Further mechanics and thermal physics: A-level Physics AQA revision notes

3.6.1

Periodic motion

Uniform circular motion 3.6.1.1

Definitions

Radian: the angle subtended at the centre of a circle by an arc equal in length to the radius; one full revolution $= 2 π$ rad.

Key results

Arc length: $s = r θ$ ( $θ$ in radians).
Angular speed: $ω = \frac{Δ θ}{Δ t} = \frac{2 π}{T} = 2 π f$ .
Linear speed: $v = ω r$ .
Centripetal acceleration: $a = \frac{v ^{2}}{r} = ω^{2} r$ .
Centripetal force: $F = \frac{m v ^{2}}{r} = m ω^{2} r$ .

Notes

An object moving in a circle at constant speed is still accelerating, because its velocity direction is constantly changing — this centripetal acceleration points towards the centre of the circle.
Centripetal force is not a new, separate kind of force — it is whatever force (tension, gravity, friction, a normal contact force) happens to be directed towards the centre in a given situation.
Common error: describing a problem as having 'a centripetal force acting outward'. Centripetal force, by definition, always points inward, and there is no separate outward 'centrifugal force' acting on the object in an inertial frame.
A conical pendulum (a mass swinging in a horizontal circle on a string) is a classic worked example combining vertical equilibrium (tension's vertical component balances weight) with horizontal centripetal force (tension's horizontal component provides $m v^{2} / r$ ).

Simple harmonic motion 3.6.1.2

Key results

SHM defining equation: $a = - ω^{2} x$ — acceleration is proportional to displacement and always directed back toward equilibrium.
Displacement (released from rest at $x = + A$ ): $x = A cos (ω t)$ ; velocity: $v = - ω A sin (ω t)$ ; acceleration: $a = - ω^{2} A cos (ω t)$ .
Maximum speed: $v_{m a x} = ω A$ (at $x = 0$ ). Maximum acceleration magnitude: $∣ a ∣_{m a x} = ω^{2} A$ (at $x = \pm A$ ).
Speed at displacement $x$ : $v = \pm ω A^{2} - x^{2}$ .

Notes

Velocity is greatest at equilibrium ( $x = 0$ ) and zero at maximum displacement; acceleration is the reverse — zero at equilibrium and greatest at maximum displacement, following directly from $a = - ω^{2} x$ .
A graph of $a$ against $x$ is a straight line through the origin with gradient $- ω^{2}$ — a direct, exam-friendly way to extract $ω$ (and so $T$ ) from experimental data without needing a full sinusoidal fit.
It is why a mass on a spring and a simple pendulum, despite looking like very different systems, share exactly the same mathematical description near equilibrium.

Mass–spring systems and simple pendulums 3.6.1.3

Key results

Mass-spring system: $T = 2 π \frac{m}{k}$ (linear spring, negligible spring mass and damping).
Simple pendulum: $T = 2 π \frac{l}{g}$ , valid for small angles ( $sin θ \approx θ$ , $θ$ in radians).

Method

Required practical 7: measure the period by timing $N$ complete oscillations (using the same point and direction each time, e.g. passing through equilibrium moving upward) and dividing, $T = \frac{Δ t}{N}$ ; repeat and average to reduce random error.
Plot $T^{2}$ against $m$ (mass-spring) or $T^{2}$ against $l$ (pendulum) — both give straight lines through the origin, with gradients $\frac{4 π ^{2}}{k}$ and $\frac{4 π ^{2}}{g}$ respectively, letting $k$ or $g$ be found from the gradient.

In practiceTwenty oscillations of a 0.25 kg mass on a spring take 14.2 s. Find the spring constant.

$T = \frac{14.2}{20} = 0.710 s$ time many oscillations
$k = \frac{4 π ^{2} m}{T ^{2}} = \frac{39.48 \times 0.25}{0.71 0 ^{2}} = 19.6 N m^{- 1}$ from $T = 2 π m / k$
With several masses, plot $T^{2}$ against $m$ : the gradient is $\frac{4 π ^{2}}{k}$ .

Notes

For the mass-spring system, $m$ must include the hanger and every moving attachment, not just the labelled mass alone.
A pendulum's period is, ideally, independent of the bob's mass — only length and $g$ (and, for larger swings, amplitude) matter, unlike the mass-spring system where mass directly sets the period.
Timing multiple oscillations and dividing (rather than timing just one) reduces the relative effect of reaction-time error on the measured period, the same principle used for other short-time measurements throughout this course.

