Internal Energy Of A Gas Mixture
A gas mixture contains two moles of a monatomic gas and four moles of a diatomic gas at 300 K; determine the total internal energy of the mixture.
Select the correct option:
Solution
3.24×104J
The internal energy of a gas mixture is the sum of the internal energies of its components, each evaluated from its own degrees of freedom at the common temperature, since the gases are non-interacting and share a single equilibrium temperature. For the monatomic gas, U1=23n1RT; for the diatomic gas, U2=25n2RT. With n1=2 mol and n2=4 mol at T=300 K, the total is U=23(2)RT+25(4)RT=(3+10)RT=13RT. Substituting R=8.314 J/mol\cdotK and T=300 K gives U=13×8.314×300≈3.24×104 J. The value 2.49×104 J wrongly treats every mole as diatomic. The value 4.99×104 J counts the diatomic component as having seven degrees of freedom by including frozen vibration. The value 1.50×104 J omits the diatomic contribution entirely. This uses the NCERT principle that energy is additive across non-interacting gas components, each obeying equipartition independently with its own characteristic share. Because the diatomic species supplies four of the six moles and carries more energy per mole, it dominates the total internal energy of the mixture. As a plausibility check, the energy per mole works out to (13/6)RT=2.17RT, lying between the monatomic value 1.5RT and diatomic value 2.5RT and skewed toward diatomic since it dominates the mixture, confirming the result.
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About This Question
- Subject
- physics
- Chapter
- kinetic theory of gases
- Topic
- internal energy of a gas mixture
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
3.24×104J
The internal energy of a gas mixture is the sum of the internal energies of its components, each evaluated from its own degrees of freedom at the common temperature, since the gases are non-interacting and share a single equilibrium temperature. For the monatomic gas, U1=23n1RT; for the diatomic gas, U2=25n2RT. With n1=2 mol and n2=4 mol at T=300 K, the total is U=23(2)RT+25(4)RT=(3+10)RT=13RT. Substituting R=8.314 J/mol\cdotK and T=300 K gives U=13×8.314×300≈3.24×104 J. The value 2.49×104 J wrongly treats every mole as diatomic. The value 4.99×104 J counts the diatomic component as having seven degrees of freedom by including frozen vibration. The value 1.50×104 J omits the diatomic contribution entirely. This uses the NCERT principle that energy is additive across non-interacting gas components, each obeying equipartition independently with its own characteristic share. Because the diatomic species supplies four of the six moles and carries more energy per mole, it dominates the total internal energy of the mixture. As a plausibility check, the energy per mole works out to (13/6)RT=2.17RT, lying between the monatomic value 1.5RT and diatomic value 2.5RT and skewed toward diatomic since it dominates the mixture, confirming the result.
This hard difficulty physics question is from the chapter kinetic theory of gases, covering the topic of internal energy of a gas mixture. It appeared in the 2025 exam.
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