Pressure variation in a capillary tube

Question

The following image shows capillary tubes placed in beakers containing water and mercury:

We know that the rise or fall in the level of liquid in a capillary tube is given by Jurin's law:

$$h=\frac{2S\cos\theta}{r\rho g}$$

where $h$ is the rise or fall in height accordingly as it's positive or negative, $S$ is the surface tension, $\theta$ is the contact angle of the liquid on the tube wall, $r$ is the radius of the capillary tube, $\rho$ is the mass density and $g$ is the local acceleration due to gravity. Contact angle for water with glass is $0^\circ$ and it is $140^\circ$ for mercury with glass. So $\cos\theta$ term is positive for water and negative for mercury, and so, water rises and mercury falls in a capillary tube.

I understood the mechanism due to which the level rises or falls in a capillary tube. But, when I tried to find the pressure variation within the fluid in the tube, I faced some problems as discussed below:

In figure $(a)$ the pressure at $A$ and $B$ is equal to the atmospheric pressure $P_{atm}$. From fluid statics, we know that pressure at a particular level is same and it differs only if there is any variation in vertical height. So, we can say that pressure inside the capillary tube in the horizontal level of $B$ is also $P_{atm}$. From this, we see that pressure at both $A$ and the point below it in the horizontal level of $B$ are same and is equal to $P_{atm}$. But from fluid statics we must expect there must a pressure difference due to the difference in the vertical height given by $\Delta P=\rho g \Delta h$. Why is there an inconsistency in the results obtained? I feel both methods are equally reasonable.

The case becomes even more interesting in $(b)$. Pressures at $A'$ and $B'$ are equal to $P_{atm}$. From fluid statics, pressure at the depth $h'$ must be same. We know pressure at $A'$ is $P_{atm}$. Now if we conclude pressure at all the points in this level is $P_{atm}$, we see pressure at two different vertical levels - one at the free surface in the beaker and the other at a depth $h'$ are same. But this result is counterintuitive and I think there must be at least some pressure difference. At the same time, I don't think my first argument is incorrect. Then why do we get contradictory results?

In short, I don't understand how pressure varies in a fluid within the capillary tube? Further, it would be great if you could explain why do we get contradictory results when we apply our familiar results from fluid statics - pressure at the same horizontal level is same and pressure difference due to difference in vertical heights is $\Delta P=\rho g \Delta h$?

^{Image Courtesy: My own work :)}

Answer 1

The pressure difference in both the cases is due to the surface tension of water. Let me explain.

Whenever a fluid(let's say $F$) has a surface which is exposed to another medium($M$), the particles on the surface experience forces due to two types of particles, namely, the particles of $F$(cohesion forces) and the particles of $M$(adhesion forces). Whenever the adhesion forces are stronger than the cohesion forces, then the fluid $F$ tends to "stick" to the medium $M$(which is the case with water and glass). And whenever the cohesion forces are stronger than the adhesion forces, then the fluid $F$ "does not wet" the medium $M$(which is the case with mercury and glass).

So when the water is rising up the capillary, adhesion forces are at work. And when it reaches the final height$\left(\frac{2S \cos\theta}{r\rho g}\right)$, it forms a concave surface. And at this point, it is being pulled up by the cohesion forces. So the water in the capillary tube is being pressed by the same atmospheric pressure that is pressing other surfaces of water in the container, but the water in the capillary tube has adhesion force(which is not present any elsewhere) exerting an upward force on the column of the water. So eventually the water in the capillary tube requires less internal pressure to balance the pressure of the atmosphere. In fact, the internal pressure at the surface of the water in the capillary tube is $P_{atm}-\rho g h$, where $h$ is the height of water in the capillary. And this pressure is clearly less than the atmospheric pressure($P_{atm})$. So there is a pressure difference of $\rho g h$ between the points in water just below $A$ and $B$.

Similarly, for mercury, the adhesion forces dominate and thus they pull the column of mercury in the downward direction. So this time, the internal pressure at the surface of the mercury column has to balance both, the force due to atmospheric pressure and the adhesion forces, so it's bound to be larger than $P_{atm}$ and again, the pressure difference($\Delta P$) between the points in the liquid just below $A'$ and $B'$ is(as you would have expected), $$\Delta P = \rho g h$$ where $h$ is the depth of the capillary drop.

Food for thought:- Try to relate the pressure difference between the two points below $A$ and $B$) with the excess pressure in a bubble. To be precise, equate both of them. You will end up discovering an alternate derivation of the Jurin's law.

CAUTION:- Be careful!! I am not saying that there's a pressure difference between the points $A$ and $B$ or $A'$ and $B'$. The pressure at all these points is $P_{atm}$. But the pressure difference is between the liquid surface on the top of the water column in the capillary and the liquid surface elsewhere.