Thermodynamics

What this chapter is about

Thermodynamics studies how energy moves between a system and its surroundings when heat flows or work is done. In earlier chapters you met kinetic and potential energy of single objects; now you deal with the internal energy of a large collection of molecules and the rules that govern how this energy can change form. The subject grew from practical questions about steam engines but now underlies chemistry, biology and every branch of engineering.

A Class 11 student meets thermodynamics here because it builds on the ideas of heat and temperature from the previous chapter and prepares you to understand heat engines, refrigerators and the direction in which natural processes run. After working through this chapter you should be able to state the first and second laws of thermodynamics, apply them to ideal-gas processes, calculate work done in expansion and compression, and explain why certain processes never run backwards on their own.

Key ideas

• A thermodynamic system is any part of the universe you choose to study; everything else is the surroundings. Systems can be isolated (no exchange of energy or matter), closed (energy can cross, matter cannot) or open (both can cross).
• The state of a gas is described by state variables such as pressure P, volume V, temperature T and internal energy U. A change between two states can follow many different paths, but U depends only on the state, not on the path.
• Heat Q is energy transferred because of a temperature difference; work W is energy transferred when a force moves through a distance (for a gas, work = force on piston × displacement).
• The first law of thermodynamics: ΔU = Q − W. Energy is conserved; the change in internal energy equals heat added minus work done by the system.
• For an ideal gas undergoing different processes, work and heat exchange take special forms: isothermal (constant T), isobaric (constant P), isochoric (constant V), and adiabatic (Q = 0).
• The second law states that heat cannot flow from a colder body to a hotter body without external work; equivalently, no heat engine can convert all absorbed heat into work. This law fixes the direction of natural processes.
• A heat engine absorbs heat Q₁ from a hot reservoir, does work W, and rejects heat Q₂ to a cold reservoir. Efficiency η = W / Q₁ = 1 − Q₂ / Q₁. The Carnot engine, an ideal reversible engine, has the maximum possible efficiency η = 1 − T₂ / T₁ (temperatures in kelvin).
• A refrigerator removes heat from a cold body and dumps it into a warmer one by using external work. Its coefficient of performance is COP = Q₂ / W.

Formulas and facts to remember

1. First law: ΔU = Q − W (sign convention: Q positive when heat enters, W positive when system does work on surroundings).
2. Work done by an ideal gas at constant pressure: W = P ΔV.
3. Work in an isothermal expansion from V₁ to V₂: W = nRT ln(V₂ / V₁).
4. For an adiabatic process with an ideal gas: PV^γ = constant and TV^(γ−1) = constant, where γ = Cₚ / Cᵥ.
5. Relation between molar heat capacities: Cₚ − Cᵥ = R (for one mole of an ideal gas).
6. Carnot efficiency: η = 1 − T_cold / T_hot (temperatures in kelvin).
7. Coefficient of performance of a refrigerator: COP = Q_cold / W = T_cold / (T_hot − T_cold) for an ideal Carnot refrigerator.
8. Second law (Kelvin–Planck form): It is impossible to construct a heat engine that, operating in a cycle, converts all absorbed heat into work.

Worked examples

Example 1 — First-law calculation

An ideal gas in a cylinder absorbs 500 J of heat and expands against a piston, doing 200 J of work on the piston. Find the change in internal energy of the gas.

Solution Use the first law: ΔU = Q − W. Q = +500 J (heat enters the gas). W = +200 J (work done by the gas). ΔU = 500 − 200 = 300 J. The internal energy of the gas increases by 300 J.

Example 2 — Work in isothermal expansion

Two moles of an ideal gas at 300 K expand isothermally from 10 L to 20 L. Calculate the work done by the gas. (R = 8.31 J mol⁻¹ K⁻¹)

Solution For isothermal expansion, W = nRT ln(V₂ / V₁). n = 2 mol, R = 8.31 J mol⁻¹ K⁻¹, T = 300 K. V₂ / V₁ = 20 / 10 = 2. ln 2 ≈ 0.693. W = 2 × 8.31 × 300 × 0.693 ≈ 3456 J. The gas does about 3.46 kJ of work on the surroundings.

Example 3 — Carnot engine efficiency

A heat engine operates between a furnace at 600 K and the atmosphere at 300 K. (a) Find the maximum possible efficiency. (b) If the engine absorbs 1000 J per cycle from the furnace, what is the maximum work output?

Solution (a) For a Carnot engine, η = 1 − T_cold / T_hot = 1 − 300 / 600 = 0.5 or 50 %. (b) W = η × Q₁ = 0.5 × 1000 = 500 J. No real engine between these reservoirs can deliver more than 500 J per cycle from 1000 J of absorbed heat.

Common mistakes

• Confusing sign conventions: forgetting that W is positive when the system does work on the surroundings and Q is positive when heat enters the system → always state your convention and apply it consistently.
• Treating internal energy as path-dependent: trying to calculate ΔU from the process instead of from the initial and final states → remember U is a state function; ΔU depends only on the endpoints.
• Using Celsius in Carnot efficiency: plugging temperatures in °C into η = 1 − T₂ / T₁ → convert temperatures to kelvin first.
• Believing a process that is quick is adiabatic: assuming any fast expansion has Q = 0 → a process is adiabatic only when heat exchange is negligible, which requires good thermal insulation, not just speed.
• Thinking the Carnot engine is a real device: assuming real engines can reach Carnot efficiency → Carnot efficiency is the theoretical upper limit; real engines have friction and irreversibility.

Quick revision

1. First law: ΔU = Q − W; energy is conserved in every thermodynamic process.
2. Work done by a gas at constant pressure: W = P ΔV; in isothermal expansion: W = nRT ln(V₂ / V₁).
3. Adiabatic means Q = 0; the relation PV^γ = constant holds.
4. Carnot efficiency η = 1 − T_cold / T_hot sets the maximum efficiency for any heat engine.
5. The second law forbids 100 % conversion of heat to work and explains why heat flows from hot to cold on its own.