Second Law of Thermodynamics

AP Physics 2· AP Physics 2 CED — Thermodynamics· 14 min read

1. Core Introduction to the Second Law★★☆☆☆⏱ 3 min

The first law of thermodynamics only requires energy to be conserved, but it cannot explain why some processes (like heat flowing from cold to hot, or a broken mug reassembling itself) never happen spontaneously. The second law fills this gap by defining the natural direction of all physical processes.

For AP Physics 2, this topic makes up 2-4% of your total exam score, and appears in both multiple-choice (conceptual questions about spontaneity) and as a sub-part of larger free-response questions about heat engines.

📘 Definition

Second Law of Thermodynamics

A fundamental physical law that defines the allowed direction of spontaneous processes, stating the total entropy of the universe never decreases for any physical process.

2. Entropy and the Entropy Form of the Second Law★★★☆☆⏱ 4 min

📘 Definition

Entropy

S

A state function that measures the number of possible microscopic arrangements (microstates) of a system's particles and energy. Intuitively, it corresponds to the disorder or randomness of the system.

For any reversible process at constant temperature (isothermal process), entropy change simplifies to:

\Delta S = \frac{Q}{T}

where $Q$ is the total heat transferred to the system, and $T$ is the constant absolute temperature. The most general statement of the second law, in terms of total entropy change of the universe (system + surroundings), is:

\Delta S_{\text{univ}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} \geq 0

Spontaneous processes have $oxed{\Delta S_{\text{univ}} > 0}$ , ideal reversible processes have $oxed{\Delta S_{\text{univ}} = 0}$ , and no process can have $\ o x e d Δ S_{univ} < 0$ . Because entropy is a state function, $D e l t a S$ only depends on start and end states, not the path taken.

📐 Worked Example

0.80 mol of an ideal gas undergoes a reversible isothermal compression at 290 K. The gas releases 1800 J of heat to the surroundings during the compression. Calculate $D e l t a S_{sys}$ , $r a c 12 D e l t a S_{surr}$ , $r a c 12 D e l t a S_{univ}$ , and state if this process violates the second law.

1
First, define signs: the system releases heat, so $Q_{sys} = - 1800$ J, and $T = 290$ K for both system and surroundings.
2
Calculate $r a c 12 D e l t a S_{sys}$ :
3
$\Delta S_{\text{sys}} = \frac{Q_{\text{sys}}}{T} = \frac{-1800}{290} \approx -6.21 \text{ J/K}$
4
The surroundings gain 1800 J, so $Q_{surr} = + 1800$ J, so:
5
$\Delta S_{\text{surr}} = \frac{+1800}{290} \approx +6.21 \text{ J/K}$
6
Calculate total entropy change:
7
$\Delta S_{\text{univ}} = -6.21 + 6.21 = 0 \text{ J/K}$
8
A total entropy change of zero is allowed for an ideal reversible process, so this process does not violate the second law.

Exam tip:

When calculating entropy change of the surroundings, always explicitly flip the sign of $Q$ from the system. If the system gains heat, the surroundings lose it, so their entropy changes in the opposite direction.

3. Heat Engines and Macroscopic Statements★★★☆☆⏱ 3 min

Two common macroscopic statements of the second law describe practical devices:

Kelvin-Planck Statement: No heat engine can convert 100% of input heat to useful work in a cycle.
Clausius Statement: Heat cannot spontaneously flow from a cooler body to a hotter body without work input.

A heat engine operates between two thermal reservoirs: a hot reservoir at $T_{H}$ that supplies heat $Q_{H}$ , and a cold reservoir at $T_{C}$ that accepts waste heat $Q_{C}$ . For a full cycle, the engine returns to its original state so $r a c 12 D e l t a U_{engine} = 0$ . By the first law, net work output is:

W = Q_H - |Q_C|

Efficiency $e$ of the engine is the ratio of useful work output to heat input:

e = \frac{W}{Q_H} = 1 - \frac{|Q_C|}{Q_H}

📐 Worked Example

A prototype heat engine absorbs 1500 J of heat from a hot reservoir at 600 K, does 500 J of work, and expels 1000 J of waste heat to a cold reservoir at 300 K. Is this engine allowed by the second law?

