Question

the diameter of a molecule of argon gas is d .find the mean free path of molecules of argon at 37°C temperature and 10^5 PA pressure.(KB = Boltzmann constant )​

Answer 1

Given :

Diameter of molecule = d

Temperature = 37°C

Pressure = 10⁵ Pa

To Find :

Mean free path of argon molecules.

Solution :

❖ The mean free path is the average distance travelled by a moving molecule between collisions.

Formula :

$\dag\:\underline{\boxed{\bf{\purple{\lambda=\dfrac{K_BT}{\sqrt2P\pi d^2}}}}}$

• $\sf{K_B}$ denotes Boltzmann constant
• T denotes temperature
• P denotes pressure
• d denotes diameter

By substituting the given values;

$\sf:\implies\:\lambda=\dfrac{K_BT}{\sqrt2P\pi d^2}$

• Temperature = 37°C = 310K

$\sf:\implies\:\lambda=\dfrac{1.38\times 10^{-23}\times 310}{\sqrt2(10^5)(3.14)d^2}$

$\sf:\implies\:\lambda=\dfrac{96.62\times 10^{-23}}{(10^5)d^2}$

$:\implies\:\underline{\boxed{\bf{\orange{\lambda=\dfrac{9.66\times10^{-27}}{d^2}\:m}}}}$

Answer 2

Given :

Diameter of molecule = d

Temperature = 37°C

Pressure = 10⁵ Pa

To Find :

Mean free path of argon molecules.

Solution :

❖ The mean free path is the average distance travelled by a moving molecule between collisions.

Formula :

$\dag\:\underline{\boxed{\bf{\purple{\lambda=\dfrac{K_BT}{\sqrt2P\pi d^2}}}}}$

$\sf{K_B}$ denotes Boltzmann constant

T denotes temperature

P denotes pressure

d denotes diameter

By substituting the given values;

$\sf:\implies\:\lambda=\dfrac{K_BT}{\sqrt2P\pi d^2}$

Temperature = 37°C = 310K

$\sf:\implies\:\lambda=\dfrac{1.38\times 10^{-23}\times 310}{\sqrt2(10^5)(3.14)d^2}$

$\sf:\implies\:\lambda=\dfrac{96.62\times 10^{-23}}{(10^5)d^2}$

$:\implies\:\underline{\boxed{\bf{\orange{\lambda=\dfrac{9.66\times10^{-27}}{d^2}\:m}}}}$