Change in fluid levels in a manometer given a pressure change

[rubytuesday]

I have a test tomorrow and I've tried this problem a million times and it won't come out! I got it right when I turned it in, but I cannot for the life of me remember what I did! Help!

Homework Statement

A manometer using oil (density 0.86 g/cm^3) as a fluid is connected to an air tank. Suddenly the pressure in the tank increases by 0.83 cm Hg.
a) How much does the fluid level rise in the side of the manometer that is open to the atmosphere?
b) What would your answer be if the manometer used mercury (density = 13.6 g/cm^3) instead?

Homework Equations

P₂= P₁ + $$\rho$$gh

The Attempt at a Solution

I've tried a couple of ways:
$$\rho$$gh = P₂ - P₁
Assuming that P₁ is atmospheric pressure, so = 10^5 Pa
(0.00086 kg)(9.8 m/s)(h) = (0.00086)(9.8 m/s)(0.0083 kg/m^2*) - 10^5
= -1.19 e7 m so -1.19 e5 cm? definitely very wrong.

*not sure about this conversion

I kind of remember using the mercury with the equation at some point, so I tried playing with that but it didn't come out then, either. Also, I think the gravity is supposed to cancel out somewhere.

P_Hg = (13.6 g/cm³)(9.8 m/s)(h) = P_oil = (0.86)(9.8)(0.83)
Gravity cancels out, so
13.6h = (0.86)(0.83)
h = 19.05 m
Then I thought maybe I could plug that in somewhere, but I can't figure out where.

I'm frustrated and freaking out. Help!

 

[anathema]

I understand your frustration and anxiety about the upcoming test. Let's take a step-by-step approach to solving this problem.

First, let's review the formula for pressure in a fluid: P = ρgh, where P is pressure, ρ is density, g is the acceleration due to gravity, and h is the height of the fluid column.

a) We are given the density of oil (ρ = 0.86 g/cm^3) and the change in pressure in the tank (ΔP = 0.83 cm Hg). We can convert the change in pressure to units of cm H2O by using the conversion factor 1 cm Hg = 13.6 cm H2O. This gives us ΔP = 0.83 cm Hg x (13.6 cm H2O/1 cm Hg) = 11.3 cm H2O.

Now, we can set up the equation P1 + ρgh = P2, where P1 is atmospheric pressure (10^5 Pa), P2 is the pressure in the tank, and h is the height of the fluid column in the manometer. Substituting in the values, we get:

10^5 Pa + (0.86 g/cm^3)(9.8 m/s^2)(h) = P2

To solve for h, we need to rearrange the equation to isolate h:

h = (P2 - 10^5 Pa)/(0.86 g/cm^3)(9.8 m/s^2)

Now, we can plug in the given value for ΔP to solve for h:

h = (11.3 cm H2O)(1 m/100 cm)(1 cm^3/0.86 g)(1 g/1000 kg)(1 kg/9.8 m/s^2) = 0.0013 m = 0.13 cm

Therefore, the fluid level in the manometer will rise by 0.13 cm.

b) To solve for the height of the fluid column if the manometer used mercury (density = 13.6 g/cm^3), we can use the same equation P1 + ρgh = P2 but with the new value for density:

10^5 Pa + (13.6 g/cm^3)(9.8 m/s^2)(h) = P2

Again

 

[thomate1]

Hello, it sounds like you have been working hard on this problem and I can understand your frustration. Let's take a step back and break down the problem.

First, let's define our variables:
P1 = initial pressure (atmospheric pressure)
P2 = final pressure (pressure in the tank)
\rho = density of the fluid (given as 0.86 g/cm^3 for oil and 13.6 g/cm^3 for mercury)
g = acceleration due to gravity (9.8 m/s^2)
h = change in fluid level in the manometer (what we are trying to find)

Now, let's look at the equation you have correctly identified:
P2= P1 + \rhogh

This is known as the hydrostatic equation and it relates pressure (P) to density (\rho), gravity (g), and height (h). We can use this equation to solve for h.

a) For the first part of the problem, we are given that the pressure in the tank (P2) increases by 0.83 cm Hg. We know that the density of oil is 0.86 g/cm^3, so we can plug these values into the equation:
P2 = (0.86 g/cm^3)(9.8 m/s^2)(h) + P1
0.83 cm Hg = (0.86 g/cm^3)(9.8 m/s^2)(h) + 10^5 Pa
Note: we need to convert cm Hg to Pa in order to have consistent units.
Solving for h, we get:
h = (0.83 cm Hg - 10^5 Pa)/(0.86 g/cm^3)(9.8 m/s^2)
h = 0.0000083 m = 0.0083 cm

Therefore, the fluid level in the side of the manometer that is open to the atmosphere will rise by 0.0083 cm.

b) For the second part of the problem, we are asked to consider what would happen if the manometer used mercury instead of oil. We can use the same equation, but with the density of mercury (13.6 g/cm^3):
P2 = (13.6 g/cm^3)(9.8 m/s^2)(h) + P1
0.83 cm Hg = (13.6 g