🔥 Thermodynamics
Thermodynamics governs energy transfer. The ideal gas law links pressure, volume and temperature across a process; the first law Q = ΔU + W is the energy balance for any closed system. Together they let you analyse heating, compression and expansion of gases in engineering systems.
2Topics
8Examples
4Calculators
12Practice Qs
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What you'll cover
Two focused topics building from gas state relations to full energy balance.
🔥Ideal Gas Laws & ProcessesCombined gas law, isobaric/isochoric/isothermal, PV=mRT
⚡First Law & Energy BalanceQ=ΔU+W, boundary work, specific heat, enthalpy
⚙️ApplicationsFour engineering calculators
🎮Quick-fire game12 randomised questions
📝Practice questions12 graded questions with full working
🔥
Ideal Gas Laws & Processes
The ideal gas law links pressure, volume and temperature for a fixed mass of gas. Different processes (constant pressure, volume or temperature) simplify the relationship in useful ways.
💡 What every symbol means, rule by rule:
- PV = nRT / PV = mRT — P is absolute pressure (Pa), V is volume (m³), T is absolute temperature (K, never °C directly). n is the amount of gas in moles (SI form); R here is the universal gas constant, 8314 J/kmol·K. m is the mass of gas in kg (engineering form); R here is the specific gas constant for that gas (J/kg·K, e.g. ≈287 for air) — specific R = universal R ÷ molar mass, so it's a different number to the R used with n.
- P₁V₁/T₁ = P₂V₂/T₂ — the subscripts 1 and 2 just mean "before" and "after": the same gas's P, V and T measured at two different moments in a process.
- Zeroth law (A, B, C) — A, B and C are just three arbitrary bodies used to state the principle. If A and C reach the same temperature, and B and C also reach the same temperature, then A and B must be equal too — even though A and B were never compared directly. This is exactly what lets a thermometer (playing the role of C) measure two different objects and let you compare them.
- Isobaric — V₁/T₁ = V₂/T₂ (Charles's Law) — pressure P stays constant ("iso" = same, "baric" = pressure), so P cancels out of the combined law, leaving only V and T linked: V and T rise and fall together in direct proportion. Sanity check — heat a gas at constant pressure and it must expand (V₂ > V₁ when T₂ > T₁); if your answer shrinks the volume on heating, a value has gone in the wrong place.
- Isochoric — P₁/T₁ = P₂/T₂ (Gay-Lussac's Law) — volume V stays constant ("choric" = volume) instead, so V cancels out, leaving only P and T linked.
- Isothermal — P₁V₁ = P₂V₂ — temperature T stays constant ("thermal" = temperature), so T cancels out, leaving only P and V linked (this is Boyle's Law). ΔU is the change in internal energy — it's zero here specifically because temperature doesn't change, and for an ideal gas internal energy depends on temperature alone, not on pressure or volume.
📚 Key vocabulary
- System
- the region of matter under analysis. Closed system: fixed mass, no mass transfer. Open system (control volume): mass may cross the boundary.
- State
- the condition of a system described by its properties (P, V, T). A process changes the system from one state to another.
- Specific gas constant R
- R = R0/M (J/kg·K), where R0 = 8314 J/kmol·K is the universal gas constant and M is molar mass. For air, R ≈ 287 J/kg·K.
- Isobaric (Charles's Law)
- constant pressure process. V₁/T₁ = V₂/T₂ — volume directly proportional to absolute temperature. W = PΔV. Used for piston-cylinder devices open to atmosphere.
- Isochoric (Gay-Lussac's Law)
- constant volume process. P₁/T₁ = P₂/T₂ — pressure directly proportional to absolute temperature. No boundary work (W = 0), so all heat added changes internal energy directly.
- Isothermal
- constant temperature process. For an ideal gas ΔU = 0, so Q = W. Work is found from W = mRT ln(V₂/V₁).
👁️ Visual aid — P–V diagram: how the three basic processes appear when plotted from the same starting state
👁️ Visual aid — Charles's Law: V against absolute temperature T
The line only passes through the origin because T is measured in Kelvin (absolute zero). Plotting the same data against °C would give a straight line that does not pass through zero — always convert to Kelvin before using V∝T.
✍️Worked example 1 — Charles's Law: isobaric heating step through it
QuestionA gas at 150 kPa, 0.4 m³ and 20 °C is heated at constant pressure to 80 °C. Find the new volume V₂.
