Calculate the value of mean free path ( λ) for oxygen molecules at temperature 270C and pressure 1.01 ×105 Pa. Assume the molecular diameter 0.3 nm and the gas is ideal. (k=1.38×10−23JK−1).

λmean =12πd2NV & PV=NKBT∴λmean =KBT2πd2P =1.38×10−23×3002×3.14×0.09×10−18×1.01×105=102nm

Calculate the value of mean free path ( λ) for oxygen molecules at temperature 270C and pressure 1.01 ×105 Pa. Assume the molecular diameter 0.3 nm and the gas is ideal. (k=1.38×10−23JK−1).

λmean =12πd2NV & PV=NKBT∴λmean =KBT2πd2P =1.38×10−23×3002×3.14×0.09×10−18×1.01×105=102nm

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Calculate the value of mean free path ( $λ$ ) for oxygen molecules at temperature $27^{0} C$ and pressure 1.01 $\times 10^{5}$ Pa. Assume the molecular diameter 0.3 nm and the gas is ideal. $(k = 1.38 \times 10^{- 23} J K^{- 1})$ .

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102 nm

86 nm

32 nm

58 nm

answer is D.

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Detailed Solution

$\begin{array}{l} λ_{mean} = \frac{1}{\sqrt{2} π d^{2} (\frac{N}{V})} & P V = N K_{B} T \\ ∴ λ_{mean} = \frac{K_{B} T}{\sqrt{2} π d^{2} P} = \frac{1.38 \times 10^{- 23} \times 300}{\sqrt{2} \times 3.14 \times 0.09 \times 10^{- 18} \times 1.01 \times 10^{5}} = 102 n m \end{array}$