B. The particulate nature of matter

Particle view

Matter is made of particles in continuous random motion. The state depends mainly on how closely particles are packed, how freely they move and how strongly intermolecular forces hold them together.

Compare the three states

Use temperature carefully

At the same temperature, particles have the same average kinetic energy in the kinetic-theory model. The different states are then distinguished by separation and intermolecular forces, not by claiming that one state automatically has hotter particles.

Common trap

Do not describe a solid as having motionless particles. “Fixed position” means the particles oscillate about equilibrium positions; it does not mean their kinetic energy is zero.

Density

Density is mass per unit volume:

$$\rho=\frac{m}{V}$$

It describes how much mass is concentrated in a given volume.

Calculation method

1. Identify the mass of the object or sample.
2. Use the volume occupied by that same sample.
3. Convert units before substituting.
4. Report density with units such as kg m⁻³ or g cm⁻³.

Useful conversion: 1 g cm⁻³ = 1000 kg m⁻³.

Interpret the result

For equal volumes, the denser sample has the greater mass. For equal masses, the denser sample occupies the smaller volume. A non-uniform object requires its total mass divided by its total external volume unless the question specifies a particular material region.

Worked example from local Question Bank row 36355

A spherical hydrogen nebula has radius $9.0\times10^{15}\,\mathrm{m}$ and number density $1.0\times10^{10}\,\mathrm{atoms\,m^{-3}}$. With $m_H=1.67\times10^{-27}\,\mathrm{kg}$, its mass density is $\rho=(1.0\times10^{10})(1.67\times10^{-27})=1.67\times10^{-17}\,\mathrm{kg\,m^{-3}}$.

$$V=\frac43\pi r^3=3.05\times10^{48}\,\mathrm{m^3}$$
$$m=\rho V=(1.67\times10^{-17})(3.05\times10^{48})=5.1\times10^{31}\,\mathrm{kg}$$

Check the boundary

Do not mix the volume of displaced fluid with the object’s mass, and do not use a material’s density formula with inconsistent units. Density is a scalar, so it has no direction.

Two temperature scales

Celsius is convenient for everyday temperature differences. Kelvin is the absolute thermodynamic scale used when temperature is linked to particle energy or radiation.

Convert between them

$$T_{\mathrm K}=\theta_{\circ\mathrm C}+273.15$$

So 0 °C = 273.15 K and 100 °C = 373.15 K. Kelvin is written without a degree symbol.

Choose the scale

Use Celsius when a question asks for a familiar temperature or a change described on the Celsius scale. Use Kelvin in equations such as $$E_k=\frac32k_BT$$, $$L=\sigma AT^4$$ and $$\lambda_{\max}T=2.9\times10^{-3}\,\mathrm{m\,K}$$.

Common trap

Never substitute a Celsius value directly into a formula that uses absolute temperature. Convert the temperature first.

Same size of change

Because the Celsius and Kelvin scales have the same interval size, a temperature change has the same numerical value in both scales:

$$\Delta T(\mathrm K)=\Delta\theta(^{\circ}\mathrm C)$$

Read a change, not an absolute value

If a sample falls from +10 °C to −10 °C, then

$$\Delta\theta=-10-10=-20^{\circ}\mathrm C$$

The same change is −20 K. The zero point shifts, but the spacing between adjacent temperatures does not.

Common trap

Do not add 273.15 when converting a temperature difference. Add 273.15 only when converting an absolute Celsius temperature to Kelvin.

Absolute temperature and motion

For particles in an ideal gas, Kelvin temperature is proportional to their average random translational kinetic energy:

$$\overline{E_k}=\frac{3}{2}k_BT$$

Here $$k_B$$ is the Boltzmann constant and T must be in kelvin.

What the equation says

If the Kelvin temperature doubles, the average translational kinetic energy doubles. A higher temperature means greater average random kinetic energy, not that every particle has exactly the same kinetic energy.

Scope of the model

The relation describes average random translational motion. It does not include the whole internal energy of a substance, which also contains intermolecular potential energy.

Worked example from local Question Bank row 31728

For helium atoms at $T=320\,\mathrm{K}$ with $m=6.6\times10^{-27}\,\mathrm{kg}$, equate mean translational kinetic energy to $\tfrac12mv^2$:

$$\frac12mv^2=\frac32k_BT\Rightarrow v=\sqrt{\frac{3k_BT}{m}}$$
$$v=\sqrt{\frac{3(1.38\times10^{-23})(320)}{6.6\times10^{-27}}}=1.4\times10^3\,\mathrm{m\,s^{-1}}$$

This is a characteristic speed derived from the average energy, not a claim that every atom has that speed.

