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

In the above case of dissociation the formula would be

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

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

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

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

Then

${\displaystyle (\eta '-\eta )dt=v{\frac {At}{M_{\text{D}}}}d\gamma _{\text{D}}.}$

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

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

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

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

This may be written

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

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 ${\displaystyle M_{\text{S}}}$ 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

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

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

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