Question Regarding Bubble in Kerosene Tank

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Homework Help Overview

The problem involves an air bubble rising in a kerosene tank, where the volume of the bubble increases as it ascends. The question asks for the depth of the tank given the atmospheric pressure and the relationship between the densities of kerosene and mercury.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of Boyle's Law and the relationship between pressure and volume. There are questions about how to incorporate the density of kerosene and the meaning of atmospheric pressure measured in centimeters of mercury. Some participants express confusion regarding the conversion between the weights of kerosene and mercury.

Discussion Status

Participants are actively engaging with the problem, exploring different approaches to find the pressure at the bottom of the tank and how to relate it to the depth. Some guidance has been offered regarding pressure units and the relationship between densities, but there is still uncertainty about certain concepts.

Contextual Notes

There is mention of a pressure unit conversion and the specific weights of kerosene and mercury, which may be relevant to solving the problem. The original poster expresses a lack of understanding regarding mercuric barometers and the implications of the given atmospheric pressure.

modulus
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1. The problem.
Okay, here's the question:
'An air bubble situated at the bottom of an open kerosene tank rises to the top surface. It is observed that at the top, the volume of the bubble is thrice it's initial volume. If the atmospheric pressure is 72cm of Hg, and mercury is 17 times heavier than kerosene, then what is the depth of the tank?'



2. Homework Equations .
As far as I think, we'll be using Boyle's Law:
P'V'=P''V''
Pressure=Density(of Kerosene)*gravitational acceleration*depth(to where the bubble is)



The Attempt at a Solution


Well, it says that at the top, the volume of the bubble triples, so the pressure must become one-by-three of the initial(and I think we're supposed to consider the pressure exerted by the kerosene on the bubble). But, I don't understand how we're supposed to use the relation between the weights of kerosene an mercury to find the the pressure exerted by kerosene. A;so, what do they mean by '72 cm of Hg'...I've never really understood the concept of mercuric barometers.
Any help will be appreciated. Thank you.
 
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Looks like you can use the first formula to find the pressure at the bottom of the tank. The second can then be used to find the height.
There is a pressure unit conversion table here:
http://en.wikipedia.org/wiki/Pressure#Units
 
modulus said:
If the atmospheric pressure is 72cm of Hg, and mercury is 17 times heavier than kerosene …

P'V'=P''V''
Pressure=Density(of Kerosene)*gravitational acceleration*depth(to where the bubble is)

But, I don't understand how we're supposed to use the relation between the weights of kerosene an mercury to find the the pressure exerted by kerosene. A;so, what do they mean by '72 cm of Hg'...I've never really understood the concept of mercuric barometers.
Any help will be appreciated. Thank you.

Hi modulus! :smile:

As you say, pressure = ρkerosenegh.

They give you ρkeroseneHg, and "72cm of Hg" means the pressure at a depth of 72cm below the surface in Hg. :wink:
 
Yeess! I got it!
You guys have no clue how easy you've made this for me!
Thanks a lot!
 

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