Fiveable
🧲AP Physics 2
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🧲AP Physics 2

FRQ 4 – Qualitative/Quantitative Translation
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Unit 9: Thermodynamics
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FRQ Types & Units

Each FRQ type tests specific skills taught in particular units. Here's why certain units appear for each question type:

This mapping reflects College Board's exam structure - each FRQ type tests specific skills that are taught in particular units.

Practice FRQ 1 of 201/20

4. A rigid, sealed cylindrical container holds 0.030 mol of an ideal monatomic gas. The gas is initially in thermal equilibrium at temperature Ti = 300 K and pressure Pi = 1.20×10^5 Pa. The container is surrounded by insulation except for a flat wall made of aluminum that has thickness L = 2.0×10^-3 m and area A = 1.5×10^-2 m^2. The thermal conductivity of the aluminum is k = 205 W/(m·K). The outer surface of the aluminum wall is suddenly brought into good thermal contact with a large thermal reservoir at temperature TR = 450 K, as shown in Figure 1. The container is rigid, so the gas volume remains constant throughout the process, and the reservoir temperature remains constant.

Figure 1. Sealed cylinder with heated monatomic gas

Black-and-white physics setup diagram with no axes and no graph grid. Use clean vector-style lines and clear printed labels.

Overall layout (left to right):
- Left region: a rigid sealed cylindrical container holding gas.
- Middle: a flat aluminum wall that is the right end-cap of the cylinder.
- Right region: a large thermal reservoir in contact with the outer surface of the aluminum wall.

Rigid sealed cylindrical container (gas chamber):
- Draw a horizontal cylinder occupying the left two-thirds of the figure width.
- Left end of the cylinder is a solid curved end-cap (sealed). The cylinder’s long axis is horizontal.
- The right end of the cylinder is not a curved end-cap; instead, it terminates at a flat vertical interface where the aluminum wall begins.
- Label centered inside the gas region (not on the wall):
  - First line: "Ideal monatomic gas"
  - Second line: "n = 0.030 mol"
  - Third line: "Initial: Ti = 300 K"
  - Fourth line: "Initial: Pi = 1.20×10^5 Pa"
- Add an external label near the cylinder wall reading: "Rigid, sealed container (V constant)" with a leader arrow pointing to the cylinder body.

Insulation (all surfaces except the aluminum wall):
- Surround the curved left end-cap and the long cylindrical sidewall (top and bottom edges of the cylinder) with a thick insulating jacket symbol (repeating zigzag/foam pattern).
- Do NOT place insulation on the aluminum wall; make the insulation stop exactly at the junction where the cylinder meets the aluminum wall.
- Add a label above the insulation pattern: "Insulation" with a leader arrow pointing to the jacket.

Aluminum wall (conduction barrier):
- Immediately to the right of the gas region, draw a rectangular slab representing the aluminum wall as a flat end-cap.
- The slab’s faces are vertical; its thickness is horizontal (left face touches the gas; right face touches the reservoir).
- Inside the slab, place the text label: "Aluminum".
- Also include (either inside the slab or immediately adjacent with a leader arrow): "k = 205 W/(m·K)".

Wall thickness L (must be an explicit dimension):
- Below the aluminum slab, draw a horizontal double-headed dimension arrow whose left arrowhead touches the inner face of the slab (gas–wall interface) and whose right arrowhead touches the outer face of the slab (wall–reservoir interface).
- Centered above that dimension arrow, print exactly: "L = 2.0×10^-3 m".

Wall area A (must be clearly the face area):
- Indicate that A is the area of the flat wall face by labeling the front face of the slab (the face separating gas and wall) with a leader arrow pointing to the face.
- The text must read exactly: "A = 1.5×10^-2 m^2".
- Ensure the leader arrow points to the wall face (not to the thickness dimension arrow).

Thermal reservoir:
- To the right of the aluminum slab, draw a large rectangular region occupying the rightmost third of the figure width, sharing a perfectly flush vertical boundary with the outer face of the aluminum wall (no gap).
- Add the label centered inside this right region: "Thermal reservoir" on one line and directly below it: "TR = 450 K".
- Add an additional small note near the reservoir (optional but included as visible text): "Large reservoir (temperature constant)".

Thermal contact statement:
- At the interface between the outer aluminum face and the reservoir, add a short label with a leader arrow to the interface line: "Good thermal contact".

Heat transfer arrow (direction must be unambiguous):
- Draw one thick arrow centered vertically, starting inside the reservoir region, crossing the aluminum slab, and ending inside the gas region.
- The arrow must point leftward (from reservoir toward gas).
- Above the arrow, print the text: "Heat transfer by conduction".

Clarity rules to enforce:
- The gas label block (including Ti and Pi) must be entirely inside the gas volume, not overlapping the wall.
- The reservoir label block (including TR) must be entirely inside the reservoir rectangle.
- L is shown only as the slab thickness (horizontal), not as a cylinder length.
- A is explicitly tied to the wall face area, not to the cylinder cross-section in general.
- No other numbers appear anywhere in the figure besides: n = 0.030 mol, Ti = 300 K, Pi = 1.20×10^5 Pa, TR = 450 K, L = 2.0×10^-3 m, A = 1.5×10^-2 m^2, k = 205 W/(m·K).
A.

A student claims: “As energy is transferred from the reservoir to the gas, the gas pressure increases because the gas atoms move faster and collide with the container walls more often and with greater impulse.”

Indicate whether the student’s claim is correct or incorrect. Without manipulating equations, justify your answer by describing (i) how the temperature of the gas is related to the motion of its atoms and (ii) how the pressure exerted by the gas is related to collisions of atoms with the container walls. Refer to the rigid container shown in Figure 1 in your reasoning.

B.

Derive an expression for the time t needed for the gas temperature to increase from Ti = 300 K to Tf = 360 K due to conduction through the aluminum wall in Figure 1. Assume the inner surface of the aluminum wall is always at the instantaneous gas temperature T (so the temperature difference across the wall is TR − T), and assume the gas is spatially uniform in temperature at all times. Express your answer in terms of n, k, A, L, TR, Ti, Tf, and physical constants as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.

C.

Indicate whether your expression from part B is or is not consistent with the following claim: “If the aluminum wall thickness L is doubled while all other given quantities remain the same, the required time t for the gas to warm from Ti to Tf doubles.” Briefly justify your answer by referencing the functional dependence of your expression on L.







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FRQ Directions
Free Response Question Practice

This practice environment simulates the AP AP Physics 2 Free Response Questions section. Here are some guidelines:

  • Read each question carefullybefore responding. Pay attention to command verbs like "identify," "explain," "analyze," or "evaluate."
  • Use the timer to practice time management. You can pause, restart, or hide the timer as needed.
  • Mark for Review if you want to come back to a question later.
  • Your responses are saved automatically as you type. You can also use the drawing tool for questions that require diagrams or graphs.
  • Use the toolbar for formatting options like bold, italic, subscript, and superscript.
  • Navigate between questions using the Previous and Next buttons at the bottom of the screen.

Tip: Answer all parts of each question. Partial credit is often available, so even if you are unsure, provide what you know.