### Video Transcript

A cylinder with a movable lid contains air with a volume of 0.042 cubic meters at 101 kilopascals pressure. The cylinder lid is pushed downward by an external force 𝐅, as shown in the diagram. The force includes the lid’s weight. As a result, the lid moves downward by 10 centimeters and the air pressure increases by 28 kilopascals. The lid then stops moving as the pressure above it and below it equalize. The pressure changes linearly as the lid is moved and the temperature of the air does not change during the compression. What is the area of the cylinder lid? Give your answer to two decimal places.

Looking at our diagram, we can imagine that our cylinder lid began here at the top of this cylinder and then moved downward a distance of 10 centimeters. We’re told the volume and pressure of the air in the cylinder before the lid moved. And we’re also told how much the air pressure changed as a result of that movement.

In this first part of our question, we want to solve for the area of the cylinder lid. To start doing that, let’s clear some space on screen and let’s consider that the air in this cylinder can be treated as an ideal gas. This means that we can describe it using the ideal gas law, which says that a gas’s pressure times its volume is equal to the number of moles of gas times a constant 𝑅 multiplied by the gas’s temperature.

Due to the movement of our cylinder lid, this gas in the cylinder changes between two conditions. In the first condition, when the lid is at the top of the cylinder and hasn’t compress the gas at all, the gas has a pressure we’ll call 𝑃 one, a volume we’ll call 𝑉 one, and a temperature we’ll call 𝑇 one. There’s no subscript on the 𝑅 because this is a constant value. And there’s no subscript on the 𝑛, the number of moles of the gas, for a reason we’ll see in a moment. This then is the ideal gas law for the first condition of our gas before the cylinder lid has descended.

But then after the lid has moved down over a distance of 10 centimeters, the air in the cylinder is at a new pressure we’ll call 𝑃 two, it has a new volume we’ll call 𝑉 two, and this is equal to 𝑛 times 𝑅 times the temperature 𝑇 two. Now the reason that 𝑛 doesn’t have a subscript on it to differentiate between the first and second conditions is because this value doesn’t change as the gas is compressed. While the density of the gas does increase while it’s compressed, the number of moles of the gas stays constant. In both conditions then, the number of moles of the gas involved is the same.

In our problem statement, we’re given a bit more information about the scenario. Specifically, we’re told that the temperature of the air in the cylinder is the same before and after compression. This means that what we’ve called 𝑇 one is equal to what we’ve called 𝑇 two. For simplicity’s sake then, let’s write both of these temperatures using the same symbol 𝑇.

When we look at these two equations, we can see that while the left-hand sides are different, the right-hand sides are identical. So if 𝑃 one 𝑉 one equals 𝑛𝑅𝑇 and 𝑃 two 𝑉 two equals 𝑛𝑅𝑇, then that means 𝑃 one 𝑉 one is equal to 𝑃 two 𝑉 two. This is helpful because both of these volumes 𝑉 one and 𝑉 two contain the area of the cylinder lid that we want to solve for.

Let’s say that the total height of this cylinder is ℎ. And let’s say further that the area of the cylinder lid that we want to solve for is 𝐴. This means that the original volume of the cylinder, its full volume 𝑉 one, is ℎ times 𝐴. We can also solve for the cylinder’s volume after the lid has descended. That’s equal to the height ℎ, whatever that is, minus 10 centimeters all multiplied by the area of the lid 𝐴.

At this point, let’s recall some information given to us in our problem statement. We’re told the volume of the cylinder before the lid has descended. We’ve called that 𝑉 one. We were also told the pressure of the air in the cylinder at this point. And we were told that after the lid has moved down 10 centimeters, that adds 28 kilopascals of pressure to the total pressure of the air in the cylinder. This means that the pressure after the lid has descended is equal to 𝑃 one plus 28 kilopascals. Since 𝑃 one is 101 kilopascals, that means that what we’ve called 𝑃 two is 129 kilopascals.

Knowing all this information, we’re now able to use this equation to solve for 𝑉 two. This will help us, as we’re on our way to solve for the area of the cylinder lid 𝐴. In our equation on the left, we’ll divide both sides by the pressure 𝑃 two, canceling that factor on the right so that we now have an expression for 𝑉 two in terms of known values. We can then substitute this ratio in for 𝑉 two in our equation on the right. That gives us 𝑃 one 𝑉 one divided by 𝑃 two being equal to the quantity ℎ minus 10 centimeters times 𝐴. And the next thing we’ll do is replace ℎ in this expression. And we’ll do it by means of this equation.

If we divide both sides of this equation by the area 𝐴, that factor cancels on the right. And we now have an expression for ℎ in terms of a known value 𝑉 one and the value we want to solve for 𝐴. We’ll use this fraction to replace ℎ in our larger expression. Next, let’s multiply through both terms in these parentheses by the value 𝐴. Doing that gives us a 𝑉 one by itself minus 10 centimeters times 𝐴.

Things are looking up here because we now have an expression all in terms of values that we know except for the one value we want to solve for. To make 𝐴 the subject of this equation, let’s first clear some space on screen. And then let’s multiply both sides of this equation by negative one. What this has done is it’s changed the signs of all three of our terms. Let’s now add the volume 𝑉 one to both sides, which causes that volume to cancel on the right. And then since the volume 𝑉 one appears on both terms on the left, we can factor it out of those terms. Finally, we divide both sides by 10 centimeters, cancelling out 10 centimeters on the right. This gives us a simplified expression for the area of the cylinder lid 𝐴.

