Pressure, Density: Water & Oil in Glass Tube

In summary, the side where the oil is sitting suggests it is a U tube open to the air at both ends, but that was not stated.
  • #1
Lui4
22
2
Homework Statement
A glass tube of cross-sectional area 0.00013 is partially filled with water. An oil with a density of 820 is slowly poured into the tube so that it does not mix with the water and floats on top of it. The height of the oil above the water's surface is 3 cm.

(a) By how much does the pressure change at a depth of 10 cm below the water's surface after the oil is added? (On the side where the oil is sitting)?
Relevant Equations
Hi I would please like some help on this question. I don't exactly know how to solve it with the information given. I have attached my work down below. Any help would be great, thanks in advance!
1B93BA5E-4E75-4065-B231-38B484D99A00.jpeg
 
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  • #2
"On the side where the oil is sitting" suggests it is a U tube open to the air at both ends, but that was not stated. Did you use that interpretation?
 
  • #3
oh thats a good point. I drew a little u tube but I don't see what to do from here. Because in this case we are only focusing on one side of the tube anyways so it wouldn't matter about the other side since they ask us about the pressure of only one side where the oil was added no? If you can see what to do could you please give me a solid advice to help me please
 
  • #4
Because I understood that we have two pressure points and they both have different heights and that is why i tried separating the two apart and I used Patm for the oil since that is what the atmospheric pressure is going to be applied to since I'd assume the u tube is open from the side the oil was put in
 
  • #5
I cannot understand your working.
You wrote an equation P2+(pressure due to 10cm water)= (atmospheric pressure)+(pressure due to 3cm oil), which is not going to give you the answer to the question.
Then you wrote and solved P2*(pressure due to 10cm water) = (atmospheric pressure)+(pressure due to 3cm oil). The + turned into *?

I cannot connect any of that with "increase in pressure at 10cm depth in the water due to adding the oil."
 
  • #6
okay yeah I see I've made a couple mistakes. If my work doesn't make sense, sure, fair. But could you then instead comment on what I should be looking to do?
 
  • #7
Lui4 said:
okay yeah I see I've made a couple mistakes. If my work doesn't make sense, sure, fair. But could you then instead comment on what I should be looking to do?
It asks for a change in pressure, so find the pressure before, the pressure after, and take the difference.
 
  • #8
I think that's pretty self-explanatory. I understand what to do I just don't see how I am supposed to do it and that's why I am asking for help on the calculation aspect of it. Lets start with the first thing you said. Lets find the pressure before. Okay great, now this is without the oil being added. I know the pressure= hpg and pressure = F/A and we are given a cross-sectional area in the question and told about the 10cm below the waters surface. What should I be thinking from here?
 
  • #9
Lui4 said:
pressure= hpg
Right, except you should add atmospheric pressure if you want absolute pressure. But here it is fine to work with gauge pressure.
Which of those variables do you know the values of?

(Note: always bear in mind you may have been given some irrelevant data.)
 
  • #10
haruspex said:
Right, except you should add atmospheric pressure if you want absolute pressure. But here it is fine to work with gauge pressure.
Which of those variables do you know the values of?
so we are told that we want the pressure below the water at 10cm for before and after so we know h and g is a constant and p we also know since its the density of the water so when you multiple these values that's where the 980 comes from no? ( 0.1 * 1000 * 9.8)= 980
 
  • #11
Use P=rho.g.h and see.
 
  • #12
haruspex said:
(Note: always bear in mind you may have been given some irrelevant data.)
Yeah I am assuming that data is the area value they gave us in this case
 
  • #13
To determine the change in pressure at a depth of 10 cm below the water's surface after the oil is added, we need to consider the hydrostatic pressure due to both the water and the oil.
The pressure in a fluid at a given depth can be calculated using the equation:

P = ρgh

where P is the pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth.

