Calculate the value of the mean free path ($$\lambda$$) for oxygen molecules at temperature 27°C and pressure $$1.01 \times 10^{5}$$ Pa. Assume the molecular diameter 0.3 nm and the gas is ideal. ($$k = 1.38 \times 10^{-23}$$ J K$$^{-1}$$)

The mean free path of a gas molecule is given by $$\lambda = \frac{kT}{\sqrt{2}\,\pi d^2 P}$$, where $$k$$ is Boltzmann's constant, $$T$$ is temperature, $$d$$ is the molecular diameter, and $$P$$ is pressure.

Substituting the given values: $$T = 27°\text{C} = 300$$ K, $$d = 0.3$$ nm $$= 0.3 \times 10^{-9}$$ m, $$P = 1.01 \times 10^5$$ Pa, and $$k = 1.38 \times 10^{-23}$$ J/K.

The numerator is $$kT = 1.38 \times 10^{-23} \times 300 = 4.14 \times 10^{-21}$$ J.

The denominator is $$\sqrt{2}\,\pi d^2 P = 1.414 \times 3.1416 \times (0.3 \times 10^{-9})^2 \times 1.01 \times 10^5$$. Computing $$d^2 = 9 \times 10^{-20}$$ m$$^2$$, so the denominator becomes $$1.414 \times 3.1416 \times 9 \times 10^{-20} \times 1.01 \times 10^5 = 1.414 \times 3.1416 \times 9.09 \times 10^{-15} = 4.037 \times 10^{-14}$$.

Therefore, $$\lambda = \frac{4.14 \times 10^{-21}}{4.037 \times 10^{-14}} = 1.026 \times 10^{-7}$$ m $$\approx 102$$ nm.