Common trap

A Celsius temperature cannot be used in this equation. Convert first; 0 °C corresponds to about 273 K, not zero particle kinetic energy.

Internal energy

The internal energy of a system is the sum of:

• random molecular kinetic energy; and
• intermolecular potential energy associated with forces between particles.

Temperature is only one part

For a fixed phase and amount of substance, raising temperature usually increases the particles’ average random kinetic energy. During a phase change, temperature can stay constant while intermolecular potential energy changes.

Do not equate heat with internal energy

Internal energy is a state property of the system. Thermal energy transfer is energy crossing the system boundary because of a temperature difference.

Temperature difference drives net transfer

When two bodies at different temperatures can exchange energy, the net thermal energy transfer is from the higher-temperature body to the lower-temperature body.

What equilibrium means

Transfer can occur in both directions microscopically, but at thermal equilibrium the opposing transfers balance and there is no net transfer. Equal temperature is the condition for zero net thermal transfer, not necessarily equal internal energy.

Apply the direction rule

First compare temperatures, then draw the net energy arrow. The arrow is independent of which object is heavier or contains more total internal energy.

Common trap

A larger object can contain more internal energy while still receiving energy from a smaller, hotter object. “Hotter” means higher temperature, not “more total energy”.

What changes in a phase change

Melting, freezing, boiling, condensing and other phase changes alter how particles are arranged and how freely they move. Energy transfer changes the balance of intermolecular potential energy.

Why temperature stays constant

During a phase change of a pure substance at constant pressure, the supplied or removed energy changes particle interactions rather than increasing the average random kinetic energy. Therefore the temperature remains constant until the phase change is complete.

Read a heating curve

A sloped section represents temperature changing within one phase. A flat section represents energy transfer during a phase change. The flat section can be long even though the thermometer reading does not change.

Common trap

“Constant temperature” does not mean “no energy transfer”. It means the transfer is not increasing average particle kinetic energy at that stage.

Temperature change within a phase

Use

$$Q=mc\Delta T$$

where c is the specific heat capacity. For a given mass, a larger c means more energy is required for the same temperature rise.

Energy during a phase change

Use

$$Q=mL$$

where L is the specific latent heat of fusion or vaporization. This energy changes particle interactions while the temperature remains constant.

Choose the equation

• temperature changes, no phase change: $$Q=mc\Delta T$$
• phase changes at constant temperature: $$Q=mL$$

If a process contains both stages, calculate the energy for each stage and add the signed or positive magnitudes consistently.

Worked example from local Question Bank row 22716

A cable receives $30\,\mathrm{W}$ and initially warms at $35\,\mathrm{mK\,s^{-1}}=3.5\times10^{-2}\,\mathrm{K\,s^{-1}}$. For copper, $c=390\,\mathrm{J\,kg^{-1}\,K^{-1}}$. Using $P=mc(\Delta T/\Delta t)$,

$$m=\frac{30}{390(3.5\times10^{-2})}=2.2\,\mathrm{kg}$$

The rate form is valid during the initial interval when losses are negligible.

Common trap

Do not use a temperature difference in $$Q=mL$$, and do not use $$Q=mc\Delta T$$ across a phase-change plateau.

Three mechanisms

Thermal energy can be transferred by conduction, convection or thermal radiation. The mechanism depends on what connects the hot and cold regions and on whether bulk matter moves.

Choose the mechanism

Real situations can combine them

A saucepan may conduct energy through its metal, transfer energy through moving water by convection and radiate energy from its surfaces. Identify the dominant mechanism being asked about rather than insisting that only one process exists.

Common trap

Radiation does not require air, and convection is not the same as “hot molecules vibrating faster through a solid”.

Conduction

In conduction, particles in a hotter region have greater average kinetic energy. Through collisions and intermolecular forces, they transfer energy to neighbouring particles in the cooler region.

What moves and what does not

Energy propagates through the material, but the material does not need to undergo bulk flow. In a solid, particles usually vibrate about fixed positions while transferring energy to neighbours.

Compare with other mechanisms

Conduction needs matter and microscopic contact. Convection transfers energy through bulk motion of a fluid. Radiation transfers energy by electromagnetic waves and can cross a vacuum.

