But for an ideal gas, internal energy is only a function of temperature and so internal energy remains constant here,no change in average kinetic energy of gas particles takes place, so where does the chaos come from to increase entropy of the system.
'Chaos' is not a very well defined term in context of statistical physics. It is not necessary to use it to understand entropy in this theory.
One way to understand this entropy is the Boltzmann-Planck definition: entropy is logarithm of number of distinct possible states that are compatible with macroscopic variables like $V$,$U$ etc. No chaos is needed in this definition.
You seem to assume that in general, increase in internal energy is necessary to increase entropy. That is not the case, for there are systems where entropy can increase just by making sudden change to the system, without supplying any energy. Consider ideal gas in a chamber that is connected to another chamber via small pipe. At first, the pipe is blocked by a closed valve, so the gas stays in the first chamber. Then the valve is opened and the gas rushes to the second chamber. In the end, the volume is twice the original value.
Now that the gas has twice the original volume, the number of accessible states increased by factor of two. But no energy was supplied or lost, so internal energy stayed the same. According to the above definition, entropy increased by a factor $\ln 2$.
A thermodynamic potential $P$ satisfies $$P(\lambda X, Y) = \lambda P(X,Y)$$ where $X$ are extensive and $Y$ are intensive parameters of state. As an example $P$ could be the internal energy of a system, it then doubles if we double all its extensive properties (volume, particle number, etc).
If we take $\lambda = \tfrac{1}{N}$, we get $$P(\tfrac{X}{N}, Y) = \tfrac{1}{N} P(X,Y)$$ Let us denote the intensive versions (such as energy per particle, volume per particle) of all extensive quantities with lower case letters: $x=\tfrac{X}{N}$, the same also applies for the potential: $p(X,Y)=\tfrac{1}{N}P(X,Y)$. We then get: $$P(x,Y) = \tfrac{1}{N} P(X,Y) = p(X,Y)$$.
Let us write this down for the Gibbs free energy $G=G(T,p,N)$. Note that its only extensive dependency is $N$, the other two, $p$ and $T$ are intensive. $$G(T,p,\tfrac{N}{N}) = G(T,p,1) = \frac{1}{N}G(T,p,N) = g(T,p,N)$$
Obviously, since the l.h.s does not depend on $N$, also the r.h.s. doesn't, so: $g=g(T,p)$. So we can rewrite the last step of the above equation as: $$G(T,p,N) = Ng(T,p)$$ Diffferentiating with respect to $N$ gives:
$$\left(\frac{\partial G}{\partial N}\right)_{T,p} = g(p,T)$$
On the other hand, from the total differential of $G$ that is given as $dG=-SdT+Vdp+\mu dN$ we know that $$\left(\frac{\partial G}{\partial N}\right)_{T,p} = \mu$$
Therefore, in total, we get: $$G(T,p,N) = Ng(T,p) = N\mu$$
Note that from this derivation you can follow Euler's relation. Indeed, since $G$ is defined as Legendre transform of $U$, $$G=U-TS+pV$$ we have, by plugging in $G=\mu N$ from above, $$N\mu = U-TS+pV$$ or $$U=TS-pV+N\mu$$
Best Answer
Your argument purports to show the chemical potential is positive. You look it up and find that it's zero. In fact, the chemical potential for an ideal gas is negative!
First, I'm not following the step where you conclude that the entropy must go down because the pressure goes up. The entropy of an ideal gas is given by the Sackur-Tetrode equation: $$ S=Nk\left[\ln\left({V\over N}\left(4\pi m U\over 3Nh^2\right)^{3/2}\right)+{5\over 2}\right] $$ Incidentally, I got this particular form of the equation from the textbook by Schroeder, which is the one I like best for this sort of thing.
Under the circumstances you describe (increasing $N$ by 1 while holding $U/N$ constant), this reduces to $$ S=Nk\left[\ln(V/Nv_Q)+ {5\over 2}\right], $$ where $v_Q$, the quantum volume per particle, is constant and $Nv_Q\ll V$. (The latter is the condition for the gas to be nondegenerate.) The derivative of this with respect to $N$ is positive: upon adding a particle at constant $T,V$, the entropy goes up by $$ \Delta S=k\left[\ln(V/nv_Q)+{3/2}\right]>0. $$
Second, where are you looking up the value for $\mu$ for argon at STP? Is it possible that the table you're looking in lists values relative to STP? It's certainly not true that the chemical potential of an ideal gas at STP is zero.
One way to get the chemical potential of an ideal gas is by imagining adding the new particle in such a way that the total energy (not the temperature) remains fixed. The relevant identity then is $$ \mu=-T\left(\partial S\over\partial N\right)_{U,V}, $$ which comes from $dU=T\,dS-P\,dV+\mu\,dN$. Taking the derivative of the above expression for $S$, we get $$ \mu=-kT\ln\left[{V\over N}\left(2\pi mkT\over h^2\right)^{3/2}\right]. $$ Again, for gases that are far from degenerate, the quantity inside the logarithm is large, so the chemical potential is negative.