Lesson 15: Thermodynamics II

Week 15 — The laws of thermodynamics and heat engines

🎯 Learning Objectives

State and apply the ideal gas law
Analyze thermodynamic processes (isothermal, adiabatic, isobaric, isochoric)
Calculate efficiency of heat engines and understand entropy

🎯 Intuition First: Why Does a Tire Puff Up?

When you pump air into a tire, the pressure rises because you're adding more molecules (more collisions). If you heat the tire, pressure rises even more — the molecules move faster and hit harder. This is the heart of thermodynamics: connecting microscopic motion to macroscopic pressure and temperature.

💡 The Big Picture:

Thermodynamics studies energy transfer, especially heat and work. The laws govern everything from car engines to your refrigerator to the stars themselves.

📋 Ideal Gas Law

PV = nRT

Where: P = pressure (Pa), V = volume (m³), n = moles, R = 8.314 J/(mol·K), T = temperature (K)

Also: PV = Nk_BT (N = number of molecules, k_B = 1.38 × 10⁻²³ J/K)

📋 Thermodynamic Processes

1. Isothermal (constant temperature):

T constant → PV = constant → W = nRT ln(V₂/V₁)

2. Adiabatic (no heat transfer):

Q = 0 → PV^γ = constant (γ = Cp/Cv, ≈1.4 for diatomic gas)

W = −ΔU = nCvΔT

3. Isobaric (constant pressure):

P constant → W = PΔV

4. Isochoric (constant volume):

V constant → W = 0 (no work done)

📋 First Law of Thermodynamics

ΔU = Q − W

Change in internal energy = heat added − work done BY the system

This is just energy conservation! Heat goes in, some does work, the rest changes internal energy.

📋 Heat Engines and Efficiency

A heat engine absorbs heat from a hot reservoir, does work, and rejects heat to a cold reservoir:

Efficiency: η = W/Q_h = 1 − Q_c/Q_h

Carnot efficiency (maximum possible): η = 1 − T_c/T_h (temperatures in Kelvin!)

📝 Key Points:

Efficiency is always less than 1 (no perfect engine)
The larger the temperature difference, the better the efficiency
No engine can violate the Second Law (entropy always increases)

📋 Second Law & Entropy

Heat spontaneously flows from hot to cold (never the reverse). Entropy measures disorder:

ΔS = Q_rev/T (for reversible processes)

The Second Law: ΔS_universe ≥ 0 (entropy never decreases)

📝 Worked Example: Heat Engine

A heat engine absorbs 500 J from a hot reservoir and rejects 300 J to a cold reservoir. Find the efficiency and work output.

Step 1 — Work output:

W = Q_h − Q_c = 500 − 300 = 200 J

Step 2 — Efficiency:

η = W/Q_h = 200/500 = 0.40 = 40%

✅ Efficiency = 40%, work = 200 J

📝 Worked Example: Carnot Engine

A Carnot engine operates between reservoirs at 500°C and 50°C. What is its maximum efficiency?

T_h = 500 + 273 = 773 K

T_c = 50 + 273 = 323 K

η = 1 − T_c/T_h = 1 − 323/773 = 1 − 0.418 = 0.582 = 58.2%

✅ Maximum efficiency = 58.2%

✅ Check Your Understanding

Q1: Why can no heat engine be 100% efficient?

Click to reveal answer

The Second Law of Thermodynamics says entropy of the universe must increase. A 100% efficient engine would convert all heat to work with no heat rejected to a cold reservoir — this would require ΔS = 0 (or negative), violating the Second Law. Some heat must always be "wasted" to a cold reservoir to increase total entropy.

Q2 (mini-problem): 0.5 moles of an ideal gas at 300 K occupies 0.02 m³. What is the pressure?