5.2 Internal energy and the ideal gas
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A2 content, including Core Practical 14. Heating calculations with \(c\) and \(L\) are in 5.1; why fusion needs very high temperatures is in 5.3.
What this page covers
7 points
- Internal energy as the random kinetic and potential energies of the molecules
- Absolute zero, the kelvin scale and the link between temperature and mean molecular kinetic energy
- The ideal gas equation \(pV = NkT\), with \(N\) the number of molecules
- Boyle's law, the pressure law and Charles's law as special cases, and their graphs
- Core Practical 14: how the pressure of a fixed mass of gas depends on its volume at constant temperature
- The kinetic theory assumptions and the derivation of \(\tfrac{1}{2}m\langle c^2\rangle = \tfrac{3}{2}kT\)
- Root-mean-square speed and how it depends on temperature and molecular mass
In one minute
6 points
- A gas obeys \(pV = NkT\) from outside and is a huge number of randomly moving molecules from inside, and putting the two views together shows that absolute temperature measures the mean kinetic energy of the molecules.
- Internal energy is the sum of the randomly distributed kinetic and potential energies of all the molecules; in an ideal gas it is all kinetic.
- Always use kelvin: \(T/\text{K} = \theta/{}^{\circ}\text{C} + 273\). At absolute zero the molecules would have their minimum possible energy.
- \(\tfrac{1}{2}m\langle c^2\rangle = \tfrac{3}{2}kT\): mean kinetic energy depends on \(T\) only, not on the type of gas.
- Fix two of \(N, T, p, V\) and you get a gas law: \(p \propto 1/V\) (Boyle), \(p \propto T\) (pressure law), \(V \propto T\) (Charles).
- Kinetic theory gives \(pV = \tfrac{1}{3}Nm\langle c^2\rangle\); comparing with \(pV = NkT\) gives \(c_{\text{rms}} = \sqrt{3kT/m}\).
Contents
12 sections
- What is internal energy?
- What is absolute zero, and how is temperature linked to molecular kinetic energy?
- What is the ideal gas equation?
- How do the three gas laws follow from pV = NkT?
- How do you investigate how the pressure of a gas depends on its volume?
- How does kinetic theory lead to ½m⟨c²⟩ = 3/2 kT?
- Exam tips for 5.2
- Common misconceptions
- How this topic connects
- Key equations
- Check your understanding
- Specification checklist
What is internal energy?
The kinetic part comes from the random motion of the molecules: vibration in a solid, vibration and sliding past each other in a liquid, free flight in a gas. The potential part comes from the forces between molecules; pulling molecules further apart against attractive forces increases it.
"Randomly distributed" matters: at any instant some molecules are fast and some slow, and collisions keep swapping energy between them. Only the total and the average are fixed by the state of the body.
- Heating during a change of state (melting, boiling) increases the potential energy of the molecules at constant temperature; heating without a change of state increases mainly their kinetic energy, so the temperature rises.
- An ideal gas has no forces between molecules except during collisions, so no potential energy: its internal energy is the total random kinetic energy of its molecules.
- The kinetic energy of the body moving as a whole, or its gravitational potential energy, is not internal energy; only random, molecular-scale energies count.
What is absolute zero, and how is temperature linked to molecular kinetic energy?
Measure the pressure of a fixed volume of gas at several temperatures and plot \(p\) against \(\theta\) in °C: the graph is a straight line. Extend it backwards and it reaches \(p = 0\) at −273 °C, for every gas and every starting pressure.
The kelvin (absolute) scale starts at absolute zero and has degrees the same size as Celsius degrees:
\[T/\text{K} = \theta/{}^{\circ}\text{C} + 273\]This is not in the formulae list (learn it). A temperature difference is the same in K and °C; a temperature ratio is only meaningful in K.
- The mean kinetic energy of the molecules is proportional to the absolute temperature: double \(T\) in kelvin and you double it (the kinetic theory section shows why).
- Only kelvin works: 10 °C to 20 °C is 283 K to 293 K, a 3.5% rise, not a doubling.
- At the same temperature, molecules of different gases have the same mean kinetic energy; heavier ones simply move more slowly.
