# What's the pressure inside a barometer?

I have the following problem I'm trying to solve. Let's say we have a barometer filled with water as in the following diagram:

The question to be answered is what's the pressure $$p$$ on the inside of the closed upper lid of the tube? The distance from the water surface to the bottom of the barometer tube is $$20$$ cm.

My reasoning is as follows. Since the hydrostatic pressure given by $$p = - \rho gh$$ comes from the weight of the fluid above some submerged body, then the pressure $$p$$ at the top of the tube should be $$0$$ since there isn't any water above it. I'm not sure if my reasoning is incorrect or if I didn't correctly understand the concepts. Could somebody tell me if my reasoning is correct? Thank you!

• The distance from the water surface to the bottom of the barometer tube is also needed... Apr 21 at 3:07
• Thank you! So I guess my reasoning is wrong? Apr 21 at 3:11
• If z is the elevation, then $p(z)=p(0)-\rho g z$, where p(0) is the pressure at elevation z = 0. Apr 21 at 14:03

Let us denote the heights $$h_1:=-2.2\;m$$ and $$h_2:=-0.2\;m=-20\;cm$$, the pressure at the point inside but at the top of the barometer by $$p$$ and the atmospheric pressure by $$p_a$$ and finally the density of water by $$\rho$$. The barometer leads into a container with water inside it exposed to the atmosphere at the height $$|h_1|$$ above the lower end of the barometer. Applying Pascal's law to obtain the hydrostatic pressure at point at the top of the barometer by equating it to that at the point at the lowest point inside the barometer which is at the height $$|h_1|$$ below the lower end of the barometer we have, $$p_a+\rho g h_1=p+\rho g h_2$$ so that $$p=p_a+\rho g (h_1-h_2).