Question

Average speed of an air molecule is 485 ms and number density of air molecules is 27 x 1025 m3 If molecule of air has spherical shape
of diameter 2 x 10-70 m. Then average distance between two successive collisions of air molecules is nearly
3.x 10-6 m
o 21 x 10-7 m
O 3 x 10-7m
5-x 10-ºm​

Answer 1

Given info : Average speed of an air molecule is 485 m/s and number density of air molecules is 27 x 10²⁵/m³ If molecule of air has spherical shape of diameter 2 x 10^-7 m.

To find : The average distance between two successive collisions of air molecules is nearly..

solution : mean free path, λ = 1/√2πd²n

where, diameter, d = 2 × 10^-7 m

no of molecules per m³, n = 27 × 10²⁵ /m³

so, λ = 1/(1.414 × 3.14 × 4 × 10¯¹⁴ × 27 × 10²⁵)

= 1/(479.51568 × 10¹¹)

≈ 0.002 × 10¯¹¹

= 2 × 10¯¹⁴

but we know, mean free path is the average distance between two successive collisions.

so, the average distance between two successive collisions is 2 × 10¯¹⁴ m

Answer 2

QUESTION

Average speed of an air molecule is 485 m/s and number density of air molecules is 27 × 10^-25 m3 .If molecule of air has spherical shape of diameter 2 ×10^-7 m. Then average distance between two successive collisions of aur molecules are nearly

SOLUTION -

Given

average speed of air mole cules = 485 m/s

number density of air molecu les, n= 27×10^25 m3

diam eter of molecule , d = 2× 10^-7 m

To Find

mean free path ( l )

l = 1 / ( √2 n π d^2 )

~ I = 1/ ( √2 × 27 × 10^25 × π × 4× 10^-14) m

~ I = 1/ ( 1.414 × 27 × 3.14 × 4 × 10^11 ) m

~ I = 1/ (479.5 × 10^11) m

~ I = 0.00208 × 10^-11 m

~ I = 0. 2 × 10^-11 cm