Question

If the mean free-path of gaseous molecule is 60 cm at a pressure of $1\times {10}^{-4}$ mm mercury, what will be its mean free-path when the pressure is increased to $1\times {10}^{-2}$ mm mercury?

• A. $6.0\times {10}^{-1}cm$
• B. $6.0cm$
• C. $6.0\times {10}^{-2}cm$
• D. $6.0\times {10}^{3}cm$

Answer 1

Answer: (A)

$6.0\times {10}^{-1}cm$

As we know,

$\lambda \propto \frac{1}{P}$

$\therefore \frac{{\lambda }_1}{{\lambda }_2}=\frac{P_2}{P_1}$

${\lambda }_2={\lambda }_1\cdot P_1/P_2=\frac{60\times {10}^{-4}}{{10}^{-2}}=0.6cm$