JR0029 : Why don't we include the adhesive and cohesive force while calculating rise in a capillary tube?

Guru Vishnu

May 15, 2020 10:34

2

Q: Why don't we include the adhesive and cohesive force while calculating rise in a capillary tube?

The contact angle of a liquid solid interface is explained by saying that the liquid surface must be perpendicular to the resultant of adhesive cohesive and gravitational forces acting on it, since it cannot sustain shear stresses. However, once the contact angle is determined, the cohesive and...

Do you have an alternate answer for the above question sir?

Particularly, I feel, I'm having the same doubt as the following comment:

I understand your point, the adhesive and cohesive forces play a role in determining the contact angle but surely that is not the end of the story. While writing Newton's Second Law for the fluid column why are we justified in omitting them even though they are at work? In the energy approach you illustrated, there would be some potential energy associated with these forces as well which has seemingly been omitted. Why? — Gerard Oct 23 '15 at 5:33

John Rennie

I'm not sure I understand what Gerard is asking

Guru Vishnu

Did you read the question sir? I think the comment alone is not self understandable.

Or let me explain:

As per Jurin's law,

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

Why aren't cohesive and adhesive forces involved in the derivation of the above formula?

How can it be self-contained in the contact angle term sir?

John Rennie

They are included. The angle $\theta$ depends on the cohesive and adhesive forces.

Guru Vishnu

Yes sir. I understand. But I don't see why is $S$ involved here. My book states the free surface attracts the tube and this is a consequence of Newton's third law.

John Rennie

Why wouldn't S be involved?

I just don't understand the problem.

Guru Vishnu

May 15, 2020 10:43

Because it's just the force within the fluid molecules. The force between the liquid surface and the tube is adhesive force and not surface tension as per my understanding.

John Rennie

Have you seen the derivation of the contact angle?

Guru Vishnu

Derivation? No sir. I thought it was experimentally determined.

John Rennie

Let me draw a diagram ...

Guru Vishnu

So is the $\cos\theta$ term inversely proportional to $S$ in some way?

John Rennie

Guru Vishnu

May 15, 2020 10:49

@JohnRennie Only thing I know about it is this - the contact angle must be in such a way that the net force (adhesive+cohesive+gravitational) must be perpendicular (normal) to the tangential plane of the free surface of the liquid.

John Rennie

All interfaces have a surface tension because the surface tension is just the interfacial energy in disguise.

So when we look at the point where the liquid, air and solid all meet there are three forces acting as I've drawn. OK so far?

Guru Vishnu

@JohnRennie Ok sir. I think I have a terminology doubt - Isn't the force between solid and liquid and liquid with non-surface liquid molecules called as adhesive and cohesive forces respectively rather than surface tension?

John Rennie

No. The terms adhesive and cohesive forces are rather vague while surface tension is precisely defined.

Or a colloid scientist might also use the term interfacial energy because the interfacial energy and the surface tension are the same thing.

Guru Vishnu

Fine. I didn't know that. This seems new to me.

John Rennie

I can prove it if you want ...

Prove that interfacial energy and the surface tension are the same thing I mean.

Guru Vishnu

May 15, 2020 10:55

@JohnRennie Another doubt before I say ok : Why are the two horizontal forces horizontal? In the explanation of contact angles, the adhesive/cohesive forces weren't normal or parallel to the interface.

John Rennie

Because surface tension always acts in the plane of the interface.

Guru Vishnu

Ok sir. Now I think cohesive/adhesive forces are much different than interfacial energy.

@JohnRennie But aren't they dimensionally inconsistent? One is force per unit length and the other is energy.

John Rennie

Dimensionally inconsistent eh? :-)

Shall we check?

Guru Vishnu

Ok sir. I think I pressed the wrong switch :-)

John Rennie

(interfacial energy is energy per unit area not just energy)

Guru Vishnu

May 15, 2020 10:59

I agree. Surface energy per area is surface tension. Yes it's consistent dimensionally. Misinterpreted energy as pure energy.

John Rennie

:-)

Guru Vishnu

Fine sir :-)
If you wish shall we proceed regarding the main question?

6 mins ago, by John Rennie

Prove that interfacial energy and the surface tension are the same thing I mean.

John Rennie

OK, consider a soap film held inside a wire frame. I'll use this as an example because it's the sort of thing we've all played with so it should be intuitive. I'll draw a diagram:

@GuruVishnu there. The light blue area is the soap film.

Guru Vishnu

Ok sir. The blue rod is movable right?

John Rennie

The blue line is a sliding wire that we can pull outwards a distance $dx$ to stretch the film.

Guru Vishnu

May 15, 2020 11:07

Ok sir. I saw this setup to find the surface energy of a soap film.

John Rennie

In that case you've done the calculation!

Guru Vishnu

Yes sir. Now I see interfacial energy you meant is same as the surface energy I referred to. However:

13 mins ago, by John Rennie

Prove that interfacial energy and the surface tension are the same thing I mean.

John Rennie

That's what I'm about to do.

Guru Vishnu

Ok sir :-)

John Rennie

Let's call the length of the wire $\ell$. Calling it $d$ was a poor choice because that's going to get confused with the $dx$.

Guru Vishnu

May 15, 2020 11:10

Ok sir. Just to simplify things - Are we going to obtain $U/A=S$?

If yes, we can definitely skip this step.

John Rennie

So if we pull the wire a distance $dx$ we create a new area of soap film $dA = \ell dx$, and the increase in energy is $EdA = E\ell dx$ where $E$ is the energy per unit area. OK so far?

Guru Vishnu

I was just wondering how could $U$ and $S$ have the same dimensions if the interfacial energy and surface energy meant the same thing.

@JohnRennie Yes sir.

John Rennie

$U$ is the total energy?

i.e. in my notation $U = EA$ ?

Guru Vishnu

Increase in surface energy. I missed a $\Delta$ sir because HCV didn't mention it. I thought it was some kind of convention.

But he uses this interchangeably.

John Rennie

OK. Anyhow the working is trivial. The increase in energy has to come from the work done $E\ell dx = dW = Fdx$

Guru Vishnu

May 15, 2020 11:13

Finally he says: We see that the surface tension of a liquid is equal to the surface energy per unit surface area.

John Rennie

And $F = S\ell$

So we end up with $E=S$

Guru Vishnu

instead of increase in surface energy per unit surface area.

@JohnRennie Ok sir. So is $E=U/A$ or in English: Interfacial energy equal to the surface energy per unit area? I see it's just surface tension then. Just a new term to be stored in memory.

John Rennie

Yes

Guru Vishnu

Apart from the forces being along the surface, especially the horizontal ones, I find it's ok to proceed sir:

26 mins ago, by John Rennie

John Rennie

It's trivial from here. The horizontal component of the total force at the junction must be zero because if it wasn't zero the liquid would flow left or right depending on the total force.

So we get:

$$ \gamma_{sa} = \gamma_{ls} + \gamma_{la}\cos\theta $$

Rearrange to get $\cos\theta$

Guru Vishnu

May 15, 2020 11:20

Ok sir. Just for confirmation: Is $\gamma_{sa}$ adhesive, $\gamma_{ls}$ cohesive and $\gamma_{la}$ surface tension sir?

John Rennie

No

$\gamma_{sa}$ is the surface tension of the solid-air interface, which is probably easier to understand as the interfacial energy of the solid-air interface.

This is related to cohesion.

Guru Vishnu

Fine sir. It also solved the mystery why it's parallel to the surface :-)

John Rennie

Imagine taking a block of the solid and splitting it along a plane of area $A$. So where we had a solid-solid interface of area $A$ we now have two solid-air interfaces of area $A$.

Guru Vishnu

So all the $\gamma$ terms are just the surface tensions of appropriate phases on either sides like solid/gas or solid/liquid, and so on.

John Rennie

Yes.

Guru Vishnu

May 15, 2020 11:24

Ok sir. It seems we're just transforming the components of cohesive and adhesive forces to surface tensions. Similar to expressing 5 m/s of velocity in the first quadrant as 3i +4j. Is this a correct inference sir?

John Rennie

Basically yes

Guru Vishnu

Ok sir. Thank you :-)

But, the following doesn't seem to hold true:

42 mins ago, by Guru Vishnu

So is the $\cos\theta$ term inversely proportional to $S$ in some way?

It needs $\gamma_{ls}=S$ to be satisfied...

John Rennie

@GuruVishnu $S$ is what I've called $\gamma_{la}$. Yes?

So our equation for the contact angle is:

$$ \gamma_{sa} = \gamma_{ls} + S\cos\theta $$

$$ \cos\theta = \frac{\gamma_{sa} - \gamma_{ls}}{S} $$

$\cos\theta$ is indeed inversely proportional to $S$ if the other two surface tensions are constant.

Guru Vishnu

Ok sir. Fine. That was an error from my side when I switched from one set of terminologies to the other. I understood this completely. And the modified Jurin's law is as per my initial expectation, independent of $S$:
$$h=\frac{2(\gamma_{sa}-\gamma_{ls})}{r\rho g}$$
and has the cohesive and adhesive terms hidden inside the surface tension terms.

Thank you very much sir :-)

John Rennie

May 15, 2020 11:43

:-)

JR0029 : Why don't we include the adhesive and cohesive force while calculating rise in a capillary tube?

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