Singly charged gas ions are accelerated from rest through a voltage of 12.5 volts. At what temperature will the average kinetic energy of gas molecules be the same as that of these ions?
In this exercise, we have two different scenarios going on. First, we have singly charged ions being accelerated across a potential difference of 12.5 volts. This description “singly charged” just means that these gas ions either have one extra proton or one extra electron. This is what gives them an overall charge and makes them ions.
Since these ions are exposed to a potential difference, that means that they already have potential energy which will be converted to kinetic energy as the ions begin to move across this distance. We can call the electrical potential energy that these ions start out with capital 𝑈. That’s energy they have just by virtue of being charged particles in the presence of a potential difference. That’s scenario one.
Then, there’s a second scenario our problem statement tells us about. In this second scenario, imagine that we have a sealed-off box that contains some number of gas molecules. These molecules move around the box at some average speed, which has to do with the temperature of their surroundings.
If we went to this temperature knob on the box and turned it so that the temperature went up, then the average speed of the molecules in the box will also go up. When we compare these two situations one with another, what we want to know is just what does the temperature reading on this knob have to be in order for the average kinetic energy of the molecules in our box to be equal to the energy of the singly charged gas ions.
In other words, we want to find a correspondence between these two scenarios. The way we’ll go about doing this is we’ll look first at what the energy is that our singly charged ions have; that is, what is 𝑈? And then, we’ll look for a mathematical relationship for the average kinetic energy of gas molecules that depends on their temperature 𝑇.
So first, the potential energy of our ions, we can recall that the electrical potential energy of a charge 𝑞 is equal to that charge multiplied by the potential difference 𝑉 that the charge is moved through. This relationship is helpful because we’re given Δ𝑉 and we can solve for 𝑞 because our gas ions remember are singly charged.
This tells us that the magnitude of our charge 𝑞 is equal to the charge of a single proton. And what is that charge? We can look it up or recollect it to be 1.6 times 10 to the negative 19th coulombs. This tells us that we have everything we need in order to calculate the initial electrical potential energy of our ions. And knowing that, let’s turn to our second scenario, the gas molecules in motion.
In this case, we want to connect the temperature of these molecules with their average kinetic energy. So we’ll look for a mathematical relationship that brings those two things together. There is such a mathematical relationship. And since we’re working in particular with gas molecules, here’s what it looks like. This relationship says that the average kinetic energy of an individual gas molecule is equal to three-halves times 𝑘, which is Boltzmann’s constant rather than Coulomb’s constant, multiplied by the temperature in kelvin 𝑇.
Since it’s a bit confusing that we sometimes use lowercase 𝑘 for Coulomb’s constant and in this case we’re using it for something different, let’s write out right away what this value 𝑘 is equal to. It’s 1.38 times 10 to the negative 23rd joules per kelvin. Having that written down should help us keep this value straight.
Let’s now reconsider the question that we’re being asked here. We want to solve for the temperature 𝑇 at which the average kinetic energy of the gas molecules, that’s 𝐾𝐸 sub avg, will be the same as the energy of the ions, that’s capital 𝑈. What we’ll do then is equate these two terms. We’ll simply set them equal to one another and enforce that as a condition.
We’ve seen that 𝑈 is equal to Δ𝑉 times 𝑞. So we rewrite that side of the equation. And we know that 𝐾𝐸 sub avg is equal to three-halves 𝑘 times 𝑇. So we make that substitution. Now, we want to solve for a temperature 𝑇. So let’s rearrange this whole equation so that 𝑇 is on one side by itself.
Making this rearrangement, we find that 𝑇 is equal to two times Δ𝑉 times 𝑞 all divided by three times 𝑘, where 𝑘 is Boltzmann’s constant rather than Coulomb’s constant. When we do solve for this temperature 𝑇, that will give us the temperature at which the average kinetic energy of our gas molecules is the same as the energy of our ions.
So let’s plug in and solve for 𝑇 now. We know that the potential difference Δ𝑉 is 12.5 volts, that the charge 𝑞 is the charge of a proton 1.6 times 10 to the negative 19th coulombs, and that Boltzmann’s constant is written as 1.38 times 10 to the negative 23rd joules per kelvin.
When we calculate this fraction and round it to two significant figures, we find a very high temperature: 97000 kelvin. That’s a temperature about as hot as some of the hottest stars in our universe. And that’s the temperature these gas molecules need in order to have the same average kinetic energy as the energy of the ions.