Question Video: Understanding Boyle’s Law | Nagwa Question Video: Understanding Boyle’s Law | Nagwa

# Question Video: Understanding Boyle’s Law Physics • Second Year of Secondary School

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A sphere of gas expands from a radius of 0.5 m to a radius of 2 m. The temperature remains constant throughout. How many times smaller is the pressure of the gas in the sphere after its expansion?

06:16

### Video Transcript

A sphere of gas expands from the radius of 0.5 meters to a radius of two meters. The temperature remains constant throughout. How many times smaller is the pressure of the gas in the sphere after its expansion?

Okay, so in this question, we’ve been told that we’ve got a sphere of gas. And this sphere of gas is expanding from a radius of 0.5 meters to a radius of two metres. In other words, here’s the sphere representing the gas before the expansion. Now, this sphere has a radius of 0.5 meters. And here’s the sphere after the gas’s expansion and this sphere has a radius of two metres. Another important point from the question is that we’ve been told that the temperature of the gas remains constant throughout the process.

Now what we’ve been asked to do is to work out how many times smaller is the pressure of the gas in the sphere after its expansion. In other words, how many times smaller is the pressure in this sphere of gas compared to this sphere of gas, at which point we consider the following. We’re talking about a sphere of gas. And because we know the shape that the gas occupies, we can work out the volume of the gas. As well as this, what we’re trying to work out is how the pressure of the gas changes from before its expansion to after its expansion. And we’ve been told that the temperature of the gas remains constant throughout this process.

So in this situation, what we need to recall is known as Boyle’s law. Now, this law tells us that the product of pressure and volume must be a constant if the temperature of a gas is constant. Now, in this case, the condition is satisfied. We’ve been told in the question that the temperature remains constant. Therefore, the product of pressure and volume must also be a constant.

In other words, if we say that the sphere of gas before expansion has a pressure 𝑃 nought and a volume 𝑉 nought and after its expansion it has a pressure 𝑃 one and a volume 𝑉 one, then because the product of pressure and volume must stay constant at any time, we can say that 𝑃 nought multiplied by 𝑉 nought must be the same as 𝑃 one multiplied by 𝑉 one. This is because we’ve got a constant temperature. So 𝑃 nought 𝑉 nought must be constant. And 𝑃 one 𝑉 one must also be that same constant because that product is not changing.

And we can use this information to work out how much smaller 𝑃 one is compared to 𝑃 nought. To do this, however, we need to know the values of 𝑉 nought and 𝑉 one first. So let’s work out the volumes. Now, we’ve been told that we’ve got two spheres of gas. And luckily, we’ve also been given the radii of those spheres. This is useful to us because we can recall that the equation that gives us the volume of a sphere is that the volume is equal to four-thirds multiplied by 𝜋 multiplied by the radius of the sphere cubed.

So we can use this equation to work out the value of 𝑉 nought and 𝑉 one. Let’s start with 𝑉 nought then. We say that the volume 𝑉 nought is equal to four-thirds multiplied by 𝜋 multiplied by 𝑟 nought cubed, where 𝑟 nought is the radius of the initial sphere. Now, we’ve been told that 𝑟 nought is equal to 0.5 meters in the question already. But we’re not gonna substitute this value in yet. We’ll see why in a second.

Instead, let’s work out an expression for 𝑉 one, the volume of the largest sphere. We said that 𝑉 one is equal to four-thirds multiplied by 𝜋 multiplied by the radius of the largest sphere cubed, where we’ve been told that 𝑟 one is equal to two meters. But again, we’re not going to sub in yet. Now that we have these expressions, let’s go back to our equation here.

Now, what we’re trying to do in this question is to work out how many times smaller is the pressure of the gas in the sphere after its expansion. In other words, we’re trying to work out how much smaller 𝑃 one is compared to 𝑃 nought. The easiest way to do this is to work out the ratio 𝑃 one divided by 𝑃 nought because this ratio gives us how many times smaller or larger 𝑃 one is compared to 𝑃 nought, which means we need to rearrange this equation so that we have 𝑃 one divided by 𝑃 nought on one of the sides.

So let’s do that then. Let’s first divide both sides of the equation by 𝑃 nought. This way we have 𝑃 one over 𝑃 nought on the right-hand side so far and the 𝑃 noughts on the left-hand side cancel. But we’ve got this spare 𝑉 one on the right-hand side. So we can just divide both sides of the equation by 𝑉 one. This way the 𝑉 ones on the right-hand side cancel. And now, on the left, we’ve got 𝑉 nought over 𝑉 one and on the right we’ve got 𝑃 one over 𝑃 nought.

So let’s tidy that up a bit and write it properly. So once again, to reiterate, we’ve got 𝑉 nought over 𝑉 one is equal to 𝑃 one over 𝑃 nought, at which one we can recall that we’ve got expressions for 𝑉 nought and 𝑉 one. So why don’t we substitute those in? So we say that 𝑉 nought is equal to four-thirds 𝜋 multiplied by 𝑟 nought cubed. And we divide this by 𝑉 one which is four-thirds 𝜋 multiplied by 𝑟 one cubed.

And now, we see why we didn’t evaluate the right-hand sides of 𝑉 nought and 𝑉 one because in our fraction we’ve got four-thirds divided by four-thirds and 𝜋 divided by 𝜋. Those all cancel. And we’re just left with 𝑟 nought cubed divided by 𝑟 one cubed on the left-hand side. So once again, tidying up this expression now, we just have 𝑟 nought cubed divided by 𝑟 one cubed on the left-hand side. And this is equal to the ratio that we’re looking for 𝑃 one divided by 𝑃 nought.

At this point, we should substitute in the values of 𝑟 nought and 𝑟 one. 𝑟 nought is equal to 0.5 meters. So we’ve got 0.5 cubed in the numerator and 𝑟 one is two meters. So we’ve got two cubed in the denominator, at which when we can evaluate the fraction on the left-hand side which gives us one over 64. What this means is that 𝑃 one the pressure afterwards in the gas is 64 times smaller than 𝑃 nought.

To see this more clearly, we can rearrange the equation by multiplying both sides by 𝑃 nought. This way the 𝑃 noughts on the right-hand side cancel. And what we’re left with is 𝑃 one on the right-hand side and 𝑃 nought multiplied by one over 64 on the left. What this means is that 𝑃 one is the same as one sixty-fourths of 𝑃 nought or 𝑃 one is 64 times smaller than 𝑃 nought or if we multiply both sides by 64, then the 64s on the left-hand side cancel. Then, we have 𝑃 nought is 64 times bigger than 𝑃 one. So 𝑃 one must be 64 times smaller than 𝑃 nought.

Whichever way you want to look at this, we arrive at the same answer. The question asks how many times smaller is the pressure of the gas in the sphere after its expansion. And the answer to that is that the pressure in the gas after its expansion is 64 times smaller than the pressure before its expansion.

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