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Ch.10 - Gases: Their Properties & Behavior

Chapter 10, Problem 108a

Two 112-L tanks are filled with gas at 330 K. One contains 5.00 mol of Kr, and the other contains 5.00 mol of O2. Considering the assumptions of kinetic–molecular theory, rank the gases from low to high for each of the following properties. (a) Collision frequency

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Hi everyone, this problem reads. You have 25 liter sealed containers at 400 kelvin, one contains three moles of helium while the other contains three moles of nitrogen gas, which gas will have a higher average velocity. So we want to know which one will have the higher average velocity and we're dealing with gasses. So the equation that we're going to need to recall in order to solve this is the equation for average velocity and average velocity is equal to the velocity of the root, mean square And that is equal to the square root of three R. T over M. And the variables that we want to pay attention to here are the root mean square and molar mass. As you can see here, they have an inverse relationship. So what that means is as the velocity increases, that means the molar mass decreases. Okay, and so in the problem we're told that the temperature is 400 kelvin for both gasses. And so let's go ahead and look at the right down the molar mass for each of our gasses. So the molar mass for helium is equal to four graham's Permal and the molar mass Of nitrogen gas is equal to 28.02 g per mole. So the question that we want to answer is which one will have the higher average velocity. So because of this inverse relationship, we know that high average velocity is going to equal lower molar mass. So out of the two answer choices, or out of the two molar mass is given, we see that the molar mass for helium is lower. Okay, so this one is the lower molar mass, which means inversely it is the higher velocity. Alright, so out of the answer choices given answer choice, a helium will have, the higher average velocity is going to be our correct answer. That's it for this problem. I hope this was helpful.