I think Pascal's law is very confusing. It says that any change in pressure in any part of a liquid is transmitted to all the parts of the liquid without being diminished in magnitude.

Consider this situation:

A vessel filled with mercury is exposed to atmospheric pressure. Then, an empty test tube having only vacuum is inverted to it so that some area of the mercury surface gets covered by it and the pressure on that part becomes zero. So the change in pressure of this part is equal to the atmospheric pressure. So, pressure at all the other parts of the liquid should decrease by 1 atm, and the normal forces on the liquid by the vessel should somewhat readjust to account for this decreased pressure.

But, if I'm correct, in this situation some mercury would itself rise into the test tube to make the pressure at the part of the surface enclosed by the test tube equal to the atmospheric pressure.

Does this law need a better statement or some modification?

And I'm also confused about when a liquid level rises and when it doesn't. For example, if we have a mercury vessel exposed to some pressure, and that pressure is decreased then no rise in mercury level happens. But if the opening of the vessel is some tube of lower cross-sectional area than the vessel, and pressure is decreased, then mercury level rises. Why?


1 Answer 1


Pascal's Law is relevant to the instantaineous case, the very moment in your example that the mercury is exposed to a vacuum, whereas you are considering the final stable state of the system after the liquid has stopped moving in response to the pressure change.
You could try to formulate a different expression of the law, but it is tricky to get right to say the least. Here, I had a go, "A pressure difference between any two areas of a liquid surface results in a net force on those areas" - how do you like that version? (me not so much).

2) for the liquid to move there has to be a pressure differential between different parts of the liquid, amounting to a total net force on it. If you just change the pressure everywhere there is no net force resultant.


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