Energy in simple harmonic motion 3.6.1.3

Key results

Potential energy at displacement $x$ : $E_{p} = \frac{1}{2} k x^{2}$ . Kinetic energy: $E_{k} = \frac{1}{2} k (A^{2} - x^{2})$ .
Total energy: $E_{total} = \frac{1}{2} k A^{2} = \frac{1}{2} m ω^{2} A^{2}$ — constant, provided there is no damping.

Notes

At the extremes ( $x = \pm A$ ): kinetic energy is zero, all energy is potential. At equilibrium ( $x = 0$ ): kinetic energy is at its greatest, potential energy is zero.
The kinetic-potential exchange completes twice per full oscillation — energy returns to being fully potential twice every period $T$ (once at each extreme), so the energy pattern itself repeats every $T /2$ , not every full $T$ .
Because total energy $\propto A^{2}$ , doubling the amplitude quadruples the total energy stored — worth remembering as a proportionality, not just a formula to substitute into.

Free oscillations and damping 3.6.1.3

Definitions

Damping: a resistive effect that removes mechanical energy from an oscillating system over time, transferring it to the surroundings.

Notes

Light damping: amplitude decays gradually over many oscillations, but the system still clearly oscillates.
Critical damping: the system returns to equilibrium in the shortest possible time without oscillating at all — the boundary case between oscillatory and non-oscillatory return.
Heavy damping: the system returns to equilibrium without oscillating, but more slowly than the critically damped case.
All three cases are released from rest from the same starting displacement — damping only changes HOW the system returns to equilibrium (or whether it oscillates on the way), not where it started.

Forced vibrations and resonance 3.6.1.4

Definitions

Natural frequency, $f_{0}$ : the frequency at which a system oscillates freely (undamped), determined by its own physical properties alone.
Resonance: the large-amplitude response that occurs when a periodic driving force's frequency matches (or nearly matches) a system's natural frequency.

Notes

A free, undamped oscillator always oscillates at its own natural frequency $f_{0}$ ; a forced oscillator, once any initial transient dies away, settles into steady motion AT the driving frequency, whatever that is.
The response amplitude is greatest when driving frequency is close to $f_{0}$ , and falls away on either side — increasing damping lowers and broadens this resonant peak, trading a sharper, taller response for a flatter, more forgiving one.
The driving force supplies exactly the energy that damping removes each cycle at resonance, sustaining a constant, maximal amplitude rather than growing without bound — real resonant systems are always at least lightly damped, which is what keeps the response finite.
A standing wave on a string is itself a resonance phenomenon: driving the string at one of its own mode frequencies ( $f_{1}$ , $2 f_{1}$ , …) produces a stable standing-wave pattern, while driving at any other frequency does not build up a clean pattern at all.

Worked examples

Worked example 3.6.1 · 4 marks

A car of mass 1200 kg rounds a flat, unbanked circular bend of radius 45 m.

The maximum coefficient of friction between the tyres and the road is 0.68.

Calculate the maximum speed at which the car can round the bend without skidding, and state what provides the centripetal force in this situation.

Show worked solution

Friction provides the centripetal force here (there is no banking to contribute a horizontal component of the normal force).

At maximum speed, friction is at its limiting value:

μ m g = \frac{m v ^{2}}{r}

so $m$ cancels and $v = μg r$ .

Substituting:

v = 0.68 \times 9.81 \times 45 \approx 300 \approx 17.3 m s^{- 1}

Mark scheme · 4 marks

States that friction provides the centripetal force in this situation 1 mark
Equates limiting friction to the centripetal force requirement, $μ m g = \frac{m v ^{2}}{r}$ , and cancels $m$ 1 mark
Rearranges to $v = μg r$ 1 mark
Calculates $v \approx 17.3 m s^{- 1}$ 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.6.1 · 5 marks

A body performs simple harmonic motion with amplitude 0.12 m and period 2.5 s.