Temperatures
T₁ = 20 + 273 = 293 K T₂ = 80 + 273 = 353 KConstant P
V₁/T₁ = V₂/T₂ (isobaric process)V₂
V₂ = V₁ × T₂/T₁ = 0.4 × 353/293 = 0.4 × 1.2048 = 0.4819 m³AnswerV₂ = 0.482 m³ ✓
Step 1 of 5
✍️Worked example 2 — ideal gas: find T₂ after compression
QuestionAir at 100 kPa, 27 °C is compressed to 400 kPa; volume changes from 0.2 m³ to 0.06 m³. Find T₂.
T₁
T₁ = 27 + 273 = 300 KCombined law
T₂ = T₁ × (P₂V₂)/(P₁V₁) = 300 × (400×0.06)/(100×0.2) = 300 × 24/20 = 360 KAnswerT₂ = 360 K = 87 °C
✍️Worked example 3 — isothermal expansion: work done
Question2 kg of air (R = 287 J/kg·K) expands isothermally at 350 K from 0.15 m³ to 0.4 m³. Find the work done.
Formula
W = mRT ln(V₂/V₁)Substitute
W = 2×287×350×ln(0.4/0.15) = 200900×ln(2.667) = 200900×0.9808AnswerW = 197.0 kJ
✍️Worked example 4 — finding the mass of a gas
QuestionA rigid tank contains air at 250 kPa, 35 °C, volume 1.2 m³. Given R = 287 J/kg·K, find the mass of air in the tank.
Temperature
T = 35 + 273 = 308 KIdeal gas law
PV = mRT → m = PV/(RT) = 250000×1.2/(287×308) = 300000/88396Answerm = 3.39 kg
✍️Worked example 5 — Charles's Law: hot air balloon envelope
QuestionA hot air balloon envelope holds 1200 m³ of air at ground temperature 15 °C. The burner heats the air at constant (atmospheric) pressure to 105 °C. Find the new volume V₂ the envelope must accommodate.
Why constant pressure?The envelope is open at the bottom, so the air inside stays at atmospheric pressure throughout — only V and T change, which is exactly Charles's Law.
Temperatures
T₁ = 15 + 273 = 288 K T₂ = 105 + 273 = 378 KCharles's Law
V₁/T₁ = V₂/T₂V₂
V₂ = V₁ × T₂/T₁ = 1200 × 378/288 = 1200 × 1.3125AnswerV₂ = 1575 m³
Sanity checkThe air was heated, so it should expand: 1575 m³ > 1200 m³ ✓. If the volume had come out smaller than 1200 m³, a temperature or ratio would have been the wrong way round.
🔨 Try it yourself
Now you can try these calculators with your own values.
🔥Ideal gas calculatorSelect calculation type
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⚡
First Law & Energy Balance
The first law of thermodynamics is a statement of energy conservation: heat added to a closed system either changes its internal energy or does work on the surroundings (or both).
📚 Key vocabulary
- Internal energy
- the sum of microscopic kinetic and potential energies of molecules. For an ideal gas: ΔU = mcvΔT, independent of the process.
- Enthalpy
- H = U + PV (J). Useful for open systems and constant-pressure processes. Q = ΔH at constant pressure.
- First law of thermodynamics
- energy cannot be created or destroyed — for a closed system, Q = ΔU + W. Sign convention: heat in is positive, work done by the system is positive.
- Boundary work
- work done by a gas as it expands (or done on it as it's compressed) by pushing against a moving boundary such as a piston: W = PΔV at constant pressure.
- Specific heat capacity
- the energy needed to raise the temperature of 1 kg of a substance by 1 K. cv (constant volume) and cp (constant pressure) differ for gases.
- Heat
- energy transferred due to a temperature difference. Not a property of the system — it's a process quantity, like work.
👁️ Visual aid — first law energy balance for a closed piston-cylinder system. Q in, W out, ΔU stored.
✍️Worked example 5 — first law: isobaric process step through it
Question2 kg of air (cv = 718 J/kg·K, cp = 1005 J/kg·K) is heated at constant pressure from 300 K to 400 K. Find W, ΔU and Q.
