### Video Transcript

A gas is heated while being kept at a constant pressure. If its temperature in kelvin increases by a factor of two, by what factor will the volume of the gas change?

Letβs say that this is our gas, which is held at a constant pressure. And weβre told that the gas is heated so that its temperature in kelvin increases by a factor of two. We can say then that if the temperature of the gas before heating is π one and the temperature of the gas after itβs heated is π two, then π two, the temperature of the gas after heating, is equal to twice π one. Similarly, weβre thinking about the volume of the gas before and after heating. And we can call these quantities π one and π two, respectively. Given the known relationship between the temperatures before and after heating, we want to know how π one and π two relate.

We can be helped in figuring this out by recalling Charlesβs law, an experimental gas law that has to do with gases that are held at a constant pressure. This law says that the volume π of a gas divided by the temperature π of that gas will always yield the same result for a gas held at constant pressure. Using this law and applying it to our scenario, we can write that the initial volume of our gas, π one, divided by the gasβs initial temperature, π one, equals the ratio of volume to temperature after heating has taken place, that is, equals π two divided by π two.

Letβs now cross multiply so that our volumes π one and π two are on the right side of this equation and the temperatures are on the left. We can bring this about by multiplying both sides of the equation by π two divided by π one. When we do this, on the left-hand side, the volume π one cancels out, while on the right-hand side, the temperature π two cancels. We are left then with this result. On the left we have a ratio of temperatures and on the right a ratio of volumes.

Letβs recall now that π two is equal to two times π one. We can make that substitution in our equation. And notice that when we do, the factor of π one cancels from numerator and denominator. The left-hand side of our expression simplifies to two. Recall that we want to know by what factor the volume of our gas changes as its temperature is increased. If we multiply both sides of our equation by π one, canceling out π one on the right, we find that π two equals two times π one. This tells us the factor by which the volume of our gas changes. The volume of the gas after heating is two times the volume of the gas before heating.