AP Physics 2 2018 Exam Scoring Guide
Apr 3, 2026
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Summary of AP Physics 2 Exam Excerpts
This document contains excerpts from an AP Physics 2 exam, covering a range of topics in electricity and magnetism, fluid mechanics, thermodynamics, and modern physics. The questions are presented in both multiple-choice and free-response formats, along with relevant tables of constants, equations, and scoring guidelines.
Section I: Multiple-Choice Questions
This section consists of multiple-choice questions designed to assess understanding of fundamental physics principles. Key topics and question types include:
- Electrostatics:
- Charge distribution on conducting and insulating spheres (Question 6).
- Electric fields and potential around charged spheres (Questions 21, 22, 23).
- Force on a charged particle in an electric field (Question 34).
- Charging by friction vs. conduction, polarization (Question 25).
- Circuits:
- Parallel resistor circuits and Ohm's Law (Question 8).
- Battery internal resistance and its effect on circuit graphs (Questions 14, 15).
- Series battery circuits (Question 31).
- Capacitors in circuits, charge storage (Question 35).
- Mechanics and Waves:
- Motion of a charged particle in an electric field (Question 7).
- Magnetic forces on current loops (Question 9).
- Wave pulse superposition (Question 13).
- Buoyant force and electrostatic forces (Question 29).
- Fluid dynamics (continuity and Bernoulli's equation) (Questions 27, 28).
- Thermodynamics:
- Work done on an ideal gas during a cycle (Question 10).
- Heat transfer mechanisms (conduction, convection, radiation) (Question 11).
- Thermodynamic processes (internal energy, work, heat) (Question 12).
- Ideal gas law and its graphical representation (Question 24).
- Gas molecule speed distribution and temperature (Questions 16, 32).
- Modern Physics:
- Radioactive decay (alpha decay, momentum, kinetic energy, energy release, half-life) (Questions 17, 18, 19).
- Wave-particle duality and diffraction (Question 36).
- Electron location probability in hydrogen atom (Question 1).
- Electromagnetic waves and frequency (Question 2).
- Gravitational vs. electrical forces (Question 3).
- Atomic energy levels and photon emission (Question 4).
- Magnetic fields from bar magnets and compass deflection (Question 5).
- Lenses and image formation (Question 20).
- Experimental Design:
- Determining thermal conductivity (Question 26).
Note: Questions 131-134 are "select two answers" questions.
Section II: Free-Response Questions
This section contains longer, more in-depth questions requiring detailed explanations, calculations, and experimental design.
- Question 1: Fluid Dynamics and Buoyancy
- Analyzes water flow in a pipe with changing diameter and elevation using the continuity equation and Bernoulli's principle.
- Assesses understanding of buoyant force and its relation to submerged and floating objects.
- Includes drawing free-body diagrams and calculating forces.
- Question 2: Circuits and Resistivity
- Focuses on experimental determination of resistivity using Ohm's Law and graphical analysis.
- Requires designing an experiment, collecting and plotting data, and calculating resistivity.
- Addresses potential sources of error, such as internal resistance and temperature effects.
- Question 3: Optics
- Investigates image formation by a converging lens using an optical bench.
- Requires drawing ray diagrams, calculating focal length and magnification, and plotting data for linear analysis.
- Explores the effect of changing the medium (air to water) on the lens's focal length and image properties.
- Question 4: Electrostatics and Capacitance
- Examines electric fields, potential, and forces due to multiple point charges.
- Covers the behavior of parallel-plate capacitors connected to a battery, including charge, electric field, and capacitance changes with varying plate separation.
- Analyzes capacitors connected in parallel.
Supporting Materials
- Table of Information: Provides fundamental physical constants and conversion factors.
- Equations: Lists relevant equations for electricity and magnetism, fluid mechanics, thermal physics, geometry, and trigonometry.
- Conventions: Outlines standard assumptions and definitions used throughout the exam.
