Question Video: Using Boyle’s Law to Calculate Volume Changes over Multiple Compressions and Expansions | Nagwa Question Video: Using Boyle’s Law to Calculate Volume Changes over Multiple Compressions and Expansions | Nagwa

# Question Video: Using Boyleβs Law to Calculate Volume Changes over Multiple Compressions and Expansions Physics • Second Year of Secondary School

A gas initially has a pressure of 800 Pa and a volume of 2 mΒ³. It is compressed at a constant temperature until its volume is half its initial value. At this point it has a pressure πβ. It is then allowed to expand again until the pressure is 0.25 Γ πβ. What is the final volume of the gas?

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### Video Transcript

A gas initially has a pressure of 800 pascals and a volume of two meters cubed. It is compressed at a constant temperature until its volume is half its initial value. At this point, it has a pressure π one. It is then allowed to expand again until the pressure is 0.25 multiplied by π one. What is the final volume of the gas?

In this problem, we have a gas which has an initial volume that we will call π zero, which the question tells us has a value of two meters cubed. The gas is initially at a pressure that we will call π zero, which the question tells us is 800 pascals. Then weβre told that the gas is compressed. At this point, we will say that the gas has a volume of π one. And we are told that this volume is half the initial volume of the gas. So we can write that π one is equal to a half multiplied by π zero.

The question tells us that the pressure of the gas at this point is π one and this value is unknown to us. Then we are told that the gas is allowed to expand. At this point, we will say that the volume of the gas is π two, and we donβt know its value. And the pressure of the gas is π two, which we are told has a value of 0.25 times π one, so π two is equal to 0.25 multiplied by π one.

The question would like us to calculate the final volume of the gas, so we must calculate π two. To answer this question, we will use a gas law known as Boyleβs law, and Boyleβs law tells us that the pressure of the gas multiplied by the volume the gas is occupying is equal to some constant πΆ. However, in order for this to be true, two things must be satisfied. The amount of gas must stay the same, and the temperature of the gas must remain constant. The question tells us that the gas is at a constant temperature, so we know that this second condition is satisfied. The question also doesnβt say anything about the gas being added or removed, so we can safely assume that the amount of gas stays constant.

So weβve established that we can apply Boyleβs law to the gas in this question. That is, the pressure of the gas multiplied by the volume of the gas at any point will be constant. And this allows us to compare the gas at two points in time. For example, we could pick two random points π΄ and π΅. The pressure of the gas multiplied by the volume of the gas at time point π΄ is equal to the pressure of the gas multiplied by the volume of the gas at point π΅. So letβs clear some space and apply Boyleβs law to this question. First, weβll use Boyleβs law to work out π one, the pressure of the gas when it has been compressed. And then weβll use Boyleβs law a second time to work out the volume of the gas after it has expanded again, π two.

To work out π one, we will consider the initial state of the gas and the compressed state of the gas. Applying Boyleβs law to this, the pressure of the gas multiplied by the volume of the gas in its initial state is equal to the pressure of the gas multiplied by the volume of the gas in its compressed state. We have an expression for π one, so letβs go ahead and substitute that in. Then expanding the brackets gives us π zero multiplied by π zero is equal to a half π one multiplied by π zero. Weβre trying to work out π one, and we have known values of π zero and π zero. So all we have to do is rearrange this equation to make π one the subject.

First, we will divide both sides by π zero, where we see that π zero in the numerator and denominator of both sides cancels. And this leaves us with the simple expression that π zero is equal to a half π one. Multiplying both sides by two, we see that the twos on the right cancel, which leaves us with π one by itself on the right. Writing this a bit more neatly with π one on the left, we get π one is equal to two π zero. Substituting our known value of π zero, π one is equal to two multiplied by 800 pascals. And evaluating this, we get π one is equal to 1600 pascals. And weβll keep a note of this over here.

Next weβll use Boyleβs law to compare the gas when it is compressed and expanded to work out π two. The pressure of the gas multiplied by the volume of the gas in its compressed state is equal to the pressure of the gas multiplied by the volume of the gas when it is expanded again. First, weβll substitute our expression for π one into this equation, and we see that π one multiplied by a half π zero is equal to π two multiplied by π two. Next, weβll substitute our expression for π two into this equation. Then expanding the brackets on each side, we see that a half π one multiplied by π zero is equal to 0.25π one multiplied by π two. We would like to calculate π two, and we have known values of π one and π zero. So all we have to do now is make π two the subject of this equation.

Dividing both sides by π one, we see that the π one in the numerator and the denominator on both sides cancel, leaving us with a half π zero is equal to 0.25π two. Another way to write 0.25 is one divided by four, so a half π zero is equal to a quarter π two. Multiplying both sides by four, we see that the fours on the right cancel and four multiplied by one-half on the left-hand side of the equation just equals two. And this leaves us with π two by itself on the right. Rewriting this a bit more neatly with π two on the left, this is our expression for π two. We can now substitute our known value of π zero into this, where we see that π two is equal to two multiplied by two meters cubed. And evaluating this, we get π two is equal to four meters cubed, so the final volume of the gas is four meters cubed.

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