amazon.co.ukA-Level Physics revision: internal energy, specific heat capacity and latent heat, the first law of thermodynamics, gas laws, ideal gas equation, kinetic theory derivation and uncertainty/percentage error calculations - AQA 7408 3.1/3.6.2, OCR A H556 5.1, Edexcel 9PH0 Topic 9.
Internal energy is the sum of the random kinetic and potential energies of all the molecules (AQA 7408 3.6.2.1; OCR A H556 5.1). Temperature measures the mean random KE per molecule - raising temperature raises internal energy and changes state. Distinguish the three: temperature is the measure, heat (Q) is energy transferred because of a temperature difference, internal energy is the store - the standard definition-matching question. Increasing the temperature of a body increases its internal energy (more molecular KE); a state change at constant temperature changes internal energy through the PE term (bonds/stretching between molecules).
Specific heat capacity: Q = mcΔθ - the energy to change the temperature of mass m by Δθ; c in J kg^-1 K^-1 (water: 4180 J kg^-1 K^-1 - large, so water is a good coolant). Note Δθ is the same in K and °C because the intervals are equal.
Specific latent heat: Q = mL - energy to change state of mass m with no temperature change (L_fusion for melting, L_vaporisation for boiling; water: L_f = 3.34 × 10^5 J/kg, L_v = 2.26 × 10^6 J/kg - vaporisation is much larger because separating molecules completely needs more energy than loosening them). Cooling curves show plateaus at state changes - the temperature stays constant while the substance changes state because the energy input goes into the latent heat, not into KE.
Find the total energy to melt 0.50 kg of ice at 0 °C and heat the water to 25 °C (L_f = 3.34 × 10^5 J/kg, c = 4180 J kg^-1 K^-1).
Line 1: Melting (no temperature change): Q_1 = mL_f = 0.50 × 3.34 × 10^5 = 1.67 × 10^5 J (M1 A1).
Line 2: Heating: Q_2 = mcΔθ = 0.50 × 4180 × 25 = 5.2 × 10^4 J (M1 A1).
Line 3: Total = 1.67 × 10^5 + 0.52 × 10^5 = 2.19 × 10^5 J ≈ 219 kJ (A1: stages added separately).
Practical (AQA RP11): determine specific heat capacity with an electrical heater: energy supplied E = IVt, temperature rise Δθ measured; c = IVt/(mΔθ). Insulate the block (reduce heat loss - the main systematic error making c too high), use a lid, stir liquids, and extrapolate the cooling curve back to correct for the heating delay.
First law: ΔU = Q + W - the increase in internal energy equals energy supplied by heating (Q, positive into the system) plus work done ON the system (W, positive when the gas is compressed) (AQA 7408 3.6.2.1). Sign conventions are the examinable trap: W = -pΔV for a gas at constant pressure (compression: ΔV negative, W positive - work done on the gas). Some boards write ΔU = Q - W with W the work done BY the gas - read the question's convention.
Work done by/on a gas: W = ∫p dV - the area under the p-V curve. Expansion does work by the gas on the surroundings (gas loses internal energy if no heat supplied); compression does work on the gas (internal energy rises).
p-V diagrams (described in words): isothermal (constant T) - a curve falling as p ∝ 1/V (Boyle); ΔU = 0 for an ideal gas, so Q = -W: all heat supplied is converted to work. Isobaric (constant p) - a horizontal line; W = pΔV. Isochoric/isovolumetric (constant V) - a vertical line; no work done, so ΔU = Q. Adiabatic (no heat exchange, Q = 0) - a steeper curve than isothermal; ΔU = W: the gas cools when it expands (work done at the expense of internal energy). The area inside a closed cycle on a p-V diagram is the net work done per cycle - the basis of engine analysis.
A gas absorbs 500 J of heat and does 200 J of work by expanding. Find ΔU, and state what happens if instead the gas is compressed by 200 J with the same heat input.
Line 1: ΔU = Q + W with W = -200 J (work done BY the gas): ΔU = 500 - 200 = 300 J (M1 A1: sign conventions).
Line 2: Compressed: W = +200 J: ΔU = 500 + 200 = 700 J (M1 A1).
Line 3: Compression adds to internal energy - the gas heats more (A1).
The empirical gas laws (AQA 7408 3.6.2.2; Edexcel 9PH0 Topic 9; OCR A H556 5.1), all for a fixed mass of ideal gas, all with temperature in kelvin (T = °C + 273): Boyle's law p ∝ 1/V at constant T (isothermal - pV = constant); Charles's law V ∝ T at constant p (isobaric); pressure law p ∝ T at constant V (isochoric). Combining: pV/T = constant, and for n moles the ideal gas equation pV = nRT (R = 8.31 J mol^-1 K^-1), or pV = NkT for N molecules (k = R/N_A = 1.38 × 10^-23 J/K).
Absolute zero: extrapolating any gas law graph to V = 0 or p = 0 gives T = -273 °C - the same intercept for every gas, defining the kelvin scale (molecules have minimum KE; pressure falls to zero when molecular motion stops).
