Theme B — The Particulate Nature of Matter
B.4.6 Thermodynamic Processes — IB Physics HL Revision Notes & Topical Questions
5 min read 10 parts Lesson 38 of 53 1 quiz
Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
Lesson summary
B.4.6 Thermodynamic Processes — IB PHYSICS HL notes: Describe isovolumetric, isobaric, isothermal and adiabatic processes of an ideal gas using the first law, p–V diagrams, entropy changes and pV5/3) = constant.Idealisation — the gas is an ideal gas and each change is slow enough…
- Monatomic gas
- Requirement
- Boyle's law
- Compression
- Pressure and volume
- Temperature and volume
Part 1 of 9The four processes
1Idealisation
- Idealisation the gas is an ideal gas and each change is slow enough to pass through equilibrium states
2Types
- Types isovolumetric (W = 0), isobaric (Δp = 0), isothermal (ΔT = 0) and adiabatic (Q = 0)
3First law
- Q = ΔU + W simplifies in each case
Part 2 of 9Isovolumetric (constant volume)
1Condition
- ΔV = 0 so the gas does no work, W = 0
2First law
- Q = ΔU
- All the heat supplied goes into internal energy
3Effect
- Effect temperature rises and pressure rises in proportion to T
4p–V diagram
- p–V diagram a vertical line
5Monatomic gas
- Q = ΔU = (3/2)nRΔT
Part 3 of 9Isobaric (constant pressure)
1Condition
- Condition gases expand or contract freely at constant pressure
2Work
- W = pΔV = nRΔT
- It is the area under a horizontal line on the p–V diagram
3First law
- Q = ΔU + pΔV, with the sign showing work done by or on the gas
4Monatomic gas
- Q = (3/2)nRΔT + nRΔT = (5/2)nRΔT
- More heat is needed for the same temperature rise than at constant volume because the gas also does work
5Volume
- Volume proportional to T during expansion
Part 4 of 9Isothermal (constant temperature)
1Condition
- T constant so ΔU = 0 for an ideal gas
2First law
- Q = W
- Heat absorbed becomes work in expansion, and work done on the gas is given out as heat in compression
3Requirement
- Requirement a slow process with good thermal contact to a reservoir
4Boyle's law
- Boyle's law pV = constant so the p–V curve is a hyperbola
5Work
- W = nRT ln(V2/V1)
Part 5 of 9Adiabatic (no heat transfer)
1Condition
- Q = 0 because the system is insulated or the change is too fast for heat to flow
2First law
- W = −ΔU
- Work is done at the expense of internal energy
3Expansion
- Expansion the gas does work and cools
4Compression
- Compression work done on the gas raises its internal energy so it warms
5p–V diagram
- p–V diagram steeper than an isotherm through the same point
Part 6 of 9Adiabatic equations (monatomic ideal gas)
Pressure and volume
- Pressure and volume pV5/3 = constant
- p1V15/3 = p2V25/3
Temperature and volume
- TV2/3 = constant
- T1V12/3 = T2V22/3
Derivation
- Derivation substitute p = nRT/V into pV5/3 = constant
Exponent
- 5/3 is the ratio of the specific heat capacities at constant pressure and constant volume for a monatomic gas
Part 7 of 9Summary of the processes
| Process | Fixed | W | ΔU | Q |
|---|---|---|---|---|
| Isovolumetric | V | 0 | = Q | = ΔU |
| Isobaric | p | pΔV | (3/2)nRΔT | ΔU + pΔV |
| Isothermal | T | = Q | 0 | = W |
| Adiabatic | Q = 0 | = −ΔU | −W | 0 |
Part 8 of 9Entropy in each process
| Process | Heat | Entropy change |
|---|---|---|
| Isothermal expansion | Q > 0, heat gained = work done by gas | ΔS > 0 |
| Isothermal compression | Q < 0, heat lost = work done on gas | ΔS < 0 |
| Isobaric expansion | Q > 0 | ΔS > 0 |
| Isobaric compression | Q < 0 | ΔS < 0 |
| Isovolumetric pressure rise | Q > 0 | ΔS > 0 |
| Isovolumetric pressure drop | Q < 0 | ΔS < 0 |
| Adiabatic expansion or compression | Q = 0 | ΔS = 0 |
Part 9 of 9More from the specification
1
Heat Engine
- A thermodynamic device that absorbs thermal energy from a high-temperature reservoir, performs mechanical work,
- and expels waste heat to a cold reservoir over a complete cyclic process where the net change in internal energy is equal to zero.
2
Thermal Efficiency
- The ratio of useful work output to total heat input, given by the efficiency formula eta = W / QH = 1 - (QC / QH);
- as the difference between reservoir temperatures increases, the maximum thermal efficiency increases.
3
Carnot Cycle
- An ideal theoretical thermodynamic cycle consisting of two reversible isothermal and two reversible adiabatic processes, representing the upper limit of efficiency achievable between two heat reservoirs.
4
First Law and Particle Energy
- Governed by Q = Delta U + W,
- where heat absorbed increases the random kinetic energy of gas particles,
- resulting in an increase in internal energy while positive work is done during volume expansion.
5
Atmospheric pressure
- Pressure of the air, about 1.01 × 105 Pa at sea level.
6
Efficiency
- Useful output divided by total input.
7
Piston
- A moving part in a cylinder that applies or receives pressure.
8
Molecule
- Tiny particle of the fluid, too small to be seen.
9
Surroundings
- Everything outside the system.
10
Carnot cycle
- A reversible cycle of two isothermal and two adiabatic processes between two reservoirs.
11
Energy transferred
- W = V Q, in joules.What the symbols mean: E or W = energy transferred (J) · Q = charge (C) · V = potential difference (V) · I = current (A) · t = time (s)
12
Kinetic energy
- K = ½mv2, energy of motion.
13
Isolated system
- System with no net external force.
Key terms
| Term | Meaning |
|---|---|
| Isovolumetric | A process at constant volume, W = 0. |
| Isobaric | A process at constant pressure, W = pΔV. |
| Isothermal | A process at constant temperature, ΔU = 0 for an ideal gas. |
| Adiabatic | A process with no heat exchange, Q = 0. |
| Quasi-static | A change so slow that the system passes through equilibrium states. |
Avoid these mistakes
1
Saying no heat transfer means no temperature change — in an adiabatic process the temperature changes because work is done.
2
Using pV = constant for an adiabatic change — for a monatomic gas pV5/3 is constant.
3
Believing ΔU = 0 in an isobaric process — the temperature changes so ΔU does not vanish.
4
Using the isothermal work formula for other processes — it applies only when T is constant.
5
Not seeing that adiabatic curves are steeper than isotherms.
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