Question:
If the mean free path of a gas molecule at $27^{\circ}\text{C}$ is $10\times10^{-7}\ \text{m}$, its mean free path at $87^{\circ}\text{C}$ is:
If the mean free path of a gas molecule at $27^{\circ}\text{C}$ is $10\times10^{-7}\ \text{m}$, its mean free path at $87^{\circ}\text{C}$ is:
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Always convert temperature to Kelvin for gas law calculations.
Updated On: Apr 28, 2026
- $12\times10^{-7}\ m$
- $8\times10^{-7}\ m$
- $6\times10^{-7}\ m$
- $10\times10^{-7}\ m$
- $14\times10^{-7}\ m$
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The Correct Option is A
Solution and Explanation
Step 1: Concept
Mean free path $\lambda$ is directly proportional to absolute temperature $T$ (at constant pressure).
Step 2: Analysis
$T_1 = 27 + 273 = 300\ K$. $T_2 = 87 + 273 = 360\ K$.
Step 3: Calculation
$\frac{\lambda_2}{\lambda_1} = \frac{T_2}{T_1} \implies \lambda_2 = 10\times10^{-7} \times \frac{360}{300} = 12\times10^{-7}\ m$. Final Answer: (A)
Mean free path $\lambda$ is directly proportional to absolute temperature $T$ (at constant pressure).
Step 2: Analysis
$T_1 = 27 + 273 = 300\ K$. $T_2 = 87 + 273 = 360\ K$.
Step 3: Calculation
$\frac{\lambda_2}{\lambda_1} = \frac{T_2}{T_1} \implies \lambda_2 = 10\times10^{-7} \times \frac{360}{300} = 12\times10^{-7}\ m$. Final Answer: (A)
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