How can the entropy of the universe increase when undergoing an irreversible process? So there’s something about entropy that I just can’t wrap my head around. So: we saw in class that when undergoing a reversible process, the entropy change of the universe (so of the system + the environment) is 0. And we saw that to calculate the entropy change of a system between two states, we forget about the original process, then we design a reversible path that links the $2$ states, and we calculate the entropy change along that path (and this entropy change of the system can be positive or negative). And then, if we want to calculate the entropy change of the universe, we imagine that our reversible process is driven by a Heat Engine or Heat Pump, and we calculate the change of entropy of the environment, which we then add to the change of entropy of the system.
==> Now, what I don’t understand is the following: when undergoing an irreversible process, the entropy of the universe increases. How is that possible? Since the whole point of calculating entropy is forgetting about the original process and designing a reversible path, and we saw at the very beginning that along a reversible path the entropy change of the universe is equal to $0$.
I know there’s something that I’ve misunderstood somewhere, but I don’t know what and this thing has been driving me mad for some time.
I hope someone will have the time to answer this!
 A: Consider an irreversible process between states a and b, 
We write $dU = dQ_{irr} - dW_{irr}$
For a reversible process between a and b,
$dU = TdS - dW_{rev}$
Sine dU is the same for both, we have
$dS = \frac{dQ_{irr}}{T} + \frac{( dW_{rev} - dW{irr})}{T}$
It is obvious that the second term is non negative at all times.
Conclusion, for any thermodynamic process
$dS \ge \frac{dQ}{T}$
Hence, the change in entropy of the universe is strictly greter than zero.
A: When you design a reversible path between initial and final state of a system to calculate entropy change of the system, what you get is ....well, entropy change of that system. But this doesn't tell you anything about the entropy change of surroundings. To calculate entropy change of surroundings, you are again free to design any path connecting its initial and final states, and this choice is independent of the particular choice of reversible path you have made for the system. To summarize, entropy change of system depends only on its initial and final states, and entropy change of surroundings depends only on its own initial and final states. This fact is not changed by the ruse of designing reversible paths between their initial and final states, separately and independently. Second law's significance lies in the fact that it shows how the two entropy changes are related.
A: The entropy of a closed system changes as a result of (a) entropy entering and leaving the system by heat flow through its boundaries and (b) entropy generated within the system as a result of process irreversibilities within the system.  These irreveribilities are the result of (a) finite rate of viscous deformation of the fluids comprising the system, (b) finite rate of heat conduction within the system, (c) finite rate of chemical species diffusion within the system, and (d) finite rate of  chemical reaction within the system.  In the absence of these irreversibilities, there is no entropy generation in the system, and the only entropy change is by heat flow through the system boundaries during all or part of the reversible process path.  The entropy flow through the boundaries of the system is given by:  $$\sum{\frac{Q_j}{T_{Bj}}}$$ where $Q_j$ is the heat flow through portion j of the boundary surface and $T_{Bj}$ is the temperature at the portion of the boundary through which $Q_j$ is occurring.  The entropy generated within a system during any process can be calculated from $$(\Delta S)_{gen}=\Delta S-\left(\sum{\frac{Q_j}{T_{Bj}}}\right)=\left(\sum{\frac{Q_j}{T_{Bj}}}\right)_{rev}-\left(\sum{\frac{Q_j}{T_{Bj}}}\right)$$where the subscript rev refers to a reversible path between the initial and final thermodynamic equilibrium states of the system.  Note that, if the actual path is reversible, $(\Delta S)_{gen}=0$.
