Page:Scientific Papers of Josiah Willard Gibbs.djvu/463

In the above case of dissociation the formula would be

$p-p_{1}=At\left({\frac {\gamma _{1}}{M_{1}}}+{\frac {\gamma _{2}}{M_{2}}}+{\frac {\gamma _{3}}{M_{3}}}\right)\cdot$

For a coexistent solid phase of the solvent we have for constant pressure

$0=\eta dt+md\mu _{\text{S}}+v{\frac {At}{M_{\text{D}}}}d\gamma _{\text{D}},$ $0=\eta 'dt+md\mu _{\text{S}},$

$m_{\text{S}}$ being for convenience taken the same in both phases.

In integrating for small values of $\gamma _{\text{D}}$ we may treat the coefficients of $dt$ and $d\gamma _{\text{D}}$ as constant. This gives

$(\eta '-\eta )\Delta t=v{\frac {At}{M_{\text{D}}}}d\gamma _{\text{D}}=At{\frac {m_{\text{D}}}{M_{\text{D}}}},$

or if we write $Q_{\text{S}}$ for (the latent heat of melting for the unit of weight of the solvent), we have

$-\Delta t={\frac {At^{2}m_{\text{D}}}{Q_{\text{S}}M_{\text{D}}m_{\text{S}}}}\cdot$

This may be written

$-\Delta t{\frac {m_{\text{S}}}{m_{\text{D}}}}{\frac {M_{\text{D}}}{M_{\text{S}}}}={\frac {At^{2}}{Q_{\text{S}}M_{\text{S}}}}\cdot$

According to Raoult, the first member of this equation has a value nearly identical for all solvents and solutes (supposed definite compounds). This would make the second member the same for all liquids of "definite" composition, when we give $M_{\text{S}}$ the value for the molecule in the liquid state. I should think it more likely that these properties should hold for the two members of the equation

$-{\frac {\Delta t}{t}}{\frac {m_{\text{S}}}{m_{\text{D}}}}{\frac {M_{\text{D}}}{M_{\text{S}}}}={\frac {At}{Q_{\text{S}}M_{\text{S}}}},$

which are pure numbers (of no dimensions in physical units). In this form it has a certain analogy with van der Waals' law of "corresponding states."

With a coexistent vapor phase of the solvent, we have

${\begin{aligned}v\,dp&=m_{\text{S}}d\mu _{\text{S}}+v{\frac {At}{M_{\text{D}}}}d\gamma _{\text{D}},\\v'\,dp&=m_{\text{S}}d\mu _{\text{S}},\\(v-v')dp&=v{\frac {At}{M_{\text{D}}}}d\gamma _{\text{D}},\\-dp&={\frac {v}{v'-v}}{\frac {At}{M_{\text{D}}}}d\gamma _{\text{D}}.\end{aligned}}$