotes the heat necessary to bring the body from the state $S$ to the state of the mixture in question under the constant temperature $t'$, while the terms of the second member denote separately the heat necessary to vaporize the part $\mu$, and to liquefy the part $\nu$ of the body. The values of $v$ and $\eta$ are such as would give the center of gravity of masses $\mu ,\nu$ and $1-\mu -\nu$ placed at the points $V,L$ and $S$.^[1] Hence the part of the diagram which represents a mixture of vapor, liquid and solid, is the triangle $VLS$. The pressure and temperature are constant for this triangle, i.e., an isopiestic and also an isothermal here expand to cover a space. The isodynamics are straight and equidistant for equal differences of energy. For ${\frac {d\epsilon }{dv}}=-p'$ and ${\frac {d\epsilon }{d\eta }}=t'$, both of which are constant throughout the triangle.

1. These points will not be in the same straight line unless

   $t'(\eta _{V}-\eta _{S}):t'(\eta _{L}-\eta _{S})::v_{V}-v_{S}:v_{L}-v_{S},$

   a condition very unlikely to be fulfilled by any substance. The first and second terms of this proportion denote the heat of vaporization (from the solid state) and that of liquefaction.