🎯⭐ INTERACTIVE LESSON

Laws of Thermodynamics

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Laws of Thermodynamics - Complete Interactive Lesson

Part 1: Internal Energy & Work

🔥 Internal Energy & Work Done by a Gas

Part 1 of 7 — The Energy of Molecules

Thermodynamics connects heat, work, and internal energy. Before we state the laws, we need to understand what these quantities mean and how gases do work.

Internal Energy ( $U$ )

The internal energy of an ideal gas is the total kinetic energy of all its molecules:

$U = \frac{3}{2} n R T \quad \text{(monatomic ideal gas)}$

Key facts:

$U$ depends only on temperature for an ideal gas
If $T$ increases → $U$ increases
If $T$ stays constant → $\Delta U = 0$
is a — it depends only on the current state, not on how the gas got there

Change in Internal Energy

$\Delta U = \frac{3}{2} n R \Delta T$

For any process where $T$ doesn't change (isothermal): $\Delta U = 0$ .

For any process where $T$ rises: $\Delta U > 0$ .

Work Done by a Gas

When a gas expands, it pushes on its surroundings and does positive work. When a gas is compressed, work is done on the gas (negative work by gas).

Isobaric Process (Constant Pressure)

$W = P \Delta V$

where $\Delta V = V_f - V_i$ .

Sign Convention Check ✅

PV Diagram Concepts 📊

Work Calculation Drill 🔧

All processes are isobaric (constant pressure). Give magnitude and correct sign.

$P = 100$ kPa, gas expands from 3.0 L to 8.0 L. Work by gas (in J)?
$P = 250$ kPa, gas compresses from 10.0 L to 4.0 L. Work by gas (in J)?
A gas at 150 kPa does 450 J of work while expanding. What is $\Delta V$ (in L)?

Part 2: First Law of Thermodynamics

⚡ The First Law of Thermodynamics

Part 2 of 7 — Energy Conservation for Thermal Systems

The First Law is simply conservation of energy applied to thermodynamic systems. It connects heat, work, and internal energy in one powerful equation.

The First Law

$\Delta U = Q - W$

or equivalently:

$Q = \Delta U + W$

Part 3: PV Diagrams

📊 PV Diagrams Deep Dive

Part 3 of 7 — Reading Work from Graphs

PV diagrams are the essential tool for analyzing thermodynamic processes. The area under the curve tells you the work, and the direction of the cycle tells you whether you have an engine or a refrigerator.

Work = Area Under the Curve

On a PV diagram, work done by the gas during any process equals the area under the path from $V_i$ to $V_f$ :

Part 4: Heat Engines & Efficiency

🏭 Heat Engines & Efficiency

Part 4 of 7 — Turning Heat into Work

A heat engine absorbs heat from a hot reservoir, converts some of it to useful work, and dumps the rest into a cold reservoir. No engine can convert ALL heat to work — this is one of the most profound results in physics.

How a Heat Engine Works

Every heat engine operates between two thermal reservoirs:

Hot reservoir (temperature $T_H$ ): supplies heat $Q_H$ to the engine

Part 5: Second Law & Entropy

🌌 The Second Law & Entropy

Part 5 of 7 — Nature's Arrow of Time

The Second Law of Thermodynamics tells us which processes can happen spontaneously and which cannot. It introduces entropy, a measure of disorder that always increases in the universe.

The Second Law of Thermodynamics

There are several equivalent statements of the Second Law:

Clausius Statement

Heat cannot spontaneously flow from a cold object to a hot object without external work being done.

Kelvin-Planck Statement

No cyclic process can convert heat entirely into work (there must always be waste heat).

Entropy Statement

The total entropy of an isolated system can never decrease; it can only increase or remain the same.

$\Delta S_{\text{universe}} \geq 0$

Part 6: Refrigerators & Heat Pumps

❄️ Refrigerators & Heat Pumps

Part 6 of 7 — Running a Heat Engine in Reverse

A refrigerator is essentially a heat engine running backward: instead of using heat to produce work, it uses work to move heat from cold to hot. Heat pumps use the same principle to heat buildings efficiently.

How a Refrigerator Works

A refrigerator uses work input to extract heat from a cold reservoir (inside the fridge) and dump it into a hot reservoir (the kitchen):

Cold reservoir ( $T_C$ ): heat $Q_C$ is absorbed from the cold interior

Part 7: Synthesis & AP Review

🎯 Synthesis & AP Review

Part 7 of 7 — Putting It All Together

Let's compare all four thermodynamic processes, review PV diagram analysis, and tackle the kinds of problems you'll see on the AP Physics 2 exam.

The Four Processes — Complete Comparison

Process	Constant	$W$	$\Delta U$	$Q$	PV Shape
Isobaric

Situation	$\Delta V$	$W$	Meaning
Gas expands	$> 0$	$> 0$	Gas does work ON surroundings
Gas compresses	$< 0$	$< 0$	Surroundings do work ON gas
Volume constant	$= 0$	$= 0$	No work done

On a Pressure vs. Volume (PV) diagram, the work equals the area under the curve between the initial and final states.

Moving right on the PV diagram → expansion → $W > 0$
Moving left → compression → $W < 0$

$\Delta U$ = change in internal energy of the gas
$Q$ = heat added to the gas (positive if heat flows IN)
$W$ = work done by the gas (positive if gas expands)

The heat added to a system goes into two places: increasing internal energy and doing work.

If you add 500 J of heat and the gas does 200 J of work expanding, then $\Delta U = 500 - 200 = 300$ J goes into raising the temperature.

Quantity	Positive means...	Negative means...
$Q$	Heat flows INTO gas	Heat flows OUT of gas
$W$	Gas expands (does work on surroundings)	Gas compressed (work done on gas)
$\Delta U$	Temperature increases	Temperature decreases

First Law Applied to Special Processes

1. Isothermal Process ( $T$ = constant)

$\Delta U = 0$ (internal energy depends only on $T$ )
$Q = W$
All heat added is converted to work (or vice versa)

2. Adiabatic Process ( $Q = 0$ , no heat exchange)

$\Delta U = -W$
If the gas expands ( $W > 0$ ): $\Delta U < 0$ → temperature drops
If the gas compresses ( $W <$ ): → temperature

3. Isochoric (Isovolumetric) Process ( $V$ = constant)

$W = 0$ (no volume change → no work)
$\Delta U = Q$
All heat goes directly into changing internal energy (temperature)

4. Isobaric Process ( $P$ = constant)

$W = P\Delta V$
$\Delta U = Q - P\Delta V$
Heat goes into both internal energy AND work

Process Identification 🔍

First Law Application ⚡

First Law Drill 🔧

Apply $\Delta U = Q - W$ to each scenario.

$Q = 1200$ J, $W = 500$ J. Find $\Delta U$ (in J).
A gas does 300 J of work while its internal energy decreases by 100 J. Find $Q$ (in J).
An adiabatic compression does 600 J of work on the gas ( $W = -600$ J by the gas). Find $\Delta U$ (in J).

$W = \int_{V_i}^{V_f} P\,dV = \text{area under curve}$

Different Processes on PV Diagrams

Process	PV Shape	Work
Isobaric ( $P$ = const)	Horizontal line	$W = P\Delta V$ (rectangle)
Isochoric ( $V$ = const)	Vertical line	$W = 0$ (no area)
Isothermal ( $T$ = const)	Curved (hyperbola: $PV = nRT$ )	Area under curve
Adiabatic ( $Q = 0$ )	Steeper curve than isothermal	Area under curve

An adiabat is steeper than an isotherm passing through the same point because $PV^\gamma = \text{const}$ (where $\gamma > 1$ ) drops off faster than $PV = \text{const}$ .

Thermodynamic Cycles

A cycle returns the gas to its original state, so $\Delta U_{\text{cycle}} = 0$ .

From the First Law for a complete cycle:

$Q_{\text{net}} = W_{\text{net}}$

Net Work = Enclosed Area

The net work done in a cycle equals the area enclosed by the cycle on a PV diagram.

Direction Matters!

Direction	Name	Net Work	Energy Flow
Clockwise	Heat engine	$W_{\text{net}} > 0$ (work OUT)	Converts heat to work
Counterclockwise	Refrigerator/heat pump	$W_{\text{net}} < 0$ (work IN)

How to Calculate Net Work for a Cycle

Find the area enclosed by the cycle
If clockwise → $W > 0$
If counterclockwise → $W < 0$

For rectangular cycles: $W_{\text{net}} = \Delta P \times \Delta V$

Cycle Direction Quiz 🔄

PV Diagram Reading 📈

PV Diagram Calculation Drill 📊

An isobaric expansion at $P = 150$ kPa from $V = 2.0$ L to $V = 6.0$ L. Work done by gas (in J)?
A rectangular clockwise cycle with $\Delta P = 100$ kPa and $\Delta V = 5.0$ L. Net work per cycle (in J)?
A gas completes a clockwise cycle with net work = 400 J. If $Q_H = 1000$ J of heat is absorbed from the hot reservoir, how much heat $Q_C$ is rejected to the cold reservoir (in J)?

