Paradox of potential energy of gas molecule We know that when water is heated at its boiling point, it will become vapors. Then because there is no change in temperature, the heat supplied will be transferred to the potential energy of the molecules. So, if molecules are far from each other so the Internal Potential energy of the substance will increase.
However, in case of ideal gas, we assume that particles are far apart, but it also states that the potential energy of gas is negligible. Is it contradicting?
 A: Unfortunately people like to teach over simplified physics, which then leads to confusion outside of the setting where the simplification wasn't over done.
So, what is temperature? Is it average kinetic energy per particle? No, it isn't.
Temperature is that number $T$ such that
$$\frac{1}{k_BT}=\frac{dS}{dE}.$$
Where $S$ is the entropy. Why? Because then when entropy increases (which it spontaneously tends to do) then energy will flow from hot to cold.
Thus we find out that things at different temperatures can increase their total combined entropy if energy flows from the hot to the cold, because they have equal and opposite $Delta E$ (conservation of energy) and yet the entropy of the cold one goes up more than the entropy of the hot one goes down.
OK. So now we know what temperature is. So when something gains energy but the temperature stays the same this just means the entropy increases at the same linear rate as you pump a bunch of energy into it.
So when you boil water you keep putting energy into the water and the entropy just increases by the same amount for each bit of energy.
The energy is going to internal energy, which is a fancy name for energy that (unlike work) you aren't specifically tracking, so the stuff that you are free to think of as spontaneously flowing since you weren't tracking exactly where it was.
So it can go to kinetic energy of the molecule. It can go to rotational energy of the molecules. It just needs to give the same increase in entropy for each bit of energy. So that it can stay at the same temperature.
A: Yes, it is a sort of contradiction. More precisely, it says that you cannot model water's liquid-to-gas phase transition with an ideal gas model, and any attempt to do so will be fraught with contradictions.
Now as to some misunderstandings that you may have: first, be careful about thinking of potential energy as infinite at infinite separation: that's true if your attractive force is something like a spring, going like $k x$ where $x$ is the separation; it's also true if the force is constant (like, if I recall correctly, the strong-force created by gluons between quarks is), and even if the force scales like $1/x$ (though the energy goes logarithmically to infinity, it does eventually get there).
However many of these forces scale even slower than that. For example if your "gas" is a cluster of astronomical particles attracted by gravity, we know that's a $1/x^2$ law, and that does not generate such an infinite potential at infinite separation. In fact, you can blast off into space with a finite energy. Or to take it closer to common gases: a very common force in quantum chemistry packages is called the "Lennard-Jones potential" and that force goes like $1/x^7$ at far distances. When these happen, there is only a finite energy difference between a separation of $x$ and a separation of $\infty.$ 
So even when we say "yes, technically the potential energy gets higher and higher" please don't think it "grows without bound." Usually it approaches some limit closer and closer, so it gets higher-and-higher, sure, but much slower-and-slower. In fact usually we set the potential energy to be 0 at infinity and look at it as "negative" when these particles are close together; it's technically true that we have to "pour" energy into this in order to separate them, but we view it as a sort of energy that we already "got for free" when they attracted themselves together into this liquid of stuff. They "want" to stick together.
So what you're comparing is the potential energy function of a non-ideal gas -- let's take the case of Lennard-Jones, $$u(x) = u_0 \left( \left(\frac{s}{x}\right)^{12} - 2~\left(\frac{s}{x}\right)^{6} \right),$$to the case of the ideal gas,$$u(x) = 0.$$
What do we see when we compare these? Yes, the potential is usually higher for the ideal gas (the LJ potential is mostly negative except if the particles are trying to bump into each other, when this $x^{-12}$ term takes over), but it's not because the ideal gas has a stronger force or anything -- in fact the ideal gas has no force and the difference is just an arbitrary additive constant that we could add to either of these.
What we see is that energy is a convenient way of talking about forces, but only energy differences (which are forces integrated over distances) matter. The ideal gas has no energy differences; the realistic potentials do have energy differences. 
