Page:EB1911 - Volume 05.djvu/272

259
CAPILLARY ACTION

${\displaystyle {\mbox{V}}}$, may be divided into two parts by considering separately the thin shell or skin extending from the outer surface to a depth ${\displaystyle \epsilon }$, within which the density and other properties of the liquid vary with the depth, and the interior portion of the liquid within which its properties are constant.

Since ${\displaystyle \epsilon }$ is a line of insensible magnitude compared with the dimensions of the mass of liquid and the principal radii of curvature of its surface, the volume of the shell whose surface is ${\displaystyle {\mbox{S}}}$ and thickness ${\displaystyle \epsilon }$ will be ${\displaystyle {\mbox{S}}\epsilon }$, and that of the interior space will be ${\displaystyle {\mbox{V}}-{\mbox{S}}\epsilon }$.

If we suppose a normal ${\displaystyle \nu }$ less than ${\displaystyle \epsilon }$ to be drawn from the surface ${\displaystyle {\mbox{S}}}$ into the liquid, we may divide the shell into elementary shells whose thickness is ${\displaystyle d\nu }$, in each of which the density and other properties of the liquid will be constant.

The volume of one of these shells will be ${\displaystyle {\mbox{S}}d\nu }$. Its mass will be ${\displaystyle {\mbox{S}}\rho d\nu }$. The mass of the whole shell will therefore be ${\displaystyle {\mbox{S}}\int _{0}^{\epsilon }\rho d\nu }$, and that of the interior part of the liquid ${\displaystyle ({\mbox{V}}-{\mbox{S}}\epsilon )\rho _{0}}$. We thus find for the whole mass of the liquid

 ${\displaystyle {\mbox{M}}={\mbox{V}}\rho _{0}-{\mbox{S}}\int _{0}^{\epsilon }(\rho _{0}-\rho )d\nu }$ (2)

To find the potential energy we have to integrate

 ${\displaystyle {\mbox{E}}=\iiint _{\rho }^{\chi }dx\,dy\,dz}$ (3)

Substituting ${\displaystyle \chi \rho }$ for ${\displaystyle \rho }$ in the process we have just gone through, we find

 ${\displaystyle {\mbox{E}}={\mbox{V}}\chi _{0}\rho _{0}-{\mbox{S}}\int _{0}^{\epsilon }(\chi _{0}\rho _{0}-\chi \rho )d\nu }$ (4)

Multiplying equation (2) by ${\displaystyle \chi _{0}}$, and subtracting it from (4),

 ${\displaystyle {\mbox{E}}-{\mbox{M}}\chi _{0}={\mbox{S}}\int _{0}^{\epsilon }(\chi -\chi _{0})\rho d\nu }$ (5)

In this expression ${\displaystyle M}$ and ${\displaystyle \chi _{0}}$ are both constant, so that the variation of the right-hand side of the equation is the same as that of the energy ${\displaystyle E}$, and expresses that part of the energy which depends on the area of the bounding surface of the liquid. We may call this the surface energy.

The symbol ${\displaystyle \chi }$ expresses the energy of unit of mass of the liquid at a depth ${\displaystyle \nu }$ within the bounding surface. When the liquid is in contact with a rare medium, such as its own vapour or any other gas, ${\displaystyle \chi }$ is greater than ${\displaystyle \chi _{0}}$, and the surface energy is positive. By the principle of the conservation of energy, any displacement of the liquid by which its energy is diminished will tend to take place of itself. Hence if the energy is the greater, the greater the area of the exposed surface, the liquid will tend to move in such a way as to diminish the area of the exposed surface, or, in other words, the exposed surface will tend to diminish if it can do so consistently with the other conditions. This tendency of the surface to contract itself is called the surface-tension of liquids.

Fig. 1.

Dupré has described an arrangement by which the surface-tension of a liquid film may be illustrated. A piece of sheet metal is cut out in the form AA (fig. 1). A very fine slip of metal is laid on it in the position BB, and the whole is dipped into a solution of soap, or M. Plateau’s glycerine mixture. When it is taken out the rectangle AACC if filled up by a liquid film. This film, however, tends to contract on itself, and the loose strip of metal BB will, if it is let go, be drawn up towards AA, provided it is sufficiently light and smooth.

