The mean free path of a molecule of diameter $5 \times 10^{- 10} m$ at the temperature $41 ° C$ and pressure $1.38 \times 10^{5} Pa$ is given as ______ m.

(Given $k_{B} = 1.38 \times 10^{- 23} J/K$ ) [2026]

Question

The mean free path of a molecule of diameter $5 \times 10^{- 10} m$ at the temperature $41 ° C$ and pressure $1.38 \times 10^{5} Pa$ is given as ______ m.

(Given $k_{B} = 1.38 \times 10^{- 23} J/K$ ) [2026]

Smart Achievers · Accepted Answer

(2)

$λ = \frac{k_{B} T}{\sqrt{2} π σ^{2} P}$

$= \frac{1.38 \times 10^{- 23} \times (273 + 41) \times 100}{\sqrt{2} \times 3.14 \times {(5 \times 10^{- 10})}^{2} \times 1.38 \times 10^{5}}$

$= 2 \sqrt{2} \times 10^{- 8}$

Solution

Q.
The mean free path of a molecule of diameter $5 \times 10^{- 10} m$ at the temperature $41 ° C$ and pressure $1.38 \times 10^{5} Pa$ is given as ______ m.

(Given $k_{B} = 1.38 \times 10^{- 23} J/K$ ) [2026]

1 $10 \sqrt{2} \times 10^{- 8}$

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

3 $2 \times 10^{- 8}$

4 $2 \sqrt{2} \times 10^{- 10}$

Ans.
(2)

$λ = \frac{k_{B} T}{\sqrt{2} π σ^{2} P}$

$= \frac{1.38 \times 10^{- 23} \times (273 + 41) \times 100}{\sqrt{2} \times 3.14 \times {(5 \times 10^{- 10})}^{2} \times 1.38 \times 10^{5}}$

$= 2 \sqrt{2} \times 10^{- 8}$

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Solution

Q. The mean free path of a molecule of diameter 5×10-10 m at the temperature 41°C and pressure 1.38×105 Pa is given as ______ m. (Given kB=1.38×10-23 J/K) [2026]

1 102×10-8

2 22×10-8

3 2×10-8

4 22×10-10

Ans. (2) λ=kBT2πσ2P =1.38×10-23×(273+41)×1002×3.14×(5×10-10)2×1.38×105 =22×10-8