Calculate the height of a water column and expressions for P

In summary, In the experiment, a tube was used to measure the density of an unknown liquid. The height of the water column was found to be twice that of the liquid, and the pressure difference between points A and C was 2.5*103Pa. There is a relationship between PCand PD, and the density of water is 1000 kg/m3.
  • #1
Richie Smash
293
15

Homework Statement



The apparatus shown may be used to determine the density of an unknown liquid X.
The tube has a uniform cross section and the height of the water column is twice that of Liquid X.
A and B are the same horizontal level. The pressure difference between points A and C is 2.5*103Pa.

(i) Calculate the height of the water column.
(ii) Express PB-PD in terms of PA and PC
[Hint: the answer can be deduced without resorting to any further calculation]
(iii)What is the relationship between PCand PD?
(Density of water = 1000 kg m-3; g = 10 m/s)

Homework Equations


Pressure =ρgh

The Attempt at a Solution


Well I know the pressure at A would be : P=ρgh and in this case the height would be the water column's height as A is at the bottom.

But they do not give me a pressure, only a pressure difference...

I could write, P=pghA+pghB...
I'm stuck i think.
 

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  • #2
Actually I did this, Pressure difference= pghA-pghC

and actually since hA-hC is the height itself, this can simply be rewritten as

2.5*103= pgh
And now I can substitute in my values for density and gravity that were given and got 0.25m is the height of the column.

I'm hoping this is correct but for parts (ii) and (iii) I'm not entirely sure.
 
  • #3
Yes, since g and rho (density) are constants, you are right to just cancel them out on LHS and RHS.

And also, since only the atmosphere is "pushing down" on points D and C, they experience about the same pressure. Hence Pc = Pd
 
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  • #4
I'm not entirely sure that's what they are asking for, because they ask write it in terms of Pa and Pc and Since D is lower than C, I don't think they are equal...

If Pa-Pc= pghA-pghC
then since the other column is half of that,
My best guess for an expression would be : (pghA)/2-(pghC)/2
= pghwater/2

can anyone confirm this??
 
  • #5
notice that they are different liquids; you cannot compare based on height alone if both liquids are different in density :) if the pressures are different at D and C, then the side of higher pressure will naturally be pushed down until there is an equilibrium. Forces must be in equilibrium.
 
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  • #6
Oh yes I understand what you mean so I know the answer for part (iii) .

But I still don't know if part (i) was correct and I don't know how to write part (ii) Pd-Pb in terms of Pa and Pc...
 
  • #7
Yes, your part (i) is correct;

so now you know that since A and C are the same height, they are at the same pressure, since the level of mercury on either side is the same (so no force "imbalance"). Since you understand now that B and D are the same pressure, you can express all of these in pressure.
 
  • #8
Umm I'm still not entirely sure on this, you are saying that the pressure of both liquids is the same, because the mercury level is constant?
 
  • #9
Richie Smash said:
Umm I'm still not entirely sure on this, you are saying that the pressure of both liquids is the same, because the mercury level is constant?
Yes. If the pressure at those points are not equal, the mercury level will be higher on one side to "compensate"
 
  • #10
Right I'm following up to this point, but this is my reasoning , if atmospheric pressure decreases as you go higher, and increases as you go lower, why is the atmospheric pressure at D and C the same if one is lower? Sorry this is my weakest area in physics I really struggle with pressure
 
  • #11
Richie Smash said:
Right I'm following up to this point, but this is my reasoning , if atmospheric pressure decreases as you go higher, and increases as you go lower, why is the atmospheric pressure at D and C the same if one is lower? Sorry this is my weakest area in physics I really struggle with pressure

Ah! The thing is, at such a small height difference, we ignore the difference in atmospheric pressure since it is rather insignificant. We will take this into account when your setup is absolutely massive (think about the size of a mountain). Otherwise, there is no need to worry as air is really not that dense to matter for pressure here.
 
  • #12
ok but I'm still confused as to this part.
They're asking for an algebraic expression of Pb-Pd in terms of Pa-Pc

I know that : Pa -Pc = pghA-pghC= 2.5x103Pa

If so the expression would be pg (hA)/2-pg (hB)/2 = 2.5x103Pa? =Pb-Pd
 
  • #13
Richie Smash said:
ok but I'm still confused as to this part.
They're asking for an algebraic expression of Pb-Pd in terms of Pa-Pc

I know that : Pa -Pc = pghA-pghC= 2.5x103Pa

If so the expression would be pg (hA)/2-pg (hB)/2 = 2.5x103Pa? =Pb-Pd
Don't worry, let's take it slow.

You know that pressure at A = pressure at B

You also know that pressure at C = pressure at D

Hence,
Pa = Pb

Pc = Pd

Hope this can give you a hint ;)
 
  • #14
Ok I just realized something, I have to find a way to get density out of the equation to express this, because the density of the water and density of liquid X are note the same.
 
  • #15
Richie Smash said:
Ok I just realized something, I have to find a way to get density out of the equation to express this, because the density of the water and density of liquid X are note the same.
Here's another hint:

The heights are different too! So my suggestion is that since you know that pressures at a couple of points are equivalent, you could sub them in directly
 
  • #16
so basically Pa-Pc= Pb-Pd?

I guess it's right but I thought it wuld have been more to it
 
  • #17
Richie Smash said:
so basically Pa-Pc= Pb-Pd?

I guess it's right but I thought it wuld have been more to it

Bingo! Fortunately, some problems have solutions simpler than expected :)
 

Related to Calculate the height of a water column and expressions for P

1. How do I calculate the height of a water column?

To calculate the height of a water column, you can use the equation h = P/ρg, where h is the height in meters, P is the pressure in Pascals, ρ is the density of water in kg/m³, and g is the gravitational acceleration in m/s². Simply plug in the values for P, ρ, and g to find the height.

2. What is the significance of the density of water in the equation?

The density of water is a crucial factor in calculating the height of a water column because it determines how much pressure the water exerts at a given depth. The higher the density, the more pressure is exerted, and thus the height of the water column will be greater.

3. How does the gravitational acceleration affect the height of the water column?

The gravitational acceleration, represented by the symbol g, plays a significant role in determining the height of the water column. The higher the value of g, the greater the force of gravity pulling the water down, resulting in a taller water column. However, g is typically considered constant for most purposes.

4. Can the equation be used for other fluids besides water?

Yes, the equation h = P/ρg can be used for other fluids besides water, as long as the appropriate values for density and gravitational acceleration are used. However, the equation is specifically derived for water, so it may not be as accurate for other fluids.

5. What are the units for pressure in the equation?

The units for pressure, P, in the equation h = P/ρg are in Pascals (Pa). However, other units such as atmospheres (atm) and millimeters of mercury (mmHg) can also be used as long as they are consistent with the units used for density and gravitational acceleration.

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