AP Physics 2 · guided topic map

Thermodynamics for AP Physics 2

Thermodynamics for AP Physics 2, organized into 1 syllabus topic and 8 mapped concept guides.

Syllabus topics: 1
Mapped concept guides: 8
Educational level: AP Physics 2: Algebra-Based

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Thermodynamics

AP Physics 2: Algebra-Based

8 guides

01Temperature and internal energy15–18%
02Kinetic theory and the particle model15–18%
03Ideal gases and gas laws15–18%
04Specific heat and latent heat15–18%
05Conduction, convection, and radiation15–18%
06The first law and gas processes15–18%
07Heat engines and the second law15–18%
08Entropy15–18%

Diagrams

Thermodynamics as AP Physics 2 draws it

The figures from the AP Physics 2 practice papers that sit on these syllabus points — the apparatus, circuits and graphs an exam question actually puts in front of you.

01Figure 3Electric PotentialAP

Figure comment

Figure 3A pressure-volume graph with a faint grid. The horizontal axis is volume V in units of 10⁻³ m³, marked from 0 to 4.0; the vertical axis is pressure P in units of 10⁵ Pa, marked from 0 to 3.0. Four states are plotted as dots at the corners of a rectangle and labelled: A at V = 1.0 and P = 2.0, B at V = 3.0 and P = 2.0, C directly below B at P = 1.0, and D directly below A at P = 1.0. Straight lines join A to B to C to D and back to A, and an arrow on each side shows the order in which the gas is taken round the closed cycle.

Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.

9 Electric Potential Open AP question · AP Physics 2: Algebra-Based Q7

Guided questions 5 parts

Reading cue. Both axes carry multipliers — A is 2.0 × 10⁵ Pa and 1.0 × 10⁻³ m³ — and the arrows run clockwise, which is why the enclosed area is net work done by the gas.

aDetermine Determine which of the four processes drawn involve no work done by or on the gas, and state why.
recall2 marks
Check answer 2 marks
1. B → C and D → A are the vertical sides of the rectangle, so ΔV = 0 for both
2. work is the area under the process line, and a vertical line encloses none, so W = PΔV = 0 in each case
bCalculate Calculate the ratio of the temperature of the gas at state C to its temperature at state A.
routine3 marks
Check answer 3 marks
1. the quantity of gas is fixed, so T ∝ PV
2. P_A V_A = 2.0 × 10⁵ × 1.0 × 10⁻³ = 200 J and P_C V_C = 1.0 × 10⁵ × 3.0 × 10⁻³ = 300 J
3. T_C/T_A = 300/200 = 1.5
cDetermine Determine which of the four plotted states is the hottest and which is the coldest, and determine the thermal energy transferred to or from the gas during the process C → D.
demanding4 marks
Check answer 4 marks
1. PV is 600 J at B and 100 J at D, against 200 J at A and 300 J at C, so B is the hottest state and D the coldest
2. work done by the gas on C → D = PΔV = 1.0 × 10⁵ × (−2.0 × 10⁻³) = −200 J
3. ΔU = (3/2)Δ(PV) = (3/2)(100 − 300) = −300 J for a monatomic gas
4. Q = ΔU + W = −300 − 200 = −500 J, so 500 J is removed from the gas
dDetermine Determine the efficiency of this cycle worked as a heat engine, and compare it with the ideal limit 1 − T_cold/T_hot set by the hottest and coldest states on the diagram.
top of the paper4 marks
Check answer 4 marks
1. heat enters only on A → B, where Q = (3/2)(600 − 200) + 400 = 1000 J, and on D → A, where Q = (3/2)(200 − 100) = 150 J with no work, giving Q_in = 1150 J
2. net work per cycle is the enclosed rectangle, 1.0 × 10⁵ × 2.0 × 10⁻³ = 200 J
3. efficiency = 200/1150 = 0.17, that is 17%
4. T ∝ PV makes T_D/T_B = 100/600, so the ideal limit is 1 − 1/6 = 83%: this cycle reaches only about a fifth of what the same two temperatures would allow

Transfer challenge

The same gas is instead expanded from state A to three times its volume while its temperature is held constant. Determine the work done by the gas, using W = PV ln(V_f/V_i) with PV evaluated at the starting state, and explain why the thermal energy supplied is equal to that work.

