Thermodynamic Processes and p-V Diagrams — B.4 Thermodynamics — IB Physics Notes

2 Lesson 2 of 5 · B.4 Thermodynamics

Four Thermodynamic Processes

The first law, $Q = \Delta U + W$, covers every possible change of state a gas can undergo. Fixing one of its four quantities at zero — $\Delta U$, $W$, $Q$ or $\Delta P$ — gives four named, extreme processes that between them describe almost every real gas change.

🚲 Hook – why a bicycle pump gets hot

Pump up a bicycle tyre quickly and the barrel of the pump gets noticeably warm — even though you haven't heated it with anything. You've compressed the air inside faster than heat can escape through the pump's metal walls, so essentially no thermal energy crosses the boundary. All the work you do on the trapped air goes straight into its internal energy, and its temperature rises. This is an adiabatic process — one of four "extreme" ways a gas can change state.

Four pressure-volume diagrams in a 2x2 grid showing isothermal, isovolumetric, adiabatic and isobaric processes, each with dotted isothermal reference curves

Fig B4.10 — the four thermodynamic processes. In every panel the dotted curves are isothermals; the solid line is the actual process.

💡 Diagnostic question (think – then read on) — if no thermal energy enters or leaves the pump's air, where does the energy that raises its temperature come from? From the work you do pushing the piston in — with $Q = 0$, the first law gives $\Delta U = -W$, so the work done on the gas is converted directly into internal energy.

🔀 Four extremes of the first law

The first law, $Q = \Delta U + W$, relates three quantities of energy transfer. Among all the changes of state a gas might undergo, it is convenient to consider four extreme cases, each obtained by keeping one variable fixed:

The four extremes $$\Delta U = 0 \quad (\text{isothermal})$$ $$W = 0 \quad (\text{isovolumetric})$$ $$Q = 0 \quad (\text{adiabatic})$$ $$\Delta P = 0 \quad (\text{isobaric})$$

(The prefix iso- means equal.) Figure B4.10 shows all four on pressure–volume ($P$–$V$) diagrams, in which the dotted lines are always isothermals — curves of constant temperature.

🌡️ Isothermal (ΔU = 0) and isovolumetric (W = 0)

In an isothermal change the gas's temperature is constant, so its internal energy cannot change: $\Delta U = 0$. The first law then reduces to:

Isothermal $$Q = W$$

In an isothermal expansion (B → A on Fig B4.10a), all the work done by the gas on the surroundings is supplied by thermal energy transferred into the gas. In an isothermal compression (A → B), the work done on the gas is all transferred away as thermal energy. For a real process to approximate this ideal, the change must be as slow as possible, giving heat time to flow in or out and keep the temperature constant. Isothermal changes obey Boyle's law: $PV = \text{constant}$.

Isovolumetric: occurring at constant volume. Since the gas can't change volume, it can do no work and none is done on it: $W = 0$. The first law reduces to:

Isovolumetric $$Q = \Delta U$$

If thermal energy is transferred into the gas, it simply gains internal energy and its temperature rises. If thermal energy is transferred away, its internal energy and temperature fall.

🧊 Adiabatic (Q = 0) and isobaric (ΔP = 0)

Adiabatic: occurring without thermal energy being transferred into or out of a closed system, so $Q = 0$. The first law reduces to:

Adiabatic $$\Delta U = -W \ \text{(compression)}$$ $$-\Delta U = W \ \text{(expansion)}$$

In an adiabatic expansion (B → A), all the work done by the gas is drawn from its own internal energy — $\Delta U$ is negative and the temperature falls. In an adiabatic compression (A → B), all the work done on the gas is transferred to its internal energy, and it gets hotter. Microscopically: gas molecules hitting an inward-moving piston gain kinetic energy (compression heats the gas); molecules hitting an outward-moving piston lose kinetic energy (expansion cools it). For a real process to approximate the adiabatic ideal, the change must be as rapid as possible in a well-insulated container.

