"Calculate the value of mean free path ($\lambda$) for oxygen molecules at temperature 27$^\circ$C and pressure 1.01 $\times$ 105 Pa. Assume the molecular diameter 0.3 nm and the gas is ideal. (k = 1.38 $\times$ 10$-$23 JK$-$1)"

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Accepted Answer

"${I_{mean}} = {{RT} \over {\sqrt 2 \pi {d^2}{N_A}P}}$

$$ = {{1.38 \times 300 \times {{10}^{ - 23}}} \over {\sqrt 2 \times 3.14 \times {{(0.3 \times {{10}^{ - 9}})}^2} \times 1.01 \times {{10}^5}}}$$

$= 102 \times {10^{ - 9}}$ m

$= 102$ nm"

Calculate the value of mean free path ($\lambda$) for oxygen molecules at temperature 27$^\circ$C and pressure 1.01 $\times$ 10⁵ Pa. Assume the molecular diameter 0.3 nm and the gas is ideal. (k = 1.38 $\times$ 10^$-$23 JK^$-$1)

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Calculate the value of mean free path ($\lambda$) for oxygen molecules at temperature 27$^\circ$C and pressure 1.01 $\times$ 105 Pa. Assume the molecular diameter 0.3 nm and the gas is ideal. (k = 1.38 $\times$ 10$-$23 JK$-$1)

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Calculate the value of mean free path ($\lambda$) for oxygen molecules at temperature 27$^\circ$C and pressure 1.01 $\times$ 10⁵ Pa. Assume the molecular diameter 0.3 nm and the gas is ideal. (k = 1.38 $\times$ 10^$-$23 JK^$-$1)