A) The mean speed of the molecules of an ideal gas is 2.0 x 103 ms. Th

1.	A) The mean speed of the molecules of an ideal gas is 2.0 x 103 ms. The radiusof 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» Given : The MEAN speed of the molecules of an ideal gas is $<klux>2</klux>.0 \times 10^3$ 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