The mean free path of a molecule is 2.25 $\times$ 10^-7 m in an ideal gas at standard temperature and pressure.What is the mean free path when the pressure of the gas is doubled and the temperature is constant? A) 0.75 $\times$ 10^-7m B) 1.13 $\times$ 10^-7m C) 2.25 $\times$ 10^-7m D) 3.38 $\times$ 10^-7m E) 3.75 $\times$ 10^-7m

Question

Quizplus · Accepted Answer

The mean free path is inversely proportional to pressure, so doubling the pressure will halve the mean free path. Therefore, the new mean free path is:
2.25*10^-7 / 2 = 1.125*10^-7 m or 1.13*10^-7 m. So, the closest option is B.

The Mean Free Path of a Molecule Is 2 ×\times× 10-7 M in an Ideal Gas at Standard Temperature and 10-7

The Mean Free Path of a Molecule Is 2 $\times$ 10^-7 M in an Ideal Gas at Standard Temperature and 10^-7