The 1.0-m-tall cylinder shown in FIGURE CP14.71 contains air at a pressure of 1 atm. A very thin, frictionless piston of negligible mass is placed at the top of the cylinder, to prevent any air from escaping, then mercury is slowly poured into the cylinder until no more can be added without the cylinder overflowing. What is the height h of the column of compressed air?
Hint: Boyle's law, which you learned in chemistry, says p₁V₁ = p₂V₂ for a gas compressed at constant temperature, which we will assume to be the case.

Question

The 1.0-m-tall cylinder shown in FIGURE CP14.71 contains air at a pressure of 1 atm. A very thin, frictionless piston of negligible mass is placed at the top of the cylinder, to prevent any air from escaping, then mercury is slowly poured into the cylinder until no more can be added without the cylinder overflowing. What is the height h of the column of compressed air?
Hint: Boyle's law, which you learned in chemistry, says p₁V₁ = p₂V₂ for a gas compressed at constant temperature, which we will assume to be the case.

Diagram of a cylinder with air and water, illustrating pressure concepts in fluid mechanics.

Hint: Boyle's law, which you learned in chemistry, says p₁V₁ = p₂V₂ for a gas compressed at constant temperature, which we will assume to be the case.

Accepted Answer

Hey, everyone. Let's go through this practice problem. If a two m long tube has one of its ends closed and the other end fitted with a thin frictionless and massless piston to trap air in the tube. And we're assuming that the two m length is initially filled with air at atmospheric pressure. Different liquids are poured on top of the piston to compress the trapped air. If the liquid used is fresh water, determine the length of the fresh water column that causes the greatest compression. And I'll give you a hint. Boyle's law states that P one multiplied by V one equals P two multiplied by V two. When a fixed amount of gas is compressed at constant temperature, assume gas, assume constant gas temperature during compression. We're given four, multiple choice options to choose from. Option A zero m, option D two m, option C 10.3 m and option D one m. All right. So first things first, it's always helpful with problems like this to define variables that might be helpful that aren't given. So we're told that the, by the diagram that the height of the column of air is given by Y. And of course, the value of Y is always going to be different depending on how, why or how high the column of water is. So I'm going to define another variable, a similar variable for the height of the column of water. And I'm just going to call it H sub W we H sub W is simply the full length of the column two m minus Y since both H sub W and Y are different values at every point in time, but we can still find a relationship between them. Using this equation. The problem asks for the length of the freshwater column that causes the greatest compression, which means we're looking for the highest possible value of H sub W that there can possibly be. One way we could do this is to alternatively find the smallest possible value that Y can be and then just subtract that from two m as the H sub W equation shows to find these extreme values of Y. Let's first create an equation that includes Y and then try to maximize or minimize it, optimize it. However, we can, we are able to, since the problem talks about Boyle's law, which gives us the pressure multiplied by the volume or different states of a gas compression, we can use that to find a relationship between the different states of two different states of the column. One thing we can do is imagine an alternate state of the column which is entirely filled with just the air. According to Boyle's law, the product of the pressure and volume of the air in the column could be the same as the pressure multiplied by the volume in the column of air. After it's been compressed by the water above it, we can use the boil law relation between these two states to see if we can find an equation that gives us a relationship for Y for the sake of labeling, I am going to use the subscript one to refer to variables relating to the column if it's just air. And I'm going to use the subscript two related to the variables of the column. When there's water above it, specifically water wear, the water is at its maximum possible height. And we know from Boyle's law that P one V one is equal to P two V two. Since we're not told anything about V recall that the volume of column of air should be equal to the cross sectional area of the column multiplied by the column's length. This will come in handy. It will also come in handy to remember that the pressure due to the a column of fluid is equal to the density of that fluid multiplied by the gravitational acceleration, G multiplied by the depth of that fluid. All right. Now let's get to work writing out some expressions for some of the variables we'll need for the Boyles Law equation. First of all P one, the pressure in the air is just going to be the same as atmospheric pressure. Since the problem tells us to assume that the two m length is initially filled with air at atmospheric pressure. So that means that the piece of one is the same as atmospheric pressure, which we should recall is equal to about 1.13 multiplied by 10. The power of five pascals also note that V one, it'd be equal to the cross sectional area of the column multiplied by the length of the column of air, which is just going to be two m. Since we're looking at the state where all two m of the column, our air. Now let's look at P two, the pressure in the air just below where all the water is. So this is going to be equal to P sub one, the pressure of the air that's already there plus the pressure due to the column of water above it. So row sub W the density of the water multiplied by G multiplied by H sub W the height of that column of water. This can also be written by substituting for H sub W the two m minus Y formula we talked about earlier. So this could also be written as P sub one plus row, sub W G multiplied by in parentheses two m minus Y. And lastly V sub two is again equal to the cross sectional area of the column, which hasn't changed multiplied by the height of that column of air, which we can see from the diagram is just Y. So B two is equal to a multiplied by Y. Now let's actually plug these expressions into Boyle's Law. So P one, V one is equal to P two V two. Let's expand this out by making subs subs some substitutions based on what we discussed earlier. So this is P one multiplied by V one which is two m multiplied by A and this is equal to P two, which we discussed earlier is P one plus row, sub W G two m minus Y. And