As mentioned in other answers the dispersion force is responsible for noble gases forming liquids. The calculation of the boiling points is now outlined after some general comments about the dispersion force.
The dispersion force (also called London, charge-fluctuation, induced-dipole-induced-dipole force) is universal, just like gravity, as it acts between all atoms and molecules. The dipole forces can be long range, > 10nm down to approx 0.2nm depending on circumstances, and can be attractive or repulsive.
  Although the dispersion force is quantum mechanical in origin it can be understood as follows: for a non-polar atom such as argon the time average dipole is zero, yet at any instance there is a finite dipole given by the instantaneous positions of the electrons relative to the nucleus. This instantaneous dipole generates an electric field that can polarise another nearly atom and so induce a dipole in it. The resulting interaction between these two dipoles gives rise to an instantaneous attractive force between the two atoms, whose time average in not zero.
  The dispersion energy was derived by London in 1930 using quantum mechanical perturbation theory. The result is $$U(r)=-\frac{3}{2}\frac{\alpha _o^2I}{(4\pi\epsilon _0)^2r^6}=-\frac{C_{disp}}{r^6}$$

where $\alpha _o$ is the electronic polarisability, $I$ the first ionisation energy, $\epsilon_0$ the permittivity of free space and $r$ the separation of the atoms. The electronic polarisability $\alpha_o$ arises from the displacement of an atom's electrons relative to the nucleus and it is the constant of proportionality between the induced dipole and the electric field $E$, viz., $\mu _{ind} = \alpha_o E$. The polarisability has units of $\ce{J^{-1}C^2m^2}$, which means that in SI units $\alpha_0/(4\pi\epsilon_)$ has units of m$^3$ and this polarisability is in effect a measure of electronic volume, or put another way $\alpha_o=4\pi\epsilon_0r_0^3$ where experimentally it is found that $r_0$ is approximately the atomic radii. The ionisation energy $I$ arises because to estimate $r_0$ a simple model of an atom is used to calculate the orbital energy and hence radius and in doing so the energy is equated to the ionisation energy since this can be measured.

If experimental values are put into the London equation then the attractive energy can be calculated. In addition the boiling point can be estimated by equating the London energy with the average thermal energy as $U(r_0)=3k_BT/2$ where $k_B$ is the Boltzmann constant and $T$ the temperature.

  The polarisability in units of $\alpha_010^{-30}/(4\pi\epsilon_0)$ m$^3$, ionisation potentials in eV, radii $r_0$ in nm and the constant $C_{disp} $ in units $10^{-70} $ Jm$^6$, temperature $T$ in K. The values in brackets are experimental values.

$~~~~~~~~~~~~~~~~~~~~ \alpha _0~~~~~~~~~~~~ I~~~~~~~~~~r_0~~~~~~~~~C_{disp}~~~~~~~~~~~~T$
$~Ne-Ne~~~0.39~~~~~~~~21.6~~~~0.308~~~~~~~3.9~(3.8)~~~22~(27) $
$~Ar-Ar~~~1.63~~~~~~~~15.8~~~~0.376~~~~~~~~50~(45)~~~~~85~(87) $
$~Xe-Xe~~~4.01~~~~~~~~12.1~~~~0.432~~~~~~233~(225)~~173~(165) $

The fit to data is very good, possibly this is fortuitous, but these are spherical atoms showing only dispersion forces and a good correlation to experiment is expected. However, there short range repulsive forces that are ignored as well as higher order attractive forces. Nevertheless it does demonstrate that dispersion forces can account for the trend in boiling quite successfully.
Source of Data. J. Israelachvilli 'Intermolecular and Surface Forces'

As mentioned in other answers the dispersion force is responsible for noble gases forming liquids. The calculation of the boiling points is now outlined after some general comments about the dispersion force.

The dispersion force (also called London, charge-fluctuation, induced-dipole-induced-dipole force) is universal, just like gravity, as it acts between all atoms and molecules. The dipole forces can be long-range, >10 nm down to approx 0.2 nm depending on circumstances, and can be attractive or repulsive. 

Although the dispersion force is quantum mechanical in origin it can be understood as follows: for a non-polar atom such as argon the time average dipole is zero, yet at any instance there is a finite dipole given by the instantaneous positions of the electrons relative to the nucleus. This instantaneous dipole generates an electric field that can polarise another nearly atom and so induce a dipole in it. The resulting interaction between these two dipoles gives rise to an instantaneous attractive force between the two atoms, whose time average in not zero. 