Common trap

Conduction is not the same as particles travelling from the hot end to the cold end. The net transfer is through local interactions.

Conduction rate

The rate of thermal energy transfer through a uniform slab is

$$\frac{\Delta Q}{\Delta t}=\frac{kA\Delta T}{\Delta x}$$

where k is the material’s thermal conductivity, A is cross-sectional area, ΔT is the temperature difference and Δx is the transfer distance.

Read the proportionalities

The rate increases with larger k, larger area and larger temperature difference. It decreases when the material is thicker, because Δx is in the denominator.

Calculation checks

Use consistent SI units: area in m², distance in m, temperature difference in K or °C, and k in W m⁻¹ K⁻¹. The rate is measured in watts, because 1 W = 1 J s⁻¹.

Worked example from local Question Bank row 127628

Ice has $k=2.3\,\mathrm{W\,m^{-1}\,K^{-1}}$, thickness $0.019\,\mathrm{m}$ and temperature difference $6\,\mathrm{K}$. Per unit area,

$$\frac{1}{A}\frac{\Delta Q}{\Delta t}=\frac{k\Delta T}{\Delta x}=\frac{(2.3)(6)}{0.019}=7.3\times10^2\,\mathrm{W\,m^{-2}}$$

The result is a heat flux; multiply by area to obtain total power.

Common trap

Use the temperature difference across the slab, not an absolute temperature. A temperature gradient is a change per distance, so do not omit Δx.

Density difference drives convection

When part of a liquid or gas is heated, it generally expands and becomes less dense. The warmer region experiences greater buoyancy and rises while cooler, denser fluid sinks.

A convection current

The rising warm fluid and sinking cool fluid form a circulation. The fluid’s bulk motion carries internal energy from the warmer region to other parts of the fluid.

What the syllabus asks

This objective is qualitative: identify the density change, the direction of motion and how that motion transfers energy. It does not require a detailed fluid-dynamics calculation.

Common trap

Convection occurs in fluids, not in a rigid solid. A solid can conduct energy even though it does not circulate as a bulk fluid.

Black-body emission

A black body is an ideal surface that emits electromagnetic radiation according to its absolute temperature. Its total emitted power, or luminosity, is modelled by

$$L=\sigma AT^4$$

Read the variables

$$A$$ is the emitting surface area, $$T$$ is absolute temperature in kelvin and $$\sigma$$ is the Stefan–Boltzmann constant. The equation gives total power emitted, not the brightness received by a particular observer.

Use proportional reasoning

At fixed area, doubling T multiplies L by $$2^4=16$$. At fixed temperature, doubling the emitting area doubles L. The fourth-power dependence makes temperature especially important.

Worked comparison from local Question Bank row 29005

Treat Mars at $200\,\mathrm{K}$ and Earth at $300\,\mathrm{K}$ as black bodies. For equal emitting area,

$$\frac{L_{Mars}}{L_{Earth}}=\left(\frac{200}{300}\right)^4=0.198\approx0.20$$

Mars emits about one fifth as much power per unit area in this ideal model.

Common trap

Do not use Celsius in the fourth-power term, and do not confuse luminosity with apparent brightness, which also depends on distance.

Apparent brightness

Apparent brightness, $$b$$, describes how much power from a distant source is received per unit area at the observer. It is an observation-dependent quantity.

Why distance matters

Radiation from an approximately point-like source spreads over larger spherical areas as it travels outward. The same emitted power is distributed over more area, so the received power per unit area decreases.

Do not confuse the quantities

Luminosity is the source’s total emitted power. Apparent brightness is what reaches a specified observer per unit area. A source can be intrinsically luminous but appear faint when it is far away.

Common trap

Apparent brightness is not simply the source’s total power. Always ask whether the question concerns emission by the source or reception at a distance.

Brightness–luminosity relation

For isotropic emission without absorption,

$$b=\frac{L}{4\pi d^2}$$

where $$L$$ is total luminosity and $$d$$ is the source–observer distance.

Use the inverse-square pattern

At fixed luminosity, doubling distance makes apparent brightness one quarter as large. At fixed distance, doubling luminosity doubles apparent brightness.

Rearrange before calculating

$$L=4\pi d^2b$$

so

$$d=\sqrt{\frac{L}{4\pi b}}$$

Keep luminosity in watts, distance in metres and brightness in W m⁻².

