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 (a) the most probable speed v_mp;

Accepted Answer

Hey everyone in this problem we are told that the di atomic nitrogen has a molar mass of 28.1 g per mole. It's an inert gas used involves to prevent evaporation and oxidation of the filament. Okay. We're asked to find the most probable speed of nitrogen at 420 Calvin. Alright, so we're talking about the most probable speed. That's what we want to find. Okay, now let's recall that the most probable speed V. M. P can be written as the square root Of two K. T. over them. Well, this is equal to because of the relationship between K over M and R over big M. We can write this as two big R. Of T over big M. Alright. So in this case we have are the gas constant which we know t the temperature we've been given and big M the molar mass that we've been given. Okay, we couldn't use the equation on the left because we didn't have the mass. Little M. Alright, so this is what we're looking for. This most probable speed V. M. P. Let's go ahead and substitute the values that we know. So we have the square root of two times R which is 8.314 jewels per mole kelvin Many Times Calvin case of the units of Calvin will cancel. We have divided by M Which is 28.01 g per mole. In this case we want to convert that into kg per mole. So that we're in our standard unit. So this is going to be 0.0- kg per mole. Okay so the per mole from the numerator will cancel with the per mole from the denominator. Alright, great. Now one thing I want to mention here is just to be careful with your molar mass when you have di atomic gasses. Okay, so here we have nitrogen which we're told is di atomic we've been given the molar mass. But if you're looking up the molar mass in the periodic table, it's only gonna be for a single atom, not for the di atomic molecules. Okay, so you're going to have to double it in order to get the molar mass of the di atomic molecules. Okay, so just be careful there. Alright so working this out, we're going to get the square root. 6983.76. Okay. The unit we have remaining in the numerator is jules recall that we can write jewel as kilogram meter squared per second squared. We're going to divide by 0.0 to kg. Okay, when we write jewel like that we can see that the unit of kilogram will cancel. We're left with meter squared per second squared. Which when we take the square root is gonna give us a unit of meters per second which is exactly what we want for a speed. Okay so taking the square root we get V M p is equal to 499.33 m per second. Okay. And so that most probable speed of the di atomic nitrogen at 420 Calvin is 499.33 m per second. Okay, So that's going to correspond with answer A when we round to 499 m per second. Thanks everyone for watching. See you in the next video.

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