Practicals: (AQA RP10; Edexcel and OCR gas-law practicals) - Boyle's law: trapped air column in a sealed tube with oil, pumped to vary pressure; plot p against 1/V - straight line through the origin confirms the law. Charles's law: capillary tube trapped air over a bead of concentrated sulfuric acid in water baths; plot V against T(°C) - extrapolate to V = 0 for absolute zero. Pressure law: fixed-volume flask in water baths; plot p against T(°C) - same extrapolation. Errors: temperature lag between bath and gas (wait for equilibrium, stir), volume/pressure reading resolution, and the assumption that the gas is dry and ideal at room temperatures.
One mole of an ideal gas occupies 0.022 m^3 at 27 °C. Find its pressure.
Line 1: T = 27 + 273 = 300 K (M1: kelvin conversion - the classic mark loss if missed).
Line 2: pV = nRT, so p = nRT/V (M1).
Line 3: p = (1 × 8.31 × 300)/0.022 = 2493/0.022 = 1.13 × 10^5 Pa (A1) - about atmospheric pressure, as expected for molar volume at room temperature.
The kinetic theory model derives the gas equation from molecular collisions (AQA 7408 3.6.2.3) - the full derivation is a standard six-mark question, so learn each step and the assumptions.
Assumptions: molecules are identical point particles; they move in random directions with a distribution of speeds; collisions (with each other and the walls) are perfectly elastic; the time of a collision is negligible compared with the time between collisions; the gas is dilute so intermolecular forces act only during collisions; Newtonian mechanics applies.
Derivation, step by step: (1) Consider one molecule of mass m moving with velocity component u_x in a box of length l, width/height giving face area A. (2) It strikes the end wall elastically, reversing u_x: change in momentum per collision Δp = 2mu_x. (3) The time between collisions with the same wall is the round trip: Δt = 2l/u_x. (4) Force on the wall from this molecule: F = Δp/Δt = 2mu_x/(2l/u_x) = mu_x^2/l. (5) Sum over all N molecules: F = m(u_1x^2 + u_2x^2 + ...)/l = Nm
Linking to temperature: compare with pV = NkT: ⅓Nm
Find the rms speed of nitrogen molecules (M = 0.028 kg/mol) at 300 K.
Line 1: c_rms = √(3RT/M) (M1: equation with molar mass in kg/mol).
Line 2: c_rms = √(3 × 8.31 × 300/0.028) = √(267,107) (M1 A1).
Line 3: c_rms = 517 m/s (A1) - comparable to the speed of sound in the gas (sound travels through molecular motion).
Measurement uncertainty (AQA 7408 3.1.2; Edexcel 9PH0 Topic 1 'Working as a physicist'; OCR Module 1): random uncertainty varies unpredictably between repeats - reduce by repeating and averaging (or a best-fit line); shown by scatter. Systematic uncertainty is a constant offset in every reading (zero error, calibration error, parallax, heat loss in thermal experiments) - repeats do not reduce it; find it by calibration or comparison with a known value. The uncertainty of a single reading is at least ± half the resolution; for repeated readings use ± half the range.
Combining uncertainties: adding or subtracting measured quantities - add the absolute uncertainties (Q = mcΔθ: absolute uncertainties in m and Δθ add; c is a constant). Multiplying or dividing - add the percentage uncertainties (E = IVt: three percentage uncertainties add). Raising to a power n - multiply the percentage uncertainty by n (V = πr^2l: the r^2 term doubles the radius's percentage uncertainty). Constant multiples - percentage uncertainty unchanged (½m
Percentage error against an accepted value: |measured - accepted|/accepted × 100% - used to conclude evaluations. Gradients and intercepts: use best-fit lines (steepest and shallowest lines through the error bars give the gradient's uncertainty); a large y-intercept where zero is expected signals a systematic error.
An aluminium block: E = IVt = 6.0 V × 2.0 A × 120 s with V ± 0.1 V, I ± 0.05 A, t ± 1 s; m = 1.000 ± 0.001 kg; Δθ = 15.0 ± 0.5 K.
Line 1: c = IVt/(mΔθ) = (6.0 × 2.0 × 120)/(1.000 × 15.0) = 96 J kg^-1 K^-1 (M1 A1).
Line 2: Percentage uncertainties: V: 0.1/6.0 = 1.7%; I: 0.05/2.0 = 2.5%; t: 1/120 = 0.8%; m: 0.1%; Δθ: 0.5/15 = 3.3% (M1: each computed).
Line 3: Total (all multiplied/divided) = 1.7 + 2.5 + 0.8 + 0.1 + 3.3 = 8.4%, so Δc = 0.084 × 96 = 8 J kg^-1 K^-1 (A1).
Line 4: c = 96 ± 8 J kg^-1 K^-1; accepted value ≈ 900 J kg^-1 K^-1 - the measured value is far too low: the dominant error is heat loss (systematic - insulation, lid, shorter heating), not random scatter (A1: evaluation - systematic vs random is the discriminator).
Exam evaluation logic for thermal practicals: identify the largest uncertainty (percentage terms), classify it random or systematic, and propose the fix (repeats for random; insulation/calibration/zeroing for systematic). The mixture method (hot metal into known water, energy conservation: m_metal c_metal Δθ_metal = m_water c_water Δθ_water) and the continuous-flow method are the alternative routes - know which errors each minimises.