Practice question

MCQ

Question

A gas expands adiabatically from 5 L to 20 L , reducing its temperature from 500 K to 250 K . If 1 mole of gas is used and R = 8.3 J mol⁻¹ K⁻¹ , what is the work done by the gas? ( gamma = 1.5 )

Options

Choose one · Correct answer highlighted

3000 J 3500 J 4150 J Correct 4500 J

Explanation

**Heat capacity** at constant pressure C_p and volume C_v, C_p = C_v + R per mole, for solids Dulong-Petit C_v≈3R≈25 J/mol·K. Specific heat and latent heat govern temperature changes and phase transitions, Q = m c ΔT for heating, Q = m L for melting/boiling at constant T. For adiabatic process: W = (μ R (T₁ - T₂))/(γ - 1) . μ = 1 , R = 8.3 , T₁ = 500 , T₂ = 250 , γ = 1.5 . W = (1 × 8.3 × (500 - 250))/(1.5 - 1) = (8.3 × 250)/(0.5) = 4150 J . Using first law ΔU =

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