Adiabatic lines on a $PV$ diagram must be steeper than isothermal lines, because in equal expansions the temperature falls during an adiabatic change but stays constant during an isothermal one. This can be quantified: an isothermal change obeys $PV = \text{constant}$ (that is, $PV^1 = \text{constant}$), but an adiabatic change of an ideal monatomic gas obeys:

Adiabatic · ideal monatomic gas $$PV^{5/3} = \text{constant}$$

Similar volume changes are therefore associated with bigger pressure changes during adiabatic processes, because there are also accompanying temperature changes. (For gases other than ideal monatomic gases, $V$ is raised to a different power — not covered in this course.)

Isobaric: any expansion or compression at constant pressure, $\Delta P = 0$. The first law keeps its full form:

Isobaric $$Q = \Delta U + W \quad (W = P\Delta V)$$

Most isobaric changes happen when a gas is allowed to expand or contract freely as its temperature changes, keeping its pressure equal to the surrounding pressure.

📈 Reading the four processes off a PV diagram

Process	Line shape	What's fixed	What to read
Isothermal	Curve following a dotted isothermal	Temperature ($\Delta U=0$)	$PV$ stays constant along the curve
Isovolumetric	Vertical straight line	Volume ($W=0$)	No area beneath the line — zero work done
Adiabatic	Curve steeper than the isothermals	Heat transfer ($Q=0$)	Steeper fall in $P$ for the same $\Delta V$ than an isothermal
Isobaric	Horizontal straight line	Pressure ($\Delta P=0$)	Area beneath = rectangle = $P\Delta V$

Real experimental graphs are read the same way. Figure B4.11 shows a student's isovolumetric experiment used to estimate absolute zero, and Figure B4.12 shows a Boyle's law (isothermal) experiment with an anomalous reading.

Scatter graph of pressure against temperature for a gas at constant volume, with a line of best fit extrapolated to intercept the temperature axis near minus 273 degrees Celsius

Fig B4.11 — isovolumetric experiment. The best-fit line, extrapolated (dashed), intercepts the temperature axis near −273 °C: absolute zero.

Volume against pressure graph for a Boyle's law experiment with a smooth inverse curve and one anomalous data point lying above the curve

Fig B4.12 — Boyle's law experiment with one anomalous reading, lying above the smooth $V \propto 1/P$ curve.

📝 Interpreting graphs in a report: discuss the spread of measurements, the quality of the graph, any relationship or trend, the significance of intercepts, gradients and areas, and any turning points. An anomalous reading should never be rejected without checking — it's usually a simple measurement or recording error. Draw a rough graph during the experiment, so an anomaly can still be re-measured before it's too late.

✍️ Worked example B4.5 – an isobaric expansion

A gas of volume 0.080 m³ and pressure $1.4\times10^{5}$ Pa expands to a volume of 0.11 m³ at constant pressure when $7.4\times10^{3}$ J of thermal energy are supplied. Name the process, and find the work done and the change in internal energy.

Identify: pressure is constant throughout, so this is an isobaric process.

Data: $P = 1.4\times10^{5}$ Pa, $\Delta V = 0.11 - 0.080 = 0.030$ m³, $Q = +7.4\times10^{3}$ J.

Equation: $W = P\Delta V$
$W = (1.4\times10^{5})(0.11-0.080)$

Answer: $W = +4.2\times10^{3}$ J. Then $Q=\Delta U+W \Rightarrow (+7.4\times10^{3}) = \Delta U + (+4.2\times10^{3})$, so $\Delta U = +3.2\times10^{3}$ J — internal energy increases.

Horizontal isobaric line on a pressure-volume diagram with the rectangular area beneath it shaded as work done

Fig — the isobaric expansion of Worked example B4.5. The shaded rectangle is the work done, $P\Delta V$.

✍️ Worked example B4.6 – an adiabatic compression

The volume of an ideal monatomic gas is reduced by a factor of 8.0 in an adiabatic compression. Determine the factor by which its pressure changes.

Identify: adiabatic monatomic ideal gas — use $PV^{5/3} = \text{constant}$.

Data: $V_1/V_2 = 8.0$ (volume reduced by this factor).

Equation: $P_1V_1^{5/3} = P_2V_2^{5/3}$
$\dfrac{P_2}{P_1} = \left(\dfrac{V_1}{V_2}\right)^{5/3} = 8.0^{5/3}$

Answer: $P_2/P_1 = 32$. The pressure increases by a factor of 32 — much more than the factor of 8.0 an isothermal compression would give, because the temperature also rises here.

