Question Video: Bulk Properties of an Ideal Gas Physics • 9th Grade

A gas cylinder holds 3.25 mΒ³ of gas at a pressure of 520 kPa and a temperature of 300 K. At what temperature would the gas pressure in the cylinder become 865 kPa? Answer to three significant figures.

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Video Transcript

A gas cylinder holds 3.25 meters cubed of gas at a pressure of 520 kilopascals and a temperature of 300 kelvin. At what temperature would the gas pressure in the cylinder become 865 kilopascals? Answer to three significant figures.

So this question describes a cylinder of gas with a volume of 3.25 meters cubed, a pressure of 520 kilopascals, which is the same as 520,000 pascals, and a temperature of 300 kelvin. We’re being asked to calculate the temperature at which the gas pressure in the cylinder would become 865 kilopascals or 865,000 pascals. So we could imagine the same cylinder of gas at sometime later where the pressure has changed to 865 kilopascals or 865,000 pascals and the temperature has taken some unknown value that we need to work out. And because it’s the same cylinder, we can assume that the volume would go unchanged.

At this point, because we have two different sets of values for the quantities volume, pressure, and temperature, let’s call the initial set of values on the left 𝑉 one, 𝑃 one, and 𝑇 one and the changed values on the right can be 𝑉 two, 𝑃 two, and 𝑇 two. So our challenge in this question is to find 𝑇 two. So to start with, let’s think of the variables that we have to deal with in this question and see if we can think of any equations that might help us.

Well, this question asks us to consider the volume, pressure, and temperature of the gas. One equation that gives us the relationship between these quantities is the ideal gas equation. This tells us that the pressure of an ideal gas multiplied by its volume is equal to some constant multiplied by its absolute temperature. We can recall that the ideal gas equation makes the assumptions that gas molecules have negligible size and don’t interact with each other. Even though this question talks about a real gas, it’s still reasonable to use the ideal gas equation as it can still give us accurate answers without being too complicated.

Now, in this question, we’re trying to find the value of 𝑇. So let’s rearrange the ideal gas equation to make 𝑇 the subject and do this by just dividing both sides of the ideal gas equation by π‘˜, giving us 𝑃𝑉 over π‘˜ equals 𝑇. Since we need to find the temperature of the gas after the pressure has increased, it seems reasonable that we can just plug these values, 𝑉 two and 𝑃 two, into this equation. And it will tell us 𝑇 two. However, while this is technically true, if we tried doing this, we’ll quickly realize that we don’t know what the value of π‘˜ is. This is because π‘˜ actually takes different values depending on the number of gas molecules that we’re dealing with. So it’s not something we can just look up.

So as it stands, we can’t substitute 𝑉 two and 𝑃 two straight into the equation and just get a value of 𝑇 two. However, because we have a set of initial conditions for the volume, pressure, and temperature of the gas, there is something else we can do with the ideal gas equation to find the answer. If we start with 𝑃𝑉 equals π‘˜π‘‡ and divide both sides of the equation by 𝑇, we get 𝑃𝑉 over 𝑇 equals π‘˜. Because the amount of gas molecules in this question is constant, this means that π‘˜ is constant. So regardless of how we try to change the pressure, volume, or temperature of a fixed amount of gas, we find that the value of 𝑃𝑉 over 𝑇 is always constant.

In other words, 𝑃 one times 𝑉 one divided by 𝑇 one is equal to 𝑃 two times 𝑉 two divided by 𝑇 two. Writing the ideal gas equation in this way is actually equivalent to its more familiar form 𝑃𝑉 equals π‘˜π‘‡. However, it makes it easy to calculate how variables change without needing to know π‘˜. And we can use this formulation for this question. Since we’re trying to find 𝑇 two, let’s first rearrange this to make 𝑇 two the subject. First, we can multiply both sides of the equation by 𝑇 two, then multiply both sides of the equation by 𝑇 one, and finally divide both sides of the equation by 𝑃 one 𝑉 one.

With the equation in this form because we know the initial temperature, initial pressure, and initial volume before the pressure increased and we know the volume and the pressure after the pressure increased. We can just substitute all of these values in and calculate 𝑇 two. Let’s just give ourselves some more space. And we know that 𝑇 one is 300 kelvin, 𝑃 two is 865,000 pascals, 𝑉 two is 3.25 meters cubed, 𝑃 one is 520,000 pascals, and 𝑉 one is 3.25 meters cubed as well. Plugging all this into our calculator, we get an answer of 499.04 which has units of kelvin since it’s a temperature. And rounding our answer to three significant figures gives us a final answer of 499 kelvin.

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