Calculate the mean free path of air molecules at 3.50 * 10^-13 at...

Question

Calculate the mean free path of air molecules at 3.50 * 10^-13 atm and 300 K. (This pressure is readily attainable in the laboratory; see Exercise 18.23.) As in Example 18.8, model the air molecules as spheres of radius 2.0 * 10^-10 m.

Accepted Answer

Welcome back everybody. We are looking at molecules inside of an evacuated container and we are told that the pressure inside is 4. times 10 to the negative 11th 80 m. We're told that the temperature inside is kelvin. And we are told to assume that the molecules are spherical with a radius of 1.6 times 10 to the negative 10th meters. And we are tasked with finding the out long. The mean free path is of the air molecules. Now the formula for the mean free path is the velocity of the molecule times the mean free time. Now we can't really work with those variables. But I'm gonna make a couple of substitution is from from our lessons here, we can say that this is equal to the volume of the container divided by four pi times the square root of two times our radius squared times the number of molecules in the container. Now this is equivalent to our bolts. Hman constant Times are absolute temperature or all divided by four pi times the square root of two times our radius squared times pressure. Which we get from our ideal gas law Now that we have this formula, we can go ahead and plug in all your values except for one thing here, we do need to make sure we're working in the correct units and our pressure needs to be in pascal's. So for 1 80 M, there is 1.13 times 10 to the fifth pascal's crossing out our units here in multiplying across. We get that the pressure inside of the evacuated container is 4.25 times 10 to the negative six pascal's. Now. We are ready to plug in our numbers into our formula for the mean free path. So we have on top are bolts, man constant of 1.38 times 10 to the negative 23rd times are absolute temperature of Calvin. All divided by four pi times the square root of two times our radius of 1.6 times 10 to the negative 10th times are pressure of 4.25 times 10 to the negative six pascal's. And when you multiply all this out, we get that are mean free path is 2127 m long, corresponding to our answer. Choice of a thank you all so much for watching. Hope this video helped. We will see you all in the next one.

Calculate the mean free path of air molecules at 3.50 * 10^-13 atm and 300 K. (This pressure is readily attainable in the laboratory; see Exercise 18.23.) As in Example 18.8, model the air molecules as spheres of radius 2.0 * 10^-10 m.

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