Why is entropy defined in the way it is?

Question

In thermodynamics course we were taught that in a reversible adiabatic process of an ideal gas the entropy change is zero.

Now what I didn't get intuitively is that if the system is compressed adiabatically and reversibly then shouldn't the randomness of the system be different from what it was earlier (as the system occupies less space). And this made me to think that why don't we define entropy change as change in internal energy per unit temperature i.e.

$$dS=\frac{dU}{T}$$

What could be wrong with the above definition ?

Answer 1

In the adiabatic reversible compression process you described, the temperature increases (entropy increase), but the volume decreases and the pressure increases (entropy decreases). The net effect is zero entropy change. In an adiabatic reversible compression or compression, it doesn't matter whether you are dealing with an ideal gas; the entropy change is zero.

To get the entropy change for an irreversible process, you need to devise an alternate reversible path between the same pair of thermodynamic end states, and determine the integral of dQ/T for this reversible path. There will actually be an infinite number of reversible paths between the two end states that you can choose from, but they all give you exactly the same value for the integral. This means that entropy is a function only of state, and not path between the states.

Answer 2

What could be wrong with the above definition ?

It’s not the definition of entropy change. The definition is a for a reversible transfer of heat as

$$dS=\frac {\delta Q_{rev}}{T}$$

Since the adiabatic process is reversible and there is no heat, $Q_{rev}=0$ and $dS=0$.

There are two potential components of entropy change, entropy transfer, which requires heat and/or entropy generated, which results from irreversible heat or irreversible work, neither of which occurs in a reversible adiabatic process.

The temperature increase in the adiabatic compression increases entropy, but the decrease in volume decreases entropy by an equal amount for a total entropy change of zero.

Hope this helps.

Answer 3

Your definition could be fine if you remember that the state of a system in thermodynamic equilibrium can be identified by a pair of thermodynamic variables (as long as there is only one chemical species and one phase).

Namely, you can define internal energy as a function of two thermodynamic variables. A common choice of this pair of variables are volume (or density) and entropy, whatever it is,

$$U(V,S) \ ,$$

and taking the differential of this function, it's possible to write

$$dU = \left(\frac{\partial U}{\partial V} \right)_S dV + \left(\frac{\partial U}{\partial S} \right)_V dS \ , $$

identifying, for a gas, the partial derivatives with pressure and temperature (physical quantities you can measure with a barometer and a thermometer),

$$P = -\left(\frac{\partial U}{\partial V} \right)_S \qquad , \qquad T = \left(\frac{\partial U}{\partial S} \right)_V \ ,$$

to be consistent with the first principle of thermodynamics

$$\begin{aligned}dU & = \delta L + \delta Q = \\ & = -P dV + T dS \ , \\ \end{aligned}$$

While the relation above holds for every transformation, only for a reversible transformation on a gas system the relations $\delta L^{rev} = -P dV$, $\delta Q^{rev} = T dS$ hold.