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QUESTION

Why does the entropy is defined by q reversible? what is the meaning of defining as q reversible not irreversible? :)

is defined as

\mathbf(DeltaS >= q/T),

where the definition separates as

DeltaS > (q_"irr")/T,

DeltaS = (q_"rev")/T,

with q_"irr" being the inefficient, irreversible heat flow, and q_"rev" being the efficient, reversible heat flow.

ENTROPY VS HEAT FLOW

By definition, entropy in the context of heat flow at a given temperature is:

The extent to which the heat flow affects the number of microstates a system can access at that temperature.

So, it's essentially a threshold for how significantly you can increase the particle randomness in a system by supplying heat into the system.

The higher the resultant entropy, the more influential the heat flow was, and the less heat you needed to impart into the system to get it to a certain amount of "randomness".

This should make sense if you compare the entropy of a gas to that of a liquid; a gas is more freely moving, so it can become more "random" more easily.

(Same with a liquid vs. a solid.)

REVERSIBLE HEAT FLOW

Adding reversible heat q_"rev" is only possible by adding differential quantities of heat delq incrementally; it is added infinitesimally slowly, in such a way that no heat is lost at all.

In other words, you're adding heat so slowly that the system has time to re-equilibrate as you add heat. This is basically the maximum amount of heat that you can add.

It's analogous to doing a full integral instead of doing an MRAM, RRAM, or LRAM approximation. You don't overestimate, underestimate, or miss anything.

You (ideally) retain all the heat that you add.

IRREVERSIBLE HEAT FLOW

Adding irreversible, inefficient heat q_"irr" results in some loss of heat.

An example is adding heat, then pausing momentarily, then adding more heat in a separate, second step. This allows the system to lose heat while you are stopping momentarily.

You could infer from what I said earlier that it's like approximating (and underestimating) the area under the curve for the T vs. S graph above using rectangles, kind of like this:

Since we know that some heat was lost when q_"irr" was supplied into the system in an attempt to increase its entropy, let's account for that heat.

This heat is represented by the unaddressed triangles in the above diagram. What we really get is:

color(blue)(q_"irr" + q_"lost" = q_"rev")

Thus, q_"irr" < q_"rev", and we call irreversible heat flow inefficient.

Now compare back to the original expressions at the top of the answer, and you should convince yourself that this holds true.

WHY IS ENTROPY EQUAL TO REVERSIBLE HEAT FLOW?

Well, let's put it this way. If you do stuff in such a way that you go from an initial state back to a final state equal to the initial state, then you've performed a cyclic process (such as the Carnot cycle).

We denote this as ointdS = 0.

This makes sense, because we know that entropy is a state function, so if S_i = S_f, then

color(blue)(ointdS = DeltaS = 0).

But if you perform an x process, and then perform a y process instead of a -x process (where x ne y), then you've performed an irreversible process, which is not necessarily cyclic...

That's not kosher, because S_i ne S_f, and thus, you've performed a path function process! That is, the path apparently does matter, when it shouldn't for a state function.

Hence, it couldn't be that DeltaS = (q_"irr")/T.

If you perform an x process and then a -x process, you've performed a reversible cyclic process. Now, you CAN say this:

color(blue)(oint_(S_i)^(S_f)dS = DeltaS = oint_(T_1)^(T_2) (delq_"rev")/TdT = 0)

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