I noticed that steam condensing on the underside of a horizontal surface formed drops which only dripped off when they reach a diameter of approximately 15mm.

Is it possible to calculate the interatomic forces between water molecules from this observation ?, (given: Avogadro's number, the molar weight of water, coordination number of liquid water, and

acceleration due to gravity).

I noticed that steam condensing on the underside of a horizontal surface formed drops which only dripped off when they reach a diameter of approximately 15mm.

Is it possible to calculate the interatomic forces between water molecules from this observation ?' date=' (given: Avogadro's number, the molar weight of water, coordination number of liquid water, and

acceleration due to gravity).[/quote']

 

 

Give it a shot. How would you proceed?

At the very least, the drop size gives some information about the surface tension involved, yes?

At the very least, the drop size gives some information about the surface tension involved, yes?

 

And that should be related to the forces, because surface tension occurs from asymmetric attraction — the surface breaks the symmetry.

Give it a shot. How would you proceed?

 

Oops, forgot to include that the shape of the drop is approximately sine-squared, (so it's volume is easy to calculate), and the distance from the horizontal surface (ceiling) to the tip of the drop is approximately 10mm,

(the circle of contact between the drop and the ceiling has diameter 15mm).

 

A cubic centimeter has side length 1cm,

a cubic centimeter of water weighs 1gram,

1 gram is 1/18th of a mole of water,

1/18th of a mole is 3.345e22 molecules,

so 1cm is the cube root of 3.345e22 => 32.2 million water molecules per centimeter.

 

The area of a circle = Pi times (radius squared),

the 15mm diameter circle of contact has 1.832 e15 water molecules...

 

Hey I'm doing all the work...

it's beginning to smell like homework

it's beginning to smell like homework

 

It has been over 20 years since I was last assigned homework.

Oh Yourdadonapogos, the correct spelling is testicles, (smells like english homework ).

I didn't spell it. Talk to IMM about spelling.

back to the original subject.

 

The volume of the sine-squared droplet is equivalent to a cylinder diameter 15mm and "height" half that of the drop i.e. 5mm.

back to the original subject.

 

The volume of the sine-squared droplet is equivalent to a cylinder diameter 15mm and "height" half that of the drop i.e. 5mm.

 

I think now you want to estimate the number of atoms in contact with the surface, and compare that to the weight of the droplet. But this will tell you about the attraction to the suface, not the attractive forces between the water molecules.

I didn't spell it. Talk to IMM about spelling.

 

Then the correct thing to do would be "testicals [sic]"

Then the correct thing to do would be "testicals [sic']"

 

It's spelled "sick":D

I think now you want to estimate the number of atoms in contact with the surface, and compare that to the weight of the droplet. But this will tell you about the attraction to the suface, not the attractive forces between the water molecules.

 

When the drop drips a small residual droplet remains stuck to the surface,

i.e. the water-water bonds have failed, not the water-surface bonds.

 

I have approximated the number of water molecules seperated when the drop drips as the same as the number in contact with the surface, (the drop begins to pinch-off close to the surface).

 

I welcome any constructive suggestions on if/how this problem can be solved.

When the drop drips a small residual droplet remains stuck to the surface' date='

i.e. the water-water bonds[b'] have failed[/b], not the water-surface bonds.

 

I have approximated the number of water molecules seperated when the drop drips as the same as the number in contact with the surface, (the drop begins to pinch-off close to the surface).

I welcome any constructive suggestions on if/how this problem can be solved.

 

Just a thought on this:

Prior to failure they "flow" into a reduced effective contact area which then "fails" or breaks off. This gives it momentum as well as weight over the reduced area. The initial "pinching off" close to the surface is a "flow" toward a somewhat stronger bond from a somewhat weaker as opposed to a failure.

When the drop drips a small residual droplet remains stuck to the surface' date='

i.e. the water-water bonds have failed, not the water-surface bonds.

 

I have approximated the number of water molecules seperated when the drop drips as the same as the number in contact with the surface, (the drop begins to pinch-off close to the surface).

 

I welcome any constructive suggestions on if/how this problem can be solved.[/quote']

 

 

A very good observation. So, the drop grows and adds mass. The water bonds hold it together, but if you look at any vertical line of molecules, each bond must, on average) support the weight beneath it (if it's cylindrical; more complicated if it's not)

 

You know the drop size, so you know the mass. You also know the mass of a water molecule, so you can estimate the size of a molecule, and how many should exist along that vertical line. The supported weight is your force.

J.C.MacSwell and Swansont, thank you for your posts on this matter.

Using the data above the water drop drips when the the interatomic spacing doubles.

Is this correct ?.