Worked example from local Question Bank row 30016

Mars is about $1.5$ times farther from the Sun than Earth. If solar intensity at Earth is $1.36\times10^3\,\mathrm{W\,m^{-2}}$,

$$b_{Mars}=b_{Earth}\left(\frac{d_E}{d_M}\right)^2=(1.36\times10^3)\frac{1}{1.5^2}=6.04\times10^2\,\mathrm{W\,m^{-2}}$$

The same solar luminosity is spread over a sphere with larger radius.

Common trap

The factor is $$d^2$$, not $$d$$. Also distinguish a source’s total emitted power from the power received per square metre.

Black-body spectrum

A black body emits a continuous spectrum of wavelengths. The wavelength at which the emitted intensity is greatest is $$\lambda_{\max}$$.

Wien’s law

The peak wavelength and absolute temperature obey

$$\lambda_{\max}T=2.9\times10^{-3}\,\mathrm{m\,K}$$

Therefore

$$T=\frac{2.9\times10^{-3}}{\lambda_{\max}}$$

Interpret the shift

A hotter black body has a smaller peak wavelength, so its spectrum shifts toward shorter wavelengths. A cooler black body peaks at a longer wavelength.

Worked example from local Question Bank row 31596

A star's spectrum peaks at $740\,\mathrm{nm}=740\times10^{-9}\,\mathrm{m}$.

$$T=\frac{2.9\times10^{-3}}{740\times10^{-9}}=3.9\times10^3\,\mathrm{K}\approx4000\,\mathrm{K}$$

The wavelength conversion is essential because Wien's constant is in metres kelvin.

Calculation checks

Use $$\lambda_{\max}$$ in metres and T in kelvin. The law identifies the peak of the spectrum; it does not say that the object emits only that one wavelength.

Treat the planet as a system

Over a long enough time, a planet at steady average temperature receives and emits radiant energy at equal rates:

$$P_{\mathrm{in}}=P_{\mathrm{out}}$$

This is conservation of energy applied to the planet–atmosphere system.

Track every pathway

Incoming solar radiation can be reflected by the planet–atmosphere system, absorbed by the atmosphere or surface, and later emitted as infrared radiation. The energy balance concerns the total absorbed input and total emitted output, not just one arrow in the diagram.

Interpret imbalance

If absorbed power exceeds emitted power, the system’s internal energy and average temperature tend to increase. If emitted power exceeds absorbed power, they tend to decrease. Equal rates mean no net long-term energy accumulation.

Boundary check

A steady temperature does not mean radiation stops. It means the net energy change is zero because input and output balance.

Emissivity

Emissivity, $$\varepsilon$$, compares the power radiated per unit area by a real surface with that radiated per unit area by an ideal black surface at the same absolute temperature:

$$\varepsilon=\frac{P/A}{\sigma T^4}$$

Use the radiation equation

For a surface of area A,

$$P=\varepsilon\sigma AT^4$$

Use T in kelvin. A black body has $$\varepsilon=1$$; real surfaces have emissivity less than or equal to 1 in this model.

What emissivity is not

Emissivity is not the fraction of incoming sunlight reflected; that is albedo. Emissivity concerns emission of thermal radiation by a surface at its temperature.

Worked example from the mapped local textbook

A planet has surface temperature $-63\,^{\circ}\mathrm{C}=210\,\mathrm{K}$, area $1.4\times10^{14}\,\mathrm{m^2}$ and luminosity $1.3\times10^{16}\,\mathrm{W}$.

$$\varepsilon=\frac{P/A}{\sigma T^4}=\frac{(1.3\times10^{16})/(1.4\times10^{14})}{(5.67\times10^{-8})(210)^4}=0.84$$

Emissivity is dimensionless, and the kelvin conversion is required by the fourth-power law.

Calculation boundary

Keep surface emission and atmospheric re-radiation as separate energy-flow terms. Do not subtract reflected solar power from the Stefan–Boltzmann emission formula.

Albedo

Albedo is the fraction of incident radiation scattered or reflected by a macroscopic system:

$$a=\frac{P_{\mathrm{scattered}}}{P_{\mathrm{incident}}}$$

Interpret the value

An albedo near 0 means most incident energy is absorbed; an albedo near 1 means most is reflected. It has no units and is often reported as a decimal or percentage.

Use ratios safely

Calculate each albedo from reflected divided by incident power before comparing surfaces. A snow surface can have a higher albedo than concrete because it reflects a larger fraction of the same incoming intensity.

