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Pressure at the Bottom

  1. Jun 10, 2009 #1
    1. The problem statement, all variables and given/known data
    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?


    2. Relevant equations
    denisty= m/v
    Pb=Pa+density*g*h


    3. 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?
     
  2. jcsd
  3. Jun 10, 2009 #2

    rock.freak667

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    Atmospheric pressure.
     
  4. Jun 10, 2009 #3

    LowlyPion

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    Well if the mass is the same, what is the pressure of ethyl alcohol at a depth of (8.53)1/3 ?
     
  5. Jun 10, 2009 #4
    pressure = F/A......F/8.53^2/3.....and the force is equal to mg?
     
  6. Jun 10, 2009 #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?
     
  7. Jun 10, 2009 #6

    LowlyPion

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    What's wrong with

    P = ρ*g*h

    where ρ in this case is the density of ethyl alcohol.
     
  8. Jun 10, 2009 #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.
     
  9. Jun 10, 2009 #8

    LowlyPion

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    I'd think that is a bit too much precision.
     
  10. Jun 10, 2009 #9
    another thing if the question is asking about water why would i be using the density of ethyl alcohol?
     
  11. Jun 10, 2009 #10

    LowlyPion

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    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: Jun 10, 2009
  12. Jun 10, 2009 #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.
     
  13. Jun 10, 2009 #12

    LowlyPion

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    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.
     
  14. Jun 10, 2009 #13
    I did and it came up wrong this is in Pascals right?
     
  15. Jun 10, 2009 #14

    LowlyPion

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    Well of course check your units. Do they want the answer as Pa or kPa?
     
  16. Jun 10, 2009 #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.
     
  17. Jun 10, 2009 #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?
     
  18. Jun 10, 2009 #17

    LowlyPion

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    That's correct. You should get something like 18,510 Pa = 1.8888*1000*9.8
     
  19. Jun 10, 2009 #18

    LowlyPion

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    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?
     
  20. Jun 10, 2009 #19

    LowlyPion

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    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.
     
  21. Jun 10, 2009 #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?
     
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