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Ch 18: A Macroscopic Description of Matter

Chapter 18, Problem 18

A diving bell is a 3.0-m-tall cylinder closed at the upper end but open at the lower end. The temperature of the air in the bell is 20°C. The bell is lowered into the ocean until its lower end is 100 m deep. The temperature at that depth is 10°C. a. How high does the water rise in the bell after enough time has passed for the air inside to reach thermal equilibrium?

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Hello, fellow physicists today, we're to solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. An open bottom container of height 2.5 m is submerged in the water of a lake to provide a temporary refuge for a scientist studying the lake's marine life. The container is initially filled with air at an initial temperature of 30 degrees Celsius. The closed upper part reaches a depth of 30 m. What is the height of the water in the container will reach once the air inside has reached a constant temperature of 20 degrees Celsius. So below the problem right here, we have a diagram to represent to help us visualize or represent the problem. So a visualization of the problem, we're also given some multiple choice answers. Let's read them off and see what our final answer might be. A is 6.5 millimeters B is 1.2 m C is centimeters and D is 1.9 m. OK. So first off, let us make the following assumptions. Let's assume that the number of moles is constant before and after the submersion of the container, also, let us assume that air behaves like an ideal gas. So we can recall and use the ideal gas Law equation which states that the pressure multiplied by the volume is equal to the number of moles multiplied by the universal gas constant multiplied by the temperature. So let's begin by writing down all of our known variables. So for, for P I, the initial pressure since the container starts out as just gas inside the container. So initially, it's just gas inside the container. It's just the atmospheric pressure which the numerical value for that is 1.13 multiplied by 10 to the fifth power pascals. And our initial temperature T I was degrees Celsius, but we need to convert degrees Celsius to Kelvin. So to do that, it's really easy, it's just the degrees Celsius. So it's just 30 plus 2 73 which is our conversion factor equals 303. Kelvin. Not too bad. And then to find our initial volume, all it is, is just the area multiplied by the length where A I should say is the cross sectional area. Let's be specific. So it's the cross sectional area of the container. And L is the initial height of the container which that is provided to us in the pro which is 2.5 m. So that's the initial height of the container. OK. So to find the final pressure, we have to do a little bit of work here. So the final pressure, we have to consider the condition because at this point, the final pressure of the container is submerged at this point. So we need to also take into account the density of water. So let's do that. So P F so the final pressure plus row, which is the density of water multiplied by gravity multiplied by lower case H which in this case, lower case H is the value is the height value that we're trying to determine. That's the unknown one is equal to the initial pressure plus plus row. The density of water multiplied by gravity multiplied by capital H which capital H is shown to us here in the diagram. So the capital H is 32.5 m and that takes into account the closed upper part reaches the depth of 30 m plus the height of the open bottom container. So it's the total height. That's what the capital H is. So when we rearrange that this equation, let's call it equation one. So when we rearrange equation one, just to solve for P F, the final pressure, we get that the final pressure is equal to the initial pressure plus the density of water multiplied by gravity multiplied by the total height minus the height value that we're trying to determine. So capital H minus lowercase H OK. And let's note here off to the side that 1000 kilograms per meters cubed is the density of water. So row equals this value. OK. OK. So now we need to determine what the final temperature is. So the final temperature was 20 degrees Celsius, but we need to convert that to Kelvin. So like before just you take it to temperature and degrees Celsius, which is 20 plus 2, 73 equals 293 Kelvin. OK. And like before, we need to find the final volume mean like like we found the initial volume, but it's slightly different. It's the cross sectional area multiplied by the length minus the height value that we're trying to find. So A is the cross sectional area of the container and L minus H or L minus lowercase H I should say is the height of the air column. OK. So we can also say really quick here that the pressure up points M and N as shown in our diagram are the same since they're on the same horizontal line. So we can say that P subscript capital M equals P equals subscript capital N. So the pressure at points M and N are equal to each other. OK. OK. So now we need to use the ideal gas law and we need to rewrite it to consider both the initial and final conditions. So that when we do that, the equation that we're gonna get is, is the initial pressure multiplied by the initial volume divided by the initial temperature is equal to the final pressure multiplied by the final volume divided by the final temperature. OK. OK. So now we could plug in all of our known variables into this equation. Let's call it equation two. OK. So let's do that. Let's plug in all of our known variables. So it's gonna be a lot of writing. So bear with me. OK. So let's start with an initial pressure, which is just the atmospheric pressure, which is 1.13, multiplied by 10 to the fifth power past scales multiplied by the volume, which is 2.5 m multiplied by a divided by the initial temperature which was 303. Kelvin. OK. So now let's do our final pressure which using our final pressure equation that we found out together is 1. three multiplied by 10 to the fifth power pascal is the initial temperature plus the density of water, which was 1000 kg per meters cubed multiplied by gravity, which the numerical value for gravity is 9. m per second squared multiplied by, by capital H minus lowercase H. So it's 32. m minus H and since we're running out of room multiplied by, let's write it up here up up 2.5 m minus H multiplied by a 293. You so OK. So for our final conditions here, it's 1.13 multiplied by 10 to the fifth power pass scales plus the density of water multiplied by gravity multiplied by 32.5 m minus H multiplied by a 2.5 m minus H multiplied by a all divided by the final temperature which is 293 Calvin. So now we must rearrange this equation to solve for H. And let's note that all the units will cancel out the, leave the H value in meters. And this is just a lot of algebra to rearrange this problem. The oldest quickly skip ahead really quick. So let's start off to, you know, to help jumpstart this problem. So let's multiply 293 Calvin to both sides and divide by a. So when you do that and you type into a calculator, you should get 244891. equals, we'll put it in brackets 420125 minus 9810 H which is the height value we're trying to determine multiplied by 2. minus H. OK. So then we can simplify this statement by using some algebra to get that H squared minus 45.326 H plus 82. equals zero. So as you know, as you can see, this is a quadratic function, so we can use the quadratic function to solve for age but it will give us two answers. But let's write it down really quick. 45.326 plus or minus square root 45.326 squared minus four eight. So four multiplied by 82.102, multiplied by one all divided by all divided by two. OK, I should say two multiplied by one because the A value is one. So just in case as a recall recall that the quadratic function equals X equals opposite B plus or minus square root B squared minus four AC all over two A. So we're called the quadratic function. OK. So when you plug that into a calculator, you'll get two values, you'll get H equals 43 0.4 m and you also get that H equals 1.9 m. So 43.4 m is incorrect. Music is not physically possible because H should be aval H I should say should be a value between zero and 2.5 m. So thus H equals 1.9 m has to be the correct answer. Music sits within the boundaries that we, that the problem needs to. So so the age value needs to reside in, needs to be, needs to be between zero and 2.5 m. Awesome. We did that was a lot of work, but we did it and isn't that awesome? OK. So that means that our final answer has to be D 1.9 m. Thank you so much for watching. Hopefully that was helpful and I can't wait to see you in the next video. Bye.
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