Thermodynamics Problem Solving

This final lesson brings together all the concepts from the course — specific heat capacity, specific latent heat, the ideal gas equation, kinetic theory, molecular kinetic energy, and the first law of thermodynamics — into multi-step problems that mirror the style and difficulty of Edexcel A-Level exam questions. Work through each example carefully, noting the problem-solving strategies.

Strategy for Multi-Step Thermal Physics Problems

graph TD
    A["Read the question"] --> B["Identify quantities\nand concepts"]
    B --> C["List knowns\nwith SI units"]
    C --> D["Choose equation(s)"]
    D --> E["Convert ALL\nunits to SI"]
    E --> F["Substitute\nand solve"]
    F --> G["Check signs\n(first law)"]
    G --> H["Evaluate: does\nthe answer make\nphysical sense?"]

Identify what is being asked and which concepts are involved.
List the known quantities with their SI units.
Choose the correct equation(s): Q = mcΔθ, Q = mL, pV = nRT, pV = NkT, pV = (1/3)Nm<c²>, (1/2)m<c²> = (3/2)kT, ΔU = Q + W, W = pΔV.
Convert all units to SI before substituting.
Check the sign conventions for the first law (Edexcel: W = work done on gas).
Evaluate your answer — does it make physical sense?

Equation Reference Table

Topic	Equation	Variables
Specific heat capacity	Q = mcΔθ	Q in J, m in kg, c in J kg⁻¹ K⁻¹, Δθ in K
Specific latent heat	Q = mL	Q in J, m in kg, L in J kg⁻¹
Ideal gas (molar)	pV = nRT	p in Pa, V in m³, n in mol, T in K
Ideal gas (molecular)	pV = NkT	N = number of molecules
Kinetic theory	pV = (1/3)Nm<c²>	m = mass of one molecule
Mean KE per molecule	(1/2)m<c²> = (3/2)kT	Independent of gas type
RMS speed	c_rms = √(3RT/M)	M in kg mol⁻¹
First law	ΔU = Q + W	W = work done ON gas
Work at constant p	W = pΔV	W = work done BY gas
Internal energy (monatomic)	U = (3/2)nRT	Ideal monatomic gas only
Gas density	ρ = pM/(RT)	Derived from pV = nRT

Worked Example 1: Combined SHC and Latent Heat (Full Phase Journey)

How much energy is needed to convert 0.40 kg of ice at −15 °C into steam at 100 °C?

Data: c_ice = 2100 J kg⁻¹ K⁻¹, c_water = 4200 J kg⁻¹ K⁻¹, L_f = 334 000 J kg⁻¹, L_v = 2 260 000 J kg⁻¹.

Step 1: Heat ice from −15 °C to 0 °C. Q₁ = mcΔθ = 0.40 × 2100 × 15 = 12 600 J

Step 2: Melt ice at 0 °C. Q₂ = mL_f = 0.40 × 334 000 = 133 600 J

Step 3: Heat water from 0 °C to 100 °C. Q₃ = mcΔθ = 0.40 × 4200 × 100 = 168 000 J

Step 4: Boil water at 100 °C. Q₄ = mL_v = 0.40 × 2 260 000 = 904 000 J

Total: Q = 12 600 + 133 600 + 168 000 + 904 000 = 1 218 200 J ≈ 1.22 MJ

Step	Process	Energy / kJ	% of total
1	Heat ice	12.6	1.0%
2	Melt ice	133.6	11.0%
3	Heat water	168.0	13.8%
4	Boil water	904.0	74.2%
Total		1218.2	100%

The vaporisation step requires far more energy than all the other steps combined. This is a common feature — always check whether latent heat dominates.

Worked Example 2: Ideal Gas Equation — Finding Temperature

A cylinder of volume 0.030 m³ contains 0.050 kg of neon gas (M = 0.020 kg mol⁻¹) at a pressure of 4.0 × 10⁵ Pa. What is the temperature of the gas?

Step 1: Find n. n = m/M = 0.050/0.020 = 2.5 mol

Step 2: Use pV = nRT. T = pV/(nR) = (4.0 × 10⁵ × 0.030)/(2.5 × 8.31) T = 12 000/20.775 = 578 K (305 °C)

Worked Example 3: Kinetic Theory — RMS Speed and Molecular KE

Calculate (a) the rms speed and (b) the mean kinetic energy of an argon atom (M = 0.040 kg mol⁻¹) at 500 K.

(a) c_rms = √(3RT/M) = √(3 × 8.31 × 500 / 0.040) = √(12 465 / 0.040) = √(311 625) = 558 m s⁻¹

(b) (1/2)m<c²> = (3/2)kT = 1.5 × 1.38 × 10⁻²³ × 500 = 1.035 × 10⁻²⁰ J

Note: the kinetic energy depends only on temperature (not on the type of gas), while the rms speed depends on both temperature and molecular mass.

