Lesson Explainer: The Liquid Column Manometer Physics

In this explainer, we will learn how to describe the process of pressure measurement using the height of a liquid column in a U-shaped tube.

A liquid column manometer is a U-shaped tube filled with a liquid that is used to measure the pressure difference of gases on either side of it. A manometer open to the atmosphere filled with water is shown below.

The water level is equal on both sides of the tube. Due to this, the pressure on the left side, 𝑃L, is equal to the pressure on the right side, 𝑃R (i.e., when the height of the water on both the left and right sides is equal, so is the pressure): 𝑃=𝑃LR when β„Ž=β„Ž.LR

If one side of the tube is attached to a pressure different from the atmosphere, then the height of the liquid can change, as seen in the diagram below.

Both the unknown gas and the atmosphere are exerting pressure on the liquid, but now the pressure is different, so the height of the water on the sides of the tube has a difference, Ξ”β„Ž.

If β„ŽR is higher than β„ŽL, then it means there is less pressure pushing down on the water; thus, 𝑃R is less than 𝑃L: 𝑃>𝑃LR when β„Ž<β„ŽLR and vice versa.

Let’s look at some examples.

Example 1: Liquid Column Manometer Unknown Gas Comparison

The diagram shows a liquid column manometer connected at one end to a gas reservoir and at the opposite end to the atmosphere. Which of the following correctly relates the pressure of the gas and the pressure of the atmosphere, 𝑃gas and 𝑃atm?

  1. 𝑃>𝑃gasatm
  2. 𝑃<𝑃gasatm
  3. 𝑃=𝑃gasatm

Answer

𝑃gas is the pressure on the right side, coming from the gas reservoir. 𝑃atm is the pressure on the left side, coming from the atmosphere.

If 𝑃gas was larger than 𝑃atm, then the height of the liquid in the manometer would be pushed up to the left. If 𝑃atm was instead larger than 𝑃atm, then it would be pushed up to the right.

Since the height is equal, then it means the pressure on both sides is equal. The correct answer is C, 𝑃=𝑃gasatm.

Example 2: Liquid Column Manometer Uneven Unknown Gas Comparison

The diagram shows a liquid column manometer connected at one end to a gas reservoir and at the opposite end to the atmosphere. Which of the following correctly relates the pressure of the gas and the pressure of the atmosphere, 𝑃gas and 𝑃atm?

  1. 𝑃<𝑃gasatm
  2. 𝑃>𝑃gasatm
  3. 𝑃=𝑃gasatm

Answer

𝑃gas is the pressure on the right side, coming from the gas reservoir. 𝑃atm is the pressure on the left side, coming from the atmosphere.

Since the liquid column heights are different in the tube, we know that 𝑃gas cannot be equal to 𝑃atm. They are different, so one must be greater than the other.

If 𝑃atm was larger, the liquid in the column would be pushed up to the right, but the liquid there is not higher; it is lower. Therefore, 𝑃gas must be greater than 𝑃atm.

The correct answer is thus B, 𝑃>𝑃gasatm.

The proportion by which the pressure is different is related to the height difference. A larger difference in height means a larger difference in pressure. We can use a modified version of the equation for pressure in a column to express this.

Equation: Change in Pressure of a Liquid Column Manometer

The equation used to relate the change in pressure to change in height in a manometer is Δ𝑃=πœŒπ‘”Ξ”β„Ž, where Δ𝑃 is the difference in pressure, 𝜌 is the density of the fluid, 𝑔 is the force due to gravity (9.81 m/s2 for Earth), and Ξ”β„Ž is the difference in height.

We can see in this equation that Δ𝑃 is proportional to Ξ”β„Ž. Twice the change in height means twice the change in pressure.

Let’s solve a problem that uses this relation. A liquid column manometer contains water, the density of which we will take as 997 kg/m3. The left side is connected to a gas of unknown pressure, and the right side is open to the atmosphere at sea level, which we will take as 101.3 kPa. We can also take gravity as 9.81 m/s2.

We are looking for the unknown gas pressure on the left side. Let’s take a look at the equation for change in pressure of a column: Δ𝑃=πœŒπ‘”Ξ”β„Ž.

It is important to note that we only care about the magnitude of change in height, so this cannot be negative.

Similarly, the difference in pressure is also a magnitude. Expressed by itself, it could look like Δ𝑃=|π‘ƒβˆ’π‘ƒ|RL or Δ𝑃=|π‘ƒβˆ’π‘ƒ|.LR

We know that, due to its liquid height being lower, 𝑃>𝑃LR. This means that we can take the second form without the absolute value, since know it will be positive.