A sample of gas is heated from 27 °C to 327 °C. (a) Calculate the ratio of the mean kinetic energy of the molecules at 327 °C to that at 27 °C. (b) Calculate the Celsius temperature at which the mean kinetic energy would be half its value at 27 °C.
(a) Convert: 27 + 273 = 300 K and 327 + 273 = 600 K. Mean kinetic energy ∝ \(T\), so the ratio is 600 ÷ 300: 2.0 (not 327 ÷ 27 = 12.1).
(b) Half the mean kinetic energy needs half the kelvin temperature: \(T = 300 \div 2 = 150\) K, so \(\theta = 150 - 273\): −123 °C.
What is the ideal gas equation?
\[pV = NkT\]\(p\) is pressure (Pa), \(V\) volume (m³), \(N\) the number of molecules (no unit), \(k\) the Boltzmann constant, 1.38 × 10⁻²³ J K⁻¹, and \(T\) the absolute temperature (K). The equation is in the formulae list and \(k\) is on the data sheet. It holds for an ideal gas; real gases behave like this at low pressures and at temperatures well above those at which they liquefy.
- \(N\) is a number of molecules, typically 10²⁰ to 10²⁷, not an amount in moles. If you are given the Avogadro constant \(N_A\) (6.02 × 10²³ mol⁻¹) and an amount \(n\) in moles, use \(N = nN_A\).
- Units check: Pa m³ = N m⁻² × m³ = N m = J, and \(kT\) is J K⁻¹ × K = J. So \(pV\) is an energy.
- Convert first: cm³ → m³ (× 10⁻⁶), litres → m³ (× 10⁻³), kPa → Pa, °C → K.
- Use the absolute pressure of the gas, not a tyre-gauge reading, which shows pressure above atmospheric.
A classroom measures 8.0 m × 6.0 m × 3.0 m. The air is at 1.01 × 10⁵ Pa and 22 °C; treat it as an ideal gas. (a) Calculate the number of molecules in the room. (b) The average mass of an air molecule is 29 u. Calculate the mass of air in the room.
(a) \(V = 8.0 \times 6.0 \times 3.0 = 144\) m³ and \(T = 22 + 273 = 295\) K. \(N = \dfrac{pV}{kT} = \dfrac{1.01 \times 10^{5} \times 144}{1.38 \times 10^{-23} \times 295}\): \(N = 3.57 \times 10^{27}\).
(b) Mass = number × mass of one molecule (1 u = 1.66 × 10⁻²⁷ kg): \(3.57 \times 10^{27} \times 29 \times 1.66 \times 10^{-27}\): 172 kg.
Given \(N_A = 6.02 \times 10^{23}\) mol⁻¹, the amount would be \(N/N_A = 5930\) mol.
How do the three gas laws follow from pV = NkT?
For a fixed mass of gas \(N\) is constant, so \(pV/T = Nk\) is constant. Hold one more quantity constant and a simpler law drops out.
| Law | Held constant | Relationship | Straight-line graph |
|---|---|---|---|
| Boyle's law | N and T | \(p \propto 1/V\), so \(p_1V_1 = p_2V_2\) | p against 1/V, through the origin |
| Pressure law | N and V | \(p \propto T\), so \(p_1/T_1 = p_2/T_2\) | p against T (in K), through the origin |
| Charles's law | N and p | \(V \propto T\), so \(V_1/T_1 = V_2/T_2\) | V against T (in K), through the origin |
A p–V curve cannot show proportionality by eye, so replot as \(p\) against \(1/V\). Since \(p = NkT \times (1/V)\), the graph is a straight line through the origin with gradient \(NkT\); a hotter isotherm is a steeper line.
A car tyre contains 0.025 m³ of air at an absolute pressure of 3.10 × 10⁵ Pa and 7 °C. After a long drive the air is at 37 °C; the volume does not change. (a) Calculate the new pressure. (b) The driver lets air out until the pressure is back to 3.10 × 10⁵ Pa, still at 37 °C. Calculate the fraction of the molecules let out, and the number this is.
