Thermodynamics 1: practice problems and exam questions

The first law, ideal gases, energy balances and the second-law limits of heat engines, refrigerators and heat pumps. Most exam mistakes come from units and temperatures, not from the physics.

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Cram pack (PDF)Formula sheets, method steps and practice problems with full solutions, ready to print.

Topics in this course

  1. Ideal gas lawP from n, V, T · from mass · state change or densitycalculations
  2. Energy balanceheating a stream · heat exchanger · exchanger with heat losscalculations
  3. Heat engines & refrigeratorsfirst law of a cycle · Carnot limit · refrigerators and heat pumpscalculations
  4. Gas power cyclesOtto cycle · Diesel cycle · Brayton cyclecalculations

Exam tips for Thermodynamics 1

  • Use absolute temperatures (K) in gas laws and in every Carnot or COP formula.
  • Write the energy balance in words first: what goes in minus what goes out equals the change.
  • Between the same two temperatures, the COP of a heat pump is always exactly 1 higher than that of a refrigerator.

Worked examples

Ideal gas lawEasy
n = 5 molV = 15 LT = 30 °CP = ?

A rigid tank holds an ideal gas. Find its pressure P.

n = 5 molV = 15 LT = 30 °CR = 8.314 J/(mol·K)

Find P (kPa)

Show hints
  1. Use PV = nRT with absolute temperature (K).
  2. Units: with V in m³ and R = 8.314, P comes out in Pa. 1 L = 0.001 m³.
  3. T = 303.1 K.
Show the answer and the step-by-step solution

Ideal gas law: PV = nRT, with absolute temperature (K) and units that match R.

Convert units

T = 30 + 273.15 = 303.1 K
V = 15 L = 0.015 m³

Ideal gas law

P = nRT/V = 5·8.314·303.1/0.015 = 840130 Pa = 840.1 kPa

Answer

P = 840.1 kPa
Energy balanceMedium
hot sidecold sideglycol mix · 150 °C120 °Cwater · 15 °Ct = ? °C

Hot glycol mix cools water in a counter-flow heat exchanger with no heat loss. Find the water outlet temperature tout.

ṁh = 1 kg/scp,h = 3.6 kJ/(kg·K)Th: 150 → 120 °Cṁc = 1 kg/scp,c = 4.18 kJ/(kg·K)tin = 15 °C

Find tout (°C)

Show hints
  1. Heat lost by the hot stream equals heat gained by the cold stream.
  2. Find Q from the hot side first, since you know both of its temperatures.
  3. Q = 108 kW.
Show the answer and the step-by-step solution

Exchanger energy balance: with no losses, ṁhcp,h(Th,in − Th,out) = ṁccp,c(tout − tin).

Heat released by the hot stream

Q = 1·3.6·(150 − 120) = 108 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 15 + 108/(1·4.18) = 40.84 °C

Answer

tout = 40.84 °C

How practising works

  • “Practise this course” mixes problems from all its topics, like a real exam.
  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Mock exams are timed and have no hints, and you get your score and a full review at the end.

Calculators for this course

Carnot efficiency and COP calculatorIdeal gas law calculatorOtto and Brayton cycle efficiency calculatorCombined gas law calculator (Boyle, Charles and Gay-Lussac)

Related courses

Physics for EngineersFE Mechanical exam practiceMaterial and Energy BalancesFE Chemical exam practiceGeneral ChemistryStaticsMechanics of MaterialsDynamics
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