Match the Process to Its PV Shape 🎯

Cold reservoir (temperature

T_C

): receives waste heat

Q_C

from the engine

Net work output:

W_{\text{net}} = Q_H - Q_C

Energy Conservation (First Law for a Cycle)

Since $\Delta U = 0$ for a complete cycle:

$Q_H = W_{\text{net}} + Q_C$

The thermal efficiency measures what fraction of input heat becomes useful work:

$e = \frac{W_{\text{net}}}{Q_H} = \frac{Q_H - Q_C}{Q_H} = 1 - \frac{Q_C}{Q_H}$

Efficiency is always between 0 and 1 (or 0% and 100%).

Since $Q_C > 0$ (you must dump some waste heat), efficiency is always less than 100%.

The Carnot Engine

The Carnot engine is a theoretical engine that achieves the maximum possible efficiency between two given temperatures. It consists of two isothermal and two adiabatic processes.

Carnot Efficiency

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

where temperatures must be in Kelvin.

Key Facts About Carnot Efficiency

No real engine can exceed Carnot efficiency
$e_{\text{Carnot}}$ increases when $T_H$ increases or $T_C$ decreases

Why 100% Efficiency is Impossible

The Second Law (coming in Part 5) forbids converting heat entirely into work in a cyclic process. There must always be waste heat $Q_C > 0$ . This is a fundamental limit of nature, not an engineering limitation.

Efficiency Concepts 🔍

Carnot Limit ⚠️

Efficiency Calculation Drill 🔧

A heat engine absorbs $Q_H = 2000$ J and does $W = 600$ J of work. What is the efficiency (in %)?
What is the Carnot efficiency of an engine operating between $T_H = 500$ K and $T_C = 200$ K (in %)?
A Carnot engine operates between 800 K and 200 K. If it absorbs 1000 J per cycle, how much work does it produce (in J)?

All three statements are equivalent — if one is violated, all are violated.

Entropy ( $S$ )

Entropy is a measure of the number of microscopic arrangements (microstates) consistent with the macroscopic state. More microstates → higher entropy → more "disorder."

Entropy Change

For a reversible process at constant temperature:

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

where $T$ is in Kelvin and $Q$ is the heat transferred.

Key Properties

Entropy is a state variable (path-independent for the system)
Units: J/K
Adding heat to a system increases its entropy
Removing heat decreases its entropy
Spontaneous processes always increase the total entropy of the universe

Reversible vs. Irreversible Processes

Type	$\Delta S_{\text{universe}}$	Example
Reversible	$= 0$	Ideal Carnot cycle
Irreversible	$> 0$

All real processes are irreversible → entropy of the universe always increases.

Heat Death of the Universe

If entropy always increases, the universe will eventually reach a state of maximum entropy — uniform temperature everywhere, no energy gradients, no ability to do work. This is called the "heat death" of the universe.

Second Law Concepts 🔍

Entropy Calculations 📐

Entropy Change Drill 🔧

Use $\Delta S = Q/T$ (with $T$ in Kelvin).

600 J of heat is added to a reservoir at 300 K. What is $\Delta S$ of the reservoir (in J/K)?
800 J of heat flows from a 400 K reservoir to a 200 K reservoir. What is $\Delta S_{\text{universe}}$ (in J/K)?
A Carnot engine absorbs 2000 J at 500 K and rejects heat at 250 K. How much heat is rejected $Q_C$ (in J)?

Work input (

W

): electrical energy drives the compressor

Hot reservoir (

T_H

): heat

Q_H

is released into the room

The heat expelled into the room is MORE than the heat removed from inside — that's why the back of your fridge feels warm!

Coefficient of Performance (COP) — Refrigerator

For a refrigerator, we want to maximize cooling per unit of work:

$\text{COP}_{\text{fridge}} = \frac{Q_C}{W} = \frac{Q_C}{Q_H - Q_C}$

Note: COP can be greater than 1 (unlike efficiency). A typical fridge has COP ≈ 3–5, meaning each joule of work removes 3–5 joules of heat.