(a) Fixed \(N\) and \(V\), so \(p/T\) is constant, in kelvin: \(p_2 = 3.10 \times 10^{5} \times \dfrac{310}{280}\): \(3.43 \times 10^{5}\) Pa.
(b) At fixed \(V\) and \(T\), \(N \propto p\): fraction remaining \(= 3.10 \div 3.432 = 0.903\), so 9.7% let out. Original number: \(N = \dfrac{3.10 \times 10^{5} \times 0.025}{1.38 \times 10^{-23} \times 280} = 2.01 \times 10^{24}\), so let out \(0.0968 \times 2.01 \times 10^{24}\): \(1.9 \times 10^{23}\) molecules.
Using °C in a gas-law ratio: 3.10 × 10⁵ × 37 ÷ 7 would give a pressure more than five times larger. Only the kelvin scale starts where pressure and molecular kinetic energy would be zero.
How do you investigate how the pressure of a gas depends on its volume?
This is Core Practical 14. A fixed mass of air is trapped and compressed in steps at constant temperature, and its pressure recorded at each volume.
- Measure the tube's internal diameter \(d\) (or use the value given), so \(V = \tfrac{\pi d^2}{4}\,l\). The area is constant, so \(V \propto l\) and the length alone tests the relationship.
- Record \(l\) and \(p\) at atmospheric pressure, then pump to raise the pressure in steps, closing the valve each time.
- Compressing a gas warms it: wait about a minute after each change so the air returns to room temperature, then read the length (eye level with the oil, to avoid parallax) and the pressure.
- Take 6–8 readings across the gauge's range; repeat while releasing the pressure and average.
- Plot \(p\) against \(1/l\) (or \(1/V\)): a straight line through the origin shows \(p \propto 1/V\) at constant temperature.
- Safety: eye protection, and stay within the working pressure of gauge and tube, since a glass tube under pressure can shatter.
The results plotted above were taken at 20 °C in a tube of internal diameter 1.20 cm.
| l / cm | 40.0 | 34.0 | 29.0 | 25.0 | 22.0 | 19.5 |
|---|---|---|---|---|---|---|
| p / kPa | 100 | 120 | 140 | 160 | 180 | 205 |
| pl / kPa m | 40.0 | 40.8 | 40.6 | 40.0 | 39.6 | 40.0 |
(a) Explain how the data support Boyle's law. (b) Use the gradient, 39.1 kPa m, to calculate the number of air molecules trapped. (c) The diameter is known to ± 0.02 cm, the gradient to ± 3.0% and the temperature to ± 1 K. Calculate the percentage uncertainty in \(N\).
(a) The products \(pl\) all lie between 39.6 and 40.8 kPa m, constant within about 2%, so \(p \propto 1/l \propto 1/V\) (and the graph is a straight line through the origin).
(b) Gradient \(= pl = pV/A = NkT/A\), so \(N = \text{gradient} \times A/kT\). \(A = \pi (0.60 \times 10^{-2})^2 = 1.131 \times 10^{-4}\) m², \(T = 293\) K: \(N = \dfrac{39.1 \times 10^{3} \times 1.131 \times 10^{-4}}{1.38 \times 10^{-23} \times 293}\): \(1.09 \times 10^{21}\).
(c) The diameter is squared, so its percentage counts twice: 0.02 ÷ 1.20 = 1.7%, so 3.3% for \(A\); \(T\): 1 ÷ 293 = 0.34%. Total \(3.3 + 3.0 + 0.34\): 6.7%, \(N = (1.09 \pm 0.07) \times 10^{21}\).
p against V is a curve. Plot p against 1/V and look for a straight line through the origin, or show that pV is the same for every reading.
How does kinetic theory lead to ½m⟨c²⟩ = 3/2 kT?
The assumptions of the kinetic model
- A gas is a very large number of identical molecules in random motion.
- The volume of the molecules is negligible compared with the volume of the container.
- There are no forces between molecules except during collisions.
- Collisions with each other and with the walls are elastic (no loss of kinetic energy).
- The duration of a collision is negligible compared with the time between collisions.
The derivation, step by step
- Change of momentum of one molecule hitting the wall: \(mu - (-mu) = 2mu\), given to the wall.