Calculate

(a) the angular frequency,

(b) the speed of the body at a displacement of 0.060 m from equilibrium, and

Show worked solution

(a):

ω = \frac{2 π}{T} = \frac{2 π}{2.5} \approx 2.51 rad s^{- 1}

(b):

v = ω A^{2} - x^{2} = 2.51 \times 0.1 2^{2} - 0.06 0^{2} = 2.51 \times 0.0108 \approx 0.26 m s^{- 1}

E_{k} = \frac{1}{2} k (A^{2} - x^{2})

and:

E_{total} = \frac{1}{2} k A^{2}

the fraction is:

\frac{A ^{2} - x ^{2}}{A ^{2}} = \frac{0.0108}{0.0144} = 0.75

so 75% of the total energy is kinetic at this displacement.

Mark scheme · 5 marks

Calculates $ω = \frac{2 π}{T} \approx 2.51 rad s^{- 1}$ 1 mark
Selects and uses $v = ω A^{2} - x^{2}$ 1 mark
Substitutes correctly and calculates $v \approx 0.26 m s^{- 1}$ 1 mark
States the kinetic fraction as $\frac{A ^{2} - x ^{2}}{A ^{2}}$ (or the equivalent ratio of $E_{k}$ to $E_{total}$ ) 1 mark
Calculates the fraction as $0.75$ (75%) 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.6.1 · 5 marks

$m$ / kg	$T$ / s
0.100	0.444
0.200	0.628
0.300	0.770
0.400	0.889
0.500	0.993

In a Required Practical 7 investigation, a student hangs a series of known masses from a spring and measures the period of oscillation (by timing 20 oscillations and dividing) for each mass.

The table shows the results.

Use the gradient of $T^{2}$ against $m$ to determine the spring constant $k$ .

Show worked solution

$T^{2}$ at $m = 0.100 kg$ :

0.44 4^{2} \approx 0.197 s^{2}

$T^{2}$ at $m = 0.500 kg$ :

0.99 3^{2} \approx 0.986 s^{2}

Gradient:

= \frac{0.986 - 0.197}{0.500 - 0.100} = \frac{0.789}{0.400} \approx 1.97 s^{2} kg^{- 1}

Since:

T^{2} = \frac{4 π ^{2}}{k} m

the gradient equals $\frac{4 π ^{2}}{k}$ , so:

k = \frac{4 π ^{2}}{gradient} = \frac{4 π ^{2}}{1.97} \approx 20.0 N m^{- 1}

Mark scheme · 5 marks

Calculates $T^{2}$ for at least two data points (e.g. $\approx 0.197 s^{2}$ and $\approx 0.986 s^{2}$ ) 1 mark
Calculates the gradient of $T^{2}$ against $m$ using two widely-separated data points 1 mark
States that the gradient equals $\frac{4 π ^{2}}{k}$ 1 mark
Rearranges to $k = \frac{4 π ^{2}}{gradient}$ 1 mark
Calculates $k \approx 20.0 N m^{- 1}$ 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.6.1 · 4 marks

A washing machine drum, containing wet clothes, is mounted on springs inside the casing.

During the spin cycle, as the drum's rotation speed increases steadily from rest, the whole machine is observed to vibrate violently at one particular speed, before smoothing out again as the spin speed rises further.

Explain why this happens, and suggest, with a reason, one design feature that would reduce the severity of the effect.

Show worked solution

As the drum's rotation speed rises, it acts as a periodic driving force on the spring-mounted casing, which has its own natural frequency $f_{0}$ set by its mass and the springs' stiffness.

Resonance occurs when the driving frequency (the rotation speed) passes through a value equal to $f_{0}$ : the amplitude of vibration becomes large because energy is transferred into the oscillating system most effectively at this matching frequency.

As the spin speed continues past $f_{0}$ , the driving frequency moves away again and the amplitude falls back down.

Increasing the damping in the mounting (e.g. a viscous damper) would reduce the severity, because damping removes energy from the system each cycle, lowering and broadening the resonant peak so the maximum amplitude at resonance is smaller.

Mark scheme · 4 marks

Identifies the spinning drum as providing a periodic driving force at the rotation frequency 1 mark
States that the casing/spring system has its own natural frequency $f_{0}$ 1 mark
Explains that resonance (large amplitude) occurs when the driving frequency equals $f_{0}$ , and that amplitude falls again as the driving frequency moves past it 1 mark
Suggests increased damping as a design feature, explaining that it lowers/broadens the resonant peak 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

3.6.2

Thermal physics

Internal energy 3.6.2.1

Definitions

Internal energy, $U$ : the sum of the random distribution of kinetic and potential energies of all the particles in a system.

Key results

First law of thermodynamics (this course's form): $Δ U = Q + W_{on}$ , where $Q > 0$ is energy transferred into the system by heating, and $W_{on} > 0$ is work done on the system (e.g. by compression).

Notes

Heating and doing work are both ways of transferring energy into or out of a system; internal energy itself is a property the system HAS, at any instant — a useful distinction when asked whether a quantity is a transfer or a stored energy.
Temperature relates to the average kinetic energy of the particles in a substance, not the total internal energy — two samples at the same temperature can have very different total internal energy if they contain different numbers of particles.

Heating and changes of state 3.6.2.1

Definitions

Specific heat capacity, $c$ : the energy required to raise the temperature of $1 kg$ of a substance by $1 K$ (equivalently $1° C$ , since a $1 K$ change equals a $1° C$ change).
Specific latent heat, $l$ : the energy required to change the phase of $1 kg$ of a substance with no temperature change — $l_{f}$ for fusion (melting), $l_{v}$ for vaporisation (boiling).

Key results

Temperature change (no phase change): $Q = m c Δ θ$ .
Phase change (no temperature change): $Q = m l$ .

Notes

The flat sections on a heating curve, where energy is supplied continuously but temperature does not rise, are exactly where a phase change is occurring.
During melting or boiling, the energy supplied increases the particles' potential energy (separating them further, breaking intermolecular bonds) while their average kinetic energy — and so temperature — stays unchanged; freezing and condensation release this same energy back to the surroundings.

Measuring specific heat capacity 3.6.2.1

Method

Electrical heating method: heat a solid (or liquid) sample of known mass $m$ electrically, measuring current $I$ , voltage $V$ and heating time $t$ , and the resulting temperature rise $Δ θ$ ; then $c \approx \frac{I V t}{m Δ θ}$ — good thermal insulation and contact reduce heat loss, and a correction is sometimes needed for energy absorbed by the heater and probe themselves.
Continuous-flow method (for a liquid): pass the liquid steadily through an electrical heater at known power $P = I V$ , and once steady inlet/outlet temperatures $T_{in}$ / $T_{out}$ are reached, measure the mass flow rate $\overset{m}{˙}$ (mass collected per unit time); then $P = \overset{m}{˙} c (T_{out} - T_{in})$ .

In practiceA 1.00 kg aluminium block is heated at 12.0 V and 4.00 A for 300 s, and warms from 18.0 °C to 34.0 °C. Find $c$ .

$E = I V t = 4.00 \times 12.0 \times 300 = 14400 J$
$c = \frac{E}{m Δ θ} = \frac{14 400}{1.00 \times 16.0} = 900 J kg^{- 1} K^{- 1}$
Heat lost to the surroundings makes this an overestimate; insulating the block reduces the error.

Notes

The continuous-flow method's key advantage is that, once steady state is reached, heat losses can be largely eliminated from the calculation entirely: repeating the experiment at a second flow rate/power and taking the difference between the two runs cancels out a constant heat-loss term without ever needing to measure it directly.
For an insulated mixing experiment (e.g. a hot solid dropped into cooler water), the underlying principle is simple conservation of energy: energy lost by the hotter material equals energy gained by the cooler material and its container, provided the system is well insulated.

Ideal gases and the gas laws 3.6.2.2

Key results

Ideal gas law (per mole): $p V = n RT$ . Ideal gas law (per molecule): $p V = N k_{B} T$ , with $R = N_{A} k_{B}$ .
Amount of substance: $N = n N_{A}$ ; $n = \frac{m}{M}$ ( $M$ : molar mass, $m$ : gas mass).
Boyle's law: $p V = constant$ (fixed $n, T$ ). Charles's law: $\frac{V}{T} = constant$ (fixed $n, p$ ). Pressure law: $\frac{p}{T} = constant$ (fixed $n, V$ ).
Combined form for a fixed amount of gas: $\frac{p _{1} V _{1}}{T _{1}} = \frac{p _{2} V _{2}}{T _{2}}$ .

Notes

The ideal gas law is a good approximation for real gases at low density, well away from the point where they would condense — where intermolecular forces and the finite volume of the molecules themselves become negligible.
Boyle's, Charles's and the pressure law are each special cases of the ideal gas law obtained by holding one variable fixed — worth recognising as consequences of a single underlying relationship rather than three separate rules to memorise independently.
Always use absolute pressure and kelvin temperature in these equations, never gauge pressure or Celsius directly — $R = N_{A} k_{B}$ links the molar gas constant to the Boltzmann constant via Avogadro's number, a conversion frequently needed when a problem mixes molecule count and moles.

Investigating gas laws 3.6.2.2

Method

Required practical 8 (Boyle's law): seal a fixed amount of air in a graduated syringe fitted with an absolute-pressure sensor; compress slowly and allow the gas to return to room temperature before each reading, so temperature stays constant; record pressure and total trapped volume (including connecting tubing) at each setting, checking for leaks.
Plot $p$ against $\frac{1}{V}$ : a straight line through the origin confirms Boyle's law, with gradient $n RT$ — if gauge pressure was measured, add atmospheric pressure first to get absolute pressure.
Required practical 8 (Charles's law): trap a fixed mass of gas as a column of length $ℓ$ behind a mobile liquid thread in a uniform-bore capillary tube, sealed at one end and open to the atmosphere at the other (so pressure stays constant); immerse in a stirred water bath and vary temperature, waiting for the gas and bath to reach the same temperature before each reading.
Plot $ℓ$ against $T$ (in kelvin): since a uniform bore means $V \propto ℓ$ , the graph is a straight line, $ℓ \propto T$ — extrapolating back gives $ℓ = 0$ at approximately $- 273° C$ , an experimental estimate of absolute zero.

In practiceBoyle: air at 101 kPa occupies 40.0 cm $^{3}$ ; find the pressure at 25.0 cm $^{3}$ . Charles: a gas column is 12.0 cm long at 20 °C; find its length at 80 °C.

$p V = constant : p_{2} = \frac{101 \times 40.0}{25.0} = 162 kPa$ temperature constant
$\frac{ℓ}{T} = constant : ℓ_{2} = 12.0 \times \frac{353}{293} = 14.5 cm$ temperatures in kelvin

Notes

The open, atmosphere-connected end in the Charles's law setup is exactly what keeps pressure constant throughout — the trapped gas column's own weight and the fixed length of liquid thread together give one constant pressure offset above atmospheric, unchanged as temperature varies.
Extrapolating the $ℓ$ against $T$ line back to $ℓ = 0$ (i.e. zero volume) gives a genuine, independent experimental estimate of absolute zero — a real measurement, not just a definition asserted in the ideal gas law.

Molecular kinetic theory 3.6.2.3

Definitions

Kinetic theory assumptions: molecules are point particles (negligible volume compared to the container), collisions are perfectly elastic, and there are no intermolecular forces except during the brief moment of a collision.

Key results

Kinetic theory pressure equation: $p V = \frac{1}{3} N m_{0} \overline{c^{2}}$ ( $m_{0}$ : mass of one molecule, $\overline{c^{2}}$ : mean square speed).

Notes

The derivation reasons about one wall directly: a molecule bouncing elastically off a wall reverses only its velocity component perpendicular to that wall, transferring momentum $2 m_{0} ∣ c_{x} ∣$ per collision, with successive collisions on the SAME wall separated by time $\frac{2 L}{∣ c _{x} ∣}$ — giving a mean force from one molecule of $\frac{m _{0} c _{x}^{2}}{L}$ .
Averaging over many molecules moving in random directions gives $⟨ c_{x}^{2} ⟩ = \frac{1}{3} ⟨ c^{2} ⟩$ (by symmetry, the three perpendicular directions share the mean square speed equally) — this is what turns the one-dimensional wall argument into the full three-dimensional pressure result $p V = \frac{1}{3} N m_{0} \overline{c^{2}}$ .
Kinetic theory derives macroscopic gas pressure entirely from the microscopic, ceaseless collisions of individual molecules with the container walls — a genuine derivation from Newtonian mechanics, not an empirical fit.

