$dQ=Tds$ and quasistatic process See Ján Lalinský's commment on Entropy $dQ=TdS$ and Work $dW = -pdV$ conditions? 
My question are:


*

*Is there a way to prove that $dQ=Tds$ is a quasistatic process? 

*What's the funcitonal description for quasistatic process? 
 A: Since it is necessary to first have a functional understanding of a quasi-static process I will attempt to answer question 2 first.
A quasi-static process is one that is carried out such that disequilibrium is minimized at each step in the process. Such a process is also a reversible process if it does not involve friction.  Such disequilibria typically involve pressure and temperature differentials.
For example, heat transfer requires a temperature difference between two things. A temperature difference is an example of disequilibrium and it is that disequilibrium which drives the heat transfer process. To reduce the disequilibrium you need to reduce the temperature difference. That of course slows down the rate of heat transfer. In order to eliminate the disequilibrium, the temperatures would have to be equal. But then you would not have heat transfer. This demonstrates that in real life all processes are necessarily irreversible. We can only approach reversible processes in the limit. 
Now regarding question 1.
Let's first rearrange the equation
$$dS=\frac {dQ}{T}$$ 
This equation does not necessarily mean a quasi-static (reversible) process is involved unless it is accompanied by an explicit statement that $dQ$ is a reversible transfer of heat. That stipulation is explicit in the following commonly used definition of $dS$:
$$ds=\frac{dQ_{rev}}{T}$$
Where $dQ_{rev}$ means a process that involves a reversible (quasi-static) transfer of heat. Since entropy is a property of the system, the difference in entropy of the system between two equilibrium states does not depend on the process being quasi-static. However the total entropy change, system + surroundings, does depend on whether or not the process is quasi-static (reversible). The total will $=0$ if the process is quasi-static (reversible) and will be $>0$ if not quasi-static (irreversible).  So I think the relevant “proof” you would want to know is how this is true. 
The following is not necessarily a “proof” but rather an example of why a process needs to be quasi-static (reversible) in order for the total entropy change (system + surroundings) to be, or approach, zero.
Consider a system H (a hot body) and its surroundings C (a cold body). Further consider both H and C to be thermal reservoirs, that is, they are so massive relative to the desired heat transfer, that heat transfer between them doesn’t change their temperatures. The temperature of body H is $T_H$ and the temperature of body C is $T_C$. We bring the bodies together and desire to transfer heat $Q$ from H to C. Since the temperature of either does not change, the heat transfer occurs isothermally.  Let’s look at the entropy changes:
For Body A (System):  $$\Delta S_A = -Q/T_H$$ (which is a drop in entropy)
For Body B (Surroundings):  $$\Delta S_C = +Q/T_C$$ (which is a rise in entropy)
The total entropy change $$\Delta S_A + \Delta S_C = -Q/T_H + Q/T_C$$
Then for any $$T_H > T_C$$
$$ Q/T_C - Q/T_H> 0$$
In order for the total entropy change to approach zero, the process must be carried out quasi-statically, meaning the temperature difference must approach zero.  This of course, results in the heat transfer rate approaching zero and the time it takes to transfer Q infinitely long. This is why in order to have quasi-static (reversible) processes those processes must be carried out very slowly
Hope this helps
A: I think, if the process is not quasi static, the properties will not be the same through out the system(i.e. the properties will change from point to point within the system).If we discuss about the pressure of the entire system or temperature of the entire system,entropy change of the system, etc.. means we are discussing about a quasi static process.This is why, quasi static processes are said to be made up of equilibrium states.These states can be plotted as points in a graph(PV or TS graphs).Since the terms T and dS in your equation are the properties of the entire system , you are describing a quasi static process.
