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Question

What is the pressure of 0.8 moles of an ideal gas in a 16-litre container at 427°C? (R = 8.31 J mol⁻¹ K⁻¹)

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Explanation

**Molecular mass and density** relation ρ = P M/(R T) allows density calculation, ideal gas law also P = n k_B T where n number density, molecular mass determines mass per molecule, density increases with pressure and decreases with temperature, inverse T dependence. PV = μ R T, P = (μ R T)/(V).T = 427 + 273 = 700 K, V = 16 × 10⁻³ m³.P = (0.8 × 8.31 × 700)/(16 × 10⁻³) = 2.90625 × 10⁵ Pa ≈ 2.91 atm (1 atm ≈ 10⁵ Pa). Substituting values gives 2.91 atm, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2

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