If the radius of an O_{2} molecule is 40 ; ma | vedclass

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(B) The relaxation time $	au$ is given by the ratio of the mean free path $\lambda$ to the root mean square velocity $V_{rms}$.$	au = \frac{\lambda}

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$10^{-12} \; 	ext{s}$

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(B) The relaxation time $	au$ is given by the ratio of the mean free path $\lambda$ to the root mean square velocity $V_{rms}$.$	au = \frac{\lambda}{V_{rms}} = \frac{1}{\sqrt{2} \pi n d^2} \cdot \frac{1}{\sqrt{3RT/M_0}} = \frac{1}{\sqrt{2} \pi n d^2} \sqrt{\frac{M_0}{3RT}}$Using the ideal gas law $PV = n_{mol}RT$,where $n = N/V = (n_{mol} N_A)/V = (P N_A)/(RT)$.Substituting $n$ into the expression:$	au = \frac{RT}{\sqrt{2} \pi P N_A d^2} \sqrt{\frac{M_0}{3RT}} = \frac{1}{\sqrt{2} \pi P N_A d^2} \sqrt{\frac{M_0 RT}{3}}$Given: $d = 2 	imes 	ext{radius} = 80 \; \mathring{A} = 8 	imes 10^{-9} \; 	ext{m}$,$P = 1.013 	imes 10^5 \; 	ext{Pa}$,$T = 300 \; 	ext{K}$,$M_0 = 32 	imes 10^{-3} \; 	ext{kg/mol}$,$N_A = 6.022 	imes 10^{23} \; 	ext{mol}^{-1}$,$R = 8.314 \; 	ext{J/(mol K)}$.Substituting these values into the formula yields $	au \approx 10^{-12} \; 	ext{s}$.

If the radius of an $O_{2}$ molecule is $40 \; \mathring{A}$,$T = 27^{\circ}C$,and $P = 1 \; \text{atm}$,find the relaxation time.

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