- Scoring Guidelines: Detailed rubrics for grading the free-response questions, including point distribution for different parts of the solution and acceptable answers/reasoning.
- Scoring Worksheet and Conversion Chart: Tools for calculating the composite score from Section I and Section II.
- Question Descriptors and Performance Data: Provides insights into student performance on specific questions.
This document serves as a comprehensive resource for understanding the scope and format of the AP Physics 2 exam, including the types of problems students are expected to solve and the criteria used for evaluation.
2016 AP Physics 2: Algebra-Based Exam Summary
This document outlines the structure, content, and scoring of the 2016 AP Physics 2: Algebra-Based exam. It includes the scoring worksheet, question descriptors, performance data, free-response questions, scoring guidelines, and multiple-choice answer keys.
Section I: Multiple Choice
- Number of Questions: 39 (Item 3 was not used for scoring, though 40 were administered).
- Scoring: The raw score (number correct) is converted to a weighted score.
- Weighted Section I Score = Number Correct × 0.9090
Section II: Free Response
- Number of Questions: 4
- Time Allotment: 90 minutes total.
- Questions 1 & 4: Short Free Response (10 points each, suggested 20 minutes each).
- Questions 2 & 3: Long Free Response (12 points each, suggested 25 minutes each).
- Scoring: Scores from each question are summed and weighted.
- Weighted Section II Score = Sum of Free Response Scores × 0.9090 (This calculation appears to be a placeholder or error in the provided text, as the scoring guidelines indicate points are awarded per question part, not a direct multiplication by a factor for the total section score).
Composite Score Calculation
- The Weighted Section I Score and Weighted Section II Score are summed.
- This composite score is then used with the AP Score Conversion Chart to determine the final AP score (1-5).
Question Descriptors and Performance Data
This section provides detailed information about the content assessed by each question and student performance data.
- Multiple-Choice Questions: Categorized by AP Physics 2 curriculum framework codes (e.g., 2-3.A.2.1) and corresponding topic areas (e.g., 3.A.2).
- Free-Response Questions: Also categorized by curriculum framework codes and topic areas. Performance data, likely in the form of average scores or percentage correct for specific parts, is presented in tables.
AP Physics 2: Algebra-Based - The College Board
- The College Board is a non-profit organization dedicated to promoting excellence and equity in education.
- It supports students' transition to college through programs like the SAT and AP.
- The AP Program aims to provide college-level coursework and exams.
Constants, Conversion Factors, and Equations
The document includes tables of:
- Constants and Conversion Factors: Essential physical constants (e.g., electron charge, speed of light, Planck's constant) and conversion factors (e.g., 1 eV to Joules, 1 atm to Pascals).
- Values of Trigonometric Functions: For common angles.
- Conventions Used in the Exam: Assumptions about inertial frames, definition of work, direction of current, ideal components, capacitor edge effects, and electric potential reference.
- AP Physics 2 Equations: A comprehensive list of equations organized by topic (Electricity and Magnetism, Fluid Mechanics and Thermal Physics, Geometry and Trigonometry).
Free-Response Questions (Examples from 2016)
The document provides the text of the 2016 AP Physics 2 Free-Response Questions, which cover a range of topics:
- Question 1: Thermodynamics (ideal gas cycles, work, internal energy, heat).
- Question 2: Optics (index of refraction, Snell's Law, experimental design, light production in hydrogen atoms).
- Question 3: Electrostatics (electric potential, charge distribution, electric field, potential energy, work, conservation of energy).
- Question 4: Circuits (RC circuits, current, potential difference, energy storage in capacitors).
Scoring Guidelines (Examples from 2016)
Detailed scoring guidelines are provided for each free-response question, outlining:
- Distribution of points: How points are awarded for specific steps, calculations, explanations, and correct answers.
- Example solutions and justifications: Demonstrating acceptable reasoning and calculations.
Multiple-Choice Answer Key
- A list of correct answers for the multiple-choice section is provided.
- Note that item 3 was not used in scoring.