Check answer 4 marks

P_A V_A = 2.0 × 10⁵ × 1.0 × 10⁻³ = 200 J, so W = 200 ln 3 = 220 J
for an ideal gas the internal energy depends only on temperature, so an isothermal process has ΔU = 0
the first law then gives Q = ΔU + W = W = 220 J, so every joule supplied leaves again as work
the isotherm falls away below the constant-pressure line A → B, so the same threefold expansion does less work: 220 J against the 400 J of the drawn isobaric step

02Fig. 7.1Electric Charges, Fields, and Gauss's Law · Electric PotentialAP

Figure comment

Fig. 7.1A large circle represents the insulating sphere. Inside it, faint concentric rings are drawn at intervals that get smaller towards the outside, indicating that the charge density increases with distance from the centre. A dimension line from the centre out to the surface, tick-marked at both ends, is labelled R. A second, shorter dimension line runs from the centre to a small marked point P inside the sphere and is labelled r. Below the sphere the density is written as a function of radius, proportional to r divided by R. No Gaussian surface is drawn.

Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.

8 Electric Charges, Fields, and Gauss's Law 9 Electric Potential Open AP question · AP Physics C: Electricity and Magnetism Q7

Guided questions 5 parts

Reading cue. The rings crowd towards the surface, so the density rises with radius: r runs from the centre out to P, and the charge inside r grows faster than r³.

aDetermine Determine the volume charge density at the centre of the sphere, at the point P marked at radius r, and at the surface, in terms of ρ₀.
recall3 marks
Check answer 3 marks
1. zero at the centre
2. ρ₀r/R at P
3. ρ₀ at the surface, which is why the rings are drawn closest together there
bDetermine Determine the fraction of the sphere's total charge that lies within a radius R/2 of the centre.
routine4 marks
Check answer 4 marks
1. q(r) = ∫₀^r (ρ₀s/R)4πs² ds = πρ₀r⁴/R
2. so q ∝ r⁴, and the total charge is Q = πρ₀R³
3. q(R/2)/Q = (1/2)⁴ = 1/16, that is 6.25%
4. much less than the 1/8 a uniform sphere would give, because the charge is pushed outwards
cSketch Sketch a graph of the magnitude of the electric field against distance from the centre, from the centre out to 3R, marking the position and the value of the maximum.
demanding5 marks
Check answer 5 marks
1. for r < R, E = q(r)/4πε₀r² = ρ₀r²/4ε₀R, drawn as a curve rising from the origin with increasing gradient, not a straight line
2. for r > R, E = Q/4πε₀r², falling as 1/r²
3. the two expressions agree at r = R, so the curve has no step in it
4. maximum at the surface, r = R, of value ρ₀R/4ε₀
5. at r = 3R the field has fallen to one ninth of the maximum
dDerive Using an energy density of ½ε₀E², derive an expression for the total electric energy stored in the field outside the sphere, and explain why the same calculation cannot be carried through for a point charge.
top of the paper5 marks
Check answer 5 marks
1. dU = ½ε₀E²(4πr² dr) with E = Q/4πε₀r² for r > R
2. U = (Q²/8πε₀)∫_R^∞ dr/r² = Q²/8πε₀R
3. substituting Q = πρ₀R³ gives U = πρ₀²R⁵/8ε₀
4. for a point charge the lower limit becomes r = 0 and the integral diverges
5. so a point charge implies infinite field energy, which is why a finite size has to be assumed in such calculations

Transfer challenge

A very long solid insulating cylinder of radius a carries a charge density that varies with distance from its axis as ρ(r) = ρ₀r/a. Derive expressions for the magnitude of the electric field inside and outside the cylinder, and verify that the two agree at the surface.

Check answer 5 marks

Gaussian surface taken as a coaxial cylinder of radius r and length ℓ, with flux only through the curved surface
charge enclosed for r < a: ℓ∫₀^r (ρ₀s/a)2πs ds = 2πρ₀ℓr³/3a
E(2πrℓ) = q_enc/ε₀ gives E = ρ₀r²/3ε₀a for r < a
for r > a the enclosed charge is fixed, giving E = ρ₀a²/3ε₀r
both expressions give ρ₀a/3ε₀ at r = a