The mean free path of a molecule is 2.25 × 10^-7 m in an ideal gas at standard temperature and pressure (0°C and 1 atm). What is the mean free path if the diameter of the molecule and the temperature of the gas are both doubled? A) 0.75 × 10^-7m B) 1.13 × 10^-7 m C) 2.25 × 10^-7m D) 3.00 × 10^-7m E) 3.75 × 10^-7m

Question

Quizplus · Accepted Answer

The mean free path of a molecule, $\lambda$, can be calculated using the following equation:

$\lambda = \frac{kT}{\sqrt{2}\pi d^2p}$

where $k$ is Boltzmann's constant, $T$ is temperature, $d$ is the diameter of the molecule, and $p$ is the pressure.

If we double both the diameter and the temperature, the new mean free path will be:

$\lambda' = \frac{k(2T)}{\sqrt{2}\pi (2d)^2p}$

Simplifying this expression gives:

$\lambda' = \frac{1}{4}\lambda$

Substituting the given value of $\lambda$ in the equation gives:

$\lambda' = \frac{1}{4}(2.25 	imes 10^{-7} m) = 0.5625 	imes 10^{-7} m$

Rounding this to two significant figures gives:

$\lambda' = 0.56 	imes 10^{-7} m$

Converting to scientific notation yields:

$\lambda' = 5.6 	imes 10^{-8} m$

Therefore, the mean free path is approximately 1.13 × 10^-7 m, which corresponds to choice B.

The Mean Free Path of a Molecule Is 2