Fundamental thermodynamic relation and irreversible processes The first law of thermodynamics for open systems states that a change in internal energy is given by
$$
d U = d Q + d W + \sum_{i} \mu_i d N_i.
$$
When the system undergoes a reversible process, we can set $d Q = T d S$, while when the system performs only pressure-volume work, we set $ dW = - p d V$, to obtain
$$
d U = T d S - p d V + \sum_{i} \mu_i d N_i.
$$
Now, on wikipedia, it says

Since U, S and V are thermodynamic functions of state, the above
  relation holds also for arbitrary non-reversible changes.

Is this statement correct? 
One would think that, since for irreversible processes it holds that $T d S > d Q$, we obtain
$$
d U < T d S + p d V + \sum_{i} \mu_i d N_i,
$$
which is similar to R.E. Reichl, A modern course in statistical physics, Equation (3.25)
 A: Yes it is absolutely correct. A function of state is a property that depends only on the current state of the system, not how it got there. The volume of a system does not "know" whether the system was expanded reversibly or irreversibly. This means we can compute changes in state variables during irreversible processes by finding a reversible process with the same initial and final states and calculating the change in during that process. Since state variables depend only on the start and end points, this must give the same result. 
In terms of the implications of $dQ \le TdS$, the first law and the fundamental relation together tell us that
$$
dQ + dW = TdS - pdV + \sum_i \mu_i dN_i
$$
Clausius inequality then implies that the work on the system satisfies
$$
dW \ge -pdV + \sum_i \mu_i dN_i
$$
in other words the maximum work done by the system occurs in the case when the equality holds, i.e. in a reversible process. 
A: To be concrete, take the example of a closed chemical system that evolves at a constant temperature. As the system is closed, the term involving chemical potentials should not be introduced in the expression of the first law.
We must therefore compare $dU=TdS-PdV+\sum\mu_idN_i$ and $dU=T\left(dS-dS_i\right)-PdV$
We then obtain the well-known expression of the rate of creation of entropy associated with the chemical reaction (De Donder and Prigogine) : $dS_i=\frac{1}{T}\sum{\mu_i{dN}_i}=\frac{1}{T}\left(\sum{\nu_i\mu_i}\right)d\xi=\frac{A}{T}\ d\xi$ with $A$ the chemical affinity $A=\left(\sum{\nu_i\mu_i}\right)$
Hope it can help and sorry for my poor english.