Temperature and molecular motion 3.6.2.3

Key results

Mean kinetic energy of a molecule: $\frac{1}{2} m_{0} \overline{c^{2}} = \frac{3}{2} k_{B} T$ .
Root-mean-square speed: $c_{rms} = \overline{c^{2}} = \frac{3 k _{B} T}{m _{0}}$ .
Internal energy of a monatomic ideal gas (intermolecular PE neglected): $U = \frac{3}{2} N k_{B} T = \frac{3}{2} n RT$ .

Notes

Combining the kinetic theory pressure equation with $p V = N k_{B} T$ shows directly that temperature is a measure of mean molecular kinetic energy — derived here, not merely asserted.
At the same temperature, different molecules share the same mean kinetic energy regardless of mass — a heavier molecule ( $4 m_{0}$ ) therefore has a lower rms speed than a lighter one ( $m_{0}$ ) at the same $T$ , specifically half the rms speed for four times the mass, since $c_{rms} \propto 1/ m_{0}$ .
$\overline{c^{2}}$ (mean square speed) is not the same as the mean speed squared — worth stating explicitly, since real gas molecules have a wide spread of individual speeds, not one shared value.
Brownian motion — the observed irregular, random motion of visible particles (e.g. smoke particles in air, viewed under a microscope) — is direct evidence for the particle model of matter: the visible particle is being struck unevenly, moment to moment, by vastly more numerous and much smaller, otherwise unseen molecules.

Worked examples

Worked example 3.6.2 · 5 marks

A 0.200 kg block of aluminium is heated electrically using a 12 V, 3.0 A heater for 100 s.

The block's temperature rises from 20.0°C to 40.0°C.

Assuming no heat is lost to the surroundings, calculate the specific heat capacity of aluminium implied by this data, and state one modification to the method that would improve the accuracy of the result.

Show worked solution

Energy supplied:

Q = I V t = 3.0 \times 12 \times 100 = 3600 J

Temperature rise:

Δ θ = 40.0 - 20.0 = 20.0 K

Using $Q = m c Δ θ$ :

c = \frac{Q}{m Δ θ} = \frac{3600}{0.200 \times 20.0} = 900 J kg^{- 1} K^{- 1}

Improvement: lag/insulate the block (e.g. with cotton wool or foam) to reduce heat loss to the surroundings, since the calculation assumes NO heat is lost, which is never fully true in practice.

Mark scheme · 5 marks

Calculates the energy supplied, $Q = I V t = 3600 J$ 1 mark
States the temperature rise, $Δ θ = 20.0 K$ 1 mark
Rearranges $Q = m c Δ θ$ to $c = \frac{Q}{m Δ θ}$ 1 mark
Calculates $c = 900 J kg^{- 1} K^{- 1}$ 1 mark
States a valid improvement (e.g. insulating the block, or correcting for the heater/thermometer's own heat capacity) with a reason linking it to reducing heat loss 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.6.2 · 6 marks

$V$ / cm³	$p$ / kPa
20	120
24	100
30	80
40	60
48	50

In a Required Practical 8 investigation of Boyle's law, a student traps a fixed mass of air in a syringe connected to an absolute pressure sensor, and records the pressure $p$ for a series of volumes $V$ at a constant room temperature of 293 K, shown in the table.

(a) Show that the results are consistent with Boyle's law.

(b) Use the gradient of $p$ against $\frac{1}{V}$ to calculate the amount of gas, in mol, trapped in the syringe. (:

R = 8.31 J mol^{- 1} K^{- 1}

)

Show worked solution

(a) Boyle's law predicts:

p V = constant

at fixed $n, T$ .

Checking the products:

120 \times 20 = 2400

100 \times 24 = 2400

80 \times 30 = 2400

60 \times 40 = 2400

50 \times 48 = 2400

(all in kPa cm³) — constant across all five readings, confirming Boyle's law.

(b) Converting the extreme points to SI units:

(V = 2.0 \times 1 0^{- 5} m^{3}, p = 1.20 \times 1 0^{5} Pa)

and:

(V = 4.8 \times 1 0^{- 5} m^{3}, p = 5.0 \times 1 0^{4} Pa)

giving:

\frac{1}{V} = 5.00 \times 1 0^{4} m^{- 3}

and $2.08 \times 1 0^{4} m^{- 3}$ .

