Calculate the mean free path in a sample of oxygen gas (M = 3.20 × 10^-2 kg/mol) at 35°C and 1.7 atm. Assume that the diameter of an O₂ molecule is 300 pm.

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Quizplus · Accepted Answer

The mean free path is given by the equation:$$\lambda = \frac{k_BT}{4\pi r^2 P N_A}$$where:* $k_B$ is the Boltzmann constant ($1.38 	imes 10^{-23}$ J/K)* $T$ is the temperature in Kelvin (35°C = 308 K)* $r$ is the radius of the molecule ($300$ pm = $3 	imes 10^{-10}$ m)* $P$ is the pressure in Pascals (1.7 atm = $1.7 	imes 101325$ Pa)* $N_A$ is the Avogadro constant ($6.022 	imes 10^{23}$ mol$^{-1}$)Substituting these values into the equation, we get:$$\lambda = \frac{(1.38 	imes 10^{-23} 	ext{ J/K})(308 	ext{ K})}{4\pi (3 	imes 10^{-10} 	ext{ m})^2 (1.7 	imes 101325 	ext{ Pa})(6.022 	imes 10^{23} 	ext{ mol}^{-1})}$$$$\lambda = 6.0 	imes 10^{-7} 	ext{ m}$$Therefore, the mean free path is $6.0 	imes 10^{-7}$ m, which is closest to choice A, $0.6 	imes 10^{-7}$ m.

Calculate the Mean Free Path in a Sample of Oxygen