Lesson Video: The Liquid Column Manometer | Nagwa Lesson Video: The Liquid Column Manometer | Nagwa

Lesson Video: The Liquid Column Manometer Physics

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

09:07

Video Transcript

In this lesson, we will learn how to describe the process of pressure measurement using the height of a liquid column in a U-shaped tube. Before we learn about this process, let’s first refresh our memory on how to find the pressure in a liquid.

We need to recall that when we’re looking for the pressure inside a fluid at different heights, we can use the formula 𝑃, the pressure of the fluid, is equal to 𝜌, the density of the fluid, times 𝑔, the acceleration due to gravity, times β„Ž, the height of the fluid above the point we are considering. Going deeper into our liquid is the same thing as saying that the height of the liquid above us is increasing, which means that the pressure of the liquid is also increasing.

We can connect this to the fact that the pressure at the bottom of a swimming pool is greater than the pressure near the top of a swimming pool. The height of the water above us when we’re at the bottom of the swimming pool is much greater than the height of the water above us when we’re near the top. Let’s apply what we just recalled about the pressure in a liquid to a manometer.

A manometer is a U-shaped tube filled with a liquid that is used to measure the pressure of a gas. How does a manometer help us measure the pressure of a gas? Let’s look at a manometer when the left tube and the right tube are both open to the atmosphere. In both tubes of the manometer, the atmosphere will apply a pressure to the liquid. In the left tube, we’ll call that pressure 𝑃 subscript 𝐿. In the right tube, we’ll call that pressure 𝑃 subscript 𝑅.

We can use the equation for the pressure of a liquid 𝑃 equals πœŒπ‘”β„Ž that we discussed earlier to compare the pressures in both of the tubes. We can see that the liquid column in both of the tubes is at the same height, which tells us that the β„Ž is the same for both the left and the right tubes, where we measure the β„Ž in both tubes from the same arbitrarily chosen vertical position. We also know that we are using a manometer on Earth. Therefore, acceleration due to gravity in both the left and the right tube is going to be the same.

And our manometer has been filled with same liquid in both the left and the right tube. Therefore, the density of the fluid is also gonna be the same in the left and right tube. Since all three of our variables happen to be the same in both the left and the right tube, this would imply that the pressure is the same in both the left and the right tube. This should make sense as both of our ends are open to the atmosphere.

In general, when we have a manometer, assuming that the liquid is the same throughout, when the height of the liquid in the left tube is equal to the height of the liquid in the right tube, then the gas pressure that the left tube is attached to is equal to the gas pressure that the right tube is attached to.

What happens if they are both not open to the atmosphere? Let’s look at an example where we attach our left tube to a gas chamber of unknown gas pressure. And our right tube will be open to the atmosphere such that the height of the left column is lower than the height of the right column with the difference between the two heights being Ξ”β„Ž. If the liquid on the right rises to a higher height than it does on the left tube, then this would imply that the liquid feels a greater push from the pressure in the left tube than it does from the pressure in the right tube.

Or another way to say that, the pressure of the gas attached to the left tube is greater than the pressure of the gas attached to the right tube. But by how much is the pressure greater? To determine that, we have to come back to our equation for the pressure of a liquid is equal to πœŒπ‘”β„Ž. The difference in the pressure of our gases will be equal to the density of the liquid times the acceleration due to gravity times the difference in height that the liquid goes in each tube.

In general, when we have a manometer that is filled with a liquid that is the same throughout, and the height of the liquid in the left tube is less than the height of the liquid in the right tube, then we can say that the pressure of the gas attached to the left tube will be greater than the pressure of the gas attached to the right tube.

In our final example, we’ll attach our left tube to a different unknown gas pressure such that the height of the liquid in the left tube is gonna be greater than the height of the liquid in the right tube with the difference in the height of the liquids in each tube being Ξ”β„Ž. Because the height of the liquid in the left tube is greater than the height of the liquid in the right tube, we can say that the pressure in the right tube is gonna be greater than the pressure in the left tube. This is because the push from the pressure in the right tube must be greater than the push from the pressure in the left tube.

