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. How much work was done to compress
the gas? Give your answer to the nearest
joule.
This question is asking us to find
the work done on the lid of this cylinder of air. There’s a lot to think about in
this question. Let’s start by breaking down all
the information we’ve been given. And then we can work out what
method we’re going to use to find the answer.
To start with, we have a cylinder
containing air with a volume of 0.042 cubic meters at a pressure of 101
kilopascals. This is the initial setup before
the lid of the cylinder moves. So we’ll label these quantities 𝑉
one and 𝑃 one. We are also told that there are 101
kilopascals of atmospheric pressure, which we’ll label 𝑃 sub a. Initially, the pressure of the gas
in the cylinder is equal to the atmospheric pressure, meaning that the lid does not
move.
Recall the formula for
pressure. Pressure is equal to the force
divided by the area over which the force is exerted. So equal pressure means there is an
equal force from the air outside the cylinder, the atmospheric pressure, and from
the air inside the cylinder. These are balanced forces, and so
the lid does not move. Then, an external force 𝐅 is
applied to the lid of the cylinder, causing it to move downwards by 10 centimeters
as the forces are no longer balanced. There is an additional downward
force. We can see that this causes the
volume of the air in the cylinder to decrease. We’ll call this new volume 𝑉
two. When the lid moves down, it causes
the pressure of the air to increase by 28 kilopascals. So the new pressure in the cylinder
𝑃 two is equal to the initial pressure of 101 kilopascals plus 28 kilopascals. This is equal to 129
kilopascals.
We also know that the lid stops
moving because the pressure above it and below it equalize. In other words, the lid stops
moving when 𝑃 two becomes equal to the sum of the atmospheric pressure and the
pressure due to the force 𝐅, which we’ll call 𝑃 sub F. We know that this force increases
the pressure of the gas 28 kilopascals. This means that the force itself
must apply a pressure of 28 kilopascals across the lid of the cylinder.
Now we’ve understood the setup,
let’s think about the question we’ve been asked. We need to find the work done by
this force 𝐅 when it moves the lid downwards. The force 𝐅 and the pressure it
applies are related by the formula 𝑃 sub F equals 𝐅 divided by 𝐴, where 𝐴 is the
surface area of the lid. Since it’s the force that we’re
really interested in, let’s rearrange this to make 𝐅 the subject. To do this, we just need to
multiply both sides of the equation by the area 𝐴. This gives us 𝐅 equals 𝑃 sub F
times 𝐴.
We need to find the work done by
the force. Recall that the work done 𝑊 is
equal to the force multiplied by the distance through which it acts 𝑑. In this case, the force acts to
move the lid downwards. So 𝑑 is equal to the distance
through which the lid moves, 10 centimeters. If we substitute our expression for
the force 𝐅 into the equation for work done, we find that the work done is equal to
the pressure 𝑃 sub F times the area of the lid times the distance moved by the
lid. We know the values of the pressure
𝑃 sub F and the distance 𝑑, so next we need to find the area 𝐴.
To find the value of 𝐴, we just
need to use some simple geometry. Recall that the volume of a
cylinder is equal to the surface area of its circular face multiplied by its
height. If we rearrange this equation to
make 𝐴 the subject, we find that 𝐴 is equal to the volume of the cylinder divided
by the height. We are not told the height of
either the initial cylinder of gas or its height after the lid is moved. However, we do know the height of
this imaginary cylinder here, 10 centimeters. We also know that the volume of
this cylinder, which we’ll call Δ𝑉, is equal to the difference between the
cylinder’s initial volume and its final volume. So Δ𝑉 is equal to 𝑉 one minus 𝑉
two. So the surface area of the cylinder
lid must be equal to 𝑉 one minus 𝑉 two all divided by the distance 𝑑.
Now, let’s return to the equation
for the work done by the force. If we substitute in our expression
for 𝐴, we find that the work done is equal to the pressure 𝑃 sub F times the
change in volume divided by 𝑑 multiplied by 𝑑. We can see that these factors of 𝑑
cancel, which simplifies the expression to 𝑊 equals the pressure 𝑃 sub F times 𝑉
one minus 𝑉 two.
Now, all that’s left to do is
calculate the value of 𝑉 two. Recall that for a gas of constant
mass, we can use the formula 𝑃 one 𝑉 one divided by 𝑇 one equals 𝑃 two 𝑉 two
divided by 𝑇 two, where the quantities labeled with a one refer to the gas’s
initial state and the quantities labeled with a two refer to the gas’s final
state. We are told that when the gas is
compressed, its temperature does not change. This means that 𝑇 one equals 𝑇
two, and we can cancel these out from our equations. This leaves us with 𝑃 one times 𝑉
one equals 𝑃 two times 𝑉 two. We can rearrange this to make 𝑉
two the subject by dividing both sides of the equation by 𝑃 two. From this, we find that 𝑉 two is
equal to 𝑃 one times 𝑉 one divided by 𝑃 two.
Now, we just need to substitute in
the values that we found at the beginning of our working. 𝑃 one is equal to 101 kilopascals,
𝑉 one is equal to 0.042 cubic meters, and 𝑃 two is equal to 129 kilopascals. If we complete the calculation, we
find that 𝑉 two is equal to 0.03288 cubic meters.
Finally, we are ready to calculate
the value of the work done by the force. 𝑊 is equal to the pressure 𝑃 sub
F multiplied by 𝑉 one minus 𝑉 two. To convert the pressure from
kilopascals into the SI unit pascals, we simply multiply by 1000, giving us 𝑃 sub F
equals 28000 pascals. We also know that 𝑉 one is 0.042
cubic meters and 𝑉 two is 0.03288 cubic meters. Substituting these values in, we
find that the work done is equal to 28000 pascals times 0.042 cubic meters minus
0.03288 cubic meters. This gives us an answer of 255.36
joules.
This means that to the nearest
joule, 255 joules of work are done by the external force to compress the gas. This is our final answer to this
question.
