What's the total work performed on the gas when it is transformed from state a to c along the path indicated?
Correct answer: A. 20 J
- A. 20 J
- B. 45J
- C. 30J
- D. 50J
Explanation
Given from the graph:a to b: Isobaric expansion (constant pressure)Pa = 2 × 10³ PaVa = 10 × 10⁻³ m³Vb = 20 × 10⁻³ m³Work done: W₁ = P × ΔV = 2×10³ × (20-10)×10⁻³ = 2×10³ × 10×10⁻³ = 20 Jb to c: Isochoric process (volume constant)Volume doesn't change → ΔV = 0So, W₂ = 0 J✅ Total Work Done from a → c:W_total = W₁ + W₂ = 20 J + 0 J = 20 J
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Thermal equilibrium and heat explain how temperature and energy transfer are related, while molar specific heats describe the heat required by a gas at constant volume or constant pressure. The first law connects heat supplied, work done and change in internal energy, with sign conventions kept distinct from temperature changes.
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