The dispersion energy was derived by London in 1930 using quantum mechanical perturbation theory. The result is

$$U(r)=-\frac{3}{2}\frac{\alpha_0^2I}{(4\pi\epsilon _0)^2r^6}=-\frac{C_{\mathrm{disp}}}{r^6}$$

where $\alpha_0$ is the electronic polarisability, $I$ the first ionisation energy, $\epsilon_0$ the permittivity of free space and $r$ the separation of the atoms. The electronic polarisability $\alpha_0$ arises from the displacement of an atom's electrons relative to the nucleus and it is the constant of proportionality between the induced dipole and the electric field $E$, viz., $\mu_{\mathrm{ind}} = \alpha_0 E$. The polarisability has units of $\pu{J-1 C2 m2}$, which means that in SI units $\alpha_0/(4\pi\epsilon_0)$ has units of $\pu{m3}$ and this polarisability is in effect a measure of electronic volume, or put another way $\alpha_0 = 4\pi\epsilon_0r_0^3$ where experimentally it is found that $r_0$ is approximately the atomic radii. The ionisation energy $I$ arises because to estimate $r_0$ a simple model of an atom is used to calculate the orbital energy and hence radius and in doing so the energy is equated to the ionisation energy since this can be measured.

If experimental values are put into the London equation then the attractive energy can be calculated. In addition the boiling point can be estimated by equating the London energy with the average thermal energy as $U(r_0)=3k_\mathrm{B}T/2$ where $k_\mathrm B$ is the Boltzmann constant and $T$ the temperature. The relevant parameters are given in the table below, with values in parentheses being experimental values:^[1]

$$\begin{array}{c|c|c|c|c|c} \text{Noble gas} & (\alpha_0/4\pi\epsilon_0)~/~\pu{10^{-30}m^3} & I~/~\pu{eV} & r_0~/~\pu{nm} & C_\mathrm{disp}~/~\pu{10^{-70} J m6} & T_\mathrm{b}~/~\pu{K} \\ \hline \ce{Ne} & 0.39 & 21.6 & 0.308 & 3.9~(3.8) & 22~(27) \\ \ce{Ar} & 1.63 & 15.8 & 0.376 & 50~(45) & 85~(87) \\ \ce{Xe} & 4.01 & 12.1 & 0.432 & 233~(225) & 173~(165) \end{array}$$

The fit to data is very good, possibly this is fortuitous, but these are spherical atoms showing only dispersion forces and a good correlation to experiment is expected. However, there short range repulsive forces that are ignored as well as higher order attractive forces. Nevertheless it does demonstrate that dispersion forces can account for the trend in boiling quite successfully.

1. Israelachvili, J. N. Intermolecular and Surface Forces, 3rd ed.; Academic Press: Burlington, MA, 2011; p 110.

As mentioned in other answers the dispersion force is responsible for noble gases forming liquids. The calculation of the boiling points is now outlined after some general comments about the dispersion force.
The dispersion force (also called London, charge-fluctuation, induced-dipole-induced-dipole force) is universal, just like gravity, as it acts between all atoms and molecules. The dipole forces can be long range, > 10nm down to approx 0.2nm depending on circumstances, and can be attractive or repulsive.
Although the dispersion force is quantum mechanical in origin it can be understood as follows: for a non-polar atom such as argon the time average dipole is zero, yet at any instance there is a finite dipole given by the instantaneous positions of the electrons relative to the nucleus. This instantaneous dipole generates an electric field that can polarise another nearly atom and so induce a dipole in it. The resulting interaction between these two dipoles gives rise to an instantaneous attractive force between the two atoms, whose time average in not zero.
The dispersion energy was derived by London in 1930 using quantum mechanical perturbation theory. The result is $$U(r)=-\frac{3}{2}\frac{\alpha _o^2I}{(4\pi\epsilon _0)^2r^6}=-\frac{C_{disp}}{r^6}$$

where $\alpha _o$ is the electronic polarisability, $I$ the first ionisation energy, $\epsilon_0$ the permittivity of free space and $r$ the separation of the atoms. The electronic polarisability $\alpha_o$ arises from the displacement of an atom's electrons relative to the nucleus and it is the constant of proportionality between the induced dipole and the electric field $E$, viz., $\mu _{ind} = \alpha_o E$. The polarisability has units of $\ce{J^{-1}C^2m^2}$, which means that in SI units $\alpha_0/(4\pi\epsilon_)$ has units of m$^3$ and this polarisability is in effect a measure of electronic volume, or put another way $\alpha_o=4\pi\epsilon_0r_0^3$ where experimentally it is found that $r_0$ is approximately the atomic radii. The ionisation energy $I$ arises because to estimate $r_0$ a simple model of an atom is used to calculate the orbital energy and hence radius and in doing so the energy is equated to the ionisation energy since this can be measured.

If experimental values are put into the London equation then the attractive energy can be calculated. In addition the boiling point can be estimated by equating the London energy with the average thermal energy as $U(r_0)=3k_BT/2$ where $k_B$ is the Boltzmann constant and $T$ the temperature.

The polarisability in units of $\alpha_010^{-30}/(4\pi\epsilon_0)$ m$^3$, ionisation potentials in eV, radii $r_0$ in nm and the constant $C_{disp} $ in units $10^{-70} $ Jm$^6$, temperature $T$ in K. The values in brackets are experimental values.

