Q26 / 125 · BITSAT 2016 · Chemistry

The average kinetic energy of an ideal gas per molecule in SI unit at $25^{\circ} \mathrm{C}$ will be

  • A
    $6.17 \times 10^{-21} \mathrm{~kJ}$
  • Correct answer
    B
    $6.17 \times 10^{-21} \mathrm{~J}$
  • C
    $6.17 \times 10^{-20} \mathrm{~J}$
  • D
    $7.16 \times 10^{-20} \mathrm{~J}$

Solution

$\overline{\mathrm{KE}}=\frac{3}{2} \mathrm{kT}=\frac{3}{2} \times \frac{8.313}{6.023 \times 10^{23}} \times 298=6.17 \times 10^{-21} \mathrm{~J}$ (Average Kinetic energy $\left.\overline{\mathrm{KE}}=\frac{3}{2} \mathrm{k} \mathrm{T}=\frac{3}{2} \frac{\mathrm{R}}{\mathrm{N}} \mathrm{T}\right)$

Hence, the correct answer is option: 'B'.

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