Question

The mean free path of a gas molecule at STP is $2.1\times 10^{-7}m$. Find the diameter of the molecule. (Boltzmann constant $=1.4\times 10^{-23}J/K$ )

• A. $5.2\mathring{A}$
• B. $4.6\mathring{A}$
• C. $3.5\mathring{A}$
• D. $2.1\mathring{A}$

Answer 1

Answer: (D)

The given, $\lambda =2.1\times 10^{-7}m$
$K_B=1.4\times 10^{-23}J/K$
For STP $T=0^{\circ }C+273=273K$
$P=1atm=1.01\times 10^{5}N/m^2$
$d=?$
$\lambda =\frac{K_{B}T}{\sqrt{2}\pi d^2\times P}$
$=\frac{1.4\times 10^{-23}\times 273}{1.41\times 3.14\times 2.1\times 10^{-7}\times 1.01\times 10^5}$
$=\frac{382.2}{9.39}\times 10^{-21}$
$d^2=40.7\times 10^{-21}=4.07\times 10^{-20}$
$d=\sqrt{4.07\times 10^{-20}}=2.01\times 10^{-10}$
$d=2.01\mathring{A}$
$d\cong 2.1\mathring{A}$