Worked example from the mapped local textbook

Radiation of intensity $610\,\mathrm{W\,m^{-2}}$ reaches water with albedo $a=0.18$. The absorbed fraction is $1-a=0.82$, so

$$I_{\mathrm{absorbed}}=(1-a)I_{\mathrm{incident}}=(0.82)(610)=5.0\times10^2\,\mathrm{W\,m^{-2}}$$

About $110\,\mathrm{W\,m^{-2}}$ is reflected; the absorbed and reflected intensities add back to the incident intensity, within rounding.

Common trap

Albedo is a ratio of powers, not the reflected power by itself. Do not compare two reflected intensities unless their incident intensities are the same or you have normalized them.

Earth’s albedo is variable

Earth’s average albedo is not a universal fixed property. It changes with the surfaces and clouds that are illuminated and with the angle at which radiation arrives.

Daily variation

Cloud cover changes with weather and the position of the Sun changes during the day. Both alter the fraction of incident radiation reflected by a region.

Latitude and incidence angle

At different latitudes, sunlight arrives at different angles to the surface normal. The effective surface and reflected fraction therefore vary; snow, ice, ocean, land and clouds also contribute different albedos.

Climate link

If snow or ice melts, a darker surface may reflect less and absorb more incoming energy. This is a consequence of changing albedo, not a change in the definition of albedo.

Solar constant

The solar constant, $$S$$, is the solar-radiation intensity received per unit area at the Earth’s orbital distance, with the surface perpendicular to the incoming rays. Its units are W m⁻².

It is not the global mean

The solar constant describes the incident intensity on a surface facing the Sun. A planet’s spherical geometry spreads the intercepted power over a larger total surface, so the planet-wide mean incoming intensity is smaller.

Use it as an input

In an energy-balance problem, S is the incoming solar intensity before accounting for the planet’s projected area, albedo, atmospheric absorption or averaging over the whole sphere.

Common trap

Do not call the solar constant the total solar power intercepted by Earth. It is an intensity: power per unit area.

Why the factor is 1/4

A planet intercepts incoming sunlight over its projected disk, area $$\pi r^2$$. The intercepted power is then averaged over the planet’s whole spherical surface, area $$4\pi r^2$$.

Mean incoming intensity

If the solar constant is S,

$$\overline I=\frac{S\pi r^2}{4\pi r^2}=\frac S4$$

This is the mean intensity before accounting for reflection or atmospheric absorption.

Add albedo when required

If the planetary albedo is a, the globally averaged absorbed intensity is

$$I_{\mathrm{abs}}=(1-a)\frac S4$$

provided the problem’s model treats the planet as a uniform system.

Worked example from local Question Bank row 30218

For $S=1400\,\mathrm{W\,m^{-2}}$ and atmospheric albedo $a=0.30$, the transmitted incident intensity is

$$I=(1-a)S=(0.70)(1400)=980\,\mathrm{W\,m^{-2}}$$

Averaging the intercepted power over the full sphere gives

$$\overline I=\frac{980}{4}=245\,\mathrm{W\,m^{-2}}$$

This result combines reflection with geometry: $(1-a)S/4$.

Common trap

Do not divide S by 4 because sunlight is four times weaker at every point. The factor comes from intercepted disk area divided by total spherical area.

Main gases

The syllabus identifies water vapour (H₂O), carbon dioxide (CO₂), methane (CH₄) and nitrous oxide (N₂O) as the main greenhouse gases.

Natural and human origins

Each of these gases has natural sources and sources affected by human activity. For example, water vapour participates in the natural water cycle, while combustion, agriculture and land-use changes can alter atmospheric concentrations of several greenhouse gases.

Abundance is not the only factor

A gas’s contribution depends on both its atmospheric abundance and how strongly it absorbs infrared radiation. Do not rank gases using concentration alone.

Common trap

O₂, N₂ and Ar are not treated as the main greenhouse gases in this syllabus objective. The question may test recognition of the listed gases, not whether a molecule is simply present in the atmosphere.

Absorb at molecular frequencies

Greenhouse-gas molecules can absorb infrared radiation when its frequency matches an allowed molecular vibration or transition. The molecule moves to a higher energy state.

Re-emit in all directions

The excited molecule soon returns to a lower energy state and emits infrared radiation. The emission is in random directions, so some radiation continues upward and some is directed back toward the surface.

Why Earth’s radiation matters

The cooler Earth emits mainly longer-wavelength infrared radiation. Greenhouse gases absorb part of this outgoing radiation; they are less likely to absorb most of the Sun’s shorter-wavelength incoming radiation.

