Can a fluids buff explain this?

[DrShoshana]

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Okay, using Nova Physics, this is the problem that I come across:

Two flasks have no ambient atmostphere, that is, they exist in a chamber in which the atmosphere has been removed. Both flasks contain mercury to a height of h. The volume in flask #2 is three times the volume of flask #1. The area at the bottom of the second flask is twice that of the first flask. Pressure P1 is the pressure at the bottom of the first and pressure P2 is at the bottom of the second flask. Which equation holds?

(A) P2=P1/2
(B) P2=P1
(C) P2=(3/2)P1
(D) P2=3P1

The answer is B. THey explain it as an application of hydrostatic equilibrium and that the pressure at the bottom of both flasks is (Density of mercury*g*h). This doesn't make any sense to me because P=F/A meaning that if you increase the area by two then the pressure should decrease by half. And these two flasks are not connected, so it doesn't make sense that they are the same pressure. Could somebody please explain to me how this works?

 

[Nevadanteater]

Okay, using Nova Physics, this is the problem that I come across:

Two flasks have no ambient atmostphere, that is, they exist in a chamber in which the atmosphere has been removed. Both flasks contain mercury to a height of h. The volume in flask #2 is three times the volume of flask #1. The area at the bottom of the second flask is twice that of the first flask. Pressure P1 is the pressure at the bottom of the first and pressure P2 is at the bottom of the second flask. Which equation holds?

(A) P2=P1/2
(B) P2=P1
(C) P2=(3/2)P1
(D) P2=3P1

The answer is B. THey explain it as an application of hydrostatic equilibrium and that the pressure at the bottom of both flasks is (Density of mercury*g*h). This doesn't make any sense to me because P=F/A meaning that if you increase the area by two then the pressure should decrease by half. And these two flasks are not connected, so it doesn't make sense that they are the same pressure. Could somebody please explain to me how this works?

let's imagine the bottom of flask 1 is a square (A) and the bottom of flask 2 is three of those squares (3A) in a line. each is filled to the exact same height.

now flask 1 is going to have a pressure equal to the weight of the fluid in it divided by that square or

(Weight Hg) / (A)

flask 2 will have exactly 3 times as much fluid in it, but also 3 times as much area distributed across the bottom of the flask

(3 Weight Hg) / (3A)

which is exactly equal to the pressure in the first flask.

the area doesn't matter, only the height of the fluid above wherever you're measuring pressure.

------------

this would be MUCH easier to explain with a whiteboard

 

[slipstream99]

...

 

[DrShoshana]

Nevadanteater, that was an unbelievably excellent explanation. I understood it very well. THank you so much for clearing this up!

 

[amar21]

EK says that Pressure (for a fluid at rest) depends only on density and depth.

 

[jash0624]

let's imagine the bottom of flask 1 is a square (A) and the bottom of flask 2 is three of those squares (3A) in a line. each is filled to the exact same height.

now flask 1 is going to have a pressure equal to the weight of the fluid in it divided by that square or

(Weight Hg) / (A)

flask 2 will have exactly 3 times as much fluid in it, but also 3 times as much area distributed across the bottom of the flask

(3 Weight Hg) / (3A)

which is exactly equal to the pressure in the first flask.

the area doesn't matter, only the height of the fluid above wherever you're measuring pressure.

------------

this would be MUCH easier to explain with a whiteboard

Maybe I'm reading it wrong because of all this late night MCAT studying...but according to the question wouldn't it be (3 Weight Hg)/(2A)?

 

[halekulani]

The answer is B. THey explain it as an application of hydrostatic equilibrium and that the pressure at the bottom of both flasks is (Density of mercury*g*h). This doesn't make any sense to me because P=F/A meaning that if you increase the area by two then the pressure should decrease by half.

you could just think about it like if you're in the pacific ocean 200m under and in the atlantic ocean 200m under water...what is the pressure? pressure at a given distance below the water is always using the equation density*gravity*depth

 

[rushrules1]

Maybe I'm reading it wrong because of all this late night MCAT studying...but according to the question wouldn't it be (3 Weight Hg)/(2A)?

If you do the math it comes out correct.

P = F/A = Weight/A

W = rho*Volume*g ; Volume = A*h

The area cancels out ---> P = rho*g*h, P is the gauge pressure.

 

[bigserve99]

EDIT: Based on jash's point below, disregard this post


As already mentioned, the static fluid pressure in this case is only a function of the height and density of the fluid.

HOWEVER, based on this observation, I would argue the answer is C, NOT B.

rho(g)h is static Pressure. rho is equal, so really only the ratio of heights matters in this problem.

We know:
(A1)*(h1) = (V1)
(A2)*(h2) = (V2)

Based on the information given in the passage:
(V2) = 3*(V1)
(A2) = 2*(A1)

so, substitution gives:
(A2)*(h2) = 3 * (A1)*(h1)

substituting again gives:
2*(A1)*(h2) = 3 *(A1)*(h1)

A1's cancel leaving:
2*(h2) = 3*(h1) or (h2) = (3/2) h1

hence, P2 = (1.5) * P1 or answer C

I would double check to make sure this isn't an error in the book. However, I do believe that C is the correct answer. Anyone else agree?

~ Peace.

 

[jash0624]

As already mentioned, the static fluid pressure in this case is only a function of the height and density of the fluid.

HOWEVER, based on this observation, I would argue the answer is C, NOT B.

rho(g)h is static Pressure. rho is equal, so really only the ratio of heights matters in this problem.

We know:
(A1)*(h1) = (V1)
(A2)*(h2) = (V2)

Based on the information given in the passage:
(V2) = 3*(V1)
(A2) = 2*(A1)

so, substitution gives:
(A2)*(h2) = 3 * (A1)*(h1)

substituting again gives:
2*(A1)*(h2) = 3 *(A1)*(h1)

A1's cancel leaving:
2*(h2) = 3*(h1) or (h2) = (3/2) h1

hence, P2 = (1.5) * P1 or answer C

I would double check to make sure this isn't an error in the book. However, I do believe that C is the correct answer. Anyone else agree?

~ Peace.

This was my original thinking, but after rereading the question, I saw that they noted that both the containers are filled to the same height (h).

 

[bigserve99]

This was my original thinking, but after rereading the question, I saw that they noted that both the containers are filled to the same height (h).

Wow, I totally missed that. Good lookin out jash......yah. It's a pretty misleading problem in my opinion. Oh well.

 

[halekulani]

the dimensions of the container are irrelevant to the question.

 

[yallo]

Yea this problem is really misleading... I guess the only explanation for both their heights being the same is that flask 2 is sort of like a pyramid cut in half and then flipped upside down. Flask 1 would be rectangular. So though flask 2 has three times as much liquid inside but only twice the area of the bottom, it rises to the same height as the liquid in flask 1 because of the extra volume of the exapanding walls. So the pressure on the bottom that we're considering to answer this question is equivalent, since the force on the bottom is the liquid above. The rest of the liquid applies pressure on the expanding walls. I don't know if this made any sense to anybody, it would help to have a diagram.

 

[halekulani]

If you do the math it comes out correct.

P = F/A = Weight/A

W = rho*Volume*g ; Volume = A*h

The area cancels out ---> P = rho*g*h, P is the gauge pressure.

nailed it on the head people

P = F/A can be re-written using density * volume * gravity as your force
rewrite volume into Area * distance and Area cancels out

 

[yallo]

it would help to have a diagram.

http://www.umiacs.umd.edu/%7Ekuijt/dba105/Castillo/Pyr16.JPG

Something like that, but with smooth outside walls. So those outside walls are what I called "expanding walls."