Question 163: Thermodynamics & KTG

Question

The mean free path of a molecule of diameter $5 	imes 10^{-10}$ m at the temperature $41^{\circ}$C and pressure $1.38 	imes 10^5$ Pa, is given as ________ m.  (Given $k_B = 1.38 	imes 10^{-23}$ J/K).

Accepted Answer

The diameter of molecule is, $d=5 	imes 10^{-10} \mathrm{~m}$
Temperature, $\mathrm{T} 41^{\circ} \mathrm{C}=(41+273) \mathrm{K}=314 \mathrm{~K}$
Pressure, $\mathrm{P}=1.38 	imes 10^5 \mathrm{~Pa}$
Boltzmann constant, $\mathrm{k}_{\mathrm{B}}=1.38 	imes 10^{-23} \mathrm{~J} / \mathrm{K}$
The standard formula for the mean free path $\lambda$ is:
$$ \lambda=\frac{\mathrm{k}_{\mathrm{B}} \mathrm{~T}}{\sqrt{2} \pi \mathrm{~d}^2 \mathrm{P}} $$
Substituting the values,
$$ \lambda=\frac{\left(1.38 	imes 10^{-23}ight) 	imes 314}{\sqrt{2} 	imes \pi 	imes\left(5 	imes 10^{-10}ight)^2 	imes\left(1.38 	imes 10^5ight)} $$
$$\Rightarrow $$ $\lambda=\frac{10^{-23} 	imes 314}{\sqrt{2} 	imes \pi 	imes 25 	imes 10^{-20} 	imes 10^5}$
$$\Rightarrow $$ $\lambda=\frac{314}{\sqrt{2} 	imes \pi 	imes 25 	imes 10^{-8}}$
Using $\pi \approx 3.14$ :
$$ \lambda=\frac{314}{\sqrt{2} 	imes 3.14 	imes 25 	imes 10^{-8}} $$
$$\Rightarrow $$ $\lambda=\frac{4}{\sqrt{2}} 	imes 10^{-8}=2 \sqrt{2} 	imes 10^{-8} \mathrm{~m}$
Hence, the correct option is D.

Answer

$2\sqrt{2} 	imes 10^{-10}$

Answer

$10\sqrt{2} 	imes 10^{-8}$

Answer

$2\sqrt{2} 	imes 10^{-8}$

Answer

$2 	imes 10^{-8}$

Thermodynamics & KTG

The mean free path of a molecule of diameter 5 \times 10^{-10} m at the temperature 41^{\circ}C and pressure 1.38 \times 10^5 Pa, is given as ________ m. (Given k_B = 1.38 \times 10^{-23} J/K).

Choose the correct answer: