Life on Earth has been made possible by the energy from the sun. The sun is a typical star belonging — Physical Chemistry — Electrochemistry Chemistry Question
Ideal gas law at the core of the sun
Life on Earth has been made possible by the energy from the sun. The sun is a typical star belonging to a group of hydrogen–burning (nuclear fusion, not oxidation) stars called main sequence stars. The core of the sun is 36 % hydrogen (1H) and 64 % helium (4He) by mass. Under the high temperature and pressure inside the sun, atoms lose all their electrons and the nuclear structure of a neutral atom becomes irrelevant. The vast space inside atoms that was available only for electrons in a neutral atom becomes equally available for protons, helium nuclei, and electrons. Such a state is called plasma. At the core of the sun, the estimated density is 158 g cm–3 and pressure 2.5×1011 atm.
Calculate the total number of moles of protons, helium nuclei, and electrons combined per cm3 at the core of the sun.
Model Answer
Protons: (158 g cm–3 × 0.36) / (1.0 g mol–1) = 57 mol cm–3
Helium nuclei: (158 g cm–3 × 0.64) / (4.0 g mol–1) = 25 mol cm–3
Electrons: 57 + (25 × 2) = 107 mol cm–3
Total: 189 mol cm–3
Calculate the percentage of space occupied by particles in hydrogen gas at 300 K and 1 atm, in liquid hydrogen, and in the plasma at the core of the sun. The density of liquid hydrogen is 0.09 g cm–3. The radius of a nuclear particle can be estimated from r = (1.4×10–13 cm) (mass number)1/3. Assume that the volume of a hydrogen molecule is twice that of a hydrogen atom, and the hydrogen atom is a sphere with the Bohr radius (0.53×10–8 cm). Estimate your answer to 2 significant figures.
Model Answer
Volume of a hydrogen molecule = 2 (4/3) π r 3
= 2 × (4/3) π × (0.53×10–8 cm)3 = 1.2×10–24 cm3
Hydrogen gas:
V = nRT / p = (1 mol × 8.314 J K–1 mol–1 × 300 K) / 101.325 kPa = 24.6 dm3 mol–1 = 4.1×10–23 dm3 / molecule
= 4.1×10–20 cm3 / molecule
(1.2×10–24 cm3) / (4.1×10–20 cm3) = 3.0×10–5 = 0.003 %
Liquid hydrogen: (2 g mol–1) / (0.09 g / cm–3) / (6×1023 molecule mol–1) = 3.7×10–23 cm3
(1.2×10–24 cm3) / (3.7×10–23 cm3) = 0.030 = 3.0 %
Solar plasma: (neglect volume of electrons)
4/3 π (1.4×10–13 cm)3 (1× 57 mol cm–3 + 4 × 25 mol cm–3) (6×1023 mol–1) = 1.1×10–12 = 1.1×10–10 %
Volume occupied is extremely small and ideal gas law is applicable.
Using the ideal gas law, estimate the temperature at the core of the sun and compare your result with the temperature required for the fusion of hydrogen into helium (1.5×107 K).
Model Answer
From 4.1 we know that there are 189 moles of particles per cm3.
T = pV / nR = (2.5×1011 atm × 1×10–3 dm3) / (189 mol × 0.082 atm dm3 mol–1 K–1) = 1.6×107 K