The 3.0-m-long pipe in FIGURE P18.49 is closed at the top end. It...

Question

The 3.0-m-long pipe in FIGURE P18.49 is closed at the top end. It is slowly pushed straight down into the water until the top end of the pipe is level with the water's surface. What is the length Lof the trapped volume of air?

Accepted Answer

Hello, fellow physicists today, we're gonna 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. A teacher placed a cylinder with length 1.0 m in cross sectional area. A in contact with a surface of a large tank filled with oil of a density 950 kg per meter cubed. The upper part of the cylinder is sealed, she slowly moves the cylinder until its upper part is level with the oil surface. The oil would start to fill the cylinder, compressing the air column inside the oil inside the cylinder stops at a distance H from its upper part as shown below. Calculate H. So our angles that calculate H the height. So here we have an illustration or a diagram depicting the cylinder at the beginning of the experiment. And then in the second picture, it's at the end of the experiment when the cylinder is partially submerged. OK. So we're given some multiple choice answers. They're all in the same units as centimeters. Let's read them up to see what our final answer might be A, is 8.0. B is 25 C is 75 D is 92. Ok. So first off, let's make the following assumptions, the temperature of the gas, which is air inside the cylinder remains constant. The temperature inside the cylinder is the same as the oil's temperature because the cylinder moves slowly, the air inside the cylinder is sealed off from the outside environment because initially the cylinder is in contact with oil. So the number of moles remains constant. Thus, we consider the air inside the cylinder to be an ideal gas. Also the process is isothermal. So that means the temperature remains constant throughout the entire process. Therefore, at this point, 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 we need to take this equation and consider the initial and final conditions of the experiment. And doing that, we can write the following, we can write that the initial pressure multiplied by the initial volume is equal to the final pressure multiplied by the final volume is equal to the number of moles multiplied by the universal gas constant multiplied by the temperature. OK. We can then go on even further to write that the initial pressure multiplied by capital H subscript I multiplied by A is equal to the final pressure multiplied by lowercase H multiplied by capital A multiplied by A where capital A in both on both sides of the equal signs is the cross sectional area where lower case H is the height of the air column inside the cylinder. At the end of the experiment where capital H subscript I is the initial height of the cylinder. And P I. In this case, let's make a note that the initial pressure is equal to atmospheric pressure, which the numerical value for atmospheric pressure is 1.13 multiplied by 10 to the fifth power pass scales. Awesome. So we can rearrange what's called this equation. One, we can rearrange equation one, the soft for P F the final pressure. So when we do that, the final pressure will equal the initial pressure multiplied by the height of the cylinder divided by the height of the air column. OK. Now, to help us further our understanding of what's going on here in this problem, let's say we have two points, we have point N inside the cylinder and we have point M outside of the cylinder in the tank with just floating around in the oil. But they both reside on the same horizontal line because they reside on the same horizontal line. We can say that the pressures at point M and N are equal to each other. So we can write the following that P subscript M. So PM is equal to, so the pressure at point M and the pressure at point N are equal. So then we can say that that is equal to the final pressure. Thus, we can write that P that PM pressure at point M is equal to the initial pressure plus plus the density which is row multiplied by gravity multiplied by h the height of the air comm. OK. So the, so the density multiplied by the gravity multiplied by the height of the air column represents the pressure due to the oil column. Awesome. OK. So we can then write that the final pressure equals the initial pressure plus the density multiplied by gravity multiplied by the height. Which if we rearrange this equation and we plug in our known value for the final pressure, which we'll call that equation two. So when we plug in the equation two and two, we'll call this equation equation three. So when we plug in two into three, we'll get that P F final pressure equals the initial pressure multiplied by the initial height of the cylinder divided by the height of the air column is equal to the initial pressure plus the density. The density multiplied by gravity multiplied by the height of the air column. OK. That was a lot of writing. OK. So now let's call that equation four. So now we need to rearrange equation four means we're trying to solve for h we're trying to break it down to solve for H. So let's rearrange it. So row multiplied by gravity multiplied by height squared, the height of the air column plus the initial pressure multiplied by the height of the air column minus the initial pressure multiplied by the height of the cylinder or the initial height of the cylinder, I should say equals zero. So this should look familiar. So this is a quadratic equation. So we need to use the quadratic function in order to solve for H. So let's make a quick little sub note just in case, we need to recall that the quadratic function is X equals opposite B plus or minus square root B squared minus four AC all over to A. OK. So when we do that sulfur H will be H equals plus or minus or actually, I should say H equals minus the initial pressure plus or minus square root. The initial pressure squared plus four multiplied by the initial pressure multiplied by density multiplied by gravity multiplied by the height of the air colum multiplied by the initial height of the cylinder all divided by two multiplied by the density multiplied by gravity. OK. So at this stage, we can plug in all of our known variables and officially solve for each. So let's do that together, shall we? So H equals So when we plug in all this is at the stage where we plug in all of our known variables. OK. So the initial pressure was the atmospheric pressure, which was 1.13 multiplied by 10 to the fifth power past scales plus or minus square root, the initial pressure which was 1.13 multiplied by 10 to the fifth power past scales squared plus four, multiplied by the initial pressure. Again, 1.13 multiplied by 10 to the fifth power past skills multiplied by gravity. Or actually, we're skipping it. We need to multiply either density the density of oil which was kg per meter cubed. And then we need to multiply it by gravity which the numerical value for gravity is 9.81 m per second squared. And we're running out of room here. So let's write it down below here. OK? And then the height was 1. m. Oh Don't. Yeah, we're trying to solve for H so let's ignore it. So then H I is 1.0 m all divided by two, multiplied by the density of oil again which was 950 kg per meter cubed multiplied by gravity which was 9.81 m per second squared. OK. So when you plug it into a calculator, you'll get two different answers. But we can't, the negative answer is physically unacceptable because the height must be a positive number. So that leaves us with H equals 0.92 m. But we need to convert that to centimeters because our final multiple choice answers are in centimeters to, to do that. We need to use dimensional analysis. So we'll remember that there's 100 centimeters and one m, the meters cancel out. So that means that our height value is 92 centimeters. So that is our final answer. Hooray, we did it. So let's scroll up to the top. So that means our final answer has to be D 92 centimeters. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

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The 3.0-m-long pipe in FIGURE P18.49 is closed at the top end. It is slowly pushed straight down into the water until the top end of the pipe is level with the water's surface. What is the length Lof the trapped volume of air?

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