This summary provides a structured overview of the 2016 AP Physics 2: Algebra-Based exam, covering its format, content areas, scoring procedures, and the types of questions students encountered.
AP Physics 2: Algebra-Based 2025 - Free-Response Questions Summary
This document contains free-response questions from the AP Physics 2: Algebra-Based 2025 exam, designed to assess students' understanding of various physics concepts. The questions cover electromagnetism, thermodynamics, circuits, and optics.
Question 1: Electromagnetism (Wires and Magnetic Fields)
This question deals with the magnetic fields and forces generated by current-carrying wires.
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Part i:
- Magnetic Field from Wire 2 at Point P: Students are asked to indicate the direction of the magnetic field produced by Wire 2 (carrying current $I$ in the +x-direction along $y=+d$) at Point P (located on Wire 1 at the origin).
- Magnetic Force on Wire 1 by Wire 2: Students are asked to indicate the direction of the magnetic force exerted on Wire 1 by Wire 2.
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Part ii:
- A third wire (Wire 3) carrying current $2I$ in the +x-direction is placed at $y=y_3$.
- The net magnetic force on Wire 1 due to wires 2 and 3 is zero.
- Students must derive an expression for $y_3$ in terms of $d$, starting with a fundamental physics principle or equation.
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Part B:
- Wire 3 is moved far away. A circular conducting loop in the xy-plane is moved at a constant speed in the -y-direction below Wire 1.
- Students must determine if there is a clockwise induced current, a counterclockwise induced current, or no induced current in the loop.
Question 2: Thermodynamics (Ideal Gas and Piston)
This question focuses on the behavior of a monatomic ideal gas contained by a movable piston in a thermally conducting container.
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Part A:
- Forces acting on the piston (mass $M$, area $A$) must be drawn and labeled. The gas inside is in thermal equilibrium with a water bath at constant temperature $T_0$. The pressure of the air above the piston is $P_{atm}$.
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Part B:
- Students must derive an expression for the internal energy of the gas in terms of $M$, $A$, $V_0$, $P_{atm}$, and physical constants, starting with a fundamental physics principle or equation.
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Part C:
- A block of mass $M$ is slowly added to the piston.
- Students must sketch the relationship between pressure ($P$) and volume ($V$) of the gas during this process, indicating the direction of the process.
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Part D:
- With the block on the piston, the water bath temperature is changed to $T_{new}$. The gas returns to its original volume $V_0$.
- Students must indicate whether $T_{new}$ is greater than, less than, or equal to $T_0$ and justify their answer by referencing parts A, B, or C.
Question 3: Circuits and Capacitors
This question involves determining the time constant of an RC circuit and the capacitance of a parallel-plate capacitor.
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Experiment 1 (Time Constant):
- Part A: Describe a procedure to determine the time constant ($\tau$) of a series circuit with an unknown resistor and an air-filled parallel-plate capacitor, using a battery, switch, ammeter, ruler, and wires. Include necessary measurements and steps to reduce uncertainty.
- Part B: Describe how the collected data would be analyzed to determine $\tau$, referencing appropriate equations and relationships.
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Experiment 2 (Capacitance):
- Part C (i): Identify two quantities (measured or calculated) from provided data (potential difference $|V|$ and charge $q$) that can be graphed to produce a straight line for determining capacitance ($C$).
- Part C (ii): Create a graph of the identified quantities, labeling axes and plotting points from a provided table.
- Part C (iii): Draw a best-fit line for the graphed data.
- Part D: Calculate an experimental value for capacitance $C$ using the best-fit line.
Question 4: Optics (Interference)
This question concerns the double-slit interference pattern produced by monochromatic light.
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Setup: Two narrow slits separated by distance $d$, a screen at distance $L$ ($L \gg d$), and monochromatic light producing bright and dark bands. Bright bands A and B are indicated, with three additional bright bands (including the central one) between them.
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Part A:
- A student claims the distance between the center of Band A and the central bright band is smaller for violet light than for red light.
