Final answer:
A molecule with double the radius and eight times the mass of another will have a shorter mean free path and a shorter mean free time under the same conditions because the mean free path is inversely proportional to the square of the radius, and mean free time is also reduced due to an increased rms speed resulting from the increase in mass.
Step-by-step explanation:
When comparing the mean free paths of two molecules under the same conditions, where one molecule has double the radius and eight times the mass of the other, there are a couple of key relationships to consider. First, the mean free path (λ), which is the average distance traveled between collisions, is inversely proportional to the cross-sectional area (which depends on the square of the radius, r). If one molecule's radius is doubled, the cross-sectional area increases by a factor of 4, reducing the mean free path by the same factor.
For the mean free time (t), this is proportional to the mean free path and inversely proportional to the root-mean-square (rms) speed, which is inversely proportional to the square root of the mass. Since the molecule with double the radius also has eight times the mass, this increases the rms speed by a factor of √2. Therefore, even though the mean free path is reduced, the increase in speed reduces the mean free time even more. Overall, the mean free time decreases by a factor of √(1/2), or divides by the square root of two (√2).
With these principles in mind, we can conclude that the molecule with double the radius and eight times the mass will have a shorter mean free path and a shorter mean free time than the other molecule under the same conditions.