Question
Liquid helium- 4 has a normal boiling point of 4.2 K . However, at a pressure of 1 mm of mercury, it boils at 1.2 K . Estimate the average latent heat of vaporization of helium in this temperature range.
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We know that liquid helium-4 has a normal boiling point of 4.2 K and at a pressure of 1 mm of mercury, it boils at 1.2 K. We need to estimate the average latent heat of vaporization of helium in this temperature range. Show more…
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At a pressure of 1 atm, liquid helium boils at 4.20 K. The latent heat of vaporization is 20.5 kJ/kg. Determine the entropy change (per kilogram) of the helium resulting from vaporization.
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The latent heat of vaporization (boiling) of helium at a pressure of 1 atm and temperature of 4.2 K is 21.8 kJ/kg. The densities at 4.2 K of the liquid and vapor are 125 kg/m^3 and 19 kg/m^3 respectively. Note that helium is monatomic. Estimate the depth of the interatomic potential energy well between two helium atoms. Take the average number of nearest neighbors to be z = 10 and note that the atomic number of helium is 4. Provide your answer as a number in exponential form in terms of Joules. Do not include the units in your answer. Hint: Be careful about units. The latent heat equation is written in terms of energy per mole, but the latent heat value given in this problem is energy per mass. So at some point, you will need to calculate the mass of a mole of helium atoms.
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