A scuba diver releases a 3.60-cm-diameter (spherical) bubble o...

Question

A scuba diver releases a 3.60-cm-diameter (spherical) bubble of air from a depth of 14.0 m. Assume the temperature is constant at 298 K, and that the air behaves as an ideal gas. Sketch a PV diagram for the process. Take the density of water to be 1000 kg/m³.

Accepted Answer

Welcome back, everyone in this problem. An underwater researcher releases an oxygen bubble with a diameter of five centimeters from a depth of about 20 m beneath sea level. The temperature remains constant at around 25 °C assume that oxygen behaves like an ideal gas draw of pressure volume diagrams for this process, using the density of sea water as 1025 kg per cubic meter and atmospheric pressure as 101,325 pascals. Now, here we have our diagram, our pressure on our Y axis or vertical axis and our volume on our horizontal axis. And we are supposed to draw a PV diagram for our process. But to do that, we need to understand then how the pressure and the volume of the oxygen bubble relate to each other. In other words, how they change as it arises from a depth of 20 m to the surface. So if we can figure out that relationship and use it to plot points, then we will be able to draw our line that represents this process. OK. So first let's make a note of all the information that we have. Now so far, we know that we are at a depth of 20 m. OK? And at that depth, the diameter of the bubble, let's call that D 20 is going to be five centimeters or 0.05 m. We also know that our temperature remains constant at 25 °C. OK. And the density of sea water is 1025 kg per cubic meter. While the atmospheric pressure, OK, is 101,325 pascals. Now how can we use this information to figure out both the pressure and the volume at varying depths? Well, let's start with the pressure. Recall that to calculate the pressure at various depths. Ph OK. Or that pressure is equal to the atmospheric pressure multiplied by the product of the density multiplied by acceleration due to gravity multiplied by or multiplied by which represents the depth. So we have a formula to calculate the pressure at varying depths. Now, how about our volume? Well, we can use the ideal gas law, we also know that the ideal gas law tells us that the product of the pressure and the volume is constant given for we have a constant temperature. Ok. So that means then that our pressure and volume at any given height is going to be equal to the product of the pressure and volume at a depth of 20 m. OK. Since we already have a formula to find the pressure at any given height, we could rearrange this formula to find the volume at any given height, which tells us then that that volume is gonna be equal to P 20 V 20 divided by ph. Now, if we're going to use this formula, we'll need to know our pressure and volume at a depth of 20 m. So let's find the both of those. Now, at 20 m, we know that our volume V 20 is going to be equal to four thirds of pi because we're talking about a bubble here. So it's we're assuming that the bubble is spherical and its volume is gonna be four thirds of pi multiplied by its radius squared. So in this case, R 20 cubed, sorry, not squared cubed. Now what's the radius of our bubble at 20 at a depth of 20 m when we know its diameter is five centimeters. So that means this radius must be 2.5 centimeters. So now we can rewrite this as four thirds of pi multiplied by 0.025 m cubed. And when we calculate here, then we should get the volume of our bubble at a depth of 20 m to be 6.54 multiplied by 10 to the negative fifth cubic meters. What about our pressure? Well, we can use our formula for pressure. Yeah. So from our formula, we also know that the pressure at a depth of 20 m is gonna be equal to 101 300 sorry 101,325 pascals plus our density 1025 kg per cubic meter multiplied by acceleration due to gravity 9.8 m per second squared multiplied by our depth of 20 m. OK? And now when we go ahead and solve here, then we get P 20 to be equal to 303,325 pascals. So now that we have our values for our volume and pressure at a depth of 20 m, we have those constants. Then we have all the information, we need to find the varying volume and pressure at different heights. So what I'm about to do now, let's quickly, let's just set up a table. Ok. Let's set up a quick table. And in this table, we're going to have our depth in meters. We're going to have our pressure, ok? In units of 10 to the fifth pascals. And we're going to have our volume in units of 10 to the negative fourth cubic meters. And now this table is going to help us to calculate our varying pressures and volumes so that we can plot our graph. So in other words, this table is going to help us find our points. OK? And now for this table, let me stop it here. Let's use different depths of increments of 5 m. So you have 0 m 5 m 10 m 15 m and 20 m now, for each of these depths, what are, what is going to be the pressure and the volume? Well, let's first start with a height of 0 m and I'm just going to do this just to show you how we can think about it. And then you can find the rest of corresponding pressures and volumes on your own. Ok. So at a height of 0 m, we can first find the pressure P zero by substituting in our formula for pressure. Remember we said it's going to be equal to the atmospheric pressure. 