Why isn't $∆S = frac{KQ_{rev}}{T}$?

Physics Asked on December 18, 2020

In my book , some sites and even on this question, I read that the change in entropy of a system in a reversible process is directly proportional to the heat added and inversely proportional to the temperature at which the heat is added.

And thus

$$∆S =frac{KQ_{rev}}{T}$$

But I want to know why don’t we have a proportionality constant here and if it is there how did we derive that it’s value is $$1$$ ?

Here is another way to think about it, If you have the first law of thermodynamics,

$$dU = dq+dw$$

You can rearrange this to become,

$$dq = dU - dw$$

Now, the above equation the heat transfer is an 'inexact differential' that means there is no function $$F$$ such that when you take the differential of $$F$$ i.e: $$dF$$ that you get $$dq$$. With this in mind, we can actually get a multiplicative factor on both sides which does make the differential exact. Turns out this multiplicative factor is temperature (*), hence we write:

$$frac{dq}{T} = dS= frac{1}{T} (dU - dw)$$

Now, the good thing is that you got an extra state function to play with whose nice property is that the integral over the whole cycle of a reversible process is zero. If you feel like getting a meaningful interpretation to this new quantity, then see this post

Edit:

On a bit more careful thought, your scaled entropy function (assuming $$k neq 0$$) is also a state function and behaves more or less like entropy. It's integral over a reversible loop is zero, is a state function etc.

However, it may not match with the entropy derived from statistical mechanics due to the scaling factors. Personally, I don't think it is a good idea to reason out entropy using dimensional analysis when it can be well understood from the differentials and integrating factors ( well at least for the most part)

A person in stackexchange actually wrote a scientific paper about derivng that integrating factor which I Found is really cool. The actual paper is in Italian but you can get a rough idea of what it is by reading his answer in this post

Answered by Buraian on December 18, 2020

Your original equation is incorrect. It should read $$Delta S=int{frac{dQ_{rev}}{T}}$$Only for a reversible process at constant temperature is the equation you wrote correct.

Answered by Chet Miller on December 18, 2020

Some background: In the thermodynamics framework, every way of adding energy to a system consists of a generalized driving force and a generalized displacement. When stretching a solid, for example, the system gains strain energy $$Delta U$$ through a mechanical force $$F$$ and an elongation $$Delta L$$. When compressing a gas, mechanical work is done through pressure $$P$$ and a shift in volume $$-V$$. In the context of electrical charge, the generalized force is a voltage $$E$$, and the charge $$C$$ is displaced.

Each conjugate pair consists of an intensive variable (i.e., one that would stay the same if two systems were pushed together) and an extensive variable (i.e., one that would double). The intensive variable is the generalized force, and the extensive variable is the generalized displacement. Each pair multiplies to give units of energy; in fact, the intensive variable is defined as a partial derivative of energy with respect to the extensive variable.

One special conjugate pair covers heating. The entropy $$S$$ is the extensive variable, representing the generalized displacement. Entropy is the "stuff" that moves when one system heats another. The temperature $$T$$ is the intensive variable; temperature gradients drive heating.

What that in mind, I hope it's clear why the differential energy change upon reversible heating is simply $$T,dS$$ and not $$KT,dS$$. There's no need for $$K$$, as the units already work out; its introduction would add needless complexity and break the definition symmetry of these conjugative variables, among various other issues.

Answered by Chemomechanics on December 18, 2020

In every equation involving physical quantities, it is always possible to add a proportionality constant to allow for different units of measurement between the left and the right side of the equation. It may look weird, but it is not wrong. Although, for clarity, I would avoid introducing it, if not necessary.

Answered by GiorgioP on December 18, 2020

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