Average Kinetic Energy Per Mole
An ideal monatomic gas is held at a uniform temperature of 350 K; calculate the average translational kinetic energy associated with one mole of the gas.
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Solution
4.36×103J
Depending only on temperature, the average translational kinetic energy of one mole of an ideal gas equals KE=23RT, since the three translational degrees of freedom each contribute 21RT per mole. Inserting R=8.314 J/mol\cdotK and T=350 K yields KE=23×8.314×350≈4.36×103 J. Importantly, this is the translational part only, and it is identical for a monatomic gas and for any other ideal gas at the same temperature, even though diatomic and polyatomic gases additionally store energy in rotation and vibration. The value 2.91×103 J uses RT without the factor 3/2. The value 7.27×103 J doubles the answer by misreading the temperature as 700 K. The value 1.45×103 J halves the correct coefficient. This is the molar form of the NCERT kinetic interpretation of temperature, which establishes translational kinetic energy as a universal function of temperature alone. The result also explains why two gases at the same temperature exchange no net translational energy on contact, which is precisely the condition that defines thermal equilibrium. As a magnitude check, multiplying the per-molecule energy 23kBT by Avogadro's number reproduces 23RT, confirming internal consistency and the value 4.36×103 J.
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About This Question
- Subject
- physics
- Chapter
- kinetic theory of gases
- Topic
- average kinetic energy per mole
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
4.36×103J
Depending only on temperature, the average translational kinetic energy of one mole of an ideal gas equals KE=23RT, since the three translational degrees of freedom each contribute 21RT per mole. Inserting R=8.314 J/mol\cdotK and T=350 K yields KE=23×8.314×350≈4.36×103 J. Importantly, this is the translational part only, and it is identical for a monatomic gas and for any other ideal gas at the same temperature, even though diatomic and polyatomic gases additionally store energy in rotation and vibration. The value 2.91×103 J uses RT without the factor 3/2. The value 7.27×103 J doubles the answer by misreading the temperature as 700 K. The value 1.45×103 J halves the correct coefficient. This is the molar form of the NCERT kinetic interpretation of temperature, which establishes translational kinetic energy as a universal function of temperature alone. The result also explains why two gases at the same temperature exchange no net translational energy on contact, which is precisely the condition that defines thermal equilibrium. As a magnitude check, multiplying the per-molecule energy 23kBT by Avogadro's number reproduces 23RT, confirming internal consistency and the value 4.36×103 J.
This medium difficulty physics question is from the chapter kinetic theory of gases, covering the topic of average kinetic energy per mole. It appeared in the 2025 exam.
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