Pressure-volume diagram comparing a steep adiabatic compression curve with a shallower dashed isothermal curve for the same volume change

Fig — the adiabatic curve (solid) rises more steeply than the isothermal (dashed) for the same volume change, in Worked example B4.6.

🌍 Real-world: writing up your investigation

Every investigation report should summarize what a chart, diagram or graph shows: the quantity and spread of measurements, the quality and reliability of the graph, any relationship between the variables, and the meaning of intercepts, gradients, areas under the graph, and any turning points. Figure B4.11 uses an intercept to estimate absolute zero; Figure B4.12 uses departure from a smooth curve to flag an anomaly.

✓ Examiner tip: if an anomalous reading wasn't checked at the time, still report it and label it as an outlier — then explain, using your judgement, whether it should affect your conclusions.

🚀 HL extension – linking PV^(5/3) to temperature

Combining $PV^{5/3} = \text{constant}$ with the ideal gas law $PV = nRT$ (so $P = nRT/V$) gives another useful adiabatic relation: substituting for $P$ shows $TV^{2/3} = \text{constant}$ along an adiabatic change.

Challenge: a monatomic ideal gas expands adiabatically from $3.13\times10^{-3}$ m³ to $3.97\times10^{-3}$ m³, reaching a final temperature of 398 K. What was its starting temperature? (Answer: using $T_1V_1^{2/3}=T_2V_2^{2/3}$, $T_1 = 398\times(3.97/3.13)^{2/3} \approx 466$ K — hotter, since the gas cools as it expands.)

⚠️ Trap corner – common mistakes

Misconception	Correct view
Isobaric means no work is done	Isobaric means constant pressure, not constant volume — the gas can still expand or contract, and $W=P\Delta V$ is not zero
Adiabatic means the temperature stays constant	That's isothermal ($\Delta U=0$). Adiabatic means $Q=0$; the temperature does change, because all the work is converted to or from internal energy
Isothermal and adiabatic lines look the same on a PV diagram	Adiabatic lines are always steeper, because temperature also changes along them, giving a bigger pressure change for the same volume change
An anomalous reading should always be deleted	Check it first — most anomalies are simple measurement errors, but a confirmed one should be kept and investigated further

⚠️ Most common slip: mixing up which variable each process holds fixed. Read the prefix literally — isothermal fixes temperature, isovolumetric fixes volume, isobaric fixes pressure; adiabatic is the odd one out, fixing $Q=0$ rather than a state variable.

ΔU=0 isothermal: Q=W W=0 isovolumetric: Q=ΔU Q=0 adiabatic: ΔU=−W ΔP=0 isobaric: Q=ΔU+W PV=constant (isothermal) PV^(5/3)=constant (adiabatic, monatomic) adiabatic lines steeper than isothermals check anomalies before rejecting them

Practice check — multiple choice

1. Which thermodynamic process has $Q = 0$?

A Isothermal

B Isovolumetric

C Adiabatic

D Isobaric

2. An adiabatic change of an ideal monatomic gas obeys

A $PV = \text{constant}$

B $PV^{5/3} = \text{constant}$

C $P/V = \text{constant}$

D $P + V = \text{constant}$

3. Why must an adiabatic line on a PV diagram be steeper than an isothermal line through the same point?

A Because the temperature also changes along the adiabatic line, adding an extra pressure change

B Because adiabatic changes always involve compression

C Because no work is done in an adiabatic change

D Because pressure cannot change in an isothermal process

4. In an isobaric expansion, the sign of the work done by the gas, $W$, is

A Positive

B Negative

C Zero

D Undefined

5. A bicycle pump gets hot during a fast compression stroke because

A Thermal energy flows in from your hand

B Gas molecules gain kinetic energy from the inward-moving piston, with no time for heat to escape

C The process is isothermal

D The gas loses internal energy as it is compressed

Practice check — fill in the blanks

Complete the statements:

A process at constant temperature is called , and obeys $Q = W$. A process at constant volume is called , and obeys $Q = \Delta U$. A process with no thermal energy transfer is called , and obeys $Q = 0$. A process at constant pressure is called , and obeys $W = P\Delta V$. An adiabatic change of an ideal monatomic gas obeys $PV^{} = \text{constant}$. On a PV diagram, adiabatic lines are always than isothermal lines through the same point.

