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Estimate the average thermal energy of a helium atom at room temperature (27 °C). - Physics

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प्रश्न

Estimate the average thermal energy of a helium atom at room temperature (27 °C).

संख्यात्मक
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उत्तर

At room temperature, T = 27°C = 300 K

Average thermal energy = `3/2` kT

Where k is Boltzmann constant = 1.38 × 10–23 m2 kg s–2 K–1

`therefore 3/2kT = 3/2 xx 1.38 xx 10^(-38) xx 300`

= 6.21 × 10–21J

Hence, the average thermal energy of a helium atom at room temperature (27°C) is 6.21 × 10–21 J.

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Gases and Its Characteristics
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