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

Question

Quizplus · Accepted Answer

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

At Atmospheric Pressure, 45 Moles of Liquid Helium Are Vaporized