Question:
The translational kinetic energy of an ideal gas containing N molecules at temperature T is (k - Boltzmann constant) ________.
The translational kinetic energy of an ideal gas containing N molecules at temperature T is (k - Boltzmann constant) ________.
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Translational kinetic energy depends only on temperature, not on the type of gas (monatomic/diatomic).
Updated On: Jun 26, 2026
- $\frac{5}{2}NkT$
- $\frac{1}{2}NkT$
- $\frac{3}{2}NkT$
- $\frac{7}{2}NkT$
- $\frac{9}{2}NkT$
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The Correct Option is C
Solution and Explanation
Step 1: Concept
According to the law of equipartition of energy, energy per molecule per degree of freedom is $\frac{1}{2}kT$.
Step 2: Meaning
Translational motion always has 3 degrees of freedom ($x, y, z$).
Step 3: Analysis
Average translational K.E. per molecule $= 3 \times \frac{1}{2}kT = \frac{3}{2}kT$.
Step 4: Conclusion
For $N$ molecules, total translational K.E. $= \frac{3}{2}NkT$. Final Answer: (C)
According to the law of equipartition of energy, energy per molecule per degree of freedom is $\frac{1}{2}kT$.
Step 2: Meaning
Translational motion always has 3 degrees of freedom ($x, y, z$).
Step 3: Analysis
Average translational K.E. per molecule $= 3 \times \frac{1}{2}kT = \frac{3}{2}kT$.
Step 4: Conclusion
For $N$ molecules, total translational K.E. $= \frac{3}{2}NkT$. Final Answer: (C)
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