# Entropy & The Second Law

In this notes sheet...

As seen in the notes sheet on the second law, we can compare a heat engine’s actual efficiency with that of its reversible counterpart. If we want to compare these in greater detail, we need to use **entropy**.

## The Clausius Inequality

In the notes sheet on the second law, we looked at heat engine connected to two thermal reservoirs, modelling two heat transfers only:

In reality, however, this is not the case. Instead, each of these heat transfers occurs in infinitely many infinitesimally small steps, dQ:

We can integrate to sum the total positive and negative heat transfers in one complete cycle:

However, this on its own is not particularly helpful. Instead, we want to be able to find the heat transfer at a specific temperature. Therefore, the Clausius inequality looks at Q/T instead:

### Reversible Cycles

The Carnot cycle is the ideal reversible cycle:

Applying the Clausius inequality:

Since the process is reversible, the two ratios must equal each other. **Therefore, for a reversible process:**

### Irreversible Cycles

We know that the efficiency of the reversible engine is greater than that of the irreversible one, so the heat output of the irreversible engine is greater than that of the reversible:

This means that

But since they are both taking heat from the same hot reservoir:

Therefore, plugging these into the equation derived above:

**We find that for an irreversible cycle:**

## Entropy

Applying Clausius’ inequality to the two reversible cycles shown above shows that the integral of dQ/T equals zero in both. This means that **the change in Q/T is path independent: it is a property.**

This property is called **entropy, S**, and the change in it is given as:

The units of entropy are J/K – Joules per Kelvin

If we take a cycle with one reversible and one irreversible process, we know the total change in entropy must be negative (see Clausius’ inequality above):

The right-hand side is the definition of change in enthalpy, so:

Whenever an irreversible process occurs, the enthalpy increases. The entropy in the system itself may decrease, but then that of the surroundings will increase significantly more.

## Temperature-Entropy Diagrams

The area under a **reversible** temperature-entropy curve is the heat transfer that takes place in that process:

### Adiabatic Process

The

**reversible process is isentropic**, as its enthalpy change equals its heat transfer, zeroThe

**irreversible**process experiences an**increase in entropy.**

### Heat Addition

The Process is not adiabatic, so there is a heat transfer to the system.

**Both**the reversible and irreversible processes**increase in entropy**.The

**irreversible process increases more**in entropy.

### Heat Rejection

There is a negative heat transfer in heat rejection

This means the change in entropy for a reversible process will also be negative

The change in entropy for the irreversible process could be either positive or negative, depending on how irreversible it is

### Carnot Cycle

The Carnot cycle is a rectangle on a T-S diagram:

Isothermal & reversible heat rejection, so reduction in entropy.

Adiabatic & reversible, so isentropic. Temperature increases.

Isothermal & reversible heat addition, so increase in entropy.

Adiabatic & reversible, so isentropic. Temperature decreases.

## Combining First & Second Laws

According to the first law, for a smaller part of a reversible process:

Taking the heat transfer as the area under the T-S graph and work as the area under the P-V diagram:

**This equation applies to both reversible and irreversible processes.**

Alternatively, we can rewrite the first law in terms of enthalpy:

**This equation also applies to both reversible and irreversible processes.**

### Isothermal, Reversible Processes

In an isothermal process, the first law becomes: