Question Video: Calculating the Volume of a Gas after Heating Using Charlesโ€™ Law Physics

The temperature of 14 mยณ of a gas is initially 350 K. It is heated to 450 K while being kept at a constant pressure. What is the volume of the gas after it is heated?


Video Transcript

The temperature of 14 cubic meters of a gas is initially 350 kelvin. It is heated to 450 kelvin while being kept at a constant pressure. What is the volume of the gas after it is heated?

Letโ€™s say that here is our gas with its initial volume of 14 cubic meters. Weโ€™ll call that volume ๐‘‰ one. The gas also has an initial temperature, weโ€™ll call it ๐‘‡ one, of 350 kelvin. But then after the gas is heated, it reaches a new temperature, weโ€™ll call ๐‘‡ two, of 450 kelvin. After itโ€™s heated, weโ€™ll say that the gas has a volume ๐‘‰ two, and this is the value we want to solve for. All throughout this process, the gas was maintained at a constant pressure. When the temperature and volume of a gas vary while its pressure is held constant, that gas is described by a law known as Charlesโ€™s law.

This law says that for a gas at constant pressure, the gasโ€™s volume ๐‘‰ divided by its temperature ๐‘‡ is equal to a constant. This means that at any instant in time, regardless of how ๐‘‰ and ๐‘‡ change, the fraction ๐‘‰ divided by ๐‘‡ has the same value for a given gas. This means that at what we could call our initial moment, when our gas has a volume ๐‘‰ one as well as a temperature ๐‘‡ one, the ratio of ๐‘‰ one to ๐‘‡ one is equal to the same volume-to-temperature ratio after the gas has been heated, that is, when the volume is now ๐‘‰ two and the temperature is now ๐‘‡ two. Since we want to solve for the volume ๐‘‰ two, we can multiply both sides of this equation by ๐‘‡ two, canceling out that temperature on the right.

And we arrive at this expression, showing us that the volume ๐‘‰ two equals the volume ๐‘‰ one times the ratio of temperatures ๐‘‡ two to ๐‘‡ one. We know all three of the values on the right-hand side of this equation. The volume ๐‘‰ one is 14 cubic meters. The temperature ๐‘‡ two is 450 kelvin, while the temperature ๐‘‡ one is 350 kelvin. As a side note, for Charlesโ€™s law to hold true, itโ€™s important that the temperatures be expressed in units of kelvin. We see our temperatures are in this case. But if they were expressed in some other units, say degrees Celsius or degrees Fahrenheit, we would need to convert them to kelvin before we could use them in this equation. Since they already are in kelvin though, weโ€™re ready to calculate ๐‘‰ two.

Notice that the units of kelvin cancel from numerator and denominator. Notice also that we could write our remaining numerator 450 as 45 times 10. Likewise, we can also write our remaining denominator 350 as 35 times 10. Weโ€™ve done this so we can see that the factors of 10 in numerator and denominator can cancel one another out, leaving our fraction as 45 divided by 35. But then 45 equals five times nine, and 35 equals five times seven. Thereโ€™s a factor of five then in both numerator and denominator that will divide out. And so we can write our fraction simply as nine-sevenths. Notice now that seven divides into 14 exactly twice. So ๐‘‰ two simplifies to two meters cubed times nine, or 18 cubic meters. This is the volume of the gas after it is heated.

So we now know that due to this heating, even though the gas was held at constant pressure, its volume has expanded.

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