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The Average Translational Kinetic Energy of Air Molecules is 0.040 Ev (1 Ev = 1.6 × 10−19j). Calculate the Temperature of the Air. Boltzmann Constant K = 1.38 × 10−23 J K−1.

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प्रश्न

The average translational kinetic energy of air molecules is 0.040 eV (1 eV = 1.6 × 10−19J). Calculate the temperature of the air. Boltzmann constant k = 1.38 × 10−23 J K−1.

योग
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उत्तर

We know from kinetic theory of gases that the average translational energy per molecule is \[\frac{3}{2}kT\].

Now,
Eavg= 0.040 eV = \[0.040 \times 1 . 6 \times  {10}^{- 19}  = 6 . 4 \times  {10}^{- 21} J\] 

\[6 . 40 \times  {10}^{- 21}  = \frac{3}{2} \times 1 . 38 \times  {10}^{- 23}  \times T\] 

\[ \Rightarrow T = \frac{2}{3} \times \frac{6 . 40 \times {10}^{- 21}}{1 . 38 \times {10}^{- 23}} = 309 . 2  \text { K }\]

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अध्याय 24: Kinetic Theory of Gases - Exercises [पृष्ठ ३४]

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एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
अध्याय 24 Kinetic Theory of Gases
Exercises | Q 14 | पृष्ठ ३४

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