# Does surface tension change with concentration?

Dec 3, 2014

Yes
${\left(\frac{\mathrm{dg} a m m a}{\mathrm{dc}}\right)}_{T} = - \frac{R T {\Gamma}_{S}}{c}$
$\gamma$=surface tension

Derivation
dG=$- S \mathrm{dT} + V \mathrm{dP} + \gamma \mathrm{ds} i g m a + \mu {\mathrm{dn}}_{s}$
At constant temperature and pressure,
dG=$\gamma \mathrm{ds} i g m a + \mu {\mathrm{dn}}_{s}$ .............. (1)
$\sigma$=surface area
n=no. of moles of the surfactant
$\mu$=chemical potential= $\left(\frac{\mathrm{dG}}{\mathrm{dn}}\right)$

so,G=$\gamma \sigma + \mu {n}_{s}$
$\Rightarrow \mathrm{dG} = \gamma \mathrm{ds} i g m a + \sigma \mathrm{dg} a m m a + \mu {\mathrm{dn}}_{s} + {n}_{s} \mathrm{dm} u$ from (Gibbs-Duhem equation)...........(2)

by comparing the two equations (1) & (2) for dG, we get
$\sigma \mathrm{dg} a m m a + n \mathrm{dm} u$=0 (Gibbs isotherm)

Now for an interface in which oil and water for example, are separated by a geometrically flat surface.The approximation implies that on the surfactant S accumulates at the surface and hence that ${\Gamma}_{o i l}$ and ${\Gamma}_{w a t e r}$ are both zero.
The equation becomes
$\mathrm{dg} a m m a = - {\Gamma}_{s} \mathrm{dm} {u}_{s}$ where ($\Gamma = \frac{n}{\sigma}$)

for dilute solutions,
$\mathrm{dm} {u}_{s} = R T \ln c$ WHERE c IS THE MOLAR CONCENTRATION OF THE SURFACTANT.

therefore at constant temperature, ${\left(\frac{\mathrm{dg} a m m a}{\mathrm{dc}}\right)}_{T} = - R T {\Gamma}_{S} / c$

I've tried to make the derivation look as simple as possible.If still not understood , leave a comment below.