Question

The mean speed of oxygen molecules is 450 ms-1. If the radius of an oxygen molecule is 1.8 Å, calculate mean time between two successive collisions and mean free path. Take .m103325

Answer 1

Answer on Question #66742-Physics-Molecular Physics-Thermodynamics

The mean speed of oxygen molecules is 450 ms-1. If the radius of an oxygen molecule is 1.8 Å, calculate mean time between two successive collisions and mean free path. Take .m103325

Solution

The mean free path is

$$l = \frac {k T}{\sqrt {2} \pi d ^ {2} P}$$ $$v _ {a v} = \sqrt {\frac {8 R T}{\pi M}} \rightarrow T = \frac {v _ {a v} ^ {2} \pi M}{8 R}$$ $$l = \frac {k}{\sqrt {2} \pi d ^ {2} P} \frac {v _ {a v} ^ {2} \pi M}{8 R} = \frac {1}{\sqrt {2} 4 r ^ {2} P} \frac {v _ {a v} ^ {2} M}{8 N _ {a}}$$ $$l = \frac {v _ {a v} ^ {2} M}{3 2 \sqrt {2} r ^ {2} P N _ {a}} = \frac {(4 5 0) ^ {2} (0 . 0 3 2)}{3 2 \sqrt {2} (1 . 8 \cdot 1 0 ^ {- 1 0}) ^ {2} (1 0 3 3 2 5) (6 . 0 2 2 \cdot 1 0 ^ {2 3})} = 7. 1 \cdot 1 0 ^ {- 8} m = 7 1 n m.$$

The time between two successive collisions is

$$t = \frac {l}{v _ {r m s}}$$ $$v _ {r m s} = \sqrt {\frac {3 R T}{M}} = \sqrt {\frac {3 \pi}{8}} v _ {a v}$$ $$t = \sqrt {\frac {8}{3 \pi}} \frac {l}{v _ {a v}} = \sqrt {\frac {8}{3 \pi}} \frac {(7 . 1 \cdot 1 0 ^ {- 8})}{(4 5 0)} = 4. 6 \cdot 1 0 ^ {- 1 0} s.$$

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