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BITSAT 2016
Exam
BITSAT
Year
2016
Subject
Chemistry
Chapter
States of Matter

BITSAT 2016 · Chemistry · States of Matter

The average kinetic energy of an ideal gas per molecule in SI unit...

The average kinetic energy of an ideal gas per molecule in SI unit at 25∘C25^{{}^{\circ}} \mathrm{C} will be

Options

  1. A6.17×10−21kJ6.17 \times 10^{-21} \mathrm{ kJ}
  2. B6.17×10−21J6.17 \times 10^{-21} \mathrm{ J}
  3. C6.17×10−20J6.17 \times 10^{-20} \mathrm{ J}
  4. D7.16×10−20J7.16 \times 10^{-20} \mathrm{ J}

Correct answer

B. 6.17×10−21J6.17 \times 10^{-21} \mathrm{ J}

Step-by-step solution

KE‾=32kT=32×8.3136.023×1023×298=6.17×10−21 J\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 KE‾=32kT=32RNT)\left.\overline{\mathrm{KE}}=\frac{3}{2} \mathrm{k} \mathrm{T}=\frac{3}{2} \frac{\mathrm{R}}{\mathrm{N}} \mathrm{T}\right)
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