Let ${\displaystyle {\mbox{T}}}$ be the surface energy per unit of area; then the energy of a surface of area ${\displaystyle {\mbox{S}}}$ will be ${\displaystyle {\mbox{ST}}}$. If, in the rectangle AACC, AA = ${\displaystyle a}$, and AC = ${\displaystyle b}$, its area is ${\displaystyle {\mbox{S}}=ab}$, and its energy ${\displaystyle {\mbox{T}}ab}$. Hence if ${\displaystyle {\mbox{F}}}$ is the force by which the slip BB is pulled towards AA,

 ${\displaystyle {\mbox{F}}={\frac {d}{db}}{\mbox{T}}ab={\mbox{T}}a,}$ (6)

or the force arising from the surface-tension acting on a length a of the strip is Ta, so that T represents the surface-tension acting transversely on every unit of length of the periphery of the liquid surface. Hence if we write

 ${\displaystyle {\mbox{T}}=\int _{0}^{\epsilon }(\chi -\chi _{0})\rho d\nu ,}$ (7)

we may define T either as the surface-energy per unit of area, or as the surface-tension per unit of contour, for the numerical values of these two quantities are equal.

If the liquid is bounded by a dense substance, whether liquid or solid, the value of χ may be different from its value when the liquid has a free surface. If the liquid is in contact with another liquid, let us distinguish quantities belonging to the two liquids by suffixes. We shall then have

 ${\displaystyle {\mbox{E}}_{1}-{\mbox{M}}_{1}\chi _{01}={\mbox{S}}\int _{0}^{\epsilon }(\chi _{1}-\chi _{01})\rho _{1}d\nu _{1},}$ (8)
 ${\displaystyle {\mbox{E}}_{2}-{\mbox{M}}_{2}\chi _{02}={\mbox{S}}\int _{0}^{\epsilon }(\chi _{2}-\chi _{02})\rho _{2}d\nu _{2}.}$ (9)

Adding these expressions, and dividing the second member by S, we obtain for the tension of the surface of contact of the two liquids

 ${\displaystyle {\mbox{T}}_{1\cdot 2}=\int _{0}^{\epsilon _{1}}(\chi _{1}-\chi _{01})\rho _{1}d\nu _{1}+\int _{0}^{\epsilon _{2}}(\chi _{2}-\chi _{02})\rho _{2}d\nu _{2}}$ (10)

If this quantity is positive, the surface of contact will tend to contract, and the liquids will remain distinct. If, however, it were negative, the displacement of the liquids which tends to enlarge the surface of contact would be aided by the molecular forces, so that the liquids, if not kept separate by gravity, would at length become thoroughly mixed. No instance, however, of a phenomenon of this kind has been discovered, for those liquids which mix of themselves do so by the process of diffusion, which is a molecular motion, and not by the spontaneous puckering and replication of the bounding surface as would be the case if T were negative.

It is probable, however, that there are many cases in which the integral belonging to the less dense fluid is negative. If the denser body be solid we can often demonstrate this; for the liquid tends to spread itself over the surface of the solid, so as to increase the area of the surface of contact, even although in so doing it is obliged to increase the free surface in opposition to the surface-tension. Thus water spreads itself out on a clean surface of glass. This shows that ${\displaystyle \int _{0}^{\epsilon }(\chi -\chi _{0})\rho d\nu }$ must be negative for water in contact with glass.

On the Tension of Liquid Films.—The method already given for the investigation of the surface-tension of a liquid, all whose dimensions are sensible, fails in the case of a liquid film such as a soap-bubble. In such a film it is possible that no part of the liquid may be so far from the surface as to have the potential and density corresponding to what we have called the interior of a liquid mass, and measurements of the tension of the film when drawn out to different degrees of thinness may possibly lead to an estimate of the range of the molecular forces, or at least of the depth within a liquid mass, at which its properties become sensibly uniform. We shall therefore indicate a method of investigating the tension of such films.