that's all multiplied by A by a multiplied by Y. And there is an A on both sides of the equation. So these can get crossed out and note that now everything left in the equation is something that's known except for the Y which we're trying to find. So let's get to work trying to isolate the Y if we can. So P one multiplied by two m is equal to. And then I'm going to distribute the Y over each term in the parentheses. So it's equal to P one Y plus row W G two m minus Y multiplied by Y. And we can even further expand out the right hand side of this equation by distributing the Y against each of the other parentheses, the two m minus Y parentheses. So that's P one Y plus two m row W G Y minus row W G Y squared. And notice that we have a Y squared in this equation. So if we're trying to solve for Y then we're gonna have to use the quadratic equation. So if we're gonna use the quadratic formula, then things might get messy. So to hopefully make things easier on ourselves, I'm just gonna start plugging in numerical values now, so that when we finally put them into the quadratic equation, things will hopefully look a lot nicer. So first off P one multiplied by two m. Well, as we, we talked earlier about what the value of P sub one is. It's 1.3 multiplied by 10 to the power of five pascals or newtons per meter squared. So I'm gonna put that into my calculator. I'm gonna put in 1.13 multiplied by 10 to the power of five pascals multiplied by two m. And we find a value of about 2600 Newton's per meter. And this is equal to P one multiplied by Y. So that's just 1.13 multiplied by 10 to the power of five nuances per meter squared multiplied by Y and then plus two m multiplied by row sub W the density of fresh water. And of course, recalled the density of fresh water is about kg per cubic meter. And also recall that the gravitational acceleration G is equal to about 9.81 m per second squared. So I'm gonna put that into a calculator. I'm going to put in two m multiplied by kg per cubic meter multiplied by 9.81 m per second squared. And that gives us a value of about 19,620 newtons per meter squared. That's another term that's being multiplied by Y. And then lastly, we minus row sub W multiplied by G multiplied by Y squared. So we take row sub W 1000 kg per cubic meter and multiply that by G and it, when we put that into a calculator, we get about 9.81 multiplied by 10 to the power of three newtons per cubic meter. And that's being multiplied by Y squared for the sake of simplification. Let's consolidate the two terms that have the Y into it into one term by adding their coefficients to one another. So you end up with a new equation slightly more simplified saying that 202,600 newtons per meter is equal to one point 2092, multiplied by 10 to the power of five newtons per meter squared multiplied by Y and then subtract the same term as before 9.81 and multiplied by 10 of the power of three nuances per cubic meter times Y squared. And to make this even neater one thing we can do is try and get the term containing the Y squared on its own by dividing all three parts of the equation by the coefficient in front of the Y squared. So we could divide all three terms by 9. multiplied by 10 to the power of three new and per cubic meter. That will make our numbers smaller and easier to write. And it'll also just make the whole thing look simpler and smaller and neater and easier to write out. So we divide all three sides of the equation by that exact term by 9.81 multiplied by 10 to the power of three new ends per cubic meter. And everything reduces down into 20.652 m squared is equal to 0.326 meters multiplied by Y minus Y squared. And the last thing I'm going to do to get this into true quadratic form is get this into a form where it's just something the Y squared plus something multiplied by Y plus the constant, all equals zero. So I'm going to add to both sides of the equation, the Y squared and then subtract from both sides of the equation. The 12.326 term. And what we find is Y squared minus 12.326 M multiplied by Y plus 20.652 m squared is equal to zero. And now we have our equation in a form where we can very easily use the quadratic equation where the coefficient in front of the Y squared is a, the coefficient in front of the Y is B, the coefficient without A Y is C. And we can then use the quadratic formula and recall that the quadratic formula states that Y is equal to the negative of B plus or minus the square root of B squared minus four multiplied by AC. And then everything is divided by two, multiplied by A. Now we'll supply this formula to the quadratic equation that we've set up. So we find that Y is equal to 0. meters plus or minus the square root of 12.326 m squared minus four multiplied by A which is just one multiplied by C which is 20.652 squared meters. And then everything, it just gets divided bye two multiplied by A which is just one. Now, if you put this into your calculator two different times, once using the plus symbol and once using the minus symbol, when we find two different values for A, we or for Y we find that Y is either equal to two m or it can also be equal to 10.3 to 6 m. Now, normally, whenever we use the quadratic equation and find two different values, then what we have to do is cross out the value which is clearly impossible for the physics of the problem we have. And if we notice if we recall from earlier, the length of the column we have is the maximum length is two m. If the column of water inside was any higher, then things would just spill out and it just is totally irrelevant. So we can ignore the 10.326 value that can just get completely crossed off. So our true value for Y is two m. And if we recall from earlier, why is the length of the air column? But the problem is actually asking us to find the maximum length of the water column. So H sub W and if we remember from earlier, H sub W is equal to two m minus the value for Y which I've just found is two m. So this means that the maximum length of a column is two minus two, which is just zero m. In other words, no matter how much water we pour onto the piston, it's not going to be able to compress the air at all. And if we look at our multiple choice options, we can see that this is exactly what option A says, it says zero m. So that is then our answer to the problem and that is it for this problem. I hope this video helped you out if it did and you're looking for more practice. Please consider checking out some of our other videos which will hopefully give you some more experience with these types of problems. But that's all for now. I hope you all have a lovely day. Bye bye.

Verified Solution

Key Concepts

Boyle's Law

Hydrostatic Pressure

Ideal Gas Behavior