This makes the equation look like π‘ƒβˆ’π‘ƒ=πœŒπ‘”Ξ”β„Ž.LR

We want 𝑃L on one side, so let’s add 𝑃R to both sides to isolate it: π‘ƒβˆ’π‘ƒ+𝑃=πœŒπ‘”Ξ”β„Ž+𝑃𝑃=πœŒπ‘”Ξ”β„Ž+𝑃.LRRRLR

𝑃R is the atmosphere at sea level, 101.3 kPa. The other values are 𝜌 as 997 kg/m3, 𝑔 as 9.81 m/s2, and Ξ”β„Ž as 10 cm. Before putting in Ξ”β„Ž, let’s convert it into metres, remembering that there are 100 cm in a metre: 11001100Γ—10=0.1.mcmmcmcmm

So, 10 cm is 0.1 m. Putting these into the equation gives 𝑃=ο€Ή997/9.81/(0.1)+(101.3).LkgmmsmkPa

The right side has the metres partially cancel each other out to become the 1/m3 given by density. Substituting the numbers gives 𝑃=978Γ—+(101.3).LkgmskPa

We need to have the units on the right side match before we can add them together. Recall that pascals are newtons per square metre Β ο€Ή/Nm and that newtons are kilogram-metres per second squared Β ο€Ήβ‹…/kgms. We can convert the units we have to newtons per square metre as follows: kgmskgmsmkgmsmNmΓ—=×1οˆο€½Γ—ο‰ο€Ό1=, making our equation 𝑃=978+(101.3).LPakPa

Let’s make the units of pascal the same. There are 1β€Žβ€‰β€Ž000 Pa in 1 kPa: 11000,11000Γ—978=0.978.kPaPakPaPaPakPa

Adding them together thus gives 𝑃=0.978+(101.3)𝑃=(102.278).LLPakPakPa

Rounding, the unknown pressure on the left side is 102.3 kPa, a small difference. Sometimes, denser fluids are used in order to observe more dramatic changes in pressure.

Let’s look at an example.

Example 3: Liquid Column Manometer Gas Pressure Calculation

The diagram shows a liquid column manometer connected at one end to a gas reservoir and at the opposite end to the atmosphere. The U-shaped tube contains mercury that has a density of 13β€Žβ€‰β€Ž595 kg/m3. The top of the mercury column in contact with the atmosphere is vertically below the top of the mercury column in contact with the gas reservoir. The vertical distance between the column tops β„Ž=25cm. Find the pressure of the gas in the reservoir. Use a value of 𝑃=101.3kPa for atmospheric pressure.

  1. 33 kPa
  2. 68 kPa
  3. 101 kPa
  4. 105 kPa
  5. 135 kPa

Answer

To find the pressure of the gas in the reservoir, we can relate the change in height to the change in pressure between the reservoir and the side open to atmosphere.

Let’s look a the change in pressure in a fluid column equation: Δ𝑃=πœŒπ‘”Ξ”β„Ž.

We have to solve for the pressure of the gas reservoir, henceforth π‘ƒοŠ¨, and we already know π‘ƒοŠ§. Looking at the liquid heights, the one open to the atmosphere is lower, so 𝑃>π‘ƒοŠ§οŠ¨. Thus, the change in pressure will look like Δ𝑃=π‘ƒβˆ’π‘ƒ.

Putting this back into the equation, π‘ƒβˆ’π‘ƒ=πœŒπ‘”Ξ”β„Ž.

We want π‘ƒοŠ¨ on one side, so let’s add it to both sides: π‘ƒβˆ’π‘ƒ+𝑃=πœŒπ‘”Ξ”β„Ž+𝑃𝑃=πœŒπ‘”Ξ”β„Ž+𝑃.

We can then subtract πœŒπ‘”Ξ”β„Ž from both sides to isolate π‘ƒοŠ¨: π‘ƒβˆ’πœŒπ‘”Ξ”β„Ž=πœŒπ‘”Ξ”β„Ž+π‘ƒβˆ’πœŒπ‘”Ξ”β„Žπ‘ƒβˆ’πœŒπ‘”Ξ”β„Ž=𝑃.

We know the density, 13β€Žβ€‰β€Ž595 kg/m3, gravity, 9.81 m/s2, and Ξ”β„Ž, 25 cm. Let’s first convert the centimetres to metres so all the units are the same: 1100Γ—25=0.25.mcmcmm

We can then put our known values into the equation: π‘ƒβˆ’ο€Ή13595/9.81/(0.25)=𝑃.kgmmsm

Multiplying together gives π‘ƒβˆ’(33341)Γ—=𝑃.kgms

The units for this recent calculation are converted to pascals()newtonspersquaremetre, or Pa (N/m2), as follows: kgmskgmsmkgmsmNmΓ—=×1οˆο€½Γ—ο‰ο€Ό1=, making the equation π‘ƒβˆ’(33341)=𝑃.Pa

The pressure for the left side, π‘ƒοŠ§, was given in kilopascals, so let’s convert the value on the right side to that as well: 11000Γ—33341=33.3.kPaPaPakPa

π‘ƒοŠ§ is given as 101.3 kPa. We put this in and solve to get 101.3βˆ’(33.3)=68.kPakPakPa

The pressure of the gas reservoir is thus B, 68 kPa.