- It crosses the box and back, distance \(2L\), before hitting the same wall again: time between hits \(= 2L/u\).
- Average force on the wall = rate of change of momentum: \(2mu \div (2L/u) = mu^2/L\).
- For \(N\) molecules use the mean of \(u^2\): force \(= Nm\langle u^2\rangle/L\), and pressure = force ÷ area \(L^2\) \(= Nm\langle u^2\rangle/V\).
- The motion is random, so \(\langle u^2\rangle = \langle v^2\rangle = \langle w^2\rangle\), and since \(c^2 = u^2 + v^2 + w^2\), \(\langle u^2\rangle = \tfrac{1}{3}\langle c^2\rangle\).
- So \(pV = \tfrac{1}{3}Nm\langle c^2\rangle\), or \(p = \tfrac{1}{3}\rho\langle c^2\rangle\) since the density \(\rho = Nm/V\).
- Compare with \(pV = NkT\): \(\tfrac{1}{3}Nm\langle c^2\rangle = NkT\), so \(m\langle c^2\rangle = 3kT\).
\(m\) is the mass of one molecule (kg), \(\langle c^2\rangle\) the mean square speed (m² s⁻²). The equation is in the formulae list, but the derivation is not supplied: learn its steps. Consequences: mean kinetic energy ∝ \(T\); the internal energy of an ideal gas is \(N \times \tfrac{3}{2}kT\); and \(c_{\text{rms}} = \sqrt{\langle c^2\rangle} = \sqrt{3kT/m}\).
Root-mean-square speed
\(\langle c^2\rangle\) is the mean of the squares of the speeds; its square root is the r.m.s. speed. Order matters: square each speed, take the mean, then the square root. The r.m.s. speed is always at least as big as the ordinary mean speed. \(c_{\text{rms}} \propto \sqrt{T}\) for a given gas and \(\propto 1/\sqrt{m}\) at a given temperature, so hydrogen and helium move much faster than nitrogen at the same temperature.
Five molecules have speeds of 200, 400, 500, 600 and 800 m s⁻¹. Calculate their mean speed and their r.m.s. speed.
Mean: \((200 + 400 + 500 + 600 + 800) \div 5\): 500 m s⁻¹.
\(\langle c^2\rangle = (200^2 + 400^2 + 500^2 + 600^2 + 800^2) \div 5 = 290\,000\) m² s⁻², so \(c_{\text{rms}} = \sqrt{290\,000}\): 539 m s⁻¹, larger than the mean because squaring weights the fast molecules more.
A nitrogen molecule has a mass of 28 u (1 u = 1.66 × 10⁻²⁷ kg). The air is at 20 °C. (a) Calculate the mean kinetic energy of a nitrogen molecule. (b) Calculate the r.m.s. speed. (c) Calculate the Celsius temperature at which the r.m.s. speed would be twice as large.
(a) \(\tfrac{3}{2}kT = 1.5 \times 1.38 \times 10^{-23} \times 293\): \(6.07 \times 10^{-21}\) J.
(b) \(m = 28 \times 1.66 \times 10^{-27} = 4.65 \times 10^{-26}\) kg; \(c_{\text{rms}} = \sqrt{\dfrac{3 \times 1.38 \times 10^{-23} \times 293}{4.65 \times 10^{-26}}}\): 511 m s⁻¹.
(c) \(c_{\text{rms}} \propto \sqrt{T}\), so doubling it needs four times the kelvin temperature: \(T = 4 \times 293 = 1172\) K: 899 °C.
A cylinder contains helium at 1.2 × 10⁵ Pa and a density of 0.20 kg m⁻³. The mass of a helium atom is 4.0 u. (a) Show that the r.m.s. speed is about 1.3 × 10³ m s⁻¹. (b) Calculate the temperature of the helium. (c) Nitrogen (28 u) at the same temperature is in a second cylinder. Calculate its r.m.s. speed.
(a) \(p = \tfrac{1}{3}\rho\langle c^2\rangle\), so \(c_{\text{rms}} = \sqrt{3p/\rho} = \sqrt{3 \times 1.2 \times 10^{5} \div 0.20} = \sqrt{1.80 \times 10^{6}}\): 1342 m s⁻¹ ≈ 1.3 × 10³ m s⁻¹.
