# Pressure on a sample of fluid (at rest)

A.T.
Is the representation in the figure right?
The molecules in a fluid are neither static nor arranged in vertical columns.

• Kaushik
Mister T
Gold Member
Molecules collide with the pressure sensor. The more frequent the collisions, and also the more energetic the collisions, the higher the reading on the pressure sensor. Let us take a monatomic ideal gas to make the explanation simpler. Moreover, let the entire sample of gas be at the same temperature so that the average kinetic energy per molecule is the same everywhere. That means that the only thing that can affect the reading on the pressure sensor is the frequency of collisions. The deeper you go in the fluid the greater the number of molecules per unit volume, and hence the greater the pressure.

It's gravity that causes the fluid to be more dense the deeper you go. To say that the pressure at some location is due to the weight of the fluid above that location is a perfectly valid macroscopic explanation. But to say that it's due to the weight of the molecules above it is a rather misleading microscopic explanation. Unless a molecule exerts a force on the pressure sensor, it cannot make a contribution to the pressure. Thus molecules located far above the pressure sensor do not contribute directly to the the reading on the pressure sensor.

• Kaushik
jbriggs444
Homework Helper
The molecules in a fluid are neither static nor arranged in vertical columns.
Note how careful @A.T. was in his phrasing here. The molecules are not static and not arranged in columns. But for Pascal's law to apply, the fluid is static. An analysis can consider a columnar arrangement since the net forces external to a column of fluid are balanced and are exactly right to keep the fluid within the column (or horizontal tube) stationary.

The point being that an analysis based on molecules is going to be difficult. But an analysis based on a continuous fluid is doable.

• Kaushik
jbriggs444
Homework Helper
It must be equal to the pressure on the top surface so that the fluid doesn't flow?
Yes.

Now consider an imaginary square horizontal tube of fluid running from the side of that cube over to another identical cube in a portion of the fluid that is hiding under a "ceiling". Is the pressure on the right end of that tube equal to the pressure at the left end?

• Kaushik
Now consider a horizontal tube running from the side of that cube over to another cube in a portion of the fluid that is hiding under a "ceiling". Is the pressure on the right end of that tube equal to the pressure at the left end?
It must be equal so that it doesn't flow.

Then for that cubical part to not flow the pressure on top surface must be equal to that of the pressure on the side surface. Hence, pressure on both the cubical part is equal. Is it?

• jbriggs444
jbriggs444
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