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JIPMER 2004
Exam
JIPMER
Year
2004
Subject
Physics
Chapter
Kinetic Theory of Gases

JIPMER 2004 · Physics · Kinetic Theory of Gases

The kinetic energy of one molecule of a gas at normal temperature...

The kinetic energy of one molecule of a gas at normal temperature and pressure will be (k=8.31 J/moleK)(k=8.31 \mathrm{~J} / \mathrm{mole} \mathrm{K})

Options

  1. A17×103 J17 \times 10^3 \mathrm{~J}
  2. B10.2×103 J10.2 \times 10^3 \mathrm{~J}
  3. C3.4×103 J3.4 \times 10^3 \mathrm{~J}
  4. D6.8×103 J6.8 \times 10^3 \mathrm{~J}

Correct answer

C. 3.4×103 J3.4 \times 10^3 \mathrm{~J}

Step-by-step solution

According to kinetic theory, K.E. of 1g-mole of an ideal gas, E=32RTE=\frac{3}{2} R T Hence, K.E. at normal temperature 0∘C0^{\circ} \mathrm{C}=273K or E=32×8.31×273=3.4×103J\begin{aligned}& =273 \mathrm{ K} \text{ or } \quad E & =\frac{3}{2} \times 8.31 \times 273=3.4 \times 10^3 \mathrm{ J} \end{aligned}$$
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