An athlete lifts a 50 kg mass through 1.5 m and lets it fall the same 1.5 m while coupled to a generator that converts 80% of the mass's lost potential energy to electrical work. This work drives

Question

An athlete lifts a 50 kg mass through 1.5 m and lets it fall the same 1.5 m while coupled to a generator that converts 80% of the mass's lost potential energy to electrical work. This work drives Hall electrolysis:

Al₂O₃(solution) + 3C(graphite) → 2 Al(l) + 3CO(g); Δ G^° = 600 kJ

How many lifts are needed to supply sufficient Gibbs energy to produce 40.5 g Al? (g = 10 m/s²; Al = 27 g/mol)

TestHub · Accepted Answer

40.5 g = 1.5 mol Al. Reaction gives 2 mol Al per 600 kJ, so 1.5 mol needs 1.5/2×600 = 450 kJ. Electrical work per lift = 0.80×50×10×1.5 = 600 J. Lifts = 450000/600 = 750. Ans = 0750.

Chemistry - Thermodynamics - II Question with Solution | TestHub

Doubts & Discussion