Sometimes, I find some websites writing laplace law as: \begin{equation} p = \gamma\frac{1}{R} \end{equation} and some \begin{equation} p = \gamma\frac{2}{R} \end{equation} I can understand that the second form is a result of spherical interfaces where curvatures are of same radii, but what about the first equation, i got confused, (knowing that the radius was not multiplied by 2 so that 2 disappeared after simplifying)? What is the interpretation of such law?
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1$\begingroup$ Could you link one example of each? $\endgroup$– stafusaCommented Apr 27, 2022 at 8:56
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$\begingroup$ @stafusa for example in this paper hal.sorbonne-universite.fr/hal-00783720/document equation (2.5), they included later that $\mu$ which is laplace pressure is $\gamma/r$ $\endgroup$– RememberCommented Apr 27, 2022 at 9:18
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2 Answers
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The Young–Laplace equation is usually written as
$$ \Delta p = -\gamma\left(\frac{1}{R_1}+\frac{1}{R_2}\right), \label{1}\tag{1} $$
where $\Delta p$ is the Laplace pressure, the pressure difference across the fluid interface (the exterior pressure minus the interior pressure), $\gamma$ is the surface tension (or wall tension) [...] and $R_{1}$ and $R_{2}$ are the principal radii of curvature.
In the given source the equation without the factor $2$ appears in a section considering a 2D system:
For simplicity we adopt the two dimensional framework introduced in Rivetti and Neukirch (2012) where the liquid-vapor interface is a cylindrical arc
where $R_2\to\infty$, rendering Eq.(\ref{1}):
$$ \Delta p = -\gamma\frac{1}{R_1}. $$
While in a (cylindrical) capillary the surface is a portion of a sphere, where $R_1=R_2=R$, so that $(1/R_1+1/R_2)$ becomes $2/R$ and Eq.(\ref{1}):
$$ \Delta p = -\gamma\frac{2}{R}. $$
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1$\begingroup$ @Aly It's just a matter of convention - of which direction is positive or whether $\Delta P$ stands for $P_{\text{out}}-P_\text{in}$ or $P_{\text{in}}-P_\text{out}$: in my link they define it as "exterior pressure minus the interior pressure", in yours, "$P_\text{inside}-P_\text{outside}$. $\endgroup$– stafusaCommented Jun 25, 2022 at 11:43
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1$\begingroup$ yeh exactly, I thought about this, but it was not clear enough in the website, I thought there was another explanation. Thanks! $\endgroup$– RememberCommented Jun 25, 2022 at 12:21
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It's because there are two radii of curvature, and there are two radii because a liquid-air interface is two dimensional. We conventionally call the two radii $R_1$ and $R_2$, and the equation for the pressure difference across the interface is:
$$ \Delta P = \gamma \left( \frac{1}{R_1} + \frac{1}{R_2} \right) $$
If we consider a cylinder this is curved in only one direction, or more formally one of the radii is infinite so its reciprocal is zero. In that case we get:
$$ \Delta P = \gamma \left( \frac{1}{R} + 0 \right) = \frac{\gamma}{R} \tag{1} $$
If we now consider a sphere it curves in both directions with equal radii of curvature $R$ so we get:
$$ \Delta P = \gamma \left( \frac{1}{R} + \frac{1}{R} \right) = \frac{2\gamma}{R} \tag{2} $$
If we look at the diagram in the paper you cited:
the drop they are considering is a cylinder so equation (1) applies.
answered Apr 27, 2022 at 9:52