Model boundary from local Question Bank row 30018

Mars's atmosphere is mainly carbon dioxide, but its pressure is less than $1\%$ of Earth's. The low pressure means far fewer absorbing molecules per unit volume, so much less outgoing infrared radiation is absorbed and re-emitted toward the surface.

The presence of a greenhouse-gas species alone does not determine the size of the effect; the amount of gas also matters.

Common trap

The greenhouse effect is not mainly reflection of incoming sunlight. The key process is absorption and re-emission of outgoing infrared radiation.

Resonance model

A greenhouse-gas molecule can absorb infrared radiation when the radiation frequency matches one of the molecule’s natural vibrational frequencies. The molecule is driven into a larger-amplitude vibration: this is the resonance description.

Molecular energy-level model

The same absorption can be described as a photon whose energy matches the gap between molecular energy levels. The molecule is excited, then returns to a lower level and emits infrared radiation.

Complete mechanism

Earth’s surface emits long-wave infrared radiation. Greenhouse gases absorb part of it and re-emit radiation in random directions; some is directed back toward the surface, raising the surface temperature compared with an atmosphere-free model.

Frequency test from local Question Bank row 31337

If a carbon dioxide vibration has period $5\times10^{-14}\,\mathrm{s}$, its natural frequency is

$$f_0=\frac1T=\frac1{5\times10^{-14}}=2\times10^{13}\,\mathrm{Hz}=20\,\mathrm{THz}$$

Earth's outgoing infrared is around $30\,\mathrm{THz}$, much closer to this molecular frequency than the cited solar infrared near $300\,\mathrm{THz}$. The resonance model therefore predicts stronger absorption of part of Earth's outgoing infrared; the energy-level model describes the same selectivity as matched photon-energy gaps.

Exam boundary

Do not write only “greenhouse gases trap heat”. Name infrared absorption, molecular resonance or energy-level matching, and re-emission in all directions.

Natural greenhouse effect

The natural greenhouse effect is the normal warming produced when atmospheric gases absorb and re-emit some of Earth’s outgoing infrared radiation. It helps keep Earth’s surface suitable for life.

Enhanced greenhouse effect

The enhanced greenhouse effect is the augmentation of that effect due to human activity. Increased concentrations of greenhouse gases can increase absorption and downward re-radiation of infrared energy.

Primary cause in the syllabus

Burning fossil fuels is identified as a primary cause because it increases atmospheric carbon dioxide and contributes to changes in the Earth–atmosphere energy balance. Other human activities can affect greenhouse-gas concentrations too.

Common trap

Do not say that the greenhouse effect itself is caused only by humans. Human activity enhances a naturally occurring effect.

Pressure

Pressure is perpendicular force distributed over area:

$$P=\frac{F_{\perp}}{A}$$

Its SI unit is the pascal, $$1\,\mathrm{Pa}=1\,\mathrm{N\,m^{-2}}$$.

Use the normal component

Only the component of force perpendicular to the surface contributes to pressure on that surface. A tangential component produces shear rather than normal pressure.

Read the proportionality

At fixed force, doubling area halves pressure. At fixed area, doubling the perpendicular force doubles pressure. Pressure is a scalar even though the force producing it has direction.

Worked example from the mapped local textbook

A $51\,\mathrm{kg}$ student stands on one foot with contact area $62\,\mathrm{cm^2}=62\times10^{-4}\,\mathrm{m^2}$. The perpendicular force is the weight, $F=mg=(51)(9.8)=5.0\times10^2\,\mathrm{N}$.

$$P=\frac{F}{A}=\frac{5.0\times10^2}{62\times10^{-4}}=8.1\times10^4\,\mathrm{Pa}$$

The result is large because the same weight acts over a small area.

Common trap

Do not use total force if the force is angled. Resolve it perpendicular to the surface first, and keep area in square metres.

Amount of substance

The amount of substance $$n$$ counts how many groups of $$N_A$$ particles are present:

$$n=\frac{N}{N_A}$$

where N is the number of particles and $$N_A$$ is the Avogadro constant.

Connect mass to moles

If molar mass is M, then

$$n=\frac{m}{M}$$

Use matching units for m and M. Once n is known, the number of particles is $$N=nN_A$$.

Use ratios efficiently

For equal numbers of particles, the samples contain equal amounts in moles even if their masses differ. For isotope or element comparisons, calculate moles before comparing particle counts.

