Pressure with Depth; understanding Apparatus

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The discussion focuses on calculating air pressure in a closed column using a setup involving water levels in a canister and a column. The pressure at point B in the column is equated to atmospheric pressure, leading to the equation P_air = P_atm - ρgh, where ρ is the density of water. Adjusting the vertical position of the open canister affects the air pressure in the column, with the pressure increasing if the canister is raised and decreasing if lowered. The conversation emphasizes the relationship between water levels and air pressure, confirming that lowering the canister decreases the water level and air pressure in the column. Understanding these principles is crucial for working with this type of scientific apparatus.
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This isn't a textbook problem, I just need some help understanding this apparatus. I attached a crude sketch...

The canister on the right is open to the air, the column on the left is closed, but the top is filled with air and the bottom is filled with water. the canister has water in it as well. How can you calculate the air pressure in the closed column using this setup? Presume you know the heights and everything...

Pressure variations with depth...

P=P_o+pgh
 

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The pressure in the closed part has to balance the pressure in the air plus the weight of the column of water.
I suspect it would hep you if you adjusted the system so the water is only in the connecting tube.
 
Thanks. That helped. I attached another image for the case where the water level in the canister is at a height "h" below the water level in the column. I also labeled the key points A, B, and C... I want to write an equation for the air pressure in the column, P_{air} which is closed to the atmosphere...

Basically I think the pressure at B must equal the pressure at C, which should be atmospheric pressure:

P_B=P_{atm}

I can also write:

P_B=P_{air}+ρgh

these equations together imply:

P_{air}=P_{atm}-ρgh

Is this true? The ρ constant is the density of water.
 

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AdkinsJr said:
Thanks. That helped. I attached another image for the case where the water level in the canister is at a height "h" below the water level in the column. I also labeled the key points A, B, and C... I want to write an equation for the air pressure in the column, P_{air} which is closed to the atmosphere...

Basically I think the pressure at B must equal the pressure at C, which should be atmospheric pressure:
P_B=P_{atm}I can also write:
P_B=P_{air}+ρghthese equations together imply:
P_{air}=P_{atm}-ρghIs this true? The ρ constant is the density of water.
You're missing the attachment.
 
That might help, it's added.
 
That looks good -- both the figure & the equations.

Don't forget, by changing the vertical position of the open canister, you will be changing the pressure of the air in the column.
 
Ok, I have another follow up question, if the water level in the canister is h meters ABOVE the water level in the column, then the air pressure in the tube is:

P_{air}=P_{atm}+pgh

If it's below, the pressure is:

P_{air}=P_{atm}-pgh

So it decreases the pressure to lower the canister? So as I lower the canister, it will cause the water level to lower as well?
 
Yes.
 
:) also see "constant volume gas thermometer".
It's a standard, and historically important, bit of scientific apparatus.
 
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