Does $\Delta U$ depend on which type of gas is considered? $\Delta U$ is a state function, hence it only depends on the 'end-points'. Let's imagine that there's a gas confined in an isolated container. For an Ideal Gas, a reversible isothermal expansion implies that $\Delta T = 0 \Rightarrow \Delta U = 0$. Would $\Delta U$ still be $0$ for the same process, for a Real Gas? If so, then why?
 A: Yes, calculating $\Delta U$ requires a knowledge of what kind of gas you're talking about.

Would $\Delta U$ still be 0 for the same process, for a Real Gas? If so, then why?

No, not in general.  An ideal gas is special because, assuming constant $N$ (or equivalently $n$, the number of moles of gas present), the internal energy is just a constant multiple of the temperature $T$.  This implies that if $\Delta T=0$, then $\Delta U=0$.  
Not all gases have this property.  In particular, consider the van der Waals gas, which has the equation of state
$$\left(P + a \frac{n^2}{V^2}\right)\left(V-nb\right) = nRT$$
Roughly speaking, the van der Waals gas is a modification of the ideal gas to account for finite molecular size as well as long-range attractive forces.  It requires a bit of work (and several additional inputs - see here) to show, but the internal energy of a monatomic van der Waals gas is
$$U = \frac{3}{2}nRT - a\frac{n^2}{V}$$
Unlike the case of the ideal gas, the van der Waals internal energy is volume dependent, and can therefore be changed even during isothermal processes.