$~~~~~~~~~~~~~~~~~~~~ \alpha _0~~~~~~~~~~~~ I~~~~~~~~~~r_0~~~~~~~~~C_{disp}~~~~~~~~~~~~T$
$~Ne-Ne~~~0.39~~~~~~~~21.6~~~~0.308~~~~~~~3.9~(3.8)~~~22~(27) $
$~Ar-Ar~~~1.63~~~~~~~~15.8~~~~0.376~~~~~~~~50~(45)~~~~~85~(87) $
$~Xe-Xe~~~4.01~~~~~~~~12.1~~~~0.432~~~~~~233~(225)~~173~(165) $

The fit to data is very good, possibly this is fortuitous, but these are spherical atoms showing only dispersion forces and a good correlation to experiment is expected. However, there short range repulsive forces that are ignored as well as higher order attractive forces. Nevertheless it does demonstrate that dispersion forces can account for the trend in boiling quite successfully.
Source of Data. J. Israelachvilli 'Intermolecular and Surface Forces'

As mentioned in other answers the dispersion force is responsible for noble gases forming liquids. The calculation of the boiling points is now outlined after some general comments about the dispersion force.
The dispersion force (also called London, charge-fluctuation, induced-dipole-induced-dipole force) is universal, just like gravity, as it acts between all atoms and molecules. The dipole forces can be long range, > 10nm down to approx 0.2nm depending on circumstances, and can be attractive or repulsive.
Although the dispersion force is quantum mechanical in origin it can be understood as follows: for a non-polar atom such as argon the time average dipole is zero, yet at any instance there is a finite dipole given by the instantaneous positions of the electrons relative to the nucleus. This instantaneous dipole generates an electric field that can polarise another nearly atom and so induce a dipole in it. The resulting interaction between these two dipoles gives rise to an instantaneous attractive force between the two atoms, whose time average in not zero.
The dispersion energy was derived by London in 1930 using quantum mechanical perturbation theory. The result is $$U(r)=-\frac{3}{2}\frac{\alpha _o^2I}{(4\pi\epsilon _0)^2r^6}=-\frac{C_{disp}}{r^6}$$

where $\alpha _o$ is the electronic polarisability, $I$ the first ionisation energy, $\epsilon_0$ the permittivity of free space and $r$ the separation of the atoms. The electronic polarisability $\alpha_o$ arises from the displacement of an atom's electrons relative to the nucleus and it is the constant of proportionality between the induced dipole and the electric field $E$, viz., $\mu _{ind} = \alpha_o E$. The polarisability has units of $\ce{J^{-1}C^2m^2}$, which means that in SI units $\alpha_0/(4\pi\epsilon_)$ has units of m$^3$ and this polarisability is in effect a measure of electronic volume, or put another way $\alpha_o=4\pi\epsilon_0r_0^3$ where experimentally it is found that $r_0$ is approximately the atomic radii. The ionisation energy $I$ arises because to estimate $r_0$ a simple model of an atom is used to calculate the orbital energy and hence radius and in doing so the energy is equated to the ionisation energy since this can be measured.

As can be seen from the formula the energy depends on the product of the square of the polarisability, i.e. volume of molecule or atom and its ionisation energy, and also on the reciprocal of the sixth power of the separation of the molecules/atoms. In a liquid of noble gases this separation may be taken to be the atomic radius, $r_0$. Thus the dependence is much more complex than just size, see table of values below. The increase in polarisability as the atomic number increases, is offset somewhat by the reduction in ionisation energy and increase in atomic radius.

If experimental values are put into the London equation then the attractive energy can be calculated. In addition the boiling point can be estimated by equating the London energy with the average thermal energy as $U(r_0)=3k_BT/2$ where $k_B$ is the Boltzmann constant and $T$ the temperature.

The polarisability in units of $\alpha_010^{-30}/(4\pi\epsilon_0)$ m$^3$, ionisation potentials in eV, radii $r_0$ in nm and the constant $C_{disp} $ in units $10^{-70} $ Jm$^6$, temperature $T$ in K. The values in brackets are experimental values.

$~~~~~~~~~~~~~~~~~~~~ \alpha _0~~~~~~~~~~~~ I~~~~~~~~~~r_0~~~~~~~~~C_{disp}~~~~~~~~~~~~T$
$~Ne-Ne~~~0.39~~~~~~~~21.6~~~~0.308~~~~~~~3.9~(3.8)~~~22~(27) $
$~Ar-Ar~~~1.63~~~~~~~~15.8~~~~0.376~~~~~~~~50~(45)~~~~~85~(87) $
$~Xe-Xe~~~4.01~~~~~~~~12.1~~~~0.432~~~~~~233~(225)~~173~(165) $

The fit to data is very good, possibly this is fortuitous, but these are spherical atoms showing only dispersion forces and a good correlation to experiment is expected. However, there short range repulsive forces that are ignored as well as higher order attractive forces. Nevertheless it does demonstrate that dispersion forces can account for the trend in boiling quite successfully.
Source of Data. J. Israelachvilli 'Intermolecular and Surface Forces'