Exam‑style questions

Exam question · Compare / Explain

Q14. Figure B4.13 represents four successive changes to the state of a gas of constant mass: B→C, C→D, D→A, A→B. (a) Name the processes shown by (i) BC and (ii) DA. (b) If CD is an adiabatic process, compare the temperatures at points C and D. (c) If AB is an isothermal process, compare the temperatures at points A and B. [5 marks]

Pressure-volume cycle diagram with four states A, B, C, D joined by isobaric, adiabatic, isovolumetric and compression curves

Fig B4.13 — four successive changes to the state of a gas.

✓ (a)(i) BC is a horizontal line at constant pressure — an isobaric expansion.
✓ (a)(ii) DA is a vertical line at constant volume — an isovolumetric change.
✓ (b) CD is adiabatic and the gas expands (volume increases from C to D), so $-\Delta U = W$: the gas does work using its own internal energy, so its temperature falls — $T_D < T_C$.
✓ (c) AB is isothermal, so by definition the temperature is constant along it — $T_A = T_B$.

Exam question · Copy and complete

Q15. Copy and complete Table B4.1 by putting 0, +, − or ± in each box, for each process listed. [8 marks]

Process	Q	ΔU	W	ΔP	ΔV	ΔT
isothermal expansion	?	?	?	?	?	?
isothermal compression	?	?	?	?	?	?
adiabatic expansion	?	?	?	?	?	?
adiabatic compression	?	?	?	?	?	?
isobaric expansion	?	?	?	?	?	?
isobaric compression	?	?	?	?	?	?
isovolumetric pressure increase	?	?	?	?	?	?
isovolumetric pressure decrease	?	?	?	?	?	?

✓ Completed table:

Process	Q	ΔU	W	ΔP	ΔV	ΔT
isothermal expansion	+	0	+	−	+	0
isothermal compression	−	0	−	+	−	0
adiabatic expansion	0	−	+	−	+	−
adiabatic compression	0	+	−	+	−	+
isobaric expansion	+	+	+	0	+	+
isobaric compression	−	−	−	0	−	−
isovolumetric pressure increase	+	+	0	+	0	+
isovolumetric pressure decrease	−	−	0	−	0	−

Exam question · Determine

Q16. An ideal monatomic gas expands adiabatically from a volume of $3.13\times10^{-3}$ m³ to $3.97\times10^{-3}$ m³. (a) If the original pressure was $2.60\times10^{5}$ Pa, determine the final pressure. (b) If the final temperature of the gas was 398 K, what was the starting temperature? [5 marks]

✓ (a) $P_1V_1^{5/3}=P_2V_2^{5/3} \Rightarrow P_2 = P_1(V_1/V_2)^{5/3} = (2.60\times10^{5})(3.13/3.97)^{5/3}$
✓ $P_2 \approx 1.75\times10^{5}\ \text{Pa}$.
✓ (b) Combining $PV^{5/3}=\text{const}$ with $PV=nRT$ gives $TV^{2/3}=\text{const}$: $T_1 = T_2(V_2/V_1)^{2/3} = 398\times(3.97/3.13)^{2/3}$
✓ $T_1 \approx 466\ \text{K}$ — hotter than the final state, consistent with cooling on adiabatic expansion.

Score: 0 pts

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First Law of Thermodynamics

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The Carnot Cycle

Process	Line shape	What's fixed	What to read
Isothermal	Curve following a dotted isothermal	Temperature (\(\Delta U=0\))	\(PV\) stays constant along the curve
Isovolumetric	Vertical straight line	Volume (\(W=0\))	No area beneath the line — zero work done
Adiabatic	Curve steeper than the isothermals	Heat transfer (\(Q=0\))	Steeper fall in \(P\) for the same \(\Delta V\) than an isothermal
Isobaric	Horizontal straight line	Pressure (\(\Delta P=0\))	Area beneath = rectangle = \(P\Delta V\)