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Volume of a liquid in a container

  • Thread starter clope023
  • Start date
  • #1
987
123
1. Homework Statement

A tall cylinder with a cross-sectional area 12.0cm is partially filled with mercury; the surface of the mercury is 6.00cm above the bottom of the cylinder. Water is slowly poured in on top of the mercury, and the two fluids don't mix.

What volume of water must be added to double the gauge pressure at the bottom of the cylinder? (express volume in cubic centimeters)


2. Homework Equations

p = p0 + (rho)gh

pgauge = p - p0

(rho)Hg = .0136g/cm^3

(rho)water = 1g/cm^3

3. The Attempt at a Solution

V = 12cm^2(6cm) = 72cm^3 - incorrect

6cmHG = 6(.0136cm(water)) = 8.2cm^3 - incorrect

I'm not even sure how to approach this problem, any help is great appreciated.
 

Answers and Replies

  • #2
alphysicist
Homework Helper
2,238
1
clope023,

They are wanting the gauge pressure to double at the bottom of the cylinder due to adding water on top of the mercury. So what is the gauge pressure at the bottom when there was just mercury, and what is the gauge pressure at the bottom when there is mercury plus some water in the cylinder?
 
  • #3
987
123
clope023,

They are wanting the gauge pressure to double at the bottom of the cylinder due to adding water on top of the mercury. So what is the gauge pressure at the bottom when there was just mercury, and what is the gauge pressure at the bottom when there is mercury plus some water in the cylinder?
gauge pressure at that height when there's only mercury would be 8.7x10^3 Pa

gauge pressure at that height with the density of both water and mercury I got the same (measured in kg/m^3)
 
  • #4
alphysicist
Homework Helper
2,238
1
I think you have a wrong value for the density of mercury; it should be something like 13.6 g/cm^3 or 13600 kg/m^3.

From the first two of your relevant equations, you have pgauge=rho g h (assuming that the pressure p0 is atmospheric pressure at the top of the mercury). However, I don't think the numerical value you got for the gauge pressure is correct. What numbers did you use to get that result?

For the second gauge pressure, we have two liquids, with water resting on top of mercury. How does the formula need to be changed to incorporate two liquids with different densities?
 
  • #5
987
123
I think you have a wrong value for the density of mercury; it should be something like 13.6 g/cm^3 or 13600 kg/m^3.

From the first two of your relevant equations, you have pgauge=rho g h (assuming that the pressure p0 is atmospheric pressure at the top of the mercury). However, I don't think the numerical value you got for the gauge pressure is correct. What numbers did you use to get that result?

For the second gauge pressure, we have two liquids, with water resting on top of mercury. How does the formula need to be changed to incorporate two liquids with different densities?
okay I'll take you through the steps I did

p = p0 + (rho)gh

p = 1.013x10^5 Pa + (13.6x10^3kg/m^3)(g)(.0600m)Pa

p = 1.093x10^5 Pa

gauge pressure = 1.093x10^5 - 1.013x10^5 = 8x10^3 Pa

for the other gauge with 2 liquids I'm not sure how to do it since I don't know how much water is being applied, the one where I received the same as my first equation I just added the densities of the 2 liquids together.
 
  • #6
alphysicist
Homework Helper
2,238
1
That gauge pressure looks right.

For the case with two liquids, we are trying to solve for the height of the water. Remember that in the equation everytime you go down a depth h in a liquid of density rho the pressure rises by (rho g h). If you go down through two different liquids you'll get two terms, etc.

You know the value of the gauge pressure for the second case (because it was twice the mercury-only value), so the only unknown will be the height of the water.

Then there's only one more step to find the volume.
 

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