Worked Example 4: First Law Application with Charles's Law

An ideal gas in a cylinder at constant pressure of 1.0 × 10⁵ Pa is heated from 300 K to 600 K. The initial volume is 0.010 m³. The gas contains 0.40 mol.

(a) What is the final volume?

At constant pressure, V/T = constant (Charles's law). V₂ = V₁ × T₂/T₁ = 0.010 × 600/300 = 0.020 m³

(b) What is the work done by the gas?

W_by = pΔV = 1.0 × 10⁵ × (0.020 − 0.010) = 1.0 × 10⁵ × 0.010 = 1000 J

(c) If 2500 J of heat is supplied, what is the change in internal energy?

Using Edexcel convention: ΔU = Q + W where W is work done ON gas = −1000 J (gas did work). ΔU = 2500 + (−1000) = 1500 J

The gas's internal energy increased by 1500 J. Of the 2500 J heat input, 1000 J went into doing work against the atmosphere, and 1500 J went into increasing the internal energy (raising the temperature).

Worked Example 5: p-V Diagram Analysis

A gas undergoes a cyclic process ABCA on a p-V diagram:

A → B: isobaric expansion at 3.0 × 10⁵ Pa from 1.0 × 10⁻³ m³ to 3.0 × 10⁻³ m³
B → C: isochoric (constant volume) pressure drop to 1.0 × 10⁵ Pa
C → A: isobaric compression at 1.0 × 10⁵ Pa back to 1.0 × 10⁻³ m³

(a) Calculate the work done by the gas in each stage.

A → B: W = pΔV = 3.0 × 10⁵ × 2.0 × 10⁻³ = 600 J (expansion, gas does work) B → C: ΔV = 0, so W = 0 J C → A: W = pΔV = 1.0 × 10⁵ × (−2.0 × 10⁻³) = −200 J (compression, work done on gas)

(b) What is the net work done by the gas per cycle?

Net W = 600 + 0 + (−200) = 400 J

This equals the area of the rectangle enclosed on the p-V diagram: Area = (3.0 − 1.0) × 10⁻³ × (3.0 − 1.0) × 10⁵ = 2.0 × 10⁻³ × 2.0 × 10⁵ = 400 J ✓

(c) What is the change in internal energy over one complete cycle?

The gas returns to its initial state, so ΔU = 0 for the complete cycle.

Worked Example 6: Pressure from Kinetic Theory

A container of volume 0.025 m³ holds 5.0 × 10²⁴ helium atoms (mass of one atom = 6.64 × 10⁻²⁷ kg) at a temperature of 400 K. Calculate the pressure using two different methods and verify they agree.

Method 1: Using pV = NkT. p = NkT/V = (5.0 × 10²⁴ × 1.38 × 10⁻²³ × 400)/0.025 p = (5.0 × 10²⁴ × 5.52 × 10⁻²¹)/0.025 p = 27 600/0.025 = 1.10 × 10⁶ Pa

Method 2: Using pV = (1/3)Nm<c²> and <c²> = 3kT/m. <c²> = 3 × 1.38 × 10⁻²³ × 400/(6.64 × 10⁻²⁷) = 1.656 × 10⁻²⁰/(6.64 × 10⁻²⁷) = 2.494 × 10⁶ m² s⁻² p = (1/3) × 5.0 × 10²⁴ × 6.64 × 10⁻²⁷ × 2.494 × 10⁶/0.025 p = (1/3) × 82.80/0.025 = (1/3) × 3312 = 1104 × 10³ ≈ 1.10 × 10⁶ Pa ✓

Both methods give the same answer, confirming the consistency of kinetic theory with the ideal gas equation.

Worked Example 7: RMS Speed Comparison at Different Temperatures

The rms speed of nitrogen molecules at 300 K is 517 m s⁻¹. At what temperature would oxygen molecules (M = 0.032 kg mol⁻¹) have the same rms speed? (M_N₂ = 0.028 kg mol⁻¹)

c_rms = √(3RT/M), so for equal rms speeds:

3RT_O₂/M_O₂ = 3RT_N₂/M_N₂

T_O₂/M_O₂ = T_N₂/M_N₂

T_O₂ = T_N₂ × M_O₂/M_N₂ = 300 × 0.032/0.028 = 300 × 1.143 = 343 K (70 °C)

Oxygen molecules need to be at a higher temperature to reach the same speed as nitrogen, because oxygen molecules are heavier.

Worked Example 8: Adiabatic Expansion Cooling

A gas at 600 K and 4.0 × 10⁵ Pa in a volume of 0.005 m³ expands adiabatically. The gas does 800 J of work during the expansion. If the gas contains 1.0 mol and behaves as an ideal monatomic gas, what is the final temperature?

