WF6 is one of the heaviest known gases. How much slower is the root-mean-square speed of WF6 than He at 300 K?

Video transcript

Hi everyone here we have a question asking us to calculate the ratio between the velocity of Xenon and nitrogen at 273 Kelvin. So we're going to use the formula velocity equals the square root as three times are gas constant times our temperature divided by our molar mass. So for zenon That is going to be three Times 8. kilograms times meters square, divided by moles times kelvin times second squared Times 273 Kelvin over our molar mass. And since our gas constant is in kilograms, we want our molar mass to be in kilograms as well. So that will be 0.1312, nine kilograms per mole And that equals 0.74 meters per second. Now for nitrogen, That will be three times 8. kilograms times meters squared, divided by moles, times kelvin times seconds squared Times 273 Kelvin over into moller mass in kilograms, Which is 0.0 to a 02 kilograms thermal. And that equals .96 m/s. And now we want to find the ratio so to do that, all we're going to do is take our 492 0. meters per second And divide it by our 227 .74 m/s And that equals 2. and that is our final answer. Thank you for watching. Bye