Skip to content

Question

A gas at 1 atm and 300 K in a 10 L container is heated isobarically to 600 K . What is the work done by the gas? ( R = 8.3 J mol⁻¹ K⁻¹ )

Options

Choose one · Correct answer highlighted

Explanation

**Latent heat** energy needed for phase change without temperature change, overcomes intermolecular forces, e.g., heating ice at 0°C to water at 0°C requires 334 kJ/kg, then heating water to 100°C requires c ΔT, then vaporization 2260 kJ/kg, illustrating two types of heat. For isobaric: W = P Δ V .Initial: P V₁ = μ R T₁ ⇒ 1 × 10 = μ × 8.3 × 300 ⇒ μ = (10)/(2490) ≈ 0.004 moles .Final: V₂ = (μ R T₂)/(P) = (0.004 × 8.3 × 600)/(1) = 19.92 L . Δ V = 19.92 - 10 ≈ 9.92 L . W = P Δ V =

Discussion

Comments

0 comments

No comments yet. Be the first to start the discussion.