Adiabatic: Q = 0, so ΔU = W_on = −800 J (the gas does work, so work ON gas is negative).

For monatomic ideal gas: ΔU = (3/2)nRΔT

−800 = (3/2) × 1.0 × 8.31 × ΔT

−800 = 12.465 × ΔT

ΔT = −800/12.465 = −64.2 K

T_final = 600 − 64.2 = 536 K

The gas cools by 64 K during the adiabatic expansion, as expected — internal energy is converted to work.

Worked Example 9: Gas Density and Ideal Gas Law

Calculate the density of carbon dioxide gas (CO₂, M = 0.044 kg mol⁻¹) at 100 °C and 2.0 × 10⁵ Pa.

Using ρ = pM/(RT):

T = 100 + 273 = 373 K

ρ = (2.0 × 10⁵ × 0.044) / (8.31 × 373)

ρ = 8800 / 3099.6 = 2.84 kg m⁻³

For comparison, CO₂ at standard conditions (20 °C, 1 atm) has a density of about 1.84 kg m⁻³. The higher pressure increases density; the higher temperature decreases it; the net effect is still an increase.

Common Exam Pitfalls

Here is a comprehensive summary of the most frequent errors in thermodynamics exam questions:

Pitfall	How to avoid it
Using °C instead of K in gas law equations	Always convert: T(K) = θ(°C) + 273
Using volume in litres or cm³ instead of m³	1 litre = 10⁻³ m³, 1 cm³ = 10⁻⁶ m³
Confusing work done BY gas with work done ON gas	Clarify sign convention before starting
Forgetting latent heat steps in heating calculations	Draw a heating curve to identify all stages
Using SHC equation during a change of state	Temperature is constant during changes of state — use Q = mL, not Q = mcΔθ
Confusing rms speed with mean speed	c_rms = √<c²> = √(3kT/m); always greater than mean speed
Applying ideal gas equation where gas is non-ideal	Check if conditions are extreme (very high p or very low T)
Saying ΔU = 0 for adiabatic (confusing with isothermal)	Adiabatic: Q = 0, ΔU = W. Isothermal: ΔU = 0, Q = −W
Forgetting to include the calorimeter in SHC experiments	Energy heats both substance AND container
Using the wrong latent heat (L_f vs L_v)	Fusion = melting/freezing. Vaporisation = boiling/condensing
Molar mass in g mol⁻¹ instead of kg mol⁻¹	Physics uses SI: M in kg mol⁻¹
Forgetting work done against atmosphere in L_v	L_v includes bond-breaking AND expansion work

Practice Strategy

For the exam, practise combining equations across topics. A typical 6-mark question might ask you to:

Use the ideal gas equation to find the number of moles.
Convert to number of molecules using Nₐ.
Calculate the rms speed using kinetic theory.
Find the mean kinetic energy using (1/2)m<c²> = (3/2)kT.
Apply the first law to determine the change in internal energy.

Each step uses a different equation, and the answer from one step feeds into the next. Practise writing out each step clearly with units.

Data Sheet Reference

In the exam, you will be given a data sheet. Make sure you know which values are provided and which you must recall:

Given on data sheet	Must be memorised
R = 8.31 J mol⁻¹ K⁻¹	Q = mcΔθ
k = 1.38 × 10⁻²³ J K⁻¹	Q = mL
Nₐ = 6.02 × 10²³ mol⁻¹	pV = nRT
SHC/latent heat values (if needed)	ΔU = Q + W
Molar masses (if needed)	W = pΔV
	(1/2)m<c²> = (3/2)kT
	pV = (1/3)Nm<c²>

Summary

Multi-step problems require combining Q = mcΔθ, Q = mL, pV = nRT, kinetic theory, and the first law.
Always convert to SI units before calculating.
Use sign conventions carefully in first law problems.
On p-V diagrams, the area under a curve is work done; the area enclosed by a cycle is net work.
For a complete cycle, ΔU = 0 (the gas returns to its initial state).
Vaporisation typically dominates the energy budget in heating calculations.
Check your answer for physical reasonableness — does the magnitude and sign make sense?

A-Level Deep Dive: Thermodynamics Problem Solving

Spec mapping

Edexcel 9PH0 specification Topic 9 — Thermodynamics treats this lesson as a synoptic capstone: candidates must combine internal energy, specific heat capacity, specific latent heat, the ideal gas equation, kinetic theory, and the first law of thermodynamics inside a single multi-stage problem (refer to the official specification document for exact wording). Topic 9 is examined principally on Paper 3 (General and Practical Principles in Physics), where the synoptic style demands that candidates select the correct model — sensible heating versus phase change versus gas process — before substituting numbers. Routine procedural fluency with each individual equation is no longer sufficient at A2 standard; the rewarded skill is recognising which physical regime applies on each step and tracking energy flow consistently through the whole problem. The Edexcel formula booklet supplies $Q = mc\Delta\theta$ , $Q = mL$ , $pV = NkT$ , $pV = nRT$ , $\tfrac{1}{2}m\langle c^2\rangle = \tfrac{3}{2}kT$ and the first law in the form $\Delta U = Q + W$ (or equivalently $Q = \Delta U + W$ ); the sign convention used in the question wording must be checked carefully.