(b) \(T = \dfrac{m\langle c^2\rangle}{3k} = \dfrac{4.0 \times 1.66 \times 10^{-27} \times 1.80 \times 10^{6}}{3 \times 1.38 \times 10^{-23}}\): 289 K.
(c) Same \(T\), same mean kinetic energy, so \(c_{\text{rms}} \propto 1/\sqrt{m}\): \(1342 \times \sqrt{4.0/28} = 1342 \times 0.378\): 507 m s⁻¹.
Check: \(p = \rho kT/m = 0.20 \times 1.38 \times 10^{-23} \times 289 \div 6.64 \times 10^{-27} = 1.20 \times 10^{5}\) Pa, consistent with the data.
Using the molar mass (4.0 or 0.0040) for \(m\). In \(\tfrac{1}{2}m\langle c^2\rangle = \tfrac{3}{2}kT\) and \(\sqrt{3kT/m}\), \(m\) is the mass of one molecule in kg: 4.0 u = 4.0 × 1.66 × 10⁻²⁷ kg.
Exam tips for 5.2
Every gas-law ratio and every use of \(pV = NkT\) or \(\tfrac{1}{2}m\langle c^2\rangle = \tfrac{3}{2}kT\) needs \(T\) in kelvin. Write the conversion as a separate line so it earns credit even if a later step slips.
Given moles and \(N_A\), multiply first. Given a mass of gas, divide by the mass of one molecule.
Two gases at the same temperature have the same mean molecular kinetic energy; their r.m.s. speeds are in the ratio \(\sqrt{m_2/m_1}\). Use ratios rather than recalculating.
Change of momentum \(2mu\) → time between collisions \(2L/u\) → force \(mu^2/L\) → sum over \(N\) molecules and divide by area → \(\langle u^2\rangle = \tfrac{1}{3}\langle c^2\rangle\) → compare with \(pV = NkT\). State which assumption you use at each step.
Pressure rises when molecules hit the walls more often or with more momentum. Smaller volume at constant \(T\): same speeds, more frequent collisions. Higher \(T\) at constant volume: faster molecules, so more frequent collisions and a larger momentum change per collision.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| Doubling the temperature from 20 °C to 40 °C doubles the mean kinetic energy. | The Celsius zero is not where molecular kinetic energy is zero. | Mean kinetic energy ∝ T in kelvin: 293 K → 313 K is a 7% rise. |
| Temperature measures the total internal energy of a body. | A large cool lake holds far more internal energy than a hot spark. | Temperature measures the mean kinetic energy per molecule. |
| When ice melts at 0 °C its internal energy does not change. | Energy is supplied while the temperature stays constant. | The molecular potential energy increases; the kinetic energy stays the same. |
| Heavier molecules have more kinetic energy at the same temperature. | Mean kinetic energy is \(\tfrac{3}{2}kT\), which contains no mass. | Same mean kinetic energy; heavier molecules have a lower r.m.s. speed. |
| Gas pressure is caused by molecules colliding with each other. | Collisions between molecules exert no net force on the container. | Pressure is the rate of change of momentum of molecules hitting the walls, per unit area. |
| The r.m.s. speed is just the average speed. | Squaring before averaging gives fast molecules more weight. | \(c_{\text{rms}} = \sqrt{\langle c^2\rangle}\) is greater than or equal to the mean speed. |
How this topic connects
5.1 Specific heat capacity and specific latent heat
Energy supplied to a body raises either the molecular kinetic energy (temperature rises, \(\Delta E = mc\Delta\theta\)) or the molecular potential energy (change of state, \(\Delta E = L\Delta m\)): the two parts of internal energy.
Momentum (Unit 1) and 5.3 Binding energy, fission and fusion
The derivation is Newton's second law as rate of change of momentum, applied to molecules bouncing elastically off a wall. Nuclei fuse only if they approach very closely, which needs a large kinetic energy; since mean kinetic energy ∝ \(T\), fusion needs temperatures of millions of kelvin.