Let ${\displaystyle {\mbox{S}}}$ be the area of the film, ${\displaystyle {\mbox{M}}}$ its mass, and ${\displaystyle {\mbox{E}}}$ its energy; ${\displaystyle \sigma }$ the mass, and ${\displaystyle e}$ the energy of unit of area; then

 ${\displaystyle {\mbox{M}}={\mbox{S}}\sigma ,}$ (11)
 ${\displaystyle {\mbox{E}}={\mbox{S}}e.}$ (12)

Let us now suppose that by some change in the form of the boundary of the film its area is changed from ${\displaystyle {\mbox{S}}}$ to ${\displaystyle {\mbox{S}}+d{\mbox{S}}}$. If its tension is ${\displaystyle {\mbox{T}}}$ the work required to effect this increase of surface will be ${\displaystyle {\mbox{T}}d{\mbox{S}}}$, and the energy of the film will be increased by this amount. Hence

 ${\displaystyle {\mbox{T}}d{\mbox{S}}=d{\mbox{E}}={\mbox{S}}de+ed{\mbox{S}}.}$ (13)

But since ${\displaystyle {\mbox{M}}}$ is constant,

 ${\displaystyle d{\mbox{M}}={\mbox{S}}d\sigma +\sigma d{\mbox{S}}=0.}$ (14)

Eliminating ${\displaystyle d{\mbox{S}}}$ from equations (13) and (14), and dividing by ${\displaystyle {\mbox{S}}}$, we find

 ${\displaystyle {\mbox{T}}=e-\sigma {\frac {de}{d\sigma }},}$ (15)

In this expression ${\displaystyle \sigma }$ denotes the mass of unit of area of the film, and ${\displaystyle e}$ the energy of unit of area.

If we take the axis of ${\displaystyle z}$ normal to either surface of the film, the radius of curvature of which we suppose to be very great compared with its thickness ${\displaystyle c}$, and if ${\displaystyle \rho }$ is the density, and ${\displaystyle \chi }$ the energy of unit of mass at depth ${\displaystyle z}$, then

 ${\displaystyle \sigma =\int _{0}^{c}\rho dz,}$ (16)

and

 ${\displaystyle e=\int _{0}^{c}\chi dz,}$ (17)

Both ${\displaystyle \rho }$ and ${\displaystyle \chi }$ are functions of ${\displaystyle z}$, the value of which remains the same when ${\displaystyle z-c}$ is substituted for ${\displaystyle z}$. If the thickness of the film is greater than ${\displaystyle 2\epsilon }$, there will be a stratum of thickness ${\displaystyle c-2\epsilon }$ in the middle of the film, within which the values of ${\displaystyle \rho }$ and ${\displaystyle \chi }$ will be ${\displaystyle \rho _{0}}$ and ${\displaystyle \chi _{0}}$. In the two strata on either side of this the law, according to which ${\displaystyle \rho }$ and ${\displaystyle \chi }$ depend on the depth, will be the same as in a liquid mass of large dimensions. Hence in this case

 ${\displaystyle \sigma =(c-2\epsilon )\rho _{0}+2\int _{0}^{\epsilon }\rho d\nu ,}$ (18)
 ${\displaystyle e=(c-2\epsilon )\chi _{0}\rho _{0}+2\int _{0}^{\epsilon }\chi \rho d\nu ,}$ (19)
 ${\displaystyle {\frac {d\sigma }{dc}}=\rho _{0},\quad {\frac {de}{dc}}=\chi _{0}\rho _{0},\quad \therefore {\frac {de}{d\sigma }}=\chi _{0}}$ (20).mw-parser-output .nowrap,.mw-parser-output .nowrap a:before,.mw-parser-output .nowrap .selflink:before{white-space:nowrap}
 ${\displaystyle {\mbox{T}}=2\int _{0}^{\epsilon }\chi \rho d\nu -2\chi _{0}\int _{0}^{\epsilon }\rho d\nu =2\int _{0}^{\epsilon }(\chi -\chi _{0})\rho d\nu ;}$ (20)

Hence the tension of a thick film is equal to the sum of the tensions of its two surfaces as already calculated (equation 7). On the hypothesis of uniform density we shall find that this is true for films whose thickness exceeds ${\displaystyle \epsilon }$.

The symbol ${\displaystyle \chi }$ is defined as the energy of unit of mass of the substance. A knowledge of the absolute value of this energy is not required, since in every expression in which it occurs it is under the