# Finding pressure of gas in u shaped tube with liquid

• JoeyBob
In summary, the conversation discusses the calculation of pressure inside a glass ball filled with gas and liquid. The initial misconception is that the pressure inside the ball would be equal to atmospheric pressure, but this is found to be incorrect. The correct equation for pressure at a certain depth below the surface of a fluid is P(h) = P0 + ρfluid * g * h. Using this equation, the correct pressure is calculated and it is found that the pressures on both sides do not equalize due to the weight of the liquid pushing against the atmospheric pressure. The conversation also briefly mentions other equations for pressure, such as P = F/A and PV = nRT, but ultimately the correct equation is found to be P(h) = P0

#### JoeyBob

Homework Statement
see attached
Relevant Equations
P=phg
The answer is suppose to be 0.9432. Initially I thought the pressure inside the glass ball would just be the same as the atmospheric pressure because these equal pressures would cancel each other out, but obviously that's not true.

I can calculate the density of the gas using the equation p1/p2=h2/h1, and get a density of 1659.8361. Now I thought I could calculate the pressure of the gas using P=pgh, where p is density, g gravity, h height of liquid. Using the height of the liquid on the gas side (0.244 m) I get 3968.9 which isn't the right answer. If I take the diffirence in height (0.169 m) I get a pressure of 2749.021. I tried subtracting this from the atmospheric pressure, because maybe the difference in height gives the difference in pressure, but this also gives the wrong answer.

The information about moles is useless because I don't know the temperature of the gas. Otherwise the question would be easy.

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Maybe you don't need the number of moles or the volume of the ball. What expression do you know that for the pressure at depth ##h## below the level of a fluid?

kuruman said:
What expression do you know that for the pressure at depth h below the level of a fluid?

Im not sure what you mean. I know P=hpg. Intuitively I thought that the liquid was higher on the side of the glass ball because it had a lower pressure than the atmospheric pressure. But I thought that the liquid would rise on the glass ball side until it had the same pressure as atmosphere outside.

It's more than ##p=\rho gh##. What does your textbook say? Will it be the same regardless of the pressure at the free surface of the fluid?

kuruman said:
It's more than ##p=\rho gh##. What does your textbook say? Will it be the same regardless of the pressure at the free surface of the fluid?

Well pressure can be P=F/A, P(h)=P(knot)-pgh, P(atm)=density*Hgh, PV=nRT

I think those sum up the equations

JoeyBob said:
P(h)=P(knot)-pgh
Can you explain this one? Are you sure you copied it correctly? Specifically what is P(knot)?
The equation I know is
$$P(h)=P_0+\rho_{\text{fluid}}~g~h$$where
##P(h)## = Pressure as a function of ##h##, the depth below the surface of a fluid.
##h## = the depth below the free surface of the fluid, a positive number.
##P_0## = the pressure at the surface of the fluid.
##\rho_{\text{fluid}}## = the density of the fluid.
##g## = the acceleration of gravity.

It says that the pressure at depth ##h## below the surface of the fluid is equal to whatever pressure is at the surface of the fluid plus the pressure exerted by the weight of the fluid to that depth.

Please remember this equation and most importantly understand and remember what the symbols stand for. Then see if you can use the equation to answer this question.

JoeyBob
kuruman said:
Can you explain this one? Are you sure you copied it correctly? Specifically what is P(knot)?
The equation I know is
$$P(h)=P_0+\rho_{\text{fluid}}~g~h$$where
##P(h)## = Pressure as a function of ##h##, the depth below the surface of a fluid.
##h## = the depth below the free surface of the fluid, a positive number.
##P_0## = the pressure at the surface of the fluid.
##\rho_{\text{fluid}}## = the density of the fluid.
##g## = the acceleration of gravity.

It says that the pressure at depth ##h## below the surface of the fluid is equal to whatever pressure is at the surface of the fluid plus the pressure exerted by the weight of the fluid to that depth.

Please remember this equation and most importantly understand and remember what the symbols stand for. Then see if you can use the equation to answer this question.

Yeah that's the equation, i transcribed it wrong. So if I am understanding this correctly, first I need to find the pressure below the surface using the atmospheric pressure. Then I need to find the pressure on the surface on the glass ball side, using the P(h) from the previous step?

I tried it now and it seems to be correct so I guess I understand the equation now.

On a side note if you don't mind, why do the pressures not equalize? For instance, if the glass ball has a lower pressure, why doesn't the atmospheric pressure push the liquid until the pressures are equal? Or is it because as the liquid is pushed up, its weight also pushes against the atmospheric pressure?

JoeyBob said:
On a side note if you don't mind, why do the pressures not equalize? For instance, if the glass ball has a lower pressure, why doesn't the atmospheric pressure push the liquid until the pressures are equal?
Who says that the atmospheric pressure does not push the liquid down util the pressures are equal? Before the bulb is connected, the fluid is at the same level on both sides. After the bulb is connected the level on the air side is lower (7.5 cm above bottom) than the bulb side (24.4 cm above bottom). Maybe you confused the distances above bottom with depths from the surface.

JoeyBob