Hello, entropy!

I think it is best to plunge right in by starting with entropy.

But, first, some preliminaries.

In thermodynamic analysis, we focus our attention on certain quantity of matter (called a system) in a box. The box walls constitute the boundary between the system and its surroundings. Interestingly, the combination of system and its surroundings is called the universe — thermodynamicists can never be accused of lacking ambition!

It is easy to see that the system has some quantities associated with it: the volume of the box V , and the amount of stuff. Since stuff can have many different species, we could use the number of moles n_{i} of each species i. Another good ‘unit’ would be the number of ‘particles’ (atoms or molecules) of each species. In classical thermo, number of moles is preferred.

It also has a certain energy, say, the sum of potential and kinetic energies of the atoms and molecules. This energy has a specific name: Internal energy, E .

At this elementary level, let us restrict changes in E to be due only to two methods of energy transfer: work and heat. Mathematically, we write:

dE = {\mathchar'26\mkern-12mu d} Q + {\mathchar'26\mkern-12mu d} W

When a small quantity of heat dQ comes into the system, it leads to an increase in the system energy by dE . It also increases the system’s entropy by dS = dQ / T. Thus, thermodynamics introduces a new quantity associated with the system.

Here, then, are the variables that we deal with: S, E, V, n_{i}

Consider now a prototype system inside a box consisting of two sub-systems, each consisting of half a mole of copper. Its external boundary is adiabatic (no heat transfer) and impervious (no stuff transfer) and rigid (no work transfer). Such a system is said to be isolated.
In other words, this is a system with a constant E, V and n_{i}.

Let the first subsystem is at 300 K and the second is at 400 K, and in the initial state, they are separated by a perfectly insulating wall (adiabatic boundary). On removing this adiabatic boundary, we expect the hotter subsystem to cool down, and the colder subsystem to heat up — a process that will proceed until the two subsystems reach the same (intermediate) temperature. If the specific heat capacity $C_{p}$ of copper is assumed to be a constant in the relevant temperature range, it’s easy to calculate the final temperature to be 350 K.

In this calculation, we make use of the fact that the total energy E of the system does not change because the system is isolated.

What about the system’s entropy?

Since we know the initial and final states of the two subsystems, we can compute their entropy changes easily.

\displaystyle S_{f,I}  - S_{i, I} = (1/2) [\textrm{mol}] \int_{300}^{350} \frac{C_{p} [\textrm{J/mol-K}] dT [\textrm{K}]} {T [\textrm{K}]}   ,


\displaystyle S_{f,II}  - S_{i, II} = (1/2) [\textrm{mol}] \int_{400}^{350} \frac{C_{p} [\textrm{J/mol-K}] dT [\textrm{K}] }{T [\textrm{K}]}  .

And therefore,

\displaystyle \Delta S = S_{f} - S_{i} = (C_{p} / 2) \left[ \ln\frac{350}{ 300}  + \ln \frac{350}{400}  \right] = C_{p} \ln  \frac{1225}{1200} [\textrm{J/K}].

In other words, the entropy has increased.

Let us unpack the argument:

We prepared a system in a special way, in the sense that it was an isolated system that had an internal constraint in its initial state. On removing this constraint, a ‘spontaneous’ process is set in motion inside the system (in this case, it is transfer of heat from the hot subsystem to the cold subsystem). The spontaneous process has the effect of increasing the entropy.

But this process does not go on forever! It stops. And the system is said to have reached its equilibrium — in this case, thermal equilibrium.

Since the process had the effect of increasing the system’s entropy, we conclude that the entropy keeps increasing, until the process stops — in other words, the equilibrium state of the system is characterized by a maximum in its entropy.

Let us look at this process in a bit more detail in the next class.

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