# The Second Law and Entropy

Physics II · Thermodynamics · https://tryals.app/learn/physics-ii/the-second-law-and-entropy

## Two Statements, One Law

The second law has two classical formulations that appear to concern different machines.

**Kelvin-Planck**: no cyclic process can take heat from a single reservoir and convert it entirely into work. **Clausius**: no cyclic process can transfer heat from a colder body to a hotter one with no other effect.

They are logically **equivalent**: assume a violation of either and you can construct a violation of the other. Couple a Kelvin-Planck violator to an ordinary refrigerator and the pair moves heat from cold to hot with no net work, breaking Clausius.

Analysing arbitrary cycles gives the **Clausius inequality**:

$$\oint \frac{\delta Q}{T} \leq 0$$

with equality exactly for reversible cycles. The equality case is the important one: if $\oint \delta Q_{rev}/T = 0$ round every reversible loop, then $\delta Q_{rev}/T$ is the differential of a state function. That function is **entropy**:

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

The subscript is essential. $S$ is a state function, so to compute $\Delta S$ for an *irreversible* process you invent any convenient reversible path between the same endpoints and integrate along that instead.

For an isolated system the inequality becomes the **entropy increase principle**, $\Delta S \geq 0$, with equality only for reversible change. Combining the first and second laws for a simple system gives the **thermodynamic identity**:

$$dU = T\,dS - P\,dV$$

which contains no $\delta$ at all, every term is a state function differential, which is why it holds for any process whatever, reversible or not.

**Mixing** two different ideal gases at the same $T$ and $P$ raises the entropy by $\Delta S = -nR\sum x_i \ln x_i$, which is strictly positive: mixing is spontaneous and unmixing costs work.

> **Common pitfall:** using $\Delta S = Q/T$ with the actual heat of an irreversible process. That underestimates the change every time. Entropy is defined through the *reversible* heat, so you must construct a reversible path between the same two states and use its heat instead.

## Practice questions

6 of this lesson's 11 practice questions, with answers. The full set is in the app.

### 1. Why are the Kelvin-Planck and Clausius statements of the second law equivalent?

A. Both can be derived mathematically from the first law of thermodynamics
B. A violation of either can be used to construct a violation of the other
C. They each demand that entropy strictly increases in reversible processes
D. Both require heat engines to operate using only a single thermal reservoir

**Answer:** B. A violation of either can be used to construct a violation of the other

**Why:** Couple a Kelvin-Planck violator to a normal refrigerator: the combination moves heat from cold to hot with no net work, violating Clausius. The reverse construction works too, so neither statement is weaker.

Page: https://tryals.app/practice/physics-ii/the-second-law-and-entropy/why-are-the-kelvin-planck-and-clausius-statements-of-the-second-law

### 2. To find the entropy change of an irreversible process, you may integrate the actual heat divided by temperature along the real path.

**Answer:** False

**Why:** False, that underestimates the change every time, since $\oint\delta Q/T < 0$ for an irreversible cycle. Because $S$ is a state function, you invent any reversible path between the same endpoints and integrate along that.

Page: https://tryals.app/practice/physics-ii/the-second-law-and-entropy/to-find-the-entropy-change-of-an-irreversible-process-you-may

### 3. The cyclic integral of actual heat exchange over temperature is non-positive, whereas entropy change is defined strictly through reversible heat. What does this distinction mean for evaluating system transformations?

A. Path heat tells us how entropy is gained, while state variables measure work
B. Actual transferred heat is only a lower bound on the true change in entropy
C. Cyclic paths destroy entropy whenever irreversible processes occur within them
D. State functions depend on the history of heat flow rather than boundary states

**Answer:** B. Actual transferred heat is only a lower bound on the true change in entropy

**Why:** Because the Clausius inequality bounds the cyclic integral below zero, the real irreversible heat transferred along a path yields less than the genuine state difference ΔS. Entropy is never destroyed, nor does a state function retain any memory of past heat paths.

Page: https://tryals.app/practice/physics-ii/the-second-law-and-entropy/the-cyclic-integral-of-actual-heat-exchange-over-temperature-is

### 4. An isolated system undergoes a process in which one part gains 12 J/K of entropy and another loses 7 J/K. Compute the total entropy change in J/K.

**Answer:** 5 (within ±0.2)

**Why:** $\Delta S_{total} = 12 - 7 = +5$ J/K. A *part* of an isolated system may lose entropy; the second law constrains only the total, which must not decrease.

Page: https://tryals.app/practice/physics-ii/the-second-law-and-entropy/an-isolated-system-undergoes-a-process-in-which-one-part-gains-12-j-k

### 5. Which statements about entropy are correct?

A. It is a state function
B. The total for an isolated system never falls
C. It is defined through the reversible heat divided by temperature
D. It cannot decrease in any part of a system

**Answer:** A. It is a state function; B. The total for an isolated system never falls; C. It is defined through the reversible heat divided by temperature

**Why:** Entropy is a state function defined by $\delta Q_{rev}/T$, and the total for an isolated system cannot fall. A *part* can lose entropy freely, a freezer does it continuously, provided the surroundings gain more.

Page: https://tryals.app/practice/physics-ii/the-second-law-and-entropy/which-statements-about-entropy-are-correct

### 6. The thermodynamic identity $dU = T\,dS - P\,dV$ holds even for irreversible processes.

**Answer:** True

**Why:** True, every quantity in it is a state function, so the relation between them holds regardless of the path taken. This is exactly why it is more powerful than the first law written in terms of heat and work.

Page: https://tryals.app/practice/physics-ii/the-second-law-and-entropy/the-thermodynamic-identity-du-t-ds-p-dv-holds-even-for