Worked example with full mark scheme

Question (12 marks):

A copper kettle contains 0.500 kg of water at an initial temperature of 15.0 °C. The heating element supplies energy at a constant rate of 2.20 kW. Assume all energy delivered by the element enters the water; ignore the kettle's heat capacity and any losses to the surroundings.

(Specific heat capacity of water $c_w = 4180\ \text{J kg}^{-1}\,\text{K}^{-1}$ ; specific latent heat of vaporisation of water $L_v = 2.26 \times 10^6\ \text{J kg}^{-1}$ ; molar mass of water $M = 0.018\ \text{kg mol}^{-1}$ ; $R = 8.31\ \text{J K}^{-1}\,\text{mol}^{-1}$ .)

(a) Calculate the energy required to raise the water from 15.0 °C to 100 °C. (2)

(b) Calculate the additional energy required to vaporise 0.100 kg of the water at 100 °C. (2)

(d) The 0.100 kg of vapour is collected at 100 °C and 1.01 × 10^5 Pa. Treating it as an ideal gas, calculate the volume it occupies. (3)

(e) State, with reasoning, whether the work done by the steam on the surrounding atmosphere during vaporisation is significant compared with the latent-heat input found in (b), and identify which term in $\Delta U = Q - W$ accounts for the rest of the energy. (3)

Solution with mark scheme:

(a) C1 — substitution into $Q_1 = mc\Delta\theta$ with $\Delta\theta = 85.0\ \text{K}$ : $Q_1 = 0.500 \times 4180 \times 85.0$ A1 — $Q_1 = 1.78 \times 10^5\ \text{J}$ (177 650 J accepted; 2 s.f. or better).

(b) C1 — substitution into $Q_2 = mL_v$ with $m = 0.100\ \text{kg}$ : $Q_2 = 0.100 \times 2.26 \times 10^6$ A1 — $Q_2 = 2.26 \times 10^5\ \text{J}$ .

(c) C1 — total energy $Q_1 + Q_2 = 1.78 \times 10^5 + 2.26 \times 10^5 = 4.04 \times 10^5\ \text{J}$ , divide by 2200 W. A1 — $t = 184\ \text{s}$ (about 3 min 4 s).

(d) C1 — convert mass to moles: $n = 0.100 / 0.018 = 5.56\ \text{mol}$ . C1 — apply $pV = nRT$ with $T = 373\ \text{K}$ and $p = 1.01 \times 10^5\ \text{Pa}$ : $V = \dfrac{nRT}{p} = \dfrac{5.56 \times 8.31 \times 373}{1.01 \times 10^5}$ A1 — $V \approx 0.171\ \text{m}^3$ (allow 0.17 m³).

(e) C1 — work done against atmosphere $W = p\Delta V \approx p V$ (initial liquid volume is roughly 10⁻⁴ m³, negligible compared with 0.171 m³): $W \approx 1.01 \times 10^5 \times 0.171 = 1.73 \times 10^4\ \text{J}$ . B1 — comparison: this is about $1.73 \times 10^4 / 2.26 \times 10^5 \approx 7.7\%$ of $Q_2$ — not negligible but smaller than the latent-heat input. B1 — apply first law: $\Delta U = Q - W \approx 2.26 \times 10^5 - 1.73 \times 10^4 \approx 2.09 \times 10^5\ \text{J}$ , the increase in internal energy associated with breaking the intermolecular bonds during the liquid–vapour transition.

Total: 12 marks (C7 A4 B1 — graded across stages). The capstone mark for AO2 is given to candidates who recognise that most of $L_v$ is the rise in internal energy and only a minority is mechanical work — a counter-intuitive result that the working makes explicit.

Specimen question modelled on the Edexcel 9PH0 Paper 3 format

Question (10 marks): A fixed mass of an ideal monatomic gas is taken around the cycle A → B → C → A.

A → B: isothermal expansion at $T_1 = 320\ \text{K}$ from $V_A = 1.0 \times 10^{-3}\ \text{m}^3$ to $V_B = 3.0 \times 10^{-3}\ \text{m}^3$ .
B → C: isochoric (constant-volume) cooling to $T_2 = 240\ \text{K}$ .
C → A: adiabatic compression back to the starting state.

The gas contains $n = 0.040\ \text{mol}$ .