Sometimes, it is not an unknown pressure that needs to be found, but an unknown height. If the pressures are given, we can find the height.

Looking at just the equation Δ𝑃=πœŒπ‘”Ξ”β„Ž.

We can isolate Ξ”β„Ž by dividing both sides by πœŒπ‘”: Ξ”π‘ƒπœŒπ‘”=πœŒπ‘”Ξ”β„ŽπœŒπ‘”, canceling the right side to give Ξ”π‘ƒπœŒπ‘”=Ξ”β„Ž.

Just like change in pressure, we then need to determine how to order the heights so as not to have a negative change. In the diagram below, for example, it can clearly be seen that β„ŽR is greater than β„ŽL.

The change in height would thus look like Ξ”β„Ž=β„Žβˆ’β„Ž,RL though typically only the change in height is needed.

Let’s look at an example.

Example 4: Liquid Column Manometer Height Difference Calculation

The diagram shows a liquid column manometer connected at opposite ends to two gas reservoirs. The pressures of the gas reservoirs are 𝑃=123.3kPa and 𝑃=110.1kPa. The U-shaped tube contains an oil with a density of 1β€Žβ€‰β€Ž080 kg/m3. What is the vertical distance β„Ž between the tops of the oil columns? Give your answer to two decimal places.

  1. 22.05 m
  2. 12.22 m
  3. 11.65 m
  4. 1.25 m
  5. 0.62 m

Answer

Let’s start by manipulating the liquid column manometer equation to isolate Ξ”β„Ž: Δ𝑃=πœŒπ‘”Ξ”β„Ž.

We can get Ξ”β„Ž by itself by dividing both sides by πœŒπ‘”: Ξ”π‘ƒπœŒπ‘”=πœŒπ‘”Ξ”β„ŽπœŒπ‘”, which becomes Ξ”π‘ƒπœŒπ‘”=Ξ”β„Ž.

The value of Δ𝑃 is the difference between the pressures π‘ƒοŠ§ and π‘ƒοŠ¨. We want it to be positive, so since 𝑃>π‘ƒοŠ§οŠ¨, the change in pressure looks like Δ𝑃=π‘ƒβˆ’π‘ƒ.

π‘ƒοŠ§ is 123.3 kPa and π‘ƒοŠ¨ is 110.1 kPa, meaning that the change in pressure is 123.3βˆ’110.1=13.2.kPakPakPa

Since we will have a final answer in terms of metres, we want kilopascals, kPa to be in pascals, Pa. There are 1β€Žβ€‰β€Ž000 Pa in a kilopascal: 1000110001Γ—13.2=13200.PakPaPakPakPaPa

Now, we have all the variables we need to find the change in height. Density is 1β€Žβ€‰β€Ž080 kg/m3 and 𝑔 is 9.81 m/s2. The equation thus looks like 13200(1080/)(9.81/)=Ξ”β„Ž.Pakgmms

We need to relate the units of pascal to kilograms, metres, and seconds. The units of pascal are newtons per square metre Β ο€Ή/Nm, and newtons are kilogram-metres per second squared. Together, the conversion looks like Nmkgmsmkgmsmkgms=×1οˆο€½Γ—ο‰ο€Ό1=Γ—.

Putting all these units together gives 13200(1080)(9.81)ο€ΎΓ—Γ—Γ—Γ—οŠ=Ξ”β„Ž.kgmsmskgm

By canceling out the units in the numerator with the units in the denominator, the units greatly simplify to just metres: 13200(1080)(9.81)=Ξ”β„Ž.m

We then simplify to give Ξ”β„Ž=1.2458.m

Given to two decimal places, the answer is thus D, 1.25 m.

Let’s summarize what we have learned in this explainer.

Key Points

  • The equation used to relate change in pressure and change in height of a liquid column manometer is Δ𝑃=πœŒπ‘”Ξ”β„Ž. where Δ𝑃 is the difference in pressure, 𝜌 is the density of the fluid, 𝑔 is the force due to gravity, and Ξ”β„Ž is the difference in height.
  • In a liquid column manometer, the ratio of change in height is equal to the ratio of the change in pressure.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.