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प्रश्न
Estimate the average thermal energy of a helium atom at room temperature (27 °C).
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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.
संबंधित प्रश्न
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Match the following:
|
|
Column A |
Column B |
|
(a) |
cm3 |
(i) Pressure |
|
(b) |
Kelvin |
(ii) Temperature |
|
(c) |
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|
(d) |
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|
(a) |
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|
|
|
(vi) PV = P1V1 |
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