Question

a) The mean speed of the molecules of an ideal gas is 2.0 x 103 ms. The radius
of a gas molecule is 1.5 x 10-10m. Calculate the (i) collision frequency, and
(ii) mean free path. It is given that n = 4 x 1024 m-3​

Answer 1

Given :

The mean speed of the molecules of an ideal gas is $2.0 \times 10^3$ ms.

i.e v = $2.0 \times 10^3$

The radius of a gas molecule is $1.5 \times 10^{-10}$ m.

i.e. r =

$n = 4 \times 10^{24} m^3$

To find:

1) Collision frequency gas of molecules of an ideal gas?

2) mean free path of an ideal gas?

Step to step explanation:

1) The collision frequency gas of molecules of an ideal :

$\displaystyle {z= 4\pi r^2nv}$

z = 4 × 3.14 × $1.5 \times 10^{-10}$ × $1.5 \times 10^{-10}$  × $4 \times 10^{24}$ × $2.0 \times 10^3$

z = 2.26 ×$\mahbf {10^9}$ $\mathbf hz$

2) The mean free path of an ideal gas :

$\displaystyle \lambda = \frac{1}{4\pi r^2n}$

$\displaystyle \lambda = \frac{1}{4 \times \pi \times 1.5 \times 10^{-10} \times 1.5 \times 10^{-10} \times 4 \times 10^{24} }$

λ = 0.88 × $\displaystyle 10^{-6}$ m