# Why does a liquids height in a Barometer depend linearly on pressure?

Here is a derivation I found relating the pressure P, gravity g, height h, and density $\rho$, of a liquid in a barometer

$$P=\frac{mg}{A}\\ V=Ah \\ P = \frac{mgh}{V}\\ \rho=\frac{m}{V}\\ P = \rho g h$$

However, this derivation seems misguided, since $P = \frac{F}{A}$, and P is the pressure of the gas, not of the liquid, as the derivation later relates it to.

Furthermore, it doesn't make sense to me that the height of the liquid and the pressure are necessarily linearly related. Wouldn't this depend on the compressibility of the liquid? Or rather, it seems like depending upon the properties of the liquid, it may be very difficult to change the height past a certain point, for example. In this sense the added energy from increasing the pressure goes into bonds or is stored in electrical interactions.

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In a barometer at equilibrium, the pressure of the mercury equals the pressure of the air (at the air/mercury interface).

The compressibility of mercury in principle does make it non-linear. The density would be 3.8 parts per million higher at the air interface than at the vacuum interface.

Certainly there are other greater sources of error: the vacuum isn't really a vacuum but contains mercury vapor, density varies with temperature, vapor pressure varies with temperature, the size of the tube varies with temperature, there are surface tension effects.

Just like any experimental technique, you need to evaluate sources of error and report a measurement with appropriate uncertainty.

Why does a liquids height in a Barometer depend linearly on pressure?

Since: $$P=\frac FA=\frac {mg}{A}=\frac {\rho Vg}A=\rho g h$$

However, this derivation seems misguided, since P=F/A, and P is the pressure of the gas, not of the liquid, as the derivation later relates it to.

No at equilibrium the pressure of both will be same.

Furthermore, it doesn't make sense to me that the height of the liquid and the pressure are necessarily linearly related. Wouldn't this depend on the compressibility of the liquid?

Since we are assuming uniform density of liquid the mass at any place of same volume will be same, so if we add some water mass adds, so force adds.

Or rather, it seems like depending upon the properties of the liquid, it may be very difficult to change the height past a certain point, for example. In this sense the added energy from increasing the pressure goes into bonds or is stored in electrical interactions.

Such a situation never arrives, as far as theoretical concerns are made, since we assume water as an in compressible fluid.