First, let's calculate the pressure at a depth of 10 cm below the water's surface without the oil:

P_water = ρ_water * g * h_water
= ρ_water * g * (h_oil + h_water)
= ρ_water * g * (0.10 m + 0.03 m)
= ρ_water * g * 0.13 m
ASK ANY QUESTION BEFORE I CONTINUE WITH CALCULATION. LET ME KNOW WHERE YOU HAVENT UNDERSTOOD
 
  • #14
Larrywriter said:
To determine the change in pressure at a depth of 10 cm below the water's surface after the oil is added, we need to consider the hydrostatic pressure due to both the water and the oil.
The pressure in a fluid at a given depth can be calculated using the equation:

P = ρgh

where P is the pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth.

First, let's calculate the pressure at a depth of 10 cm below the water's surface without the oil:

P_water = ρ_water * g * h_water
= ρ_water * g * (h_oil + h_water)
= ρ_water * g * (0.10 m + 0.03 m)
= ρ_water * g * 0.13 m
ASK ANY QUESTION BEFORE I CONTINUE WITH CALCULATION. LET ME KNOW WHERE YOU HAVENT UNDERSTOOD
Yeah this is what I did too but the only difference is, if this is before the oil then we wouldn't add the 0.03m right? since that is the height contributing to when the oil is added?
 
  • #15
Lui4 said:
Yeah this is what I did too but the only difference is, if this is before the oil then we wouldn't add the 0.03m right? since that is the height contributing to when the oil is added?
Apologies for the confusion. You are correct. The height of the oil, 0.03 m, should only be considered when calculating the pressure after the oil is added. Therefore, when calculating the pressure without the oil, we should not include the height of the oil.

Let's recalculate the pressure at a depth of 10 cm below the water's surface without the oil:

P_water = ρ_water * g * h_water= ρ_water * g * h_water= ρ_water * g * 0.10 m

Now, let's calculate the change in pressure at the depth of 10 cm below the water's surface after the oil is added:
To be continued.......
 
  • #16
Larrywriter said:
Apologies for the confusion. You are correct. The height of the oil, 0.03 m, should only be considered when calculating the pressure after the oil is added. Therefore, when calculating the pressure without the oil, we should not include the height of the oil.

Let's recalculate the pressure at a depth of 10 cm below the water's surface without the oil:

P_water = ρ_water * g * h_water= ρ_water * g * h_water= ρ_water * g * 0.10 m

Now, let's calculate the change in pressure at the depth of 10 cm below the water's surface after the oil is added:
To be continued.......
No worries and okay sure I will be patiently waiting for your response, thank you!
 
  • #17
where P is the pressure, ρ is the density of water, g is the acceleration due to gravity, and h is the depth;

ΔP = P_water - P_oil= ρ_water * g * 0.10 m - ρ_oil * g * 0.03 m

Substituting the given values:

ρ_water = 1000 kg/m³ (density of water)ρ_oil = 820 kg/m³ (density of oil)g = 9.8 m/s² (acceleration due to gravity)

ΔP = (1000 kg/m³) * (9.8 m/s²) * (0.10 m) - (820 kg/m³) * (9.8 m/s²) * (0.03 m)= 980 kg/(m·s²) - 240.12 kg/(m·s²)= 739.88 kg/(m·s²)

Therefore, the pressure decreases by 739.88 Pascal (Pa) at a depth of 10 cm below the water's surface after the oil is added.

Hope i helped you
 
  • #18
I appreciate your efforts and all the time you took to write that out and make sure it was nicely written in a way that's easy to understand and all those units you put. But for whatever reason that is not the correct answer. Thank you for trying to be of help, its greatly appreciated!
 
  • #19
Trial and error, ! which is the answer?
your are welcome though
 
  • #20
When the question says "10cm below the water level", does it mean 10cm below a(n imaginary) line drawn on the tube where the water level is before the oil is poured in ? or are we talking a consistent 10cm below the water level, regardless of whether there's oil on top, or not.

(ps: in your diagram, you've written what looks like "241.08", buried in the middle of a bunch of other numbers : yes, that would be the answer for a straight tube, closed on the bottom).
 
  • #21
Lui4 said:
so we are told that we want the pressure below the water at 10cm for before and after so we know h and g is a constant and p we also know since its the density of the water so when you multiple these values that's where the 980 comes from no? ( 0.1 * 1000 * 9.8)= 980
Yes, I get that's where the 980 came from, but it seemed you multiplied that by P2(?!)
 
