Photons of light scatter off molecules, and the distance you can see through a gas is proportional to the mean free path of photons through the gas. Photons are not gas molecules, so the mean free path of a photon is not given by Equation 20.3, but its dependence on the number density of the gas and on the molecular radius is the same. Suppose you are in a smoggy city and can barely see buildings 500 m away. b. How far would you be able to see if the temperature suddenly rose from 20°C to a blazing hot 1500°C with the pressure unchanged?

Video transcript

Hey, everyone. Let's go through this practice problem. A group of students was conducting experiments on how far a green light can travel in the water before being completely scattered by water molecules. The student presumed that the mean free path of photon water molecule collisions is equal to the distance that light can travel through water before it scatters completely using a very thin walled transparent container. And at 15 °C, the distance the green light could travel through the water was 20.3 m. If the water is being heated, increasing its temperature to 90 °C while keeping its pressure constant, how far can green light travel before being completely scattered? Hint the mean free path of the photons is inversely proportional to the product of the number density and the square of the radius of the water molecules. Option A 13.6 m, option B 20.3 m, option C 25.6 m and option D 31.2 m. All right. So a few things to know for this problem. First of all, this problem is nice in that it very clearly tells us pretty much everything we need to know first off, we're assuming that the mean free path of the photon molecule. So mean free path is equal to the distance that the light can travel before scattering. So that basically tells us that the distance we're looking for is equal to the weight to the mean free path we wanna find. So as long as we can solve for the mean free path, then we'll be able that that'll just be the answer. The easiest way we can find this mean free path is by using the temperatures that the problem gives us, we're given the temperature for two different cases. And we're given the mean free path for one of those two cases rather the distance it could travel in one of those cases. So if we can consider how the mean free path is proportional to temperature, what the relationship is, we can compare the two different scenarios. And what's convenient is that the hint of the problem reminds us that the mean free path is inversely proportional. I'm going to use the proportional symbol inversely proportional to the product of the number density. So the number of particles per unit volume and the square of the radius of the water molecules. Now, it's important to realize is that since we're only looking, since we're trying to find a relationship here, and since the radius of the water molecules is constant, as far as we know for this problem, we can ignore the R squared relationship there. So what's really only important to us is the inversely proportional to the number density. The one divided by N divided by V, which can be more simply written as V divided by N if we bring the V into the numerator. Now, this probably still doesn't look very helpful right now because this doesn't tell us anything about the temperature. But we, we can relate temperature to the volume and the number of particles by using the ideal gas law. And recall that the ideal gas law states, the pressure multiplied by the volume is equal to the number of particles multiplied by the Boltzmann constant multiplied by the temperature. So if we divide both sides of the equation by N, so, so that we can get a V divided by N term, we can see that if we have V divided by N on one side of the equation, then the temperature T is in the numerator on the other side of the equation. So what this means is that the mean free path is not only proportional to V divided by N, but it's also proportional to the temperature. So let's use this proportionality to concoct a relationship between the temperatures and the free paths for both cases. So the mean free path for one case divided by the mean free path. For another case, I'll call this Lambda prime. This will have the same ratio as the temperatures. So lambda divided by Lambda prime is going to be equal to T divided by T prime. Let's say that the unprimed Lambda is the one we're solving for. So, Lambda represents the mean free path when our temperature T is 90 degrees. So we'll say that the primed Lambda is the 20.3 m, which is the mean free path for when the temperature is 15 °C. So the temperature ratio is equal to and then we're going to convert to Calvin sphere. So we'll take the 90 °C and add 273 Kelvins to that. So our tea becomes 363 Kelvins and our T prime is the other temperature. So 15 °C plus 273 Kelvins, which we find as about 288 Kelvins, we put this into a calculator. We find that this ratio is equal to about 1.26. So in other words, lambda, we multiply both sides of the equation by lambda prime is equal to 1.26 multiplied by Lambda prime or 1.26 multiplied by the mean free path for that case. So 20.3 m put that into a calculator and we find a new mean free path for the 90 degrees case of about 25 0.6 m. And as we established earlier, the distance that we're looking for should be equal to that same mean free path. So the distance that the particle travels is equal to 25.6 m. So that is our answer to the problem. And if we look at our multiple choice options, this agrees with option C so option C is the answer to the problem and that is it for this problem. I hope this video helped you out. If you need more practice, please check out some of our other videos which will give you more experience with these types of problems, but that's all for now. I hope you all have a lovely day. Bye bye.