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Thermodynamics

Six topics on what temperature actually measures and where the energy goes. Kinetic theory, which makes temperature the average kinetic energy of one particle and pressure a rate of momentum delivery, the ideal gas law and the constraint that earns you each shortcut, thermal transfer and what equilibrium does and does not switch off, the first law with its sign convention nailed down, specific heat and conduction rate as separate questions, and entropy, which supplies the direction that energy conservation never does.

AP exam 15-18%6 topics
Topics
Key forms For every problem in this unit
Pressure
P = F⊥ / A, in pascals. Defined at every point INSIDE the gas, not only where it touches a wall
Average kinetic energy
K(avg) = (3/2) k_B T, per PARTICLE. k_B = 1.38 × 10⁻²³ J/K
Root-mean-square speed
v(rms) = √(3k_B T / m) = √(3RT / M). Speed goes as √T, so 4× the temperature doubles it
Same T, two gases
equal K(avg) per particle, NOT equal speed. Heavier particles move slower by √(m₂/m₁)
Ideal gas law
PV = nRT = N k_B T. R = 8.31 J/(mol·K). T is ALWAYS in kelvins
Two states, fixed n
P₁V₁ / T₁ = P₂V₂ / T₂. Cancel only what the setup pins: rigid tank cancels V, free piston cancels P
Kelvin conversion
T(K) = T(°C) + 273. A ratio in °C means nothing
Model assumptions
particle volume negligible, no appreciable force BETWEEN collisions, collisions elastic and frequent. Those wall collisions ARE the pressure
Speed distribution
Maxwell-Boltzmann: a spread, not one speed. Heating moves population right and flattens the curve; the area is the fixed particle count
First law
ΔU = Q + W, with W the work done ON the gas
Work on the gas
W = −PΔV at constant pressure. Compression gives W > 0. On a sloped path use the AREA under the curve
Internal energy
U = (3/2)nRT for a monatomic ideal gas. Depends on TEMPERATURE alone, so ΔU is path independent
Named processes
isothermal: ΔU = 0, Q = −W · adiabatic: Q = 0, ΔU = W · isovolumetric: W = 0, ΔU = Q · isobaric: W = −PΔV
Closed cycle
ΔU = 0, so Q(net) = −W(net). The enclosed area sets how big both are
Heating a substance
Q = mcΔT, so ΔT = Q / (mc). Larger c means a SMALLER temperature change
Mixing, insulated
m₁c₁(T(f) − T₁) + m₂c₂(T(f) − T₂) = 0. T(f) lands nearer the larger mc, not at the midpoint
Conduction rate
Q / Δt = kAΔT / L, in watts. A RATE, not an amount and not a temperature. Energy = rate × time
Entropy
ΔS = Q / T. The total for an ISOLATED system cannot decrease; a system trading energy with its surroundings certainly can
Direction
energy moves spontaneously from higher TEMPERATURE to lower, whatever the totals are. The first law permits both directions; the second picks one
Unit 9 tools
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60 open-ended problems.

Read the question, work it out, then flip the card to compare your reasoning to the worked solution. Mark each card so you can return to the ones that still bite.

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Twenty mixed items drawn from across all 6 topics, with guaranteed misconception-code coverage. Identifies which misconceptions still bite when you cannot see which topic the question came from.

20questions
6topics
22codes covered
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