- Students must indicate if the claim is correct or incorrect and justify it by referencing the difference in path length without using equations.
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Part B:
- Derive an expression for the distance between the centers of bands A and B when light of frequency $f$ is used. The expression should be in terms of $d$, $L$, $f$, and physical constants, starting with a fundamental physics principle or equation.
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Part C:
- Indicate whether the expression derived in Part B is consistent with the answer in Part A and provide a brief justification.
General Exam Instructions
- Time: Section II has 4 questions and lasts 1 hour and 40 minutes. Suggested times are ~25 minutes for Q1 & Q3, ~30 minutes for Q2, and ~20 minutes for Q4.
- Tools: Calculators, rulers, and straightedges are allowed. Reference information (equations) is accessible.
- Answering: Write answers in the provided booklet, labeling parts clearly. Show work for credit. Use pencil or black/dark blue ink. Graphs should use one color.
- Format: The exam was originally digital and is presented here for classroom use. Students have access to reference information on AP Central.
Summary of AP Physics 2 Practice Exam Content
This document contains a practice exam for AP Physics 2, covering a range of topics in electricity and magnetism, fluids and thermal physics, and modern physics. It includes multiple-choice questions and free-response questions with scoring guidelines.
Section I: Multiple-Choice Questions
This section assesses understanding of various physics concepts through 40 questions, with four answer choices each. Key topics covered include:
- Radioactive Decay and Half-Life: Questions involve calculating remaining amounts of radioactive substances and determining half-lives.
- Thermodynamics and Ideal Gas Law: Concepts such as pressure-volume relationships, work done on/by a gas, internal energy, heat transfer, and the ideal gas law are tested.
- Electricity and Magnetism: This broad category includes:
- Electric Fields and Potentials: Understanding equipotential lines, electric field strength, and charge distribution.
- Electric Forces and Fields: Coulomb's Law, forces between charges, and electric fields generated by charges.
- Capacitance: Circuits with capacitors, stored charge, and factors affecting capacitance.
- Magnetic Fields and Forces: Forces on moving charges in magnetic fields, magnetic fields produced by currents, and induced currents.
- Electromagnetic Waves: Properties of waves, including electric field amplitude and wavelength.
- Optics: Refraction, reflection, lenses (converging and diverging), and mirrors (concave and convex).
- Fluid Mechanics: Buoyant force and its dependence on volume.
- Wave Properties: Diffraction and its dependence on wavelength.
- Modern Physics: Mass-energy equivalence, radioactive decay products, and relativistic effects (time dilation).
Key Features of Section I:
- Questions 1-36 have four answer choices (A-D).
- Questions 131-134 require selecting two correct answers.
- A table of physical constants and equations is provided.
- Calculators, rulers, and straightedges are allowed.
Section II: Free-Response Questions
This section consists of four free-response questions, designed to assess deeper understanding and application of physics principles.
Question 1: Heater Design (10 points, ~20 minutes)
- Topic: Electrical power, resistance, heat transfer, thermodynamics.
- Part (a): Calculate the length of wire needed for an electric heater delivering 10 kW of power from a 240 V supply, given wire resistivity and cross-sectional area.
- Part (b):
- i. Identify and describe the primary energy transfer processes through electrical insulation (conduction) and water (convection).
- ii. Explain, using the first and second laws of thermodynamics, why a heater must remain on to maintain a constant water temperature higher than the surrounding air, considering heat loss.
- Part (c): Sketch a thermodynamic process for air warming in a closed room and compare the heat transferred (Q) to the change in internal energy (ΔU).
- Part (d): Calculate the force exerted by air on a door, using the ideal gas law to find the pressure at a new temperature.
Question 2: Concave Mirror Optics (12 points, ~25 minutes)
- Topic: Reflection, concave mirrors, image formation.