Worked comparison from local Question Bank row 28984

For $40\,\mathrm{g}$ of argon-40, $n_{Ar}=40/40=1.0\,\mathrm{mol}$. For $8\,\mathrm{g}$ of helium-4, $n_{He}=8/4=2.0\,\mathrm{mol}$. Since $N=nN_A$,

$$\frac{N_{Ar}}{N_{He}}=\frac{1.0N_A}{2.0N_A}=\frac12$$

The larger argon mass does not mean more atoms; its molar mass is also larger.

Common trap

Do not confuse N (number of particles) with $$N_A$$ (particles per mole) or n (amount in moles).

Ideal-gas model

An ideal gas is a kinetic-theory model: particles are in constant random motion, occupy negligible volume compared with the container, and interact negligibly except during collisions. Collisions are treated as elastic.

Connect microscopic and macroscopic quantities

Temperature is related to average translational kinetic energy. Pressure comes from momentum transfer when particles collide with the container walls. More energetic or more frequent collisions can increase pressure.

It is an approximation

Real gases have finite-size particles and intermolecular forces. The ideal model is most reliable when particles are far apart and interactions are relatively unimportant.

Common trap

The model does not say every particle has the same speed. It uses a distribution of speeds and averages over many particles.

One relation for a fixed amount of gas

The constant-pressure, constant-volume and constant-temperature observations combine to give

$$\frac{PV}{T}=\text{constant}$$

provided the amount of gas is unchanged.

Compare two states

$$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$$

Temperatures must be absolute. If pressure and temperature stay fixed, volume is fixed; if volume and temperature stay fixed, pressure is fixed.

Use the constraint first

Name what is constant before choosing a simplified law: Boyle-type behaviour uses constant T, Charles-type behaviour uses constant P, and pressure–temperature behaviour uses constant V.

Worked example from the mapped local textbook

A fixed gas changes from $P_1=1.1\times10^5\,\mathrm{Pa}$, $V_1=0.27\,\mathrm{m^3}$ and $T_1=289\,\mathrm{K}$ to $V_2=0.35\,\mathrm{m^3}$ and $T_2=423\,\mathrm{K}$.

$$P_2=\frac{P_1V_1T_2}{T_1V_2}=\frac{(1.1\times10^5)(0.27)(423)}{(289)(0.35)}=1.2\times10^5\,\mathrm{Pa}$$

The amount of gas must remain fixed for this two-state relation.

Common trap

Do not apply $$PV/T=\text{constant}$$ after gas has been added or removed unless the problem explicitly tracks the changed amount of gas.

Molar form

For an ideal gas,

$$PV=nRT$$

Use n in mol, T in kelvin and $$R=8.31\,\mathrm{J\,K^{-1}\,mol^{-1}}$$.

Particle form

Using the number of particles N, the same law is

$$PV=Nk_BT$$

where $$k_B$$ is the Boltzmann constant. The forms are equivalent because $$R=N_Ak_B$$ and $$N=nN_A$$.

Choose the form from the data

Use the molar form when amount of substance is given. Use the particle form when a question asks for the number of molecules or gives microscopic quantities. Rearrange before substituting, for example $$N=PV/(k_BT)$$.

Worked example from local Question Bank row 22708

$N=2.7\times10^{15}$ helium atoms occupy $V=1.3\times10^{-5}\,\mathrm{m^3}$ at $18\,^{\circ}\mathrm{C}=291\,\mathrm{K}$. The particle form matches the data:

$$P=\frac{Nk_BT}{V}=\frac{(2.7\times10^{15})(1.38\times10^{-23})(291)}{1.3\times10^{-5}}=0.83\,\mathrm{Pa}$$

The very small pressure is consistent with the tiny number of atoms compared with a macroscopic mole.

Common trap

Do not use Celsius in either equation, and do not mix n with N. Pressure must be in pascals and volume in cubic metres for SI results.

Pressure from collisions

Gas particles collide with a wall and change momentum. The wall exerts a force on the particles; by Newton’s third law, the particles exert an equal and opposite force on the wall. Pressure is this normal force per unit area.

Kinetic-theory relation

For an ideal gas,

$$P=\frac13\rho\overline{v^2}$$

where ρ is gas density and $$\overline{v^2}$$ is the mean square molecular speed. The speed in this equation is not simply the square of the average speed.

Read the trends

Greater molecular speed increases momentum change per collision and collision rate, increasing pressure. At fixed speed, greater density means more mass per unit volume and therefore greater pressure.

