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Physics · Kinetic Theory of Gases

JEE Main 2024 — 27 January, Shift 2 — Question 39

The total kinetic energy of 1 mole of oxygen at 27∘C27^{\circ} \mathrm{C} is : [Use universal gas constant (R)=8.31 J/moleK](\mathrm{R})=8.31 \mathrm{~J} / \mathrm{mole} \mathrm{K]}

  1. Option A:

    6845.5 J

  2. Option B:

    5942.0 J

  3. Option C:

    6232.5 J

    Correct
  4. Option D:

    5670.5 J

Answer: C

Step-by-step solution

Kinetic energy =f2nRT=\frac{\mathrm{f}}{2} \mathrm{nRT}

=52×1×8.31×300 J=6232.5 J\begin{aligned}& =\frac{5}{2} \times 1 \times 8.31 \times 300 \mathrm{~J} &=6232.5 \mathrm{~J} \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
27 January, Shift 2
Subject
Physics
Chapter
Kinetic Theory of Gases
Topic
Energy of Gas and Gas Laws and Miscellaneous Problems
The total kinetic energy of 1 mole of oxygen at 27 ° C is : [Use… | JEE Main 2024 PYQ with Solution · DhiX AI