# Question Video: Using Gay-Lussac’s Law to Find the Pressure of a Gas Physics • 9th Grade

A gas is heated from 19°C to 80°C, while the volume is kept constant. If the initial pressure was 2000 Pa, what is the pressure of the gas after it has been heated? Give your answer to 3 significant figures.

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### Video Transcript

A gas is heated from 19 degrees Celsius to 80 degrees Celsius, while the volume is kept constant. If the initial pressure was 2000 pascals, what is the pressure of the gas after it has been heated? Give your answer to three significant figures.

Alright, so in this example, we have a gas which starts out at 19 degrees Celsius, but then it’s heated and its temperature goes up to 80 degrees Celsius. We’re told when the temperature of the gas is 19 degrees Celsius, it has a pressure of 2000 pascals. Then as the temperature of the gas changes, we want to see what impact — if any — this has on the pressure of the gas. As we consider what kind of method would help us figure out this final gas pressure, we find a clue in the problem statement: this gas was heated and the volume was kept constant.

When the volume of a gas is held constant over some process, that means that Gay-Lussac’s law applies to that process. This law tells us that for a gas held at constant volume, the pressure of that gas divided by its temperature is equal to a constant value. In other words, even as the pressure or the temperature of the gas will change, so long as the volume of the gas is constant, that ratio 𝑃 divided by 𝑇 is always the same.

Another way of writing this is to consider some change that happens in a gas and to say that the pressure divided by the temperature before the change 𝑃 one divided by 𝑇 one is equal to the pressure divided by temperature after the change. This second way of writing the law is especially helpful for us because we indeed have a gas that has gone through a change. It’s been heated. So we can say the initial pressure of our gas divided by its initial temperature is equal to its final pressure — that’s what we want to solve for — divided by its final temperature.

Since it’s 𝑃 two, the final gas pressure that we want to solve for, let’s rearrange this equation to let us do that. If we multiply both sides of the equation by 𝑇 two, the temperature of the gas after it’s heated, then that term cancels on the right-hand side and we have 𝑃 two all by itself. We can write the resulting equation this way. We can say that 𝑃 two, the final pressure of the gas, equals 𝑃 one times the temperature ratio 𝑇 two divided by 𝑇 one.

Now at this point in our solution, it’s important to notice something. See the temperatures we’ve been given in the problem statement, 19 degrees Celsius; 80 degrees Celsius. That is these temperatures are given as values on the Celsius scale. But that’s not the SI unit of temperature. The SI temperature unit is kelvin. In other words, before we plug in 𝑇 two and 𝑇 one for our calculation, we want to convert these given temperatures in to temperatures on the Kelvin scale. If we forgot to do that, our answer would be off.

So we want to know how to convert a temperature from degrees Celsius to kelvin. And roughly speaking, a temperature in kelvin is equal to that same temperature in degrees Celsius plus 273. So for example, if we had a temperature in Celsius of zero degrees, that is freezing, then the equivalent temperature in kelvin would be zero plus 273 or 273 kelvin. Using this relationship, let’s write out the initial and final temperatures of our gas in kelvin. 𝑇 one, the initial temperature of the gas, will be equal to 19 plus 273 and all this will be in kelvin. All that together is 292 kelvin. This value in kelvin is the value we’ll use to substitute in for 𝑇 one in our equation.

Next, what about 𝑇 two, the temperature of our gas after it’s been heated? That’s equal to 80 plus 273 all in kelvin. And this equals 353 kelvin. That’s the temperature we’ll use for 𝑇 two in our equation. With these values plugged in, notice that the units kelvin cancel out. But be careful. This doesn’t mean that if we had used degrees Celsius instead of kelvin, the units would also have cancelled and therefore wouldn’t have made a difference. It’s true that if we had forgotten to convert our initial temperatures in degrees Celsius into kelvin and simply plug them into the equation, then those units would have also cancelled.

But notice what our fraction would have been. Instead of 353 divided by 292, we would have had a fraction of 80 divided by 19. These two fractions are not the same thing. All that to say whenever we’re working within the SI system as we usually are, then whenever an equation involves temperature, we’ll want to make sure to use the Kelvin temperature scale in our calculation. Okay, we’re almost ready to solve for 𝑃 two, the final pressure of this gas.

In order to do so, let’s plug in for 𝑃 one, the initial pressure. And that’s given as 2000 pascals. This unit, pascals, is the base unit for pressure in the SI system. As the unit of pressure, a pascal is equal to a newton of force divided by a square metre of area. So when we see 2000 pascals, that’s another way of saying 2000 newtons of force per square metre of area. And that value can now go in for 𝑃 one in our equation. We’re now ready to calculate 𝑃 two, the pressure of the gas after it’s been heated. We see the units of this result will be pascals. And when we calculate this expression and round it to three significant figures, we find a result of 2420 pascals. This is the pressure of the gas after it’s been heated.