- Part (a): Describe a procedure to collect data for determining the focal length of a concave mirror using a lightbulb, screen, and meterstick. This includes drawing a diagram, labeling axes for a graph (e.g., 1/s₀ vs. 1/sᵢ), and outlining the measurement process. Explain how to determine the focal length from the graph.
- Part (b): Draw a ray diagram for a given object position relative to the focal point of a concave mirror to illustrate image formation.
- Part (c): Analyze data from an experiment where a concave mirror focuses sunlight. Determine the image distance of the Sun based on a temperature vs. distance graph and relate this distance to the mirror's focal length using the mirror equation.
Question 3: Electrostatic Systems (12 points, ~25 minutes)
- Topic: Electric potential energy, work, conservation of energy, electric fields, electric potential, charge distribution, induction.
- Part (a):
- i. Write an expression for the work done to move two charged spheres apart, considering only the spheres (Earth external).
- ii. Explain how the total energy of the two-sphere system changes when they swing apart due to electrostatic repulsion, considering forces.
- Part (b): Calculate the work done on the system including Earth for the processes in (a)(i) and (a)(ii), explaining the reasoning.
- Part (c): Draw an electric field vector and isolines of electric potential at a specific point near two spheres with opposite charges.
- Part (d): Explain, in terms of charge distribution, why a charged plastic sphere and an uncharged metal sphere attract each other when allowed to pivot.
Question 4: Nuclear Decay (10 points, ~20 minutes)
- Topic: Radioactive decay, half-life, conservation of momentum, mass-energy equivalence, magnetic forces on charged particles.
- Part (a): Determine the approximate half-life of Uranium-231 from provided data.
- Part (b):
- i. Calculate the velocity of the center of mass of the decay products (alpha particle and Thorium nucleus) immediately after decay, using conservation of momentum.
- ii. Calculate the energy released during the decay using the mass difference and Einstein's mass-energy equivalence (E=mc²).
- Part (c): Explain how the path of an alpha particle in a uniform magnetic field can be used to verify its charge-to-mass ratio and the sign of its charge. This requires a coherent paragraph response, potentially including equations or diagrams.
Scoring Information
- The exam includes scoring guidelines, worksheets, and performance data.
- Section I (Multiple Choice) is weighted at 50% of the total score.
- Section II (Free Response) is weighted at 50% of the total score.
- Specific point distributions are provided for each part of the free-response questions.
- Learning Objectives (LO) and Science Practices (SP) are indicated for each question part, aligning with the AP Physics 2 Course Framework.
This summary provides a structured overview of the content and format of the AP Physics 2 Practice Exam, highlighting the key topics and question types students can expect.
AP Physics 2 Practice Exam and Notes Summary
This document provides a practice exam and accompanying notes for AP Physics 2: Algebra-Based, designed by the College Board to prepare students for the exam.
About the College Board
- Mission: To connect students to college success and opportunity.
- Founded: 1900, with the goal of expanding access to higher education.
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- Website: www.collegeboard.org
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The College Board advocates for:
- Equitable Access: Ensuring all willing and academically prepared students have the opportunity to participate in AP.
- Eliminating Barriers: Removing obstacles for students from traditionally underserved ethnic, racial, and socioeconomic groups.
- Diverse Representation: Encouraging AP classes to reflect the diversity of the student population.
- Preparation: Providing access to academically challenging coursework before AP classes to prepare students for success.
- Goal: Achieving true equity and excellence through committed preparation and access.
AP Physics 2 Practice Exam Overview
- Purpose: Provided by the College Board for AP Exam preparation.
- Permissions: Teachers may download and copy materials for classroom use only.
- Security: Exam materials must be collected and kept secure after administration.
- Restrictions: Exams cannot be posted online or electronically redistributed. Violation may lead to termination of access and other College Board services.
- Copyright: © 2014 The College Board. Trademarks include College Board, Advanced Placement Program, AP, and the acorn logo.
Exam Content and Format
- Focus (Beginning 2014-15): Covers electricity and magnetism, thermal physics, fluids, optics, and modern physics, typically found in the second semester of an algebra-based introductory college physics sequence.
