A diver 50 m deep in 10°C fresh water exhales a 1.0-cm-diameter b...

Question

A diver 50 m deep in 10°C fresh water exhales a 1.0-cm-diameter bubble. What is the bubble's diameter just as it reaches the surface of the lake, where the water temperature is 20°C?

Accepted Answer

Hey, everyone. Let's go through this practice problem. A group of researchers prepared a spherical shaped inflatable weather balloon filled with helium gas at a temperature of 15 degrees Celsius and a pressure of 150.98 atmospheres making a radius of three m. As the balloon reaches the highest altitude, the temperature drops to negative 60 degrees Celsius and the atmospheric pressure is at 600.10 atmospheres determine the radius of the weather balloon at that altitude. We're given four multiple choice options. Option, a 50 m, option B 13 m, option C six m and option D three m. First off know that since we're looking for the radius of the weather balloon, the spherical weather balloon, you want to have a relationship on hand between the radius of the balloon and its geometry, we know it's a sphere. So we can recall that the volume of the sphere is equal to four thirds multiplied by pi multiplied by the cube of the radius of the sphere. So that's the relationship we're gonna want to use to ultimately solve for the radius. However, since solving for the radius, using this equation requires that we first know the volume. We first need to find what volume the balloon expands to before we can solve for the radius. R. Let's first find the initial volume of the balloon using the volume equation. I'm gonna call that V sub one in the sequel to four thirds multiplied by pi multiplied by the cube of the first radius. The initial radius the balloon has which we're told is three m. So four thirds pi multiplied by the cube of three m. If we put that into a calculator, you find a volume of about 113.97 cubic meters. Now, in order to find the final volume that the volume expands to, we can compare the variables of the initial pressure and the initial temperature and the initial volume to the final versions of those variables using the combined ideal gas law which states that the pressure of a gas multiplied by its volume divided by its temperature are always constant. In other words, P one V one divided by T one is equal to P two V two, divided by T two. So we're looking for the final volume V two. So we're going to algebraically solve this for the final volume. This is a pretty simple algebraic task. And what we find is that V sub two is equal to P sub one V, sub one multiplied by T sub two, all divided by P sub two T sub one. So all we have to do now is plug in the variables that we have that were given the problem. So for P sub one, we're told that the initial pressure is 10.98 atmospheres and the V sub one, the volume is what we've just found as 13 0.97 cubic meters. And then this is multiplied by T sub two, the final temperature. And we're told that the final temperature is negative 60 degrees Celsius. But in order to get temperature to work with the ideal gas law, we need to convert this into Kelvins. So, and to convert from Kelvins, we just add 273. So negative 60 degrees plus 273 and all that's Kelvin's. We take this whole thing and then divide it by the final pressure which we're told by the problem is 0.10 atmospheres multiplied by the initial temperature, which we're told is 15 degrees Celsius. Once again, we have to add 273 to this for it to be converted into Kelvins. And what we find from a calculator is that the final volume is about 97.2 cubic meters. And now that we have the final volume, we can use the volume of the sphere formula that we discussed earlier to solve for the radius. So once again, let's go back to our volume equation. The volume is equal to four thirds pi are cubed we're going to solve this for R. So let's first isolate R cubed on its own algebraically by dividing both sides of the equation by four thirds. Pi we find that R cubed is equal to three V three multiplied by the volume divided by four pi. And I should add subscripts to make it clear that we're now looking at the final variables because we're solving for R two in terms of V two. And finally, to solve for our sub two completely on its own, we take the cubed root of both sides of the equation. So R sub two is equal to the cubed root of three V sub two divided by four pi. So all we have to do is plug into this equation. The value or V sub two that we found earlier. So R sub two is equal to the cube root of three multiplied by 8197.2 cubic meters divided by four pi. And if we put this into a calculator, then we find a radius of about 12. m. However, if we look at the multiple choice options that were given to us, all of them are rounded to only two significant figures. So if we round this answer to two significant figures, then the radius is approximately 13 m. So that then is our answer to this problem. And if we look at the options we've given, we can see that Option B agrees with what we wrote. Option B says 13 m. So that means that option B is the correct answer to this problem and that's all for now. I hope this video helped you out. If you need more practice. Please check out some of our other videos which will give you more experience with these types of problems, but that's all for now and I hope you all have a lovely day. Bye bye.

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A diver 50 m deep in 10°C fresh water exhales a 1.0-cm-diameter bubble. What is the bubble's diameter just as it reaches the surface of the lake, where the water temperature is 20°C?

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