The first ionization energy of polonium, Po, is 812 kJ/mol. Is the first ionization energy of Lv expected to be greater than, less than, or equal to that of Po? Justify your answer in terms of Coulomb’s law.

The first ionization energy comparison between Lv and Po can be justified using Coulomb’s law.

Explanation:

Following Coulomb's law, the first ionization energy of Lv is expected to be less than that of Po.

According the Coulomb's law, the magnitude of attractive/repulsive force between charged particles is directly proportional to he product of the charges and inversely proportional to the square of the distance between the charges. It is an inverse square law.

Hence;

The first ionization energy is affected by the size of the nuclear charge as well as the distance of the outermost electron from the nucleus.

As we descend from Po to Lv, the increase in size of the nuclear charge is counteracted by the addition of more shells hence the distance of the outermost electrons increases.

According to Coulomb's law, as the distance between the charges increases, the first ionization energy of Lv decreases.

Hence, first ionization energy of Lv is less than that of Po.

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the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.