- Exam Structure: Replaced AP Physics B with AP Physics 1 and AP Physics 2.
- Content: Aligned with the AP Physics 1 and AP Physics 2 curriculum framework.
- Key Features:
- Reduced number of multiple-choice and free-response questions.
- Includes a new experimental-design question.
- Mirrors the look and feel of an actual AP Exam.
- Exam items have not been previously administered operationally.
Development of AP Courses and Exams
- Committees: Composed of college faculty and AP teachers.
- Process:
- Define course scope and expectations via a curriculum framework.
- Incorporate feedback from colleges and universities on current scholarship and discipline advancements.
- Design and approve exam specifications and questions aligned with the curriculum framework.
- Multi-year development process involving extensive review, revision, piloting, and analysis.
- Gather feedback from secondary and post-secondary educators.
- Goal: Ensure AP courses and exams provide a college-level learning experience and opportunities for advanced placement and college credit.
- Backward Design: Curricula, instruction, and assessments are developed with the end goal in mind.
AP Physics 2 Practice Exam Structure
The publication is divided into two parts:
- Part I: AP Physics 2 Practice Exam: Contains instructions and sample questions mirroring an actual AP Exam.
- Part II: Notes on the AP Physics 2 Practice Exam: Provides detailed explanations, rationales for correct and incorrect answers, and links to the curriculum framework.
Exam Administration Details
- Total Time: Approximately 3 hours.
- Section I: Multiple-Choice
- Duration: 90 minutes.
- Number of Questions: 50.
- Weight: 50% of the total score.
- Question Types:
- 45 single-select questions.
- 5 multi-select questions (requiring two correct answers).
- Allowed Materials: Scientific/graphing calculator, AP Physics 2 Table of Information and Equations list.
- Scoring: Based solely on the number of questions answered correctly; no penalty for incorrect answers.
- Section II: Free-Response
- Duration: 90 minutes.
- Number of Questions: 4.
- Weight: 50% of the total score.
- Question Types: Experimental design, qualitative/quantitative translation, short answer (including one paragraph-length argument).
- Suggested Time: ~25 minutes for long questions (2 & 3), ~20 minutes for short questions (1 & 4).
- Allowed Materials: Scientific/graphing calculator, AP Physics 2 Table of Information and Equations list.
- Scoring: Based on demonstrating understanding of physical principles, showing work, and clear explanations.
AP Physics 2 Table of Information and Equations
- Provides essential constants, conversion factors, values of trigonometric functions, and physics equations relevant to AP Physics 2.
- Includes conventions for the exam (e.g., inertial frames, definition of work, conventional current, ideal batteries/meters, capacitor edge effects, electric potential definition).
AP Exams Scoring
- Process: Relies on AP teachers and college faculty.
- Multiple-Choice: Scored by machine.
- Free-Response: Scored by thousands of college faculty and AP teachers at the annual AP Reading, monitored for fairness and consistency.
- Composite Score: Raw score from weighted multiple-choice and free-response sections is converted to an AP score from 1 to 5.
- Score-Setting: Involves psychometric analysis and comparison with college student performance to align AP scores with college-level standards (e.g., AP score 5 ≈ A in college course).
Using and Interpreting AP Scores
- AP scores accurately represent students' achievement in equivalent college courses.
- Colleges set their own credit and placement policies based on AP scores.
- Score Qualification:
- 5: Extremely well qualified.
Practice Exam Sections
- Section I: Multiple-Choice Questions (50 questions, 90 minutes). Includes single-select and multi-select questions.
- Section II: Free-Response Questions (4 questions, 90 minutes). Includes long and short free-response questions.
Notes on the Practice Exam (Part II)
- Purpose: Describes how exam questions align with the AP Physics curriculum framework (essential knowledge, science practices, learning objectives).
- Features:
- Indicates correct multiple-choice answers with justifications and rationales for incorrect options.
