Entropy

The thermodynamic definition of entropy

concentrates on the change in entropy, dS, that occurs as a result of a physical or chemical change (in general, as a result of .a ‘process’)

The definition is motivated by the idea that a change in the extent to which energy is dispersed depends on how much energy is transferred as heat.

THERMODYNAMIC DEFINITION OF ENTROPY

THERMODYNAMIC DEFINITION OF ENTROPY

The thermodynamic definition of entropy is based on the expression:

For a measurable change between two states i and f this expression integrates to:

That is, to calculate the

difference in entropy between any two states of a system, we find a reversible path between them, and integrate the energy supplied as heat at each stage of the path divided by the temperature at which heating occurs.

THERMODYNAMIC DEFINITION OF ENTROPY

Furthermore, because the temperature of the

surroundings is constant whatever the change, for a measurable change.

That is, regardless of how the change is brought about in the system, reversibly or irreversibly, we can calculate the change of entropy of the surroundings by dividing the heat transferred by the temperature at which the transfer takes place.

THERMODYNAMIC DEFINITION OF ENTROPY

Calculating the entropy change in the

surroundings.

To calculate the entropy change in the surroundings when 1.00 mol H2O(l) is formed from its elements under standard conditions at 298 K, we use ΔHѲ = −286 kJ from Table 2.7. The energy released a heat is supplied to the surroundings, now regarded as being at constant pressure, so qsur = +286 kJ.

Calculating the entropy change in the surroundings

Therefore,

Δssur = 2.86 × 105 J

298 KΔssur = +960 J

K−1

This strongly exothermic reaction results in an increase in the entropy of the surroundings as energy is released as heat into them.

dStotal = dSsystem +

dSsurroundings After complete cycle: Total heat supplied is:

Total work output is: ΔU = 0, so Q = W

Other way is to define entropy change of the

system alone.

So: Either W and Q must be negative: work is done to the system and heat is extracted from it or W and Q must be zero.Thus: W = Q ≤ 0, or Hence:

What if the cycle is reversible?So: W = Q ≥ 0, So:

Hence: In order to be consistent with the first result,

Take a room and an ice cube as an

example. Let's say that the room is the isolated system. The ice will melt and the total entropy inside the room will increase. This may seem like a special case, but it's not. All what I'm really saying is that the room as whole is not at equilibrium meaning that the system is exchanging heat, etc inside itself increasing entropy.

Example of increasing entropy