Answer:
83.72 K
Explanation:
[tex]\alpha[/tex] = Polarizability of argon = [tex]1.66\times 10^{-30}\ m^3[/tex]
I = First ionization = 1521 kJ/mol
r = Distance between atoms = 3.8 A
R = Gas constant = 8.314 J/mol K
T = Boiling point
Potential energy due to dispersion of gas is given by
[tex]P=-\frac{3}{4}\frac{\alpha^2I}{r^6}\\\Rightarrow P=-\frac{3}{4}\frac{(1.66\times 10^{-30})^2\times 1521\times 10^3}{(3.8\times 10^{-10})^6}\\\Rightarrow P=-1044.01\ J/mol[/tex]
Kinetic energy is given by
[tex]K=\frac{3}{2}RT[/tex]
The potential and kinetic energy will balance each other
[tex]P=\frac{3}{2}RT\\\Rightarrow 1.04401\times 10^{-33}=\frac{3}{2}RT\\\Rightarrow T=\frac{1044.01\times 2}{3\times 8.314}\\\Rightarrow T=83.72\ K[/tex]
The boiling point of argon is 83.72 K
