For diatomic carbon dioxide gas (CO2, molar mass 44.0 g/mol) at T...

Question

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

Accepted Answer

Hey everyone in this problem we have oxygen. Which is a di atomic gas with the molar mass of 32 g per mole. And were asked to determine its average speed at 273 Calvin. Okay. Alright, so average speed, let's recall. We can write the average is equal to the square root of eight R. T. Divided by pi. But yeah our is our gas constant which we know t. Is the temperature which we've been given pie and mm is a molar mass which we've also been given. Okay, 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. So we're gonna get the swear route of eight times r which is 8.314 jewels per mall Calvin Times the temperature 273 Calvin divided by pi there's a molar mass m in this case it's 32 g from all we want to put that into kilograms from. Also that we have our standard units. So this is going to be 0.32 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 di atomic gasses, okay, oxygen is di atomic were given the molar mass. But if you were to look this up in the periodic table would show a molar mass of 16 g per mole. Okay, For one atom you would need to multiply that by two for that. Di atomic molecules. Okay, so just be careful when you're working with Mueller masses of di atomic gasses. Alright so working this out, we are going to have well let's look at units first. Okay this per mole. We have an n 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 18,157 0.776 kg meters squared per second squared divided by pi time. 0.32 kg in the denominator. Okay 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. We're gonna be left with meters squared per second squared. Which when we take the square root is going to be meters per second. Just like we want for a speed. So the average will be 424.99 m per second. Okay we can round to 425 m per second. The average speed of the di atomic oxygen molecule at the temperature we were given 273 Calvin is going to be 425 m per second. That's answer. C thanks everyone for watching. I hope this video helped see you in the next one

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