But how much bigger? Once again, we apply the pressure in a liquid is equal to πœŒπ‘”β„Ž. And just as we saw last time, the difference in the pressure of the gases is equal to the density of the liquid times the acceleration due to gravity times the change in height of the liquid in each of the tubes. In general, when we have a manometer that is filled with a liquid that is the same throughout, and the height of the liquid in the left tube is greater than the height of the liquid in the right tube, then we can say that the pressure of the gas attached to the left tube is less than the pressure of the gas attached to the right tube.

In a manometer, the preferable liquid is mercury. This is because mercury is very dense and we don’t need to use tall columns to measure significant pressure differences. From a practical point of view, mercury also has a low evaporation rate, which means that it’s easy to work with without having to lose any of the material. If we don’t have access to mercury or we’re working with gases that have low pressure differences, we can use either oil or water as our liquid in the manometer. Manometers are widely used because they’re very simple.

Now that we’ve learned how a manometer can be used to find differences in pressure and compare our pressures to each other, let’s go over two examples.

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, 𝑃 gas and 𝑃 atmosphere? (A) 𝑃 gas is equal to 𝑃 atmosphere. (B) 𝑃 gas is greater than 𝑃 atmosphere. (C) 𝑃 gas is less than 𝑃 atmosphere.

We need to remember that a manometer is a U-shaped tube filled with a liquid that is used to measure the pressure of a gas. For this problem, one end of the tube is opened up to the atmosphere, and the other end is connected to a gas reservoir. The atmosphere applies a pressure 𝑃 atmosphere onto the liquid in one side of the tube. On the other side of the tube, the gas from the gas reservoir also applies a pressure onto the liquid. We can compare the pressure of the atmosphere to the pressure of the gas by looking at the height that the liquid goes to on either end of the tube.

We need to recall that the pressure of a fluid 𝑃 is equal to the density of the fluid 𝜌 times the acceleration due to gravity 𝑔 times the height of the fluid β„Ž. If we wanna compare the two pressures, we need to analyze the height that the liquid goes to on either end of the tube. Because the liquid rises to the same height on either end of the tube, this implies that both ends of the tube apply the same amount of pressure. Or put another way, the pressure that the gas applies on the liquid is gonna be equal to the pressure that the atmosphere applies on the liquid. Therefore, we can say answer choice (A) is correct. The pressure of the gas from the gas reservoir is equal to the pressure from the atmosphere.

Next, we’ll try an example where our liquid is not going to the same height in either end of the tube.

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, 𝑃 gas and 𝑃 atmosphere? (A) 𝑃 gas equals 𝑃 atmosphere. (B) 𝑃 gas is less than 𝑃 atmosphere. (C) 𝑃 gas is greater than 𝑃 atmosphere.

We need to recall that a manometer is a U-shaped tube filled with a liquid that is used to measure the pressure of a gas. In a manometer, the pressure on one side, in this case, the pressure of the atmosphere, pushes down on the liquid on one end of the tube. And the pressure on the other end of the tube, in this case, the pressure from the gas reservoir, pushes down on the liquid on the other end. The relative pressure of the gases will determine what height the liquid will go to on either end of the tube.

We can see in our diagram that we have a difference in height between the liquid on one end of the tube to the liquid on the other end of the tube. In this case, the height of the liquid that’s in the tube attached to the atmosphere is lower than the height of the liquid in the tube that’s attached to the gas reservoir. This implies that the pressure the atmosphere puts on the liquid provides a greater push than the pressure the gas reservoir puts on the liquid. Or another way to put it would be that the pressure of the atmosphere is greater than the pressure of the gas in the gas reservoir. Therefore, we can say that when we compare the pressure of the gas in the gas reservoir to the pressure of the atmosphere that the pressure of the gas is going to be less than the pressure of the atmosphere or answer choice (B).

Key Points

We saw that a liquid manometer works on the principle of different pressures being exerted on the liquid in each side of the tube. The ratio of the column heights in the U-shaped tube is equal to the ratio of the pressures on top of the two columns.

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