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Question

What is the total internal energy of 2 moles of a monatomic gas at 400 K? (R = 8.31 J mol⁻¹ K⁻¹)

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Explanation

**Ideal gas equation** combines Boyle, Charles, Avogadro laws, P V = N k_B T, N number of molecules, k_B Boltzmann constant, for 1 mole N_A=6.022×10²³, R = N_A k_B, enabling calculation of volume from P,T,n. For monatomic gas, U = (3)/(2) μ R T.U = (3)/(2) × 2 × 8.31 × 400 = 9972 J ≈ 9.97 kJ . Substituting values gives 9.97 kJ, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and v_rms = √(3 R T/M) relations.

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