For diatomic carbon dioxide gas (CO2, molar mass 44.0 g/mol) at T = 300 K, calculate (b) the average speed v_av;

Video transcript

Hey, everyone in this problem, we have oxygen which is a diatomic gass. And with the molar mass of 32 g per mole, and we're asked to determine its average speed at Calvin. OK. All right. So average speed. Let's recall, we can write the average is equal to the square root of eight RT divided by pi. Yeah. OK. R is our gas constant which we know T is the temperature which we've been given. Pi and mm is a molar mass which we've also been given. OK. So you know everything, all of these values we can go ahead and plug them in and find the average what we're looking for. Yeah. So we're gonna get the square root of eight times R which is 8.314 jewels per mole. Kelvin times the temperature 273 Calvin divided by pi. It was a molar mass M. In this case, it's 32 g per mole. We wanna put that into kilograms per mole so that we have our standard units. And so this is gonna be 0.032 kg per mole and just be careful here when you're given molar masses or when you're looking at molar masses, when you're working with diatomic gasses, OK. Oxygen is diatomic. We're given the molar mass. But if you were to look this up in the periodic table, it would show a molar mass of 16 g per mole. OK. For one atom, you would need to multiply that by two for that diatomic molecule. OK. So just be careful when you're working with molar masses of diatomic gasses. All right. So working this out, we are going to have, well, let's look at units first. OK. This per mole, we have an numerator and denominator that will cancel. We have this Calvin divided by Calvin and the numerator that will cancel. So we're gonna be left with the square root of 18, 0.7763 kg meters squared per second squared divided by pi times 0. kg in the denominator. OK? And I've written this stuff so I can show you we can write jewels here as kilogram meter squared per second squared. The unit of kilogram will cancel and we're gonna be left with meter squared per second squared which when we take the square root is gonna be meters per second just like we want for a speed. OK? So the average will be 424.99 m per second. OK? We can round to 425 m per second. So the average speed of the diatomic oxygen molecule at the temperature we were given 273. Kelvin is gonna be 425 m per second. That's answer. C thanks everyone for watching. I hope this video helped see you in the next one.