Maximum COP (Carnot Refrigerator)

$\text{COP}_{\text{Carnot, fridge}} = \frac{T_C}{T_H - T_C}$

Heat Pumps

A heat pump is the same device as a refrigerator, but the goal is reversed: we want to heat the hot reservoir (the building) by extracting heat from the cold reservoir (outside air).

COP — Heat Pump

$\text{COP}_{\text{heat pump}} = \frac{Q_H}{W} = \frac{Q_H}{Q_H - Q_C}$

Maximum COP (Carnot Heat Pump)

$\text{COP}_{\text{Carnot, HP}} = \frac{T_H}{T_H - T_C}$

Useful Relationship

$\text{COP}_{\text{heat pump}} = \text{COP}_{\text{fridge}} + 1$

This makes sense: $Q_H/W = Q_C/W + W/W = Q_C/W + 1$ .

Real-World Applications

Air conditioners: refrigerators that cool indoor air
Heat pumps: heat buildings by extracting heat from outdoor air (even cold air has thermal energy!)
Geothermal heat pumps: use the stable ground temperature as a reservoir — very efficient
A heat pump with COP = 4 delivers 4 kW of heating for every 1 kW of electricity — far better than a space heater (COP = 1)

Refrigerator & Heat Pump Concepts 🔍

Engine vs. Refrigerator ⚖️

COP Calculation Drill 🔧

A refrigerator removes 600 J from the cold reservoir using 200 J of work. What is its COP?
What is the maximum COP of a refrigerator operating between $T_C = 250$ K and $T_H = 300$ K?
A Carnot heat pump operates between $T_C = 270$ K and $T_H = 300$ K. What is its COP?

Quick Decision Guide

Temperature doesn't change? → Isothermal → $\Delta U = 0$
No heat exchange? → Adiabatic → $Q = 0$
Volume doesn't change? → Isochoric → $W = 0$
Pressure doesn't change? → Isobaric → $W = P\Delta V$

For any complete cycle: $\Delta U = 0$ , so $Q_{\text{net}} = W_{\text{net}} = \text{enclosed area on PV diagram}$ .

Common AP Mistakes to Avoid

❌ Mistake 1: Using °C in Carnot efficiency

Carnot efficiency requires Kelvin: $e = 1 - T_C/T_H$ . Always convert: $T(K) = T(°C) + 273$ .

❌ Mistake 2: Confusing $W$ sign conventions

Some textbooks define $W$ as work done ON the gas. AP Physics 2 typically uses $W$ = work done BY the gas. Check which convention the problem uses!

❌ Mistake 3: Thinking $\Delta U = 0$ means $T$ is constant

For an ideal gas, $\Delta U = 0 \Leftrightarrow \Delta T = 0$ (true). But this is specific to ideal gases.

❌ Mistake 4: Assuming adiabatic = isothermal

Adiabatic ( $Q = 0$ ) means no heat exchange. Isothermal ( $\Delta T = 0$ ) means constant temperature. An adiabatic expansion causes temperature to DROP.

❌ Mistake 5: Forgetting that COP > 1 is normal

Unlike efficiency (always < 1), COP of a refrigerator or heat pump can exceed 1. A COP of 5 does not violate any law.

❌ Mistake 6: Confusing engine efficiency with refrigerator COP

Engine: $e = W/Q_H < 1$ . Fridge: $\text{COP} = Q_C/W$ (can be > 1). They measure different things.

Process Identification Mastery 🔍

AP-Style Analysis 📝

Process Property Match 🎯

For each scenario, identify the correct result.

Final Mastery Drill 🏆

A Carnot engine operates between 600 K and 300 K with $Q_H = 1200$ J. Find the work output $W$ (in J).
A refrigerator has COP = 4 and does 150 J of work per cycle. How much heat is removed from the cold reservoir $Q_C$ (in J)?
900 J of heat flows irreversibly from a 450 K reservoir to a 300 K reservoir. Find $\Delta S_{\text{universe}}$ (in J/K).

e_{\text{Carnot}} = 1

(100%) only if

T_C = 0

K — impossible to reach!

Real engines have friction, turbulence, and other irreversibilities, so

e_{\text{real}} < e_{\text{Carnot}}