1
Calculate entropy change of each reservoir: the hot reservoir loses 1500 J, so:
2
$\Delta S_H = \frac{-1500}{600} = -2.5 \text{ J/K}$
3
The cold reservoir gains 1000 J, so:
4
$\Delta S_C = \frac{+1000}{300} \approx +3.33 \text{ J/K}$
5
The engine completes a full cycle, so its entropy change $r a c 12 D e l t a S_{engine} = 0$ , because entropy is a state function.
6
Total entropy change of the universe:
7
$\Delta S_{\text{univ}} = -2.5 + 3.33 + 0 = +0.83 \text{ J/K}$
8
Since $r a c 12 D e l t a S_{univ} > 0$ , this engine is allowed by the second law.

Exam tip:

For any full-cycle process, the entropy change of the engine itself is always zero—don't accidentally add a non-zero entropy change for the engine to the total.

4. Maximum Carnot Efficiency★★★☆☆⏱ 3 min

The most efficient possible heat engine operating between two fixed temperatures $T_{H}$ and $T_{C}$ is the ideal reversible Carnot engine. The Carnot efficiency sets an absolute upper limit that no real engine can exceed, per the second law. The formula is:

e_{\text{Carnot}} = 1 - \frac{T_C}{T_H}

Critical note: $T_{H}$ and $T_{C}$ must be absolute temperatures in Kelvin, not Celsius. 100% efficiency ( $e = 1$ ) is only possible if $T_{C} = 0$ K, which is physically impossible, consistent with the second law.

📐 Worked Example

A natural gas power plant has a combustion chamber temperature of 1100°C and discharges waste heat to cooling towers at 40°C. What is the maximum possible efficiency this plant can achieve?

1
Convert temperatures to Kelvin:
2
$T_H = 1100 + 273 = 1373 \text{ K}, \quad T_C = 40 + 273 = 313 \text{ K}$
3
Substitute into the Carnot efficiency formula:
4
$e_{\text{max}} = 1 - \frac{313}{1373}$
5
Calculate the final result:
6
$\frac{313}{1373} \approx 0.228, \quad e_{\text{max}} = 1 - 0.228 = 0.772 = 77.2\%$
7
This matches expectations: real plants typically achieve 50-60% efficiency, which is below the Carnot limit.

Exam tip:

Always convert Celsius temperatures to Kelvin before plugging into the Carnot efficiency formula. Using Celsius directly will often give you a negative or impossible efficiency, which is an immediate red flag.

5. AP Style Concept Check★★★★☆⏱ 4 min

✓ Quick check

Test your understanding of core second law rules:

Which of the following processes is allowed by the second law of thermodynamics?
- A process with $r a c 12 D e l t a S_{sys} = - 10$ J/K and $r a c 12 D e l t a S_{surr} = + 8$ J/K
- A heat engine operating between 500 K and 200 K with an efficiency of 65%
- A process that decreases the total entropy of the universe
- An ice cube freezing in a 0°C freezer, where the freezer uses work input to move heat from the ice to the room
Reveal answer
An ice cube freezing in a 0°C freezer, where the freezer uses work input to move heat from the ice to the room —
Option A has total $r a c 12 D e l t a S_{univ} = - 2 < 0$ (violates second law). Option B's maximum Carnot efficiency is 60%, so 65% exceeds the allowed limit. Option C directly violates the second law. Option D is allowed: the entropy increase of the surroundings (room) outweighs the entropy decrease of the ice, so total $r a c 12 D e l t a S_{univ} > 0$ .

📐 Worked Example

1.00 mol of an ideal gas undergoes a spontaneous irreversible free expansion into a vacuum, doubling its volume at constant temperature 300 K. (a) What is the change in internal energy of the gas? Justify. (b) Calculate the entropy change of the system. (c) Calculate total entropy change and state if the process is spontaneous.

