Calculating Pressure at the Bottom of a Tank Using Density and Volume

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In summary, to find the volume of water with the same mass as 8.58 m3 of ethyl alcohol, the density of water needs to be multiplied by the volume and divided by the density of ethyl alcohol. To find the pressure at the bottom of the cubic tank, the equation Pb=Pa+density*g*h can be used, with the height being the cube root of the volume. The atmospheric pressure can be used for Pa.
  • #1
talaroue
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Homework Statement


What volume of water has the same mass as 8.58 m3 of ethyl alcohol?

If this volume of water is in a cubic tank, what is the pressure at the bottom?


Homework Equations


denisty= m/v
Pb=Pa+density*g*h


The Attempt at a Solution



I found the first part by setting denisty=m/v solving for m=Vrho. Then plugged that in for water in the same equation and got 6.7782 m^3. For the second part though, I thought I could use Pb=Pa+density*g*h. and the height i found by taking the 6.7782^(1/3). But would the Pa be zero since its at the bottom if not what do I use for the Pa?
 
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  • #2
Atmospheric pressure.
 
  • #3
talaroue said:

Homework Statement


What volume of water has the same mass as 8.58 m3 of ethyl alcohol?

If this volume of water is in a cubic tank, what is the pressure at the bottom?

Homework Equations


denisty= m/v
Pb=Pa+density*g*h

The Attempt at a Solution



I found the first part by setting denisty=m/v solving for m=Vrho. Then plugged that in for water in the same equation and got 6.7782 m^3. For the second part though, I thought I could use Pb=Pa+density*g*h. and the height i found by taking the 6.7782^(1/3). But would the Pa be zero since its at the bottom if not what do I use for the Pa?

Well if the mass is the same, what is the pressure of ethyl alcohol at a depth of (8.53)1/3 ?
 
  • #4
pressure = F/A...F/8.53^2/3...and the force is equal to mg?
 
  • #5
I don't understand this step at all. So it wants the pressure at the bottom of the tank of water. Since its a cube you can find out that the sides are 1.89x1.89x1.89. since its at the bottom that gives the height 1.89, density of 1000, and gravity is 9.8. So that gives yous Pb=Pa+19404 correct?

But then Pa=Fa/A. The area is 1.89*1.89, but then is the force just mg?
 
  • #6
talaroue said:
pressure = F/A...F/8.53^2/3...and the force is equal to mg?

What's wrong with

P = ρ*g*h

where ρ in this case is the density of ethyl alcohol.
 
  • #7
That woudl me that...

P=790kg/m^3*9.8 m/s^2*2.04 m
P=15818.65341 Pa, which is wrong according to the program.
 
  • #8
talaroue said:
That woudl me that...

P=790kg/m^3*9.8 m/s^2*2.04 m
P=15818.65341 Pa, which is wrong according to the program.

I'd think that is a bit too much precision.
 
  • #9
another thing if the question is asking about water why would i be using the density of ethyl alcohol?
 
  • #10
talaroue said:
another thing if the question is asking about water why would i be using the density of ethyl alcohol?

OK. Use water of a different dimension cube.
I think I was just cutting out the middleman.

Ooops. I wasn't I forgot about the cube root of the dimension. Sorry. Not a short cut at all.
 
Last edited:
  • #11
so use the volume of water to find the dimension? and then use density of water to find out the pressure?

I tried what you gave me and i rounded to the tenth and it still didn't work.
 
  • #12
talaroue said:
so use the volume of water to find the dimension? and then use density of water to find out the pressure?

I tried what you gave me and i rounded to the tenth and it still didn't work.

You're right. I corrected myself. I forgot that the dimension is cube rooted not linear.

So your method of

P = ρgh

using water and the 1.8888 depth is the way to go.
 
  • #13
I did and it came up wrong this is in Pascals right?
 
  • #14
talaroue said:
I did and it came up wrong this is in Pascals right?

Well of course check your units. Do they want the answer as Pa or kPa?
 
  • #15
It doesn't specifiy which one the want but we when work out this problem the way we did we get N/m^2 which is Pa.
 
  • #16
A 1.03 m diameter vat of liquid is 5.47 m deep. The pressure at the bottom of the vat is 1.79 atm. What is the mass of the liquid in the vat?

Pb=density*gravity*height
V*Pb/density*g*h
density can be put in m/v and Pb can be turned to Pa. then just solve for the mass right? Is there something wrong with the way I am rearranging this equation for both of these problems?
 
  • #17
talaroue said:
It doesn't specifiy which one the want but we when work out this problem the way we did we get N/m^2 which is Pa.

That's correct. You should get something like 18,510 Pa = 1.8888*1000*9.8
 
  • #18
talaroue said:
A 1.03 m diameter vat of liquid is 5.47 m deep. The pressure at the bottom of the vat is 1.79 atm. What is the mass of the liquid in the vat?

Pb=density*gravity*height
V*Pb/density*g*h
density can be put in m/v and Pb can be turned to Pa. then just solve for the mass right? Is there something wrong with the way I am rearranging this equation for both of these problems?

In this one you know the depth and gravity and pressure in atm that you convert to Pa, then solve directly for ρ don't you and then figure the mass from there?
 
  • #19
You can save a step by noting that it is a column so ...

F = m*g = P*A

So m = P * A / g

where A is the area of the bottom.
 
  • #20
That anwser isn't coming out right. and neither is this one... Why does it seem i am doing all the work right but the anwser is wrong?
 
  • #21
still having a bit of a problem
 
  • #22
talaroue said:
still having a bit of a problem

Show how you are calculating it.
 
  • #23
talaroue said:
A 1.03 m diameter vat of liquid is 5.47 m deep. The pressure at the bottom of the vat is 1.79 atm. What is the mass of the liquid in the vat?

Pb=density*gravity*height
V*Pb/density*g*h
density can be put in m/v and Pb can be turned to Pa. then just solve for the mass right? Is there something wrong with the way I am rearranging this equation for both of these problems?
What about atmospheric pressure?
 
  • #24
I attached it

EDIT: that answer i came up with is wrong
 

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  • #25
Doc Al said:
What about atmospheric pressure?

P(below)= P(above)+density*gh

P(above) would that be considered atmospheric pressure?
 
  • #26
Am i wrong in assuming that volume is of cylinder?
 
  • #27
talaroue said:
P(below)= P(above)+density*gh

P(above) would that be considered atmospheric pressure?
That's what I would assume.
 
  • #28
Sorry I confused things earlier by dropping the atmospheric pressure. It is needed here for a static liquid.
 
  • #29
How would I calculate that? Force=Pressure/Area
 
  • #30
LowlyPion said:
Sorry I confused things earlier by dropping the atmospheric pressure. It is needed here for a static liquid.

No worries, you have helpped me a lot! I apperiate it.
 
  • #31
talaroue said:
How would I calculate that? Force=Pressure/Area
Calculate what? You are given Pb and you know Pa.
 
  • #32
how do i know Pa already?<<<<<<This would solve all my problems?
 
  • #33
talaroue said:
how do i know Pa already?<<<<<<This would solve all my problems?
See post #27. That's what we've been talking about.
 
  • #34
I understand the equation, but i don't know how to come up with a number for Pa, to plug into the equation.
 
  • #35
talaroue said:
I understand the equation, but i don't know how to come up with a number for Pa, to plug into the equation.
Hint: What's atmospheric pressure in units of atm? :-p

(You can also express atmospheric pressure in standard units.)
 

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