Entropy: At atmospheric pressure, 45 moles of liquid helium are vaporized at its boiling point of 4.22 K. The heat of vaporization of helium, at atmospheric pressure, is 2.09 × 10⁴ J/kg, and the atomic weight of helium is 4.00 g/mol. The change in the entropy of the helium, as it vaporizes, is closest to A) 890 J/K. B) 14,000 J/K. C) 18,000 J/K. D) -9400 J/K. E) -14,000 J/K.

Question

Quizplus · Accepted Answer

The change in entropy (ΔS) is calculated using ΔS = Q/T, where Q is the heat absorbed and T is the temperature. First, convert moles to mass: 45 moles × 4.00 g/mol = 180 g = 0.18 kg. Then, calculate Q: 0.18 kg × 2.09 × 10^4 J/kg = 3762 J. Finally, ΔS = 3762 J / 4.22 K ≈ 890 J/K.

Entropy: at Atmospheric Pressure, 45 Moles of Liquid Helium Are