Entropy Change in Reversible Processes

Starting with the Best: The Carnot Cycle

Let's circle back to our hero of efficiency, the Carnot cycle. Remember, it's a completely reversible cycle, and it sets the universal speed limit for any heat engine.

We found two different ways to express its efficiency:

1. From the definition of any heat engine's efficiency:
   ${\eta }_{th}=1-\frac{Q_L}{Q_H}$
   Where $Q_H$ is the heat added from the hot reservoir and $Q_L$ is the heat rejected to the cold reservoir.

2. From the special properties of the Carnot cycle, we proved that its efficiency depends only on the absolute temperatures of the reservoirs:
   ${\eta }_{th,rev}=1-\frac{T_L}{T_H}$

Since the Carnot cycle is the engine that achieves this ideal efficiency, we can set these two expressions equal to each other:
$1-\frac{Q_L}{Q_H}=1-\frac{T_L}{T_H}$
A little bit of algebra gives us:
$\frac{Q_L}{Q_H}=\frac{T_L}{T_H}$
And rearranging this, we get a fascinating result:
$\frac{Q_H}{T_H}=\frac{Q_L}{T_L}$
Let's think about what this means for the entire cycle. We use the standard sign convention: heat entering the system is positive ($+Q_H$), and heat leaving the system is negative ($-Q_L$). So, we can write $Q_L$ as $-Q_{out}$. The relationship is then:
$\frac{Q_H}{T_H}=\frac{|Q_L|}{T_L}$
If we sum these quantities around the cycle, using our sign convention, we get:
$\frac{Q_H}{T_H}+\frac{-Q_L}{T_L}=0$
Or, more generally, for any reversible cycle:
${\sum }_{cycle}\frac{Q_{rev}}{T}=0$
This is a huge deal! Whenever we have a quantity whose net change is zero over a complete cycle, it means we're dealing with a property, or a state function. Think back to internal energy ($U$). No matter what you do, if you end up back in the same state, the change in internal energy, $\Delta U$, is always zero.

Work ($W$) and heat ($Q$) are not properties; their values depend on the path you take. But this special quantity, $Q_{rev}/T$, acts like a property. This gives us the motivation to define a new property. We call it Entropy, and we give it the symbol $S$.

The Mathematical Definition of Entropy Change

Based on the discovery from the Carnot cycle, the Austrian physicist Rudolf Clausius defined the change in entropy for a system as follows.

For a process that happens at a constant temperature (isothermal), the change in entropy is:
$\Delta S=\frac{Q_{rev}}{T}$
Where $Q_{rev}$ is the heat transferred during a reversible process, and $T$ is the absolute temperature in Kelvin.

But what if the temperature changes during the process? Since you're comfortable with calculus, you'll see this is a natural extension. We just look at an infinitesimally small step of the process, where the temperature is essentially constant. The tiny change in entropy, $dS$, is:
$dS={\left(\frac{\delta Q}{T}\right)}_{rev}$
(We use $\delta Q$ instead of $dQ$ as a reminder that heat is a path function, not a property.)

To find the total entropy change for a process from state 1 to state 2, we just add up all these little changes by taking the integral:
$\Delta S=S_2-S_1={\int }_1^2{\left(\frac{\delta Q}{T}\right)}_{rev}$

This short video gives a great, clear summary of this definition.

The units of entropy are energy per temperature, so in the SI system, that's Joules per Kelvin (J/K) or kilojoules per Kelvin (kJ/K). More often, we'll talk about specific entropy (entropy per unit mass), which has units of kJ/kg·K.

But Why Reversible?

This is the most critical and often confusing part of the definition. The subscript "rev" is not optional. The entropy change between two states is a fixed value (because it's a property), but to calculate it using this formula, you MUST use the heat transfer that would occur during an imaginary reversible path between those states.

Why? Because only for a reversible process is the quantity $Q/T$ guaranteed to be a property that returns to zero over a cycle.

Let's build the intuition for this.

This is the key takeaway: Entropy change is path-independent, but its calculation using heat transfer is path-dependent.

