Question Video: Calculating the Height of a Rising Air Bubble Physics • Second Year of Secondary School

Video Transcript

A rising air bubble in a column of water has a volume at the base of the column and a greater volume near the top of the column, as shown in the diagram. Find the height of the bubble above the base of the column when it has a volume 1.55 times greater than that at the base of the column. Use a value of 101 kilopascals for atmospheric pressure and a value of 1000 kilograms per meter cubed for the density of water. The water temperature is uniform. Give your answer to one decimal place.

In this question, we’re thinking about a bubble of air that rises from the bottom of a column of water. We need to calculate the height of the bubble above the base of the column when its volume is 1.55 times larger than its initial volume. To answer this question, we will need to use the ideal gas law for a gas of constant mass. 𝑃 one 𝑉 one divided by 𝑇 one is equal to 𝑃 two 𝑉 two divided by 𝑇 two, where 𝑃 one, 𝑉 one, and 𝑇 one are the initial pressure, volume, and temperature of the gas and 𝑃 two, 𝑉 two, and 𝑇 two are the final pressure, volume, and temperature of the gas.

Since we’re told that the temperature of the water is uniform, we know that 𝑇 one must be equal to 𝑇 two. This means that the temperature terms cancel, leaving us with a simpler expression. 𝑃 one 𝑉 one is equal to 𝑃 two 𝑉 two. Let’s start by working out these quantities for the bubble.

First, let’s think about the bubble’s initial pressure, 𝑃 one. When the bubble is at the bottom of the water column, it will experience pressure from two sources. It’ll experience a pressure due to the weight of the water above it, which we’ll call 𝑃 𝑤. And it will also experience atmospheric pressure, 𝑃 𝑎, which pushes down on all of the water in the column. So, the value of 𝑃 one will equal 𝑃 𝑎 plus 𝑃 𝑤. We are told that atmospheric pressure, 𝑃 𝑎, has a value of 101 kilopascals.

Recall that the pressure due to a fluid is given by the formula 𝑃 is equal to 𝜌𝑔ℎ, where 𝜌 is the density of the fluid, 𝑔 is the gravitational field strength, and ℎ is the height of the fluid above the point at which the pressure is being calculated. So, the pressure exerted by the water on the air bubble, 𝑃 𝑤, is equal to the density of the water, 𝜌 𝑤, multiplied by the gravitational field strength 𝑔 multiplied by the height of the water above the bubble. Since the bubble is at the base of the column, this is equal to the column’s height, which we’ll label with a capital 𝐻. So, the pressure exerted on the bubble when it is at the base of the column is given by 𝑃 one is equal to 𝑃 𝑎 plus 𝜌 𝑤 𝑔𝐻.

We aren’t told the volume of the bubble at this point, so we’ll just label it as 𝑉 one. As the bubble rises upwards through the column, the height of water above it decreases. This means that the pressure exerted by the water on the bubble decreases. And so, the volume of the bubble increases. The diagram shows us the bubble when it has reached the very top of the column, which means there is a negligible amount of water above it. So, the only significant pressure that the bubble experiences is atmospheric pressure, 𝑃 𝑎. So, we know that 𝑃 two is equal to 𝑃 𝑎. Although we aren’t given an exact value for the bubble’s volume, we do know that it’s 1.55 times greater than its volume at the base of the column. So, we can once again note that 𝑉 two is equal to 1.55 times 𝑉 one.

Now we have worked out expressions 𝑃 one, 𝑃 two, and 𝑉 two, we can substitute these into our simplified ideal gas equation to get 𝑃 𝑎 plus 𝜌 𝑤 𝑔𝐻 all multiplied by 𝑉 one is equal to 𝑃 𝑎 multiplied by 1.55 𝑉 one.

To answer this question, we need to find the height of the bubble above the base of the water column. Since the bubble is so near the top of the column, this is equivalent to finding the height of the column, 𝐻. To do this, we need to rearrange this equation to make 𝐻 the subject. First, we can divide both sides of the equation by 𝑉 one and cancel out these terms. This leaves us with the equation 𝑃 𝑎 plus 𝜌 𝑤 𝑔𝐻 is equal to 1.55𝑃 𝑎. Next, we can subtract 𝑃 𝑎 from both sides. This leaves us with 𝜌 𝑤 𝑔𝐻 is equal to 1.55𝑃 𝑎 minus 𝑃 𝑎, which simplifies to 𝜌 𝑤 𝑔𝐻 is equal to 0.55𝑃 𝑎. Finally, we can divide both sides of the equation by 𝜌 𝑤 𝑔 to get our expression for 𝐻. 𝐻 is equal to 0.55𝑃 𝑎 divided by 𝜌 𝑤 𝑔.

Now, we just need to substitute in the values of 𝑃 𝑎, 𝜌 𝑤, and 𝑔. We know that atmospheric pressure, 𝑃 𝑎, is equal to 101 kilopascals. We have to be careful with units here. This value is given to us in units of kilopascals, but we need it in the SI unit of pascals. To convert from kilopascals to pascals, we simply multiply the value by 1000. So, our atmospheric pressure 𝑃 𝑎 is equal to 101000 pascals. We are told that the density of water, 𝜌 𝑤, is equal to 1000 kilograms per meter cubed, and we can recall that the gravitational field strength of Earth, 𝑔, is equal to 9.8 meters per second squared. Substituting these values into our expression for 𝐻, we find that 𝐻 is equal to 0.55 times 101000 pascals divided by 1000 kilograms per meter cubed times 9.8 meters per second squared.

Completing this calculation, we find that 𝐻 is equal to 5.668 meters. Rounded to one decimal place, this is equal to 5.7 meters. So, we have found that the bubble reaches a height of 5.7 meters above the base of the column. This is the final answer to this question.