What is the change in entropy of helium gas with total mass 0.130 kg at the normal boiling point of helium when it all condenses isothermally to liquid helium? Assume that the normal boiling point of helium is 4.216 K and the heat of vaporization of helium is 2.09×104 J/kg .

ΔS = - 645 J/K

Explanation:

The entropy change for a phase change is given by

ΔS = ΔH phase change / T

For our question we will use this expression for the condensation of Helium:

ΔScondensation = ΔH condensation / Tb

where ΔH condensation is the change in enthaly of condensation which is the negative of the enthalpy of vaporization, and  Tb is the boiling temperature in Kelvin which is the same for condensation.

We can see we are  in position to plug the values in  and solve for ΔScondensation :

For 0.130 kg Helium we will have

ΔHcondensation = - 0.130 Kg  x 2.09 x 10⁴ J/Kg = -2.72 x 10³ J

ΔH is negative since heat is released (exothermic) during condensation.

ΔS = ΔH phase change / T = -2.72 x 10³ J / 4.216 K = - 645 J/K

it is to be noted that an alpha particle is an helium nucleus. so it contains 2 protons 2 neutrons and has a mass number of 4. when alpha decay particles are ejected from the nucleus it gets transformed into an atom which has atomic number less than 2 and a mass number less than 4.

an equation that could represent alpha decay is radium 226 decay of alpha particle forms radon 222.

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