Question Regarding Bubble in Kerosene Tank

  • Thread starter Thread starter modulus
  • Start date Start date
  • Tags Tags
    Bubble Tank
AI Thread Summary
An air bubble in a kerosene tank rises and triples in volume as it reaches the surface, prompting a question about the tank's depth. Boyle's Law is relevant for relating pressure and volume changes of the bubble. The atmospheric pressure is given as 72 cm of Hg, which needs to be converted to understand the pressure exerted by kerosene. The discussion clarifies that the density ratio of kerosene to mercury is essential for calculating the pressure at the bubble's initial depth. Participants express gratitude for the clarity provided in understanding the pressure concepts related to the problem.
modulus
Messages
127
Reaction score
3
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.
 
Physics news on Phys.org
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!
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top