Gradient:

= \frac{1.20 \times 1 0 ^{5} - 5.0 \times 1 0 ^{4}}{5.00 \times 1 0 ^{4} - 2.08 \times 1 0 ^{4}} \approx 2.40 J

Since:

p = n RT \times \frac{1}{V}

this gradient equals $n RT$ , so:

n = \frac{gradient}{RT} = \frac{2.40}{8.31 \times 293} \approx 9.86 \times 1 0^{- 4} mol

Mark scheme · 6 marks

Calculates the $p V$ product for at least two data points (e.g. both equal 2400 kPa cm³) 1 mark
States that $p V$ is constant across the data, confirming Boyle's law 1 mark
Converts at least one $V$ value to m³ and the corresponding $p$ value to Pa correctly 1 mark
Calculates the gradient of $p$ against $\frac{1}{V}$ using two data points ( $\approx 2.40 J$ ) 1 mark
States that this gradient equals $n RT$ (from rearranging $p V = n RT$ ) 1 mark
Calculates $n \approx 9.86 \times 1 0^{- 4} mol$ 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.6.2 · 5 marks

$θ$ / °C	$ℓ$ / mm
20	117.2
40	125.2
60	133.2
80	141.2
100	149.2

In a Required Practical 8 investigation of Charles's law, a student traps a column of air of length $ℓ$ in a uniform capillary tube sealed at one end, and immerses the tube in a water bath at a series of temperatures $θ$ , obtaining the results in the table.

Determine the gradient of $ℓ$ against $θ$ , and use it together with one data point to extrapolate an estimate of absolute zero, in °C.

Show worked solution

Gradient:

= \frac{149.2 - 117.2}{100 - 20} = \frac{32.0}{80} = 0.400 mm ° C^{- 1}

Using the point $(20, 117.2)$ to find the intercept:

ℓ = 0.400 θ + c \Rightarrow 117.2 = 0.400 \times 20 + c \Rightarrow c = 109.2 mm

Extrapolating to $ℓ = 0$ :

0 = 0.400 θ + 109.2 \Rightarrow θ = - \frac{109.2}{0.400} \approx - 273 ° C

This is a genuine experimental estimate of absolute zero, obtained by extrapolation rather than assumed in advance.

Mark scheme · 5 marks

Calculates the gradient of $ℓ$ against $θ$ using two data points ( $0.400 mm ° C^{- 1}$ ) 1 mark
Uses the gradient and one data point to find the intercept, $c \approx 109.2 mm$ (or an equivalent method to find the line equation) 1 mark
Sets $ℓ = 0$ in the line equation to extrapolate 1 mark
Solves to find $θ \approx - 273 ° C$ 1 mark
States that this is an independent experimental estimate of absolute zero, not an assumed value 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.6.2 · 5 marks

A sample of helium gas, treated as an ideal monatomic gas, contains $2.5 \times 1 0^{23}$ atoms at a temperature of 350 K.

Calculate

(a) the mean kinetic energy of one helium atom, and

(b) the total internal energy of the sample.

k_{B} = 1.38 \times 1 0^{- 23} J K^{- 1}

)

Show worked solution

(a):

\overline{E_{k}} = \frac{3}{2} k_{B} T = 1.5 \times 1.38 \times 1 0^{- 23} \times 350 \approx 7.25 \times 1 0^{- 21} J

(b):

U = N \times \overline{E_{k}} = 2.5 \times 1 0^{23} \times 7.25 \times 1 0^{- 21} \approx 1.81 \times 1 0^{3} J

(c) The kinetic theory model assumes there are no intermolecular forces except during the brief moment of a collision, so an ideal gas has no intermolecular potential energy — its internal energy is therefore purely the kinetic energy of its molecules.

Mark scheme · 5 marks

Selects and uses $\overline{E_{k}} = \frac{3}{2} k_{B} T$ 1 mark
Calculates $\overline{E_{k}} \approx 7.25 \times 1 0^{- 21} J$ 1 mark
States $U = N \times \overline{E_{k}}$ (or equivalently $U = \frac{3}{2} N k_{B} T$ ) 1 mark
Calculates $U \approx 1.81 \times 1 0^{3} J$ 1 mark
States that an ideal gas has no intermolecular potential energy (no intermolecular forces except during collisions), so its internal energy is purely kinetic 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.