At what temperature is the kinetic energy of a gas molecule half its value at $327^{\circ} C$ ?

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Solution

The correct option is B

KE per molecule $= \frac{3}{2} k T$

$\frac{E_{2}}{E_{1}} = \frac{(\frac{3}{2} T_{2})}{(\frac{3}{2} T_{1})} = \frac{T_{2}}{T_{1}} E_{1} = E E_{2} = (\frac{E}{2}) T_{1} = 600 K T_{2} = ? [\frac{\frac{E}{2}}{E}] = \frac{T_{2}}{600} ∴ T_{2} = 300 K (o r) 27^{\circ} C$

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327^{\circ} C

327^{\circ} C

At what temperature KE of a gas molecules is half of its value at $327^{\circ} c ?$

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