Identify process
Constant pressure (isobaric) → W = PΔV = m(cp−cv)ΔTWork
W = 2×(1005−718)×100 = 2×287×100 = 57400 JΔU
ΔU = mcvΔT = 2×718×100 = 143600 JFirst law
Q = ΔU + W = 143600 + 57400 = 201000 JCheck
Q = mcpΔT = 2×1005×100 = 201000 J ✓AnswerQ = 201 kJ, ΔU = 143.6 kJ, W = 57.4 kJ
Step 1 of 7
✍️Worked example 6 — first law: isochoric (constant volume) process
Question1.5 kg of nitrogen (cv = 742 J/kg·K) is heated at constant volume. Q = 45 kJ. Find ΔT and ΔU.
Isochoric
W = 0 (no boundary work when volume is constant)First law
Q = ΔU + 0 = ΔU → ΔU = 45000 JΔT
ΔT = ΔU / (mcv) = 45000 / (1.5×742) = 45000/1113 = 40.43 KAnswerΔU = 45 kJ ΔT = 40.4 K
✍️Worked example 7 — first law: isothermal process
Question3 kg of air (R = 287 J/kg·K) expands isothermally at 400 K from 0.2 m³ to 0.5 m³. Find Q.
Isothermal
ΔU = 0 for an ideal gas at constant temperature → Q = WWork
W = mRT ln(V₂/V₁) = 3×287×400×ln(0.5/0.2) = 344400×ln(2.5) = 344400×0.9163AnswerQ = W = 315.6 kJ
✍️Worked example 8 — enthalpy change at constant pressure
Question1.5 kg of air is heated at constant pressure from 20 °C to 150 °C. Given cp = 1005 J/kg·K, find ΔH and Q.
ΔT
ΔT = 150 − 20 = 130 KEnthalpy change
ΔH = mcpΔT = 1.5×1005×130 = 195975 JAnswerΔH = 196.0 kJ Q = ΔH at constant pressure = 196.0 kJ
🔨 Try it yourself
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⚡First law & heat transfer calculatorSelect calculation type
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⚙️
Engineering Applications
Four real engineering scenarios where ideal gas laws and the first law are used to size, check or predict equipment behaviour.
🔥 Compressed gas cylinder heating in storage
A sealed industrial gas cylinder is moved from a cold outdoor store into a heated workshop. As its temperature rises, the pressure inside climbs too — engineers must check the new pressure stays below the cylinder's rated limit before it's brought inside. This uses the combined gas law, since the cylinder is rigid (fixed volume) but both P and T change.
P₁V₁/T₁ = P₂V₂/T₂ (T in Kelvin)
🧮 Calculator
Select what to find and fill known values.
📋 Slow, cooled air compression for a pneumatic system
A workshop's air compressor runs slowly enough, and is cooled well enough, that the compressed air stays at roughly constant temperature throughout the stroke — an isothermal process. Engineers need to know the work input the compressor motor must supply to compress the air from its intake volume down to the storage volume.
W = mRT ln(V₂/V₁)
🧮 Calculator
Enter mass, R, T, V1 and V2.
⚡ Energy balance across an engine cylinder
During one stroke of an engine cylinder, heat is released by burning fuel, some of that energy leaves as useful work pushing the piston, and the rest changes the internal energy of the gas trapped inside. Given any two of Q, ΔU and W from test data, an engineer can find the third to check the cylinder's energy balance closes correctly.
Q = ΔU + W +Q = heat in, +W = work out
🧮 Calculator
Enter any two of Q, ΔU, W.
🌡️ Sizing a process water heater
A factory needs to heat a batch of water (or another fluid) from its intake temperature to a target process temperature. Before specifying the heater's power rating, an engineer must first calculate the total heat energy the batch actually requires, using its mass and specific heat capacity.
Q = mcΔT
🧮 Calculator
Enter mass, specific heat and temperature change.
🎮
Thermodynamics Challenge
12 randomised questions — ideal gas laws, first law and heat transfer.
🔥 Quick-fire round
Question 1 / 12Score: 0Streak: 0 🔥
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Practice Questions
12 graded questions — type your answer, check, then reveal full working.
🔥 Always convert temperatures to Kelvin. State equations used.
Q1 · Ideal gas — V₂2 marks
Gas at 200 kPa, 0.3 m³, 30 °C is compressed to 500 kPa at 60 °C. Find V₂.