1
(a) For an ideal gas, internal energy depends only on temperature. The expansion is constant temperature, so $r a c 12 D e l t a T = 0$ , meaning $r a c 12 D e l t a U = 0$ . No work is done in free expansion, so the first law confirms $Q = 0$ .
2
(b) Entropy is a state function, so $r a c 12 D e l t a S$ equals the change for a reversible isothermal expansion between the same states. For reversible expansion, $Q = n R T ln (V_{2} / V_{1})$ , so $r a c 12 D e l t a S_{sys} = Q / T = n R ln (V_{2} / V_{1})$ :
3
$\Delta S_{\text{sys}} = (1.00)(8.31)(\ln 2) \approx 5.76 \text{ J/K}$
4
(c) No heat is exchanged with the surroundings, so $Q_{surr} = 0$ , $r a c 12 D e l t a S_{surr} = 0$ . Total entropy change:
5
$\Delta S_{\text{univ}} = 5.76 + 0 = +5.76 \text{ J/K}$
6
Since $r a c 12 D e l t a S_{univ} > 0$ , the process is spontaneous, as expected.

6. Common Pitfalls

Wrong move:

Using Celsius temperatures directly in the Carnot efficiency formula instead of converting to Kelvin

Why:

Most problems give temperatures in Celsius, and students forget the formula requires absolute temperature

Correct move:

Always add 273 to any Celsius temperature before plugging into the Carnot efficiency formula

Wrong move:

Claiming a process with $r a c 12 D e l t a S_{sys} < 0$ is impossible per the second law

Why:

Students memorize "entropy increases" but forget this applies to the total entropy of the universe, not just the system

Correct move:

Always calculate $rac{1}{2}\\Delta S_{\text{univ}} = rac{1}{2}\\Delta S_{\text{sys}} + rac{1}{2}\\Delta S_{\text{surr}}$ before judging if a process is allowed

Wrong move:

Calculating heat engine efficiency as $e = W / Q_{C}$ instead of $e = W / Q_{H}$

Why:

Students mix up input heat from the hot reservoir and waste heat to the cold reservoir

Correct move:

Remember: efficiency is work you get out divided by heat you put in, and input heat always comes from the hot reservoir ( $Q_{H}$ )

Wrong move:

Forgetting to flip the sign of $Q$ when calculating $r a c 12 D e l t a S_{surr}$

Why:

Students carry the system's $Q$ sign directly over to the surroundings, flipping the sign of the total entropy change

Correct move:

Explicitly write $Q_{surr} = - Q_{sys}$ before calculating $r a c 12 D e l t a S_{surr}$ for any process

Wrong move:

Claiming a reversible Carnot engine can have 100% efficiency

Why:

Students confuse reversibility with zero total entropy change, but forget the temperature requirement

Correct move:

Remember even ideal Carnot engines only reach 100% efficiency if $T_{C} = 0$ K, which is physically impossible, so all real engines have $e < 1$

7. Quick Reference Cheatsheet

Category	Formula	Notes
Entropy change (reversible isothermal)	$Δ S = \frac{Q}{T}$	$Q$ = heat added to system, $T$ = absolute temperature (K)
Second Law (entropy form)	$Δ S_{univ} = Δ S_{sys} + Δ S_{surr} \geq 0$	$> 0$ = spontaneous, $= 0$ = ideal reversible
Heat engine net work	$W = Q_{H} - ∣ Q_{C} ∣$	Full cycle: $r a c 12 D e l t a U_{engine} = 0$
Heat engine efficiency	$e = \frac{W}{Q _{H}} = 1 - \frac{∣ Q _{C} ∣}{Q _{H}}$	Always $e < 1$ for real engines
Maximum (Carnot) efficiency	$e_{Carnot} = 1 - \frac{T _{C}}{T _{H}}$	Temperatures must be in Kelvin, upper limit for all engines
Entropy change for full cycle	$Δ S_{cycle} = 0$	Entropy is a state function, returns to starting value
Clausius Statement	N/A	No device can transfer heat cold to hot with no work input

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Spontaneity concept multiple choice
2022 · FRQ
Carnot efficiency calculation sub-part

What's Next

The second law of thermodynamics is a foundational concept that extends beyond thermodynamics to all areas of physics, explaining why all real processes are irreversible and why perpetual motion machines are impossible. After mastering this sub-topic, you have completed the core thermodynamics content for AP Physics 2 Unit 2, and are ready to move on to applied topics. Understanding the second law also helps you contextualize real-world energy conversion systems like power plants and refrigerators, which are common FRQ contexts on the AP exam, so be sure to practice calculating entropy changes and Carnot efficiency before test day.