Practical skills (Unit 6)
Core Practical 14 is a common context for planning, graph plotting and combining percentage uncertainties in the practical skills paper.
Key equations
| Equation | Use | In the formulae list? |
|---|---|---|
| \(T/\text{K} = \theta/{}^{\circ}\text{C} + 273\) | convert to the absolute temperature scale | no — learn it |
| \(pV = NkT\) | any calculation on an ideal gas; N = number of molecules (p in Pa, V in m³, k = 1.38 × 10⁻²³ J K⁻¹) | yes |
| \(\dfrac{p_1V_1}{T_1} = \dfrac{p_2V_2}{T_2}\) | a fixed mass of gas changing state (Boyle, pressure law, Charles as special cases) | no — derive it from pV = NkT with N constant |
| \(p = F/A\) | pressure on a wall in the kinetic theory derivation | no — learn it |
| \(pV = \tfrac{1}{3}Nm\langle c^2\rangle\) and \(p = \tfrac{1}{3}\rho\langle c^2\rangle\) | pressure from molecular motion | no — derive it |
| \(\tfrac{1}{2}m\langle c^2\rangle = \tfrac{3}{2}kT\) | mean kinetic energy of a molecule at temperature T | yes |
| \(c_{\text{rms}} = \sqrt{3kT/m}\) | r.m.s. speed of the molecules of a gas | no — derive it from ½m⟨c²⟩ = 3/2kT |
Check your understanding
1. Water at 100 °C turns to steam at 100 °C. State what happens to the mean kinetic energy and the potential energy of its molecules, and to its internal energy.
Mean kinetic energy unchanged (same temperature). Potential energy increases, because the molecules are separated against the attractive forces between them. So the internal energy increases.
2. Calculate the ratio of the mean kinetic energy of air molecules at 20 °C to that at −40 °C.
20 °C = 293 K and −40 °C = 233 K, so the ratio is 293 ÷ 233 = 1.26.
3. A fixed mass of gas is compressed slowly to half its volume at constant temperature. State and explain what happens to its pressure.
The pressure doubles (Boyle's law). The molecules have the same speeds, but in half the volume they hit each unit area of the walls twice as often, so the rate of change of momentum per unit area doubles.
4. A balloon holds 2.0 litres of gas at 300 K. It is heated to 450 K at constant pressure. Calculate the new volume.
\(V \propto T\), so \(V = 2.0 \times 450 \div 300 = 3.0\) litres.
5. By what factor does the r.m.s. speed of the molecules of a gas change when its temperature rises from 300 K to 1200 K?
\(c_{\text{rms}} \propto \sqrt{T}\) and \(T\) is multiplied by 4, so the r.m.s. speed doubles (factor 2).
6. Calculate the total internal energy of the 1.09 × 10²¹ air molecules in Worked example 4, treating the air as an ideal gas at 293 K.
\(U = N \times \tfrac{3}{2}kT = 1.09 \times 10^{21} \times 1.5 \times 1.38 \times 10^{-23} \times 293 = 6.61\) J.
Specification checklist
Wording is ours; numbers refer to the specification points.
- (128) I can describe internal energy as the sum of the randomly distributed kinetic and potential energies of the molecules, and say which part changes during heating and during a change of state.
- (129) I can explain what absolute zero is, convert between °C and K, and use the idea that the mean kinetic energy of the molecules is proportional to the absolute temperature.
- (130) I can use \(pV = NkT\), with \(N\) as the number of molecules (converting from moles with \(N_A\) when it is given).
- (130) I can show that Boyle's law, the pressure law and Charles's law are special cases of \(pV = NkT\), and sketch and interpret their graphs.
- (131) I can describe how to investigate the relationship between the pressure and volume of a fixed mass of gas at constant temperature, analyse the results with a graph of p against 1/V, and combine the percentage uncertainties.
- (132) I can state the assumptions of the kinetic model, derive \(pV = \tfrac{1}{3}Nm\langle c^2\rangle\) and then \(\tfrac{1}{2}m\langle c^2\rangle = \tfrac{3}{2}kT\), and calculate r.m.s. speeds.