To solve this equation, we’ll substitute in the values we know for 𝑉 one, 𝑃 one, and 𝑃 two. And we’re also going to convert our distance of 10 centimeters so that it has units of meters. Because 100 centimeters equals a meter, we convert 10 centimeters to meters by moving the decimal place two spots to the left. 10 centimeters then is 0.10 meters. And if we substitute in for 𝑉 one, 𝑃 one, and 𝑃 two, we get this overall expression.

Before we calculate, let’s look at the units inside the parentheses. Kilopascals divided by kilopascals will cancel out. So there will be no units inside these parentheses. And then outside the parentheses, we have meters cubed divided by meters. That will make one factor of meters cancel out, giving us a final answer in units of meters squared, an area. Computing this expression and then rounding our result so that we only keep two decimal places, we find that the area of the lid of the cylinder is 0.09 meters squared.

Let’s store this result off to the side and now move on to part two of our question.

What is the cylinder’s height? Give your answer to two decimal places.

In part one, we labeled the height of the cylinder ℎ. And now we want to solve for that value. Since we now know the area of the cylinder lid 𝐴 as well as the volume of the cylinder before the lid is compressed 𝑉 one, we’re able to use these values along with the fact that 𝑉 one equals 𝐴 times ℎ to solve for ℎ. Dividing both sides of this equation by the area 𝐴, we find that ℎ equals 𝑉 one over 𝐴. 𝑉 one is 0.042 cubic meters, and 𝐴, we found, is approximately 0.09 meters squared. When we perform this division, two factors of meters will cancel from numerator and denominator. We’ll be left with final units of meters. To two decimal places, ℎ is 0.46 meters. This is the height of the cylinder.

Let’s look now at the next part of our question.

What was the magnitude of the force applied to the lid? Give your answer to the nearest newton.

In this part of our question, we want to solve for the magnitude of this force labeled 𝐅. This, we recall, is the force that pushes down on the lid causing it to compress 10 centimeters. To begin solving for 𝐅, let’s recall Pascal’s relationship that pressure is equal to a force divided by an area. In words, we could say that the pressure is equal to a force spread out over an area. In our case, that force 𝐅 is spread out over the area of the cylinder lid. Multiplying both sides of this equation by the area 𝐴, we find that force equals pressure times area. And this area 𝐴 is the area of our lid, the one we solved for earlier.

To use the right value for pressure 𝑃 here, we’ll want to be a bit careful. We’ve seen that initially the air in our cylinder was at a pressure 𝑃 one. And then in the end, after the lid compressed, it was at a pressure 𝑃 two. The pressure 𝑃 in this equation is specifically that associated with the force 𝐅. That is, 𝑃 is equal to the difference between 𝑃 one and 𝑃 two. The pressure 𝑃 then is equal to 129 kilopascals minus 101 kilopascals, or 28 kilopascals. This is how much pressure is added to the original air pressure 𝑃 one due to the application of the force 𝐅.

If we substitute in the pressure 𝑃 and the area 𝐴 to our equation for the force 𝐅, there are a couple of steps we’ll want to take before we calculate this force. First, we’ll want to make sure the units agree. And to do that, we’ll change this value from units of kilopascals to units of pascals. 1000 pascals is equal to one kilopascal. So what we’ll do is multiply this number by 1000. 28 times 1000 is 28000.

In terms of units, we can now recall that a pascal is equal to a newton per meter squared. If we replace pascals with newtons per meter squared in this expression, we see then that the meters squared in the denominator here will cancel with the meters squared here. This will leave us with units of newtons, which is just what we want.

Before we multiply these values together though, notice that this is a very large value and we’re using it to multiply an approximate value. Recall that we got 0.09 meters squared by an approximation, by rounding. Because we’re multiplying this number by such a large number, 28000, even small errors due to rounding will be magnified. What we’ll want to do then is to use an exact value for the area of our cylinder lid rather than this approximate one.

The original expression we used to calculate the area looked like this. We had our volume 𝑉 one divided by 0.10 meters. And this was multiplying the quantity one minus our pressure 𝑃 one divided by our pressure 𝑃 two. When we compute this, we get an exact value for the area of our cylinder 𝐴. Multiplying the pressure 𝑃 times the area 𝐴, we get 2552.55 and so on newtons. We’re told to give our answer rounded to the nearest newton, and so because the digit immediately following the decimal place is greater than or equal to five, the final answer we report is 2553 newtons. This is the magnitude of the force pressing down on the cylinder lid.

Let’s now record this value off to the side and move on to the last part of our question.

How much work was done to compress the gas? Give your answer to the nearest joule.

The work done on an object 𝑊 is equal to the force acting on that object multiplied by that object’s displacement. In our scenario, we have a force of approximately 2553 newtons moving the lid of the cylinder a distance of 10 centimeters. We’ve seen that to convert 10 centimeters into meters involves moving the decimal point two spots to the left. Making this conversion, we’ll get a final answer in units of newton meters, or joules. Rounding this product to the nearest joule, we get 255. The work done to compress the gas was 255 joules.