# I have a somewhat silly question about pressure and atmosphere [duplicate]

The very basic calculation of pressure is <<$\rho gh$>>. So now if I keep myself in a totally enclosed room, then the force exerted by all the air above me should be cut out by the structure of the room and not affect me, in other words the pressure that I feel in the chamber should be very low compared to when I am exposed in the atmosphere.

Perhaps pressure is also(or merely?) affected by the density of fluid(air) in the room, that's why I am feeling the same atmospheric pressure as outside because the moment I close the room, the pressures inside and outside are equal. But this doesn't sound so true when I look at <<$\rho gh$>>, the height is still in the equation. One might argue that the deeper you get in a container full of fluid (like in the ocean) the density increases, which is very true when huge depth is concerned, but clearly in the basic <<$\rho gh$>>, density is assumed to be constant.

My thought looks kinda messy, hopefully someone would be able to clear it up a bit, thank you. Btw when I asked this question to my physics teachers, somehow they managed to shut me up or what, I don't remember.

• Possible duplicate of Pressure inside a closed room Commented Nov 19, 2017 at 10:05
• Commented Nov 19, 2017 at 10:05
• "But this doesn't sound so true when I look at <<ρgh>>, the height is still in the equation" $\to$ That's the $h$ for when the air-tight room was sealed. "the deeper you get in a container full of fluid (like in the ocean) the density increases" $\to$ Not by much, actually. Water is almost incompressible over a very large range of pressures. Commented Nov 19, 2017 at 12:53
• @stafusa more or less yes: once you break ~10 MPa it starts diverging, so the bottoms of seas are only a few percent denser than above sea level. Commented Nov 20, 2017 at 11:08
• @KyleKanos, Thanks, I had not seen this plot before. Commented Nov 20, 2017 at 11:22

$\rho g h$ gives the extra pressure at a depth $h$ in a fluid of constant density $\rho$, compared with the pressure above the fluid. So for an open tank of liquid, the pressure at depth $h$ is given by $$p=p_0+\rho g h$$ in which $p_0$ is the atmospheric pressure at the exposed surface.
$\rho g h$ also gives the increase in pressure when we go down a distance $h$ in a gas, such as the atmosphere, provided that $h$ is small enough for $\rho$ not to change significantly over that distance.
In a typical room the difference in air pressure between floor and ceiling is given pretty accurately by $\rho g h$, because $\rho$ will change so little. Putting $\rho= 1.3\ \text{kg m}^{-3}$, $h=3 \text m$, $g=9.8\ \text{N kg}^{-1}$, gives $p=38 \text{Pa}$, which is pretty small (negligible for most purposes) compared with atmospheric pressure (approximately 100kPa)
But you're interested in the mean pressure in the room. This pressure is determined by $n$, the number of moles of air in the room and the temperature, according to $$pV=nRT.$$ However, if the room communicates with the atmosphere, or did before you closed the doors and windows, then $n$ has been fixed automatically, without consultation, by the mechanical equilibrium condition that $$p_{room}=p_{atmos}$$ in which $p_{atmos}$ is the pressure of the atmosphere in the vicinity of the room.