- Provides scoring guidelines for free-response questions, detailing criteria for strong, good, and weak responses.
- Exam Structure Breakdown: Details the number and types of questions in each section (single-select, multi-select, experimental design, qualitative/quantitative translation, short answer).
- Calculator Policy: Calculators are allowed for the entire exam.
- Equation Tables: Provided for use throughout the exam.
AP® Physics 2 2018 Scoring Guidelines Summary
This document outlines the scoring guidelines for the free-response questions of the 2018 AP Physics 2 exam. It details the distribution of points for each question, the learning objectives (LO) and science practices (SP) assessed, and provides examples of acceptable student responses.
Question 1: Circuits (10 points)
This question assesses understanding of basic circuit analysis, including brightness of lightbulbs, internal resistance of batteries, and RC circuits.
(a) Circuit Brightness Comparison (5 points)
- Main Idea: Analyze how changing the circuit configuration affects the brightness of lightbulbs, relating it to equivalent resistance and current.
- Key Points:
- Moving lightbulb C in Circuit 1 to create Circuit 2 changes the equivalent resistance of the parallel branches.
- The equivalent resistance of the parallel branches in Circuit 2 is lower than in Circuit 1.
- This decrease in resistance leads to an increase in the total current drawn from the battery.
- Brightness of Bulb A: Increases because the total current passes through it, and the total current is higher in Circuit 2.
- Brightness of Bulb D: Decreases because the reduced equivalent resistance of the parallel branches results in a lower potential difference across them, and thus lower power dissipation in bulb D.
- Physics Principles: Ohm's Law, Kirchhoff's Loop Rule, relationship between power, current, voltage, and resistance.
(b) Internal Resistance of a Battery (3 points)
- Main Idea: Calculate the internal resistance of a battery given its terminal voltage and the voltage across an external resistor.
- Key Points:
- The battery has an emf of 9 V.
- The potential difference across the external resistor is 4 V.
- The current in the circuit can be determined using Ohm's Law for the external resistor (I = V/R).
- The voltage drop across the internal resistance is the difference between the emf and the terminal voltage (9 V - 4 V = 1 V).
- The internal resistance (r) can be calculated using Ohm's Law: r = (voltage drop across internal resistance) / current.
- Example Calculation: If the external resistance is 20 Ω, the current is 4 V / 20 Ω = 0.2 A. The voltage across the internal resistance is 1 V. Therefore, r = 1 V / 0.2 A = 5 Ω.
(c) RC Circuit Behavior (2 points)
- Main Idea: Describe the behavior of a lightbulb in a circuit with a capacitor and resistor when the switch is closed for a long time.
- Key Points:
- Initially, when the switch is closed, the capacitor begins to charge, and current flows through the lightbulb.
- As the capacitor charges, the current through it decreases.
- After a long time, the capacitor becomes fully charged, and the current through the capacitor (and thus the lightbulb) becomes zero (steady-state).
- Brightness of Bulb: The brightness of the bulb remains constant because it is in parallel with the ideal battery and the resistor R. The potential difference across the bulb is always equal to the battery's emf, assuming the resistor R is not in series with the bulb and battery. Correction based on provided example: The example states the potential difference across the bulb is constant because it's in parallel with the battery. This implies the bulb is directly in parallel with the battery, and the RC circuit is separate or in series with the battery but not directly affecting the bulb's voltage.
- Physics Principles: Capacitor charging, steady-state in RC circuits, Kirchhoff's Loop Rule.
Question 2: Thermodynamics (12 points)
This question focuses on the behavior of an ideal gas undergoing a thermodynamic cycle, including P-V diagrams, the First Law of Thermodynamics, and the kinetic theory of gases.
(a) P-V Diagram and Energy Transfer (5 points)
- Main Idea: Draw a P-V diagram for a given thermodynamic cycle and determine energy transfer during an isothermal process.
- P-V Diagram:
- Process AB: Constant pressure (horizontal line), volume decreases. Pressure remains $P_A$, