A rigid vessel contains 0.80 mol monatomic ideal gas at 300 K and is heated to 450 K. Use R = 8.31 J/(mol K), Nₐ = 6.02 × 10²³ mol⁻¹. (a) Find ΔU using U = 3nRT/2. (b) Find molecule number. (c) Find rms-speed factor. (d) Find pressure factor and explain molecularly. (e) Explain zero boundary work and contrast constant-pressure heating.

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ΔU = 1.50 kJ; N = 4.82 × 10²³; rms-speed factor 1.225; pressure factor 1.50; rigid-volume boundary work is zero. ΔU = 3nRΔT/2 = 1.5 × 0.80 × 8.31 × 150 = 1496 J = 1.50 kJ. Molecule number is nNₐ = 0.80 × 6.02 × 10²³ = 4.82 × 10²³. Mean translational kinetic energy is proportional to T, so rms speed changes by √(450/300) = √1.5 = 1.225. At fixed amount and volume, p ∝ T, giving factor 1.50. Faster molecules collide more often and transfer more momentum to vessel walls. Rigid volume has ΔV = 0, so boundary work is zero and heating equals ΔU. A constant-pressure piston expands, does work and needs additional heat for the same temperature rise.

Molecular interpretation of a heated gas

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