Chemistry - Thermodynamics - I Question with Solution | TestHub

P = k·d, k = P₁/d₁ = 1 bar m⁻¹. Final diameter 3 m ⇒ final pressure 3 bar. V = πd³/6 ⇒ dV = (πd²/2) dd. W = ∫₁³ k d·(πd²/2) dd = (kπ/8)(d₂⁴ − d₁⁴) = (10⁵·π/8)(81 − 1) = 10⁵·π·10 = π×10⁶ ≈ 3.14×10⁶ J ≈ 3 in units of 10⁶ J. (The trap: pressure varies with diameter, so integrate P dV rather than using a constant pressure; also final pressure must be read off the proportionality, not assumed constant at 1 bar.)