Think of it like calculating the change in your elevation when hiking between two points on a mountain. The change in elevation ($\Delta z$) is fixed—it only depends on your start and end points. But if you tried to calculate it by measuring the length of your path, the answer would be different if you took a winding trail versus a steep, direct one. The formula for entropy is like this: it only gives the right "elevation change" if you use the specific "path" of a reversible process.

Tying the Math Back to Energy Dispersal

Okay, so we have a formula: $dS=\delta Q_{rev}/T$. How does this strange-looking fraction connect to our beautiful, intuitive idea of "energy dispersal"?

This is where the $T$ in the denominator becomes the star of the show.

Let's break it down with an analogy. Imagine you're in a completely silent library. If someone whispers ($\delta Q$), it's a huge disturbance. The change in the "acoustic environment" is massive. The "dispersal" of that sound is very noticeable. This is like adding heat to a very cold system (low $T$). The entropy change is large.
$\Delta S=\frac{\delta Q}{T_{low}}\to \text{Large change}$
Now, imagine you're at a loud rock concert. If someone whispers that same whisper ($\delta Q$), you won't even notice. The change is negligible compared to the existing noise. This is like adding the same amount of heat to a very hot system (high $T$). The molecules are already buzzing like crazy. The entropy change is small.
$\Delta S=\frac{\delta Q}{T_{high}}\to \text{Small change}$

This simple idea explains the Second Law! It tells us why heat spontaneously flows from hot to cold.

• When a quantity of heat $Q$ leaves a hot object ($T_H$), the object's entropy decreases by $Q/T_H$.
• When that same heat $Q$ enters a cold object ($T_L$), the object's entropy increases by $Q/T_L$.

Since $T_H>T_L$, the fraction $Q/T_L$ is bigger than $Q/T_H$. The entropy gain of the cold object is greater than the entropy loss of the hot object. The total entropy of the universe has increased. The energy has become more dispersed overall, and our mathematical definition perfectly captures this.

The T-S Diagram: A New Tool

Our definition $dS=\delta Q_{rev}/T$ can be rearranged to $\delta Q_{rev}=TdS$.
Integrating this gives us:
$Q_{rev}={\int }_1^{2}TdS$
This gives us an amazing graphical tool. Just like the area under the curve on a P-V diagram represents work, the area under the curve on a Temperature-Entropy (T-S) diagram represents the heat transferred during a reversible process.

This is an incredibly useful way to visualize heat transfer in cycles, as we'll see many times in the future.

• An isothermal (constant temperature) process is a horizontal line on a T-S diagram.
• An adiabatic and reversible (no heat transfer) process has $Q_{rev}=0$, which means $\Delta S=0$. This is a vertical line on a T-S diagram. We call this special process isentropic (constant entropy).

Let's Do Some Calculations

Time to make this concrete with some problems.

Key Takeaways

Let's pause and recap. This was a math-heavy but crucial step.

• We discovered from the Carnot cycle that for any reversible cycle, the sum of $Q/T$ is zero. This motivated the definition of a new property: Entropy ($S$).
• The mathematical definition for the change in entropy is $\Delta S=\int (\delta Q/T{)}_{rev}$.
• This formula is a calculation tool. Entropy is a property, so its change between two states is fixed. We use an imaginary reversible path to calculate that change.
• The $T$ in the denominator is key: adding heat to a cold system increases entropy more than adding the same heat to a hot system. This is the mathematical reason heat flows from hot to cold.
• On a T-S diagram, the area under the curve of a reversible process represents the heat transferred.

Preview of the Next Lesson

We've successfully defined and calculated the entropy change for a system. But what about the entropy change of the surroundings? And most importantly, what happens to the total entropy of the universe (system + surroundings) when we deal with real, irreversible processes like friction, or heat transfer across a finite temperature difference?

In our next lesson, we will answer that question by introducing the Increase of Entropy Principle. This powerful principle is the ultimate statement of the Second Law of Thermodynamics and will finally give us the tool to determine whether a process is possible, impossible, or ideal. See you then