2bThermal Processes and Heat Engines
Analyze isothermal, adiabatic, isobaric, and isochoric processes. Apply the Carnot efficiency formula and interpret PV diagrams.
Heat engines power our civilization — from car engines to power plants. Understanding thermodynamic processes and the Carnot limit is essential for designing efficient energy systems and understanding why 100% efficiency is physically impossible.
What is the theoretical maximum efficiency of any heat engine — and why can no real engine ever reach it?
Lesson Overview
Thermodynamic processes describe how a gas changes state. In an isothermal process, temperature is constant (ΔU = 0, Q = W). In an adiabatic process, no heat is exchanged (Q = 0, ΔU = −W). In an isobaric process, pressure is constant (W = PΔV). In an isochoric process, volume is constant (W = 0, ΔU = Q). On a PV diagram, work done by the gas equals the area under the curve. The Carnot engine is the most efficient possible heat engine operating between two temperatures: efficiency e = 1 − T_C/T_H. Real engines are always less efficient due to irreversibilities.
Key Equations
Worked Examples
A Carnot engine operates between T_H = 800 K and T_C = 400 K. Find its efficiency.
A heat engine absorbs Q_H = 2000 J per cycle and rejects Q_C = 1400 J. Find (a) the net work output, (b) the efficiency.
A gas expands isobarically at P = 2 × 10⁵ Pa from V₁ = 0.01 m³ to V₂ = 0.04 m³. Find the work done by the gas.
A gas undergoes an adiabatic compression. W = −800 J (work done on gas). Find ΔU and state what happens to the temperature.
A Carnot engine has efficiency e = 40% and absorbs Q_H = 5000 J per cycle. Find (a) W_net, (b) Q_C, (c) T_C if T_H = 600 K.
Guided Problems
A Carnot engine operates between T_H = 500 K and T_C = 300 K. What is its maximum efficiency?
Hint: Use e = 1 − T_C/T_H. Both temperatures must be in Kelvin.
A heat engine has efficiency e = 25% and produces W_net = 500 J per cycle. Find Q_H and Q_C.
Hint: Use e = W_net/Q_H to find Q_H. Then Q_C = Q_H − W_net.
Identify the thermodynamic process: (a) A gas is heated in a rigid container. (b) A gas expands slowly in a thermally insulated cylinder. (c) A gas expands at constant temperature. (d) A gas is compressed at constant pressure.
Hint: Rigid container → constant volume (isochoric). Insulated → no heat exchange (adiabatic). Constant temperature → isothermal. Constant pressure → isobaric.
On a PV diagram, a gas expands from (P = 3×10⁵ Pa, V = 0.02 m³) to (P = 3×10⁵ Pa, V = 0.05 m³). What process is this, and what is the work done?
Hint: Constant pressure = isobaric. Work = area under PV curve = P × ΔV.
A steam engine operates between 200 °C and 50 °C. What is the maximum (Carnot) efficiency? Why does the actual efficiency fall short of this value?
Hint: Convert to Kelvin first: T_H = 473 K, T_C = 323 K. Real engines have friction, heat losses, and irreversible processes.
Key Vocabulary
Isothermal Process
A thermodynamic process at constant temperature. For an ideal gas, ΔU = 0, so Q = W. The gas must exchange heat with its surroundings to maintain constant T.
Example: A gas expanding slowly in contact with a large heat reservoir at constant temperature is approximately isothermal.
Adiabatic Process
A process with no heat exchange (Q = 0). From the First Law: ΔU = −W. Compression raises temperature; expansion lowers it. Occurs in thermally insulated systems or very rapid processes.
Example: Rapid compression in a diesel engine is nearly adiabatic — the air temperature rises enough to ignite fuel without a spark.
Isobaric Process
A process at constant pressure. Work done by the gas: W = PΔV. Heat added goes partly to work and partly to increasing internal energy.
Example: Heating a gas in a cylinder with a freely moving piston occurs at constant (atmospheric) pressure.
Isochoric Process
A process at constant volume (also called isovolumetric). No work is done (W = 0), so all heat goes into changing internal energy: ΔU = Q.
Example: Heating a gas in a rigid sealed container is isochoric — pressure rises but volume stays fixed.
Carnot Engine
A theoretical heat engine operating on the Carnot cycle (two isothermal and two adiabatic processes). It achieves the maximum possible efficiency between two temperatures: e = 1 − T_C/T_H. It is reversible and produces no entropy.
Example: A Carnot engine between 600 K and 300 K has e = 50% — no real engine can exceed this.
PV Diagram
A graph of pressure vs. volume for a thermodynamic process. The area under the curve equals the work done by the gas. Different processes (isothermal, adiabatic, isobaric, isochoric) have characteristic shapes.
Example: An isobaric process is a horizontal line on a PV diagram; an isochoric process is a vertical line.
Workbook Check — Interactive Quiz
Interactive Practice — 5 Questions
A Carnot engine operates between T_H = 600 K and T_C = 200 K. What is its efficiency?
In an adiabatic process, Q = 0. If a gas is compressed (W = −500 J done on gas), what is ΔU?
Which process is represented by a horizontal line on a PV diagram?
A heat engine absorbs Q_H = 3000 J and does W_net = 900 J. What is its efficiency?
For an isothermal process of an ideal gas, which is true?
Independent Practice
A Carnot engine operates between T_H = 900 K and T_C = 300 K. Find (a) the efficiency, (b) the work output if Q_H = 3000 J, and (c) the heat rejected Q_C.
A gas expands isobarically at P = 1.5 × 10⁵ Pa from V = 0.02 m³ to V = 0.06 m³. It absorbs Q = 10,000 J. Find W and ΔU.
Describe what happens to the temperature of a gas during (a) adiabatic expansion and (b) adiabatic compression. Explain using the First Law.
A heat engine has efficiency e = 35% and rejects Q_C = 650 J per cycle. Find Q_H and W_net.
★ A gas undergoes a cycle on a PV diagram: (A→B) isobaric expansion at P = 2×10⁵ Pa from V = 0.01 to 0.03 m³; (B→C) isochoric cooling; (C→A) isothermal compression back to start. Calculate the net work done per cycle and sketch the cycle on a PV diagram.
ChallengeCommon Mistakes
Using Celsius temperatures in the Carnot efficiency formula
Always use Kelvin: e = 1 − T_C/T_H. Using Celsius gives wrong answers. Convert: K = °C + 273.
Thinking a Carnot engine is a real, buildable device
The Carnot engine is a theoretical ideal. It sets the upper limit on efficiency. Real engines are always less efficient due to friction, heat losses, and irreversible processes.
Confusing the work done ON the gas with work done BY the gas
In physics, W in ΔU = Q − W is work done BY the gas. If the gas is compressed, W < 0 (gas does negative work), so ΔU increases.
Thinking the area under a PV curve always equals work regardless of direction
Work = area under PV curve, but sign matters: expansion (moving right) → positive work; compression (moving left) → negative work.
Math Tips
Carnot efficiency: e = 1 − T_C/T_H (Kelvin only). This is the MAXIMUM efficiency — real engines are always less efficient.
Engine energy balance: Q_H = W_net + Q_C. Efficiency e = W_net/Q_H = 1 − Q_C/Q_H.
Process summary: Isothermal → ΔU = 0; Adiabatic → Q = 0; Isobaric → W = PΔV; Isochoric → W = 0.
PV diagram: area under curve = work. Clockwise cycle → net positive work (engine). Counterclockwise → net negative work (refrigerator).