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