The mean free path of a molecule is $2.25 \times 10^{-7} \mathrm{~m}$ in an ideal gas at standard temperature and pressure $\left(0^{\circ} \mathrm{C}\right.$ and $1 \mathrm{~atm}$ ). What is the mean free path if the diameter of the molecule and the temperature of the gas are both doubled?
A) $1.13 \times 10^{-7} \mathrm{~m}$
B) $2.25 \times 10^{-7} \mathrm{~m}$
C) $3.00 \times 10^{-7} \mathrm{~m}$
D) $3.75 \times 10^{-7} \mathrm{~m}$
E) $0.75 \times 10^{-7} \mathrm{~m}$

Question

Quizplus · Accepted Answer

Doubling the diameter of the molecule increases the collision cross-section, reducing the mean free path by a factor of 4. Doubling the temperature increases the mean free path by a factor of $\sqrt{2}$, due to increased kinetic energy. Overall, the mean free path is halved: $2.25 \times 10^{-7} \mathrm{~m} / 2 = 1.13 \times 10^{-7} \mathrm{~m}$.

The Mean Free Path of a Molecule Is 2.25×10−7 m2.25 \times 10^{-7} \mathrm{~m}2.25×10−7 m

The Mean Free Path of a Molecule Is $2.25 \times 10^{-7} \mathrm{~m}$