Question

Calculate the value of mean free path $\left(\lambda \right)$ for oxygen molecules at temperature ${27}^{\circ }\text{C}$ and pressure $1.01\times {10}^5\text{P}a$ . Assume the molecular diameter $0.3\text{nm}$ and the gas is ideal. $(k=1.38\times {10}^{-23}{\text{JK}}^{-1})$

• A. $102\text{nm}$
• B. $32\text{nm}$
• C. $58\text{nm}$
• D. $86\text{nm}$

Answer 1

The correct option is B $32\text{nm}$

Given: $P=1.01\times {10}^5\text{Pa}$
$d=0.3\text{nm}=0.3\times {10}^{-9}\text{m}$
$k=1.38\times {10}^{-23}{\text{JK}}^{-1}$
$T={27}^{\circ }\text{C}=300\text{K}$

We know that:
$${\lambda }_{mean}=\frac{kT}{\sqrt{2}\pi d^{2}P}$$$\Rightarrow {\lambda }_{mean}=\frac{1.38\times {10}^{-23}\times 300}{\sqrt{2}\times 3.14\times 0.09\times {10}^{-18}\times 1.01\times {10}^5}$
$\therefore {\lambda }_{mean}=102\text{nm}$