Worked example from the mapped local textbook

Nitrogen at pressure $1.0\times10^5\,\mathrm{Pa}$ has density $1.17\,\mathrm{kg\,m^{-3}}$. Rearranging the kinetic-theory relation gives

$$v_{\mathrm{rms}}=\sqrt{\overline{v^2}}=\sqrt{\frac{3P}{\rho}}=\sqrt{\frac{3(1.0\times10^5)}{1.17}}=5.1\times10^2\,\mathrm{m\,s^{-1}}$$

This is the root-mean-square speed, not the ordinary arithmetic mean speed.

Common trap

A single particle’s collision force is not the total gas force. Pressure is a statistical average over many collisions on the surface.

Internal energy model

For an ideal monatomic gas, internal energy is the total random translational kinetic energy of its particles:

$$U=\frac32Nk_BT=\frac32nRT$$

What is included

The model includes translational kinetic energy only. It neglects intermolecular potential energy and does not include rotational or vibrational molecular energy.

Read the dependence

At fixed amount of gas, U is proportional to T. At fixed temperature, U is proportional to N or n. Particle mass does not appear directly in $$U=\frac32Nk_BT$$.

Worked example from the mapped local textbook

For $1.0\,\mathrm{mol}$ of an ideal monatomic gas at $300\,\mathrm{K}$,

$$U=\frac32nRT=\frac32(1.0)(8.31)(300)=3.7\times10^3\,\mathrm{J}$$

This is the total random translational kinetic energy in the model. At the same temperature, doubling the amount of gas doubles $U$.

Common trap

Equal mass samples of different monatomic gases do not necessarily have equal internal energy: compare their number of particles or moles at the same temperature.

Good approximation

A real gas is closest to the ideal-gas model at relatively low pressure and low density, where particles are far apart and intermolecular forces and particle volume are small compared with the container volume.

Temperature matters

Higher temperature gives particles more kinetic energy, making attractive interactions less important. Low temperature increases the importance of intermolecular attractions and can bring the gas closer to condensation.

Where it fails

At high pressure or high density, particles are crowded: their finite size and interactions matter. Near phase changes, the ideal model is especially unreliable.

Use the full condition

Do not state only “high temperature”. The reliable region is generally high temperature together with low pressure or low density.

A cell as an energy source

A cell transfers energy from a non-electrical source, such as chemical or solar energy, to charge carriers. The energy source establishes an electromotive force (emf) that can drive charge around a circuit.

Meaning of emf

The emf is the energy supplied by the source per unit charge when charge passes through the source. Its unit is the volt, $$1\,\mathrm V=1\,\mathrm{J\,C^{-1}}$$.

Follow the energy

The cell is not a reservoir of charge that gets used up. Charge circulates; the cell supplies energy that is transferred in circuit components such as lamps, motors and resistors.

Common trap

Emf is not the same as current. Emf is energy per charge supplied by the source; current is charge flow per unit time.

Two ways to supply emf

Chemical and solar cells both supply energy per unit charge, but they obtain that energy differently.

Boundary

Compare the energy source and operating conditions, not just the external circuit. A separate battery may store energy produced by a solar cell.

Resistance

Resistance is the ratio of potential difference across a component to current through it:

$$R=\frac{V}{I}$$

Its SI unit is the ohm, Ω.

Conductors, insulators and the origin of resistance

A conductor has mobile charge carriers that can drift when an electric field is applied. In a metal these carriers are electrons. In an insulator, charge carriers are not sufficiently mobile for a sustained current under ordinary conditions. Resistance arises because moving carriers interact with the material's lattice and transfer energy to it.

Interpret the ratio

For a given current, a larger potential difference means larger resistance. Resistance describes how strongly a component opposes charge flow under the stated operating conditions.

Worked example from the mapped local textbook

A component carries $0.78\,\mathrm{A}$ when the potential difference across it is $4.4\,\mathrm{V}$.

$$R=\frac{V}{I}=\frac{4.4}{0.78}=5.6\,\Omega$$

This is its resistance at that operating point; it should not be assumed constant unless the component is ohmic under fixed conditions.

Unit check

From $$R=V/I$$, $$1\,\Omega=1\,\mathrm{V\,A^{-1}}$$. Use the voltage across the component, not the emf of the whole source unless they are equal in the circuit.

Common trap

Resistance is not the same as current. A component can have high resistance and a small current for a given voltage.