T₁=303K T₂=333K
V₂ = 200×0.3×333/(303×500) = 19980/151500 = 0.1319 m³
V₂ = 200×0.3×333/(303×500) = 19980/151500 = 0.1319 m³
Q2 · Ideal gas — T₂2 marks
Gas at 300 kPa, 0.5 m³, 100 °C. Constant volume heating raises pressure to 450 kPa. Find T₂ in °C.
T₁=373K, isochoric: T₂=T₁×P₂/P₁=373×450/300=559.5K = 286.5 °C
Q3 · First law3 marks
A closed system receives 80 kJ of heat. The gas expands doing 25 kJ of work. Find ΔU.
Q = ΔU + W → 80 = ΔU + 25 → ΔU = 55 kJ
Q4 · Isobaric3 marks
3 kg of air (cv = 718, cp = 1005 J/kg·K) heated at constant pressure from 290 K to 390 K. Find Q, W and ΔU.
Q = mcpΔT = 3×1005×100 = 301500 J
ΔU = mcvΔT = 3×718×100 = 215400 J
W = 301500−215400 = 86100 J
ΔU = mcvΔT = 3×718×100 = 215400 J
W = 301500−215400 = 86100 J
Q5 · Isochoric2 marks
2 kg of gas (cv = 650 J/kg·K) is heated at constant volume by 78 kJ. Find ΔT.
Isochoric: W = 0, Q = ΔU = mcvΔT
ΔT = 78000/(2×650) = 60 K
ΔT = 78000/(2×650) = 60 K
Q6 · Combined multi-part4 marks
Gas initially at 100 kPa, 0.6 m³, 27 °C. Heated at constant pressure to 177 °C. Find V₂. Then, given cv = 720 J/kg·K, mass = 1.2 kg, find W, ΔU and Q.
T₁=300K T₂=450K
V₂=0.6×450/300 = 0.9 m³
W = PΔV = 100000×0.3 = 30000 J
ΔU = 1.2×720×150 = 129600 J
Q = 129600+30000 = 159600 J
V₂=0.6×450/300 = 0.9 m³
W = PΔV = 100000×0.3 = 30000 J
ΔU = 1.2×720×150 = 129600 J
Q = 129600+30000 = 159600 J
Q7 · Isothermal work3 marks
1.8 kg of air (R = 287 J/kg·K) expands isothermally at 320 K from 0.1 m³ to 0.3 m³. Find the work done.
W = mRT ln(V₂/V₁) = 1.8×287×320×ln(3) = 165312×1.0986 = 181614 J ≈ 181.6 kJ
Q8 · Finding mass2 marks
Gas at 180 kPa, 0.9 m³, 45 °C. R = 260 J/kg·K. Find the mass.
T = 45+273 = 318K
m = PV/(RT) = 180000×0.9/(260×318) = 162000/82680 = 1.96 kg
m = PV/(RT) = 180000×0.9/(260×318) = 162000/82680 = 1.96 kg
Q9 · Enthalpy change2 marks
4 kg of gas is heated at constant pressure through ΔT = 85 K, with cp = 1005 J/kg·K. Find ΔH.
ΔH = mcpΔT = 4×1005×85 = 341700 J = 341.7 kJ
Q10 · Heat transfer2 marks
0.8 kg of water (c = 4186 J/kg·K) is heated from 20 °C to 80 °C. Find the heat required.
Q = mcΔT = 0.8×4186×60 = 200928 J ≈ 200.9 kJ
Q11 · First law — find W2 marks
In a process, Q = −30 kJ (heat rejected) and ΔU = −45 kJ. Find W.
Q = ΔU + W → −30 = −45 + W → W = 15 kJ
Q12 · Combined ideal gas + first law4 marks
Gas at 120 kPa, 0.5 m³, 25 °C is heated at constant pressure to 125 °C. Find V₂. Then, with mass = 1.0 kg and cv = 718 J/kg·K, find W, ΔU and Q.
T₁=298K T₂=398K
V₂ = 0.5×398/298 = 0.6678 m³
W = PΔV = 120000×0.1678 = 20136 J
ΔU = 1.0×718×100 = 71800 J
Q = 71800+20136 = 91936 J
V₂ = 0.5×398/298 = 0.6678 m³
W = PΔV = 120000×0.1678 = 20136 J
ΔU = 1.0×718×100 = 71800 J
Q = 71800+20136 = 91936 J