Is the entropy change the same for both closed reversible and irreversible processes? It is said that entropy is a state function and doesn't depend on path. 
Also
$$S_2-S_1=\int \frac{1}{T}dQ$$
for reversible process,
$$S_2-S_1>\int \frac{1}{T}dQ$$
for irreversible process.


*

*If the same amount of heat đQ enters in both reversible and irreversible process, then shouldn't $S_2-S_1$ be different in reversible and irreversible processes?

*Is $\int dQ/T$ equal in both reversible and irreversible processes since the same amount of heat is transferred at same boundary temperature? Please explain this.
Also,
$$dS=d_eS+d_iS=\frac{dQ}{T}+d_iS$$
"dS" for a reversible process is only equal to "đQ/T" and no entropy generation.
But an irreversible process, working between same conditions(heat supplied at same T) is accompanied by entropy generation. 


*So shouldn't "$\int dS$" for an irreversible process greater than "$\int dS$" in a reversible process? Or, is $\int dQ/T$ for an irreversible process less than $\int dQ/T$ for a reversible process and together with entropy generation in an irreversible process, the total entropy change is the same in both an reversible and irreversible process? Please clarify.

 A: Let me start by saying I agree with what both @Barbaud Julien and @Chester Miller said. So my answer is only intended to provide a different perspective.
First, it is true that the entropy is a state function of a system, and therefore the change of entropy of the system between any two given equilibrium states will be the same whether or not the process between the states is reversible. However, the change in entropy of the system plus surroundings will depend on the process. It will be zero if the process is reversible, and greater than zero if irreversible. 
Consider a cycle consisting of two isothermal and two adiabatic processes. If the processes were all reversible, we would obviously have a Carnot cycle.  Since your question focuses on the effect of heat transfer on entropy, let’s assume the adiabatic processes are reversible adiabatic (isentropic) and consider only the impact of the heat transfer processes being reversible or irreversible.
Let’s take the example of a heat addition expansion process of an ideal gas between two equilibrium states.  Let the initial and final temperatures be equal. Consequently, from the ideal gas law, the initial and final pressure-volume products are equal. Clearly the process between the states may be a reversible isothermal process. However, it also need not be. We’ll consider both.
Let the temperature of the system be $T_{sys}$ and the temperature of the surroundings be $T_{sys}+dT$. Consider the system and surroundings as thermal reservoirs, i.e., heat transfer between them does not alter their temperatures so that the heat transfer occurs isothermally.
Let a specific quantity of heat, $Q$, transfer from the surroundings to the system. The resulting entropy changes are:
$$\Delta S_{sys}=\frac{+Q}{T_{sys}}$$
$$\Delta S_{surr}=\frac{-Q}{T_{sys}+dT}$$
The net change in entropy (system + surroundings) is thus:
$$\Delta S_{net}=\frac{+Q}{T_{sys}}+\frac{-Q}{T_{sys}+dT}$$
Now, note that if $dT\to 0$, then $\Delta S_{net}\to 0$ and the process is a reversible isothermal process.  
However, for any finite temperature difference, $dT>0$, $\Delta S_{net}>0$ and the process is irreversible.
For both the reversible and irreversible processes, the change in entropy of the system is the same. However, for the irreversible process the change in entropy of the system is greater than the change in entropy of the surroundings. The excess entropy that is created in the irreversible expansion process means more heat must be rejected to the cold temperature reservoir (surroundings) during the isothermal compression in order for the cycle entropy to be zero.  That results in less energy to do work in the cycle. 
Although in this example the same amount of heat is transferred reversibly and irreversibly, clearly the rate of heat transfer will be greater for the irreversible than the reversible process owing to the finite temperature differential for the irreversible process. So for the amount of heat transfer to be the same, the product of the very slow heat transfer rate times the very long time for the reversible process would need to equal the product of the higher heat transfer rate times the shorter time duration for the irreversible process. 
Hope this helps.
A: 1 -> First, Q is not the same in both cases. Q (like W) is a path-dependant function. So whether you take the reversible or the non-reversible path to get to your final step, you are going to end up with different values of Q and W (think about the Carnot cycle for example : you get the most efficeint output W when it is reversible, but as soon as you introduce some irreversibilities, W is reduced)
I did not get your points 2 and 3, they seem to be mostly repeating the same question. You can enlight me more on what you don't understand
EDIT : I would like to illustrate the fact that heat can be different according to reversibility by taking the example of an isothermal gas compression. Unfortunately, Idon't have time to upload pictures to make it more clear, so I will just assume that you can visualize this classical problem by yourself.
If you have a reversible isothermal compression, (with P=Pext all along), you will get that the work perforomed is Wreversible = nRTln(V1/V2) >0 (basic thermo exercise). The heat will be given by Qrev=deltaU-Wrev
Now, if you take another isothermal transformation [from the same P1,T1V1 to the same P2,T1,V2 states], with a contant outside pressure P0. In this cqse, it correpsonds to a pressure brutally applied all at once, which will cause irreversibilities. The work performed will be W = P0(V1-V2) (<*Wrev)
deltaU is a state funtcion so it remains the same in both cases, and you get Q = deltaU-W > Qrev
The irreversible process betweeen the same states caused more heat to be transfered. what happened ? Well the violence of the second process caused some additional tranfers to occur (e.g. friction between materials moving fast relative to each other...). this is what creates irreversibilities.
So the irreversibility is indeed associated with different transfer processes.
The "benefit" of a reversible process, physically speaking, is that you can get more work from your system, if you're studying heat engines for example. You will get the best efficiency out of a reversible engine, as stated by 2nd law. But the idea of "benefit" depends of course on the context and what you want.
On a problem-solving point of view, the benefit of considering reversible process is obvious. If you are faced with an irreversible process, just forget it, and consider a reversible one that takes you to the same point instead. In this newly made up process, you can use dS=deltaQ/T, and compute S2-S1 this way. And then, you use the fact that s is a state function to say that the change of S is the sxame regardless of the path, so your result S2-S1 holds true even for that irreversible transformation you are studying
A: I would like to expand a little on what @Barbaud Julien said (even though what he said was perfectly correct).  
If you want to determine the change in entropy for an irreversible process, what you need to do is devise (dream up) a reversible path between the same two end states.  This reversible path does not have to bear any resemblance whatsoever to the actual irreversible path.  For example, even if the irreversible path is adiabatic, the reversible path does not need to be.  So comparing the heat transferred and boundary temperatures in the reversible path with that of the irreversible path is not too useful.  
But let's talk about that comparison that Barbaud did for an "isothermal" process.  In the reversible path, the interior temperatures of the gas are all essentially at the boundary temperature throughout the process.  In the irreversible path however, the interior temperatures depend on position throughout the gas, and are less than the boundary temperature during the expansion.  So, even though we are calling the process isothermal (because the boundary is held at fixed temperature), the process is not really isothermal because the interior temperatures are different.
In the irreversible expansion, entropy is being generated within the gas both as a result of temperature gradients within the gas and as a result of viscous friction within the gas.  So the physical processes taking place within the gas for reversible and irreversible paths between the same to end states are very different.
ADDENDUM
What would you think if I told you that there is not just one reversible path between the initial and final state of the "isothermal" irreversible process discussed by @Barbaud Julien.  
What would you think if I told you that there are an infinite number of reversible paths, and that some of these paths absorb more heat and do more work than the irreversible process, and some of these paths absorb less heat and do less work than the irreversible process.  But, all of these paths have exactly the same entropy change.  Thoughts?
