# Change in internal energy and reversible paths between states

As I understand it, the relation (1) $$dU = TdS - pdV + \sum_{i} \mu_idN_i$$ always holds, even for irreversible paths. The justification seems to be that (2) "$$U, S, V, N_1, N_2, ...$$ are all state variables." I sort of intuitively see how (1) follows from (2) but I would like a more rigorous explanation of it. I was thinking that, IF it is possible to find a reversible path with the same starting point (a) and endpoint (b) as any arbitrary (possibly non-reversible) path, then we have $$\int_a^b TdS - pdV + \sum_{i} \mu_idN_i = \int_a^bdQ +dW = \int_a^b dU$$. In the limit, for infinitesimal processes, we should have $$TdS - pdV + \sum_{i} \mu_idN_i = dU$$.

The problem is that I'm not sure that there always is a reversible path corresponding to any arbitrary non-reversible path (this related question doesn’t seem to have a satisfying answer). For instance, if $$\Delta G < 0$$ between point (a) and (b), then every path between (a) and (b) has a negative Gibbs free energy and is presumably not reversible (I may be wrong about this since $$\Delta G < 0$$ implies non-reversibly under the assumption of constant p and T).

A. Does there always exist a reversible path between two states? And if so, does my reasoning about why (2) implies (1) seem correct?

B. If not are there any other (rigorous) ways of explaining why (2) implies (1)?