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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.
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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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संबंधित प्रश्न
Figure shows two vessels A and B with rigid walls containing ideal gases. The pressure, temperature and the volume are pA, TA, V in the vessel A and pB, TB, V in the vessel B. The vessels are now connected through a small tube. Show that the pressure p and the temperature T satisfy `Ρ/T = 1/2 ({P_A}/{T_A}+{P_B}/{T_B))` when equilibrium is achieved.
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