  • #22
haruspex said:
Yes, I get that's where the 980 came from, but it seemed you multiplied that by P2(?!)
yeah I didn't mean to do that but I knew that number is correct. Could you please tell me how to calculate the pressure after adding the oil? Because then you just take the difference of those two and I think the problem is solved.
 
  • #23
hmmm27 said:
When the question says "10cm below the water level", does it mean 10cm below a(n imaginary) line drawn on the tube where the water level is before the oil is poured in ? or are we talking a consistent 10cm below the water level, regardless of whether there's oil on top, or not.

(ps: in your diagram, you've written what looks like "241.08", buried in the middle of a bunch of other numbers : yes, that would be the answer for a straight tube, closed on the bottom).
I don't have a clarifying answer. We are just given these questions as homework. So Im not too sure. I would assume its 10cm below the water regardless of the oil or not

Update. The correct answer was 241 but how?? Like do we just not consider before the oil was added. since to get 241 as the answer its just as simple as p =hpg and solve it using h= 0.03 p=820 and g=9.8
 
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  • #24
Lui4 said:
Because in this case we are only focusing on one side of the tube anyways so it wouldn't matter about the other side since they ask us about the pressure of only one side where the oil was added no?
No. What's happening to the other side might affect what's happening on this side. Try making a sketch of the situation. That's where I would start.
 
  • #25
Lui4 said:
The correct answer was 241 but how??
The contribution from the atmosphere does not change, and the contribution from the water (10cm in both cases) does not change, so….
 
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  • #26
Larrywriter said:
ΔP = P_water - P_oil
This is wrong I guess.
##\Delta p=p_f-p_i##
##p_f :## Pressure in the mentioned depth after adding oil.
##p_i :## Pressure in the mentioned depth before adding oil.
So we should find pressure at a depth of 10cm below water's surface, before and after adding oil.
Larrywriter said:
Therefore, the pressure decreases
We've added oil so pressure should increase.
 
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  • #27
Lui4 said:
I don't have a clarifying answer. We are just given these questions as homework. So Im not too sure. I would assume its 10cm below the water regardless of the oil or not

Update. The correct answer was 241 but how?? Like do we just not consider before the oil was added. since to get 241 as the answer its just as simple as p =hpg and solve it using h= 0.03 p=820 and g=9.8
There is certain value of static pressure at a point that is submerged 0.1 m from the surface of the body of water.
There is certain value of static pressure at a point that is submerged 0.03 m from the surface of the body of oil.
Calculate those two values first.

The weight of 0.03 m of oil added on the surface of the water only adds up to that calculated value of static pressure at a point that is still submerged 0.1 m from the surface of the body of water.

Considering 0.13 m from the surface of the water-oil column only creates confusion, as you are dealing with a mix of densities and heights.

The added oil increases static pressure in our summerged point in 241 N/m2 or a 124 %.
 
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1. What is pressure?

Pressure is the amount of force applied per unit area. In other words, it is the measure of how much force is pushing on a certain area.

2. How is pressure related to density?

Pressure and density are directly related. As the density of a substance increases, the pressure it exerts also increases. This is because there are more particles in a given area, resulting in a greater force being applied.

3. How does water and oil behave in a glass tube?

Water and oil have different densities, with water being more dense than oil. When placed in a glass tube, the water will sink to the bottom while the oil will float on top. This is due to the principle of buoyancy, where less dense substances float on top of more dense substances.

4. Why does the level of water and oil change when pressure is applied to the glass tube?

When pressure is applied to the glass tube, the volume of the tube decreases. This causes the water and oil to become more compressed, resulting in a decrease in their volume as well. As a result, the level of the water and oil in the tube will rise.

5. How does the behavior of water and oil in a glass tube relate to real-life applications?

The behavior of water and oil in a glass tube is a demonstration of how substances with different densities behave when placed in a confined space. This principle is used in many real-life applications, such as in hydraulic systems, where liquids with different densities are used to create pressure and move objects.

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