# Why is entropy equal to #q# for a reversible process? What is the difference between reversible and irreversible?

##### 1 Answer

**Entropy** is defined as

#\mathbf(DeltaS >= q/T),#

where the definition separates as

#DeltaS > (q_"irr")/T,#

#DeltaS = (q_"rev")/T,#

with *inefficient*, *irreversible* heat flow, and *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 dispersion of particle energy 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 its energy distribution to a certain amount of dispersiveness.

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

(Same with a liquid vs. a solid.)

**REVERSIBLE HEAT FLOW**

Adding reversible heat **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 **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 *rectangles*, kind of like this:

Since we know that **some heat was lost** when

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, **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

This makes sense, because we know that entropy is a state function, so if

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

But if you perform an *irreversible* process, which is not necessarily cyclic...

That's not kosher, because **path function** process! That is, the path apparently *does* matter, when it *shouldn't* for a state function.

Hence, it couldn't be that

If you perform an

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