# Pressure in a closed system using an inclined manometer

• Dylan.Wallett
In summary: Pa = (n - Δn)RT/VSolving for n, we get:n = (100 kPa V)/(RT)Now, we can plug in this value of n into our previous equation for Δn, and solve for the change in air density (Δρair):Δρair = Δn/V Plugging in the values we know, we get:Δρair = (100 kPa - 130 kPa)/(RT)Now, we can use this value of Δρair to find the new positions of the mercury menisci. We can use the equation PA - PB = ρmerc g (ΔL) sin σ, but this time, we will use the new
Dylan.Wallett

## Homework Statement

In the figure attached the pressure at A and B are the same 100 kPa. If water is introduced at A to increase p_A to 130 kPa find the new positions of the mercury menisci. The connecting tube has a diameter of 10 mm. Assume no change in Liquid density.

## Homework Equations

P A - P B = ρmerc g (Δ L) sin σ

## The Attempt at a Solution

30 000 = (13600)(9.81) (Δ L) (0.26)
ΔL = 0.868
h = 0.225 m

Intuitively this answer should be wrong because if there is an increase in pressure at A it should transfer through out the system to create an equilibrium. There would be no change in the mercury menisci if the air did not compress.

I think I need to work out a change in air density to get the correct answer. I am also not sure why they mentioned the diameter of the connecting tube.

Would you mind helping me get on the correct track.

Thank you

#### Attachments

• Static Pressure problem.jpg
15.1 KB · Views: 994
for your post! From what I understand, the system described in the problem is a U-tube manometer, with a liquid (mercury) on one side and air on the other. When water is introduced at A, it will cause the pressure at A to increase, but the pressure at B will remain the same. This will result in a pressure difference between A and B, causing the mercury levels to change.

To solve this problem, you can use the equation you mentioned: PA - PB = ρmerc g (ΔL) sin σ. However, in this case, the density of the liquid (ρmerc) will remain constant, as no change in liquid density is specified in the problem. The important thing to note is that the pressure difference (PA - PB) is now equal to 30 kPa, since PA has increased from 100 kPa to 130 kPa.

Now, let's consider the air on the other side of the U-tube. The air is compressed due to the increase in pressure at A, and this will result in a change in air density. To determine the new positions of the mercury menisci, we will need to consider the new pressure and density of the air at both points A and B.

To find the new air pressure at A, we can use the ideal gas law: PA = nRT/V, where n is the number of moles of air, R is the ideal gas constant, T is the temperature, and V is the volume. Since the volume of air at A is the same as before (the connecting tube diameter is given, so we can calculate the volume), and the temperature remains constant, the only variable that changes is the number of moles of air. This number will decrease due to the compression, but we don't know the exact amount. However, we do know that the pressure at A has increased by 30 kPa, so we can set up an equation:

130 kPa = (n - Δn)RT/V

where Δn is the change in the number of moles of air. We can rearrange this equation to solve for Δn:

Δn = n - (130 kPa V)/(RT)

Now, we can use this value of Δn to find the new air pressure at B. Since the pressure at B is the same as before (100 kPa), we can set up a similar equation:

100 k

## 1. What is pressure in a closed system?

Pressure in a closed system refers to the amount of force exerted by a gas or liquid on the walls of a container. It is measured in units of pressure, such as Pascals (Pa) or pounds per square inch (psi).

## 2. How is pressure measured in a closed system?

In a closed system, pressure can be measured using a device called a manometer. This device consists of a tube filled with a liquid, typically mercury, and the pressure is determined by the height of the liquid in the tube.

## 3. What is an inclined manometer?

An inclined manometer is a type of manometer that uses a U-shaped tube filled with a liquid to measure pressure. The tube is inclined at an angle, which allows for more accurate measurements of small pressure differences.

## 4. How does an inclined manometer work?

The inclined manometer works by comparing the difference in the height of the liquid in the two arms of the tube, which is caused by the pressure difference between the two ends. By measuring this difference, the pressure in the closed system can be determined.

## 5. Why is an inclined manometer used instead of a regular manometer?

An inclined manometer is used instead of a regular manometer because it provides more precise measurements for small pressure differences. The angle of the tube allows for a larger change in height for a smaller change in pressure, making it more sensitive and accurate for measuring pressure in closed systems.

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