101,325 pascals plus our density 1025 kg per cubic meter multiplied by our acceleration due to gravity multiplied by our height of 0 m. So we replace it with zero. When we go ahead and solve the pressure at 0 m is gonna be 101 325 sorry 101,325 pascals which again makes sense because at 0 m, we are at the surface or we are at sea level. Now we're going to write it to two significant figures. So that's going to be one multiplied by 10 to the fifth pascals approximately. OK. Now what about our volume at this point? Well, now remember we said that our volume at any particular height equals the pressure at 20 m multiplied by the volume at 20 m divided by the pressure at that height. OK. So in this case, I don't want to write out our values for P 20 V 20 you already know what they are, you know, just to save some time, but we're going to divide both of those by this pressure of 101,325 pascals. And when we do that, we get it to equal 0.0001957 or uh up to written to two significant figures and in units of 10 to the negative fourth cubic meters, it would be approximately two multiplied by 10 to the negative fourth cubic meters. OK. So this is going to be our volume and pressure at 0 m. So now we can go ahead and put those into our table. So that's gonna be one multiplied by 1/10 of the fifth pascals and two multiplied by 1/10 of the negative fourth cubic meters. Now to find the rest of corresponding pressures and volumes for our varying depth, we can repeat the process, substitute our value for into the formula for pressure to find the pressure at that depth. And then substitute that value for the pressure into the formula for the volume to find the volume at that depth. So in other words, at H equals 5 m, again, we're gonna substitute five anywhere we see H and go ahead and solve and get our values. Now, I'd recommend you pause this video and you try to figure out what those values are on your own. Ok? But I'm gonna go ahead and write what I found, ok? And when you calculate these are the values that you should get, ok? Because at a depth of 1.5 m, you should find that the pressure is 1.5 multiplied by 10 to the fifth pascals and the volume is 1.3 multiplied by 10 to the negative fourth cubic meters at 10 m. It should be two and one at 15 m. It should be 2.5. And let me just fix that here. It should be 2.5 and 0.8. And then at 20 it should be three and 0.7. So these are the values that you've probably found or you should find for these depths know that we have our corresponding pressures and volumes, we can plot our points. So if I come up here, then as we said, for a pressure of one multiplied by 10 to the fifth pascal, the volume is two multiplied by 10 to the negative fourth cubic meters at 5 m. For 1.5 the volume is 1.3. So it's probably going to be somewhere here at two. The volume is going to be one. OK. Oh, sorry. At a pressure of two, the volume is going to be one at a pressure of 2.5. The volume is gonna be 0.8. So it's approximately gonna be somewhere here. And then for a volume, a pressure of three, the volume is 0.7. So now we have our points, we can go ahead and plot. So now if we go ahead and draw our curve here, you notice that it forms a curve, then it should look something like this. I'm gonna try and draw my curve a bit better. OK. But it should look something like that. Let me give it one more try. Sometimes when you draw your curve, it might be a little off, but it should look something like that. I remember it. It might not be exact because I'm just estimating where the points should be. But this is the diagram of the pressure, the, the PV diagram. Sorry for our process. Thanks for watching everyone. I hope this video helped.

A scuba diver releases a 3.60-cm-diameter (spherical) bubble of air from a depth of 14.0 m. Assume the temperature is constant at 298 K, and that the air behaves as an ideal gas. Sketch a PV diagram for the process. Take the density of water to be 1000 kg/m³.

Key Concepts

Ideal Gas Law

Hydrostatic Pressure

PV Diagram

Watch next

A scuba diver releases a 3.60-cm-diameter (spherical) bubble of air from a depth of 14.0 m. Assume the temperature is constant at 298 K, and that the air behaves as an ideal gas. Sketch a PV diagram for the process. Take the density of water to be 1000 kg/m3.

Key Concepts

Ideal Gas Law

Hydrostatic Pressure

PV Diagram

Watch next

A scuba diver releases a 3.60-cm-diameter (spherical) bubble of air from a depth of 14.0 m. Assume the temperature is constant at 298 K, and that the air behaves as an ideal gas. Sketch a PV diagram for the process. Take the density of water to be 1000 kg/m³.