Flashcards · IB Diploma Programme Physics · Topic 2

B. The particulate nature of matter flashcards

48 cards on Topic 2, subtopics B.1 to B.5. Pick a subtopic or study them all.

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B.1 Thermal energy transfers

Internal energy: The sum, over all molecules, of the intermolecular potential energy and the random kinetic energy.
Temperature and average kinetic energy: Temperature (in kelvin) measures the average random kinetic energy per particle: Ek = (3/2)kB T. It is not a measure of total internal energy.
Average kinetic energy of a gas particle at 620 K? (kB = 1.38 × 10⁻²³ J K⁻¹): Ek = 1.5 × 1.38 × 10⁻²³ × 620 = 1.28 × 10⁻²⁰ J
Specific heat capacity and specific latent heat equations: Q = mcΔT (temperature change, no change of phase) Q = mL (change of phase at constant temperature)
What happens to the particles during a phase change?: The temperature and average kinetic energy stay constant; the energy supplied increases the potential energy as the particles separate.
Evaporation vs boiling: Evaporation: at the surface, at any temperature. Boiling: throughout the liquid, at the boiling point.
Rate of thermal conduction: ΔQ/Δt = kA ΔT/Δx, where k is the thermal conductivity (W m⁻¹ K⁻¹), A the area and ΔT/Δx the temperature gradient.
Convection: Heated fluid expands and becomes less dense, so it rises while cooler, denser fluid sinks to replace it, forming a convection current. It needs a fluid.
Black body: An ideal body that absorbs all radiation incident on it and emits the maximum power for its temperature.
Stefan–Boltzmann law, and the effect of doubling T: L = σAT⁴ (σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴, T in kelvin). Doubling T multiplies L by 16.
Apparent brightness and Wien’s law: b = L / (4πd²), in W m⁻². λmax T = 2.9 × 10⁻³ m K: a hotter body peaks at a shorter wavelength.

B.2 Greenhouse effect

Emissivity: The power radiated per unit area by a surface divided by that of a black body at the same temperature (σT⁴). No unit. P = eσAT⁴.
Albedo: Total scattered (reflected) power divided by total incident power. No unit, between 0 and 1. Fraction absorbed = 1 − α.
Solar constant, and why the mean incoming intensity is S/4: S: the intensity of the Sun’s radiation at the planet’s distance, on a surface facing the Sun at the top of the atmosphere. The planet intercepts S × πR² but has surface area 4πR².
Equilibrium temperature of a planet with no atmosphere: Absorbed = emitted: S(1 − α)/4 = eσT⁴, so T = [S(1 − α)/(4eσ)]^(1/4).
S = 1800 W m⁻², α = 0.50, e = 1. Equilibrium temperature?: Absorbed = 1800 × 0.50 ÷ 4 = 225 W m⁻². T⁴ = 225 ÷ 5.67 × 10⁻⁸ = 3.97 × 10⁹, so T = 251 K.
Four main greenhouse gases: Carbon dioxide (CO₂), methane (CH₄), nitrous oxide (N₂O) and water vapour (H₂O).
Why do greenhouse gases absorb infrared but not sunlight?: Their molecules have vibrational energy gaps (or natural frequencies) matching infrared photons. Sunlight peaks near 500 nm and passes through; the surface’s radiation peaks near 10 μm and is absorbed.
How do greenhouse gases warm the surface?: They absorb infrared from the surface and re-emit it in all directions, so part returns to the surface. The surface warms until the power escaping to space equals the absorbed solar power.
Enhanced greenhouse effect: The extra warming caused by greenhouse gases added by human activity (fossil fuels, deforestation, livestock, fertilisers).

B.3 Gas laws

Pressure and amount of substance: P = F/A, F perpendicular to the surface (Pa). n = N/NA, with NA = 6.02 × 10²³ mol⁻¹.
The gas laws for a fixed mass of gas: PV/T = constant (T in kelvin). Constant T: PV = constant. Constant P: V/T = constant. Constant V: P/T = constant.
Ideal gas equation in two forms: PV = nRT (R = 8.31 J K⁻¹ mol⁻¹) PV = NkBT (kB = 1.38 × 10⁻²³ J K⁻¹)
Amount of gas in 3.1 × 10⁻³ m³ at 1.5 × 10⁵ Pa and 280 K?: n = PV/RT = (1.5 × 10⁵ × 3.1 × 10⁻³) ÷ (8.31 × 280) = 0.20 mol
Pressure from kinetic theory, and mean molecular kinetic energy: P = (1/3)ρv², where v² is the mean of the squared speeds. ½mv² = (3/2)kBT per molecule.
Internal energy of a monatomic ideal gas: U = (3/2)NkBT = (3/2)nRT: only random kinetic energy, because there are no intermolecular forces.
The temperature of a gas is quadrupled. What happens to the rms speed of its molecules?: It doubles, because rms speed ∝ √T.
When does a real gas behave most like an ideal gas?: At low pressure, low density and high temperature (well above its condensation point), where molecular volume and intermolecular forces are negligible.
A graph of V against temperature at constant pressure meets the temperature axis where?: At 0 K, which is about −273 °C.

B.4 Thermodynamics (HL only)

First law of thermodynamics, with the sign convention: Q = ΔU + W. Q positive for heat supplied to the gas; W positive for work done by the gas (expansion); ΔU positive when the temperature rises. (HL only)
Closed system: Matter cannot cross the boundary, but energy can cross it in either direction, as heat or as work. (HL only)
Adiabatic: No thermal energy is transferred to or from the gas, so Q = 0 and ΔU = −W. (HL only)
Work done by a gas, and change of internal energy of a monatomic ideal gas: W = PΔV at constant pressure; in general, the area under the P–V graph. ΔU = (3/2)nRΔT. (HL only)
Isothermal vs adiabatic change of a monatomic ideal gas: Isothermal: ΔU = 0, so Q = W. Adiabatic: Q = 0, temperature changes, PV^(5/3) = constant; its curve is steeper on a P–V graph. (HL only)
Entropy: two expressions: ΔS = ΔQ/T (T in kelvin). S = kB ln Ω, where Ω is the number of microstates. (HL only)
Second law of thermodynamics: Clausius, Kelvin and entropy forms: Clausius: heat does not flow by itself from cold to hot. Kelvin: no cyclic device can turn all the energy from one reservoir into work. Entropy of an isolated system never decreases. (HL only)
750 J flows from a reservoir at 360 K to one at 250 K. Total entropy change?: −750/360 + 750/250 = −2.08 + 3.00 = +0.917 J K⁻¹. Positive, so the flow is allowed. (HL only)
Heat engine efficiency and the Carnot limit: η = W/Qh, with W = Qh − Qc. η_Carnot = 1 − Tc/Th (kelvin). For Th = 900 K and Tc = 300 K, η_Carnot = 0.667. (HL only)

B.5 Current and circuits

Current, potential difference and emf: Current: rate of flow of charge, I = Δq/Δt. P.d.: work done per unit charge, V = W/q. Emf: energy transferred to each coulomb of charge by the cell.
Resistance, Ohm’s law and resistivity: R = V/I. Ohm’s law: at constant temperature, current is directly proportional to p.d. R = ρL/A, where ρ (Ω m) is a property of the material.
A wire of resistance 8.0 Ω has area 2.0 × 10⁻⁷ m² and length 1.5 m. Resistivity?: ρ = RA/L = 8.0 × 2.0 × 10⁻⁷ ÷ 1.5 = 1.1 × 10⁻⁶ Ω m
Origin of resistance in a metal: Electrons driven by the p.d. collide with vibrating lattice ions and transfer energy to them. Hotter means larger vibrations, more collisions and higher resistance.
Electrical power, and resistors in series and parallel: P = IV = I²R = V²/R Series: Rs = R₁ + R₂ + … Parallel: 1/Rp = 1/R₁ + 1/R₂ + …
Real cell: emf, internal resistance and terminal p.d.: ε = I(R + r), so terminal p.d. V = ε − Ir. A graph of V against I has intercept ε and gradient −r.
A 1.5 V cell with r = 0.60 Ω supplies 0.50 A. Terminal p.d.?: Lost volts Ir = 0.30 V, so V = 1.5 − 0.30 = 1.2 V
I–V graphs of a filament lamp and a thermistor: Filament lamp: bends towards the V axis; heating raises R. Thermistor (common type): bends towards the I axis as it heats; more charge carriers are freed, so R falls.
Resistance of an ideal ammeter and an ideal voltmeter: Ammeter: zero (connected in series). Voltmeter: infinite (connected in parallel).
Chemical cell vs solar cell: Both are sources of emf. A chemical cell converts stored chemical energy; a solar cell converts the energy of absorbed light, so its output depends on light intensity.