The mean free path of a molecule of diameter 5 10-10 m at the temperature 41 C and pressure 1.38 105 Pa, is given as m. (Given kB = 1.38 10-23 J/K).
- 2 2 10-10
- 10 2 10-8
- 2 2 10-8
- 2 10-8
Answer
(3) 2 2 10-8
Solution
= kB T 2 d2 P, T = 41 + 273 = 314 K
kB TP = 1.38 10-23 3141.38 105 = 3.14 10-26 m3
d2 = 3.14 25 10-20 m2
= 3.14 10-26 2 3.14 25 10-20
= 4 10-8 2 = 2 2 10-8 m
Answer: (3)