Thermodynamics
'Thermodynamics is the only physical theory of universal content which, within the framework of the applicability of its basic concepts, I am convinced will never be overthrown.' — Albert Einstein
1. Chapter Overview
THERMODYNAMICS studies the RELATIONSHIPS between HEAT, WORK, and the INTERNAL ENERGY of systems. Unlike the thermal properties chapter (which focuses on specific phenomena), thermodynamics provides UNIFYING LAWS that apply to ALL macroscopic systems. This chapter covers the ZEROTH, FIRST, and SECOND LAWS of thermodynamics, thermodynamic PROCESSES, and the CARNOT ENGINE.
2. Thermodynamic Systems
| Type | Exchange with Surroundings | Examples |
|---|---|---|
| Open | Both matter and energy | Boiling water in an open pot |
| Closed | Energy only, no matter | Gas in a sealed cylinder with a movable piston |
| Isolated | Neither matter nor energy | Gas in a perfectly insulated rigid container |
State Variables and Equation of State
- Intensive: Independent of system size (T, P, ρ)
- Extensive: Depend on system size (V, U, S, n)
- For an ideal gas: PV = nRT (equation of state)
- Thermodynamic Equilibrium: Mechanical, thermal, and chemical equilibrium SIMULTANEOUSLY
3. Zeroth Law of Thermodynamics
- Statement: If TWO systems are each in thermal equilibrium with a THIRD system, they are in thermal equilibrium WITH EACH OTHER.
- Significance: Provides the basis for TEMPERATURE measurement — if thermometer reads T, any object in contact at equilibrium has the SAME temperature.
4. First Law of Thermodynamics
- Statement: Energy can neither be created nor destroyed; it can only change forms.
- Mathematical Form: ΔU = Q — W
- ΔU = Change in INTERNAL ENERGY of the system
- Q = Heat ADDED to the system
- W = Work done BY the system
- Sign convention: Q > 0 (heat added), W > 0 (work by system), ΔU > 0 (energy increase)
For Different Processes
| Process | Condition | First Law |
|---|---|---|
| Isothermal | ΔT = 0, ΔU = 0 | Q = W |
| Adiabatic | Q = 0 | ΔU = -W |
| Isochoric | ΔV = 0, W = 0 | ΔU = Q |
| Isobaric | P = constant | ΔU = Q — PΔV |
Worked Problem
Q: In a process, a gas absorbs 500 J of heat and does 200 J of work. Find ΔU. A: ΔU = Q — W = 500 — 200 = 300 J (internal energy increases).
5. Thermodynamic Processes
Isothermal Process (Constant Temperature)
- For ideal gas: PV = constant (Boyle's law)
- Work done: W = nRT ln(V₂/V₁)
- Internal energy does NOT change for ideal gas (U depends only on T)
Adiabatic Process (No Heat Exchange)
- PV^γ = constant (γ = C_p/C_v)
- For monatomic gas: γ = 5/3, for diatomic: γ = 7/5
- Work done: W = (P₁V₁ — P₂V₂)/(γ — 1) = nR(T₁ — T₂)/(γ — 1)
- Temperature CHANGES during adiabatic compression/expansion
Isochoric Process (Constant Volume)
- Work done = 0 (no volume change)
- Heat added = Change in internal energy
- P/T = constant (Gay-Lussac's law)
Isobaric Process (Constant Pressure)
- Work done W = PΔV
- V/T = constant (Charles' law)
- Heat added = ΔU + PΔV
Cyclic Process
- System returns to INITIAL state (ΔU = 0)
- Net work done = Net heat exchanged (Q_net = W_net)
- Efficiency η = W_net/Q_in
6. Specific Heat Capacities
- Molar specific heat at constant volume (C_v): Q = nC_vΔT at constant V
- Molar specific heat at constant pressure (C_p): Q = nC_pΔT at constant P
- Relation: C_p — C_v = R (Mayer's formula)
- For monatomic gas: C_v = 3R/2, C_p = 5R/2
- γ (adiabatic index) = C_p/C_v
7. Second Law of Thermodynamics
Statements
- Kelvin-Planck Statement: It is IMPOSSIBLE to construct an engine that converts ALL heat from a single source into work (no 100% efficient engine)
- Clausius Statement: It is IMPOSSIBLE for heat to flow spontaneously from a COLD body to a HOT body without external work
Reversible and Irreversible Processes
| Reversible | Irreversible |
|---|---|
| Can be REVERSED without any change in system/surroundings | Cannot be reversed |
| Infinitely SLOW (quasi-static) | Natural processes (friction, diffusion) |
| No dissipation | Energy dissipates |
| Idealisation | REAL processes |
8. Heat Engines and the Carnot Engine
Heat Engine
- Absorbs Q₁ from hot reservoir, does work W, rejects Q₂ to cold reservoir
- Efficiency: η = W/Q₁ = 1 — Q₂/Q₁
Carnot Engine
- The MOST EFFICIENT heat engine operating between two temperatures
- Carnot cycle: Isothermal expansion → Adiabatic expansion → Isothermal compression → Adiabatic compression
- Efficiency: η_carnot = 1 — T₂/T₁ (T₁ = hot, T₂ = cold, in KELVIN)
- Key: Carnot efficiency depends ONLY on temperatures, NOT on the working substance
Worked Problem
Q: A Carnot engine operates between 500 K and 300 K. Find efficiency. A: η = 1 — T₂/T₁ = 1 — 300/500 = 1 — 0.6 = 0.4 = 40%.
9. Common Mistakes
- ΔU = Q — W, NOT Q + W: Always check the sign convention (some textbooks use Q + W with W as work done ON the system)
- Isothermal ≠ adiabatic: In isothermal, ΔT = 0 so ΔU = 0; in adiabatic, Q = 0 so ΔU = -W
- C_p — C_v = R applies ONLY to ideal gases: Real gases have different relationships
- 100% efficiency is impossible (2nd Law): No engine can convert all heat into work
- Carnot efficiency is the MAXIMUM possible: Real engines have lower efficiency
10. CBSE Exam Focus
- First law of thermodynamics — sign convention, numerical problems
- Isothermal and adiabatic processes — PV diagrams, work done
- Carnot cycle and efficiency derivation (5-mark)
- Second law statements — Kelvin-Planck and Clausius
- Heat engine numericals (3-mark)
- Mayer's formula C_p — C_v = R
11. Key Formulas
- ΔU = Q — W
- W_isothermal = nRT ln(V₂/V₁)
- TV^(γ-1) = constant; PV^γ = constant (adiabatic)
- C_p — C_v = R
- γ = C_p/C_v
- η = 1 — Q₂/Q₁ = 1 — T₂/T₁ (Carnot)
- For monatomic: C_v = 3R/2, C_p = 5R/2
12. Self-Test (5+ Q&A)
Q1: 2 moles of ideal gas expand isothermally at 300 K from 5 L to 15 L. Find work done (R = 8.314 J/mol·K). A: W = nRT ln(V₂/V₁) = 2×8.314×300×ln(3) = 4988.4×1.099 = 5480 J.
Q2: A Carnot engine has 40% efficiency. If sink temperature is 300 K, find source temperature. A: η = 1 — T₂/T₁ → 0.4 = 1 — 300/T₁ → 300/T₁ = 0.6 → T₁ = 500 K.
Q3: In a cyclic process, the system absorbs 1000 J and rejects 600 J. Find work done and efficiency. A: W = Q₁ — Q₂ = 1000 — 600 = 400 J. η = W/Q₁ = 400/1000 = 40%.
Q4: What is the difference between isothermal and adiabatic processes? A: Isothermal: T constant, ΔU = 0, heat exchanged. Adiabatic: No heat exchange (Q = 0), temperature changes due to work.
Q5: Why can't a heat engine have 100% efficiency? A: Second Law of Thermodynamics (Kelvin-Planck statement). Some heat MUST be rejected to a cold reservoir; Q₂ cannot be zero.
13. Conclusion
Thermodynamics provides UNIVERSAL LAWS governing energy transformations. The First Law quantifies energy conservation including heat. The Second Law establishes the DIRECTION of natural processes and limits efficiency. The Carnot engine represents the IDEAL — a standard against which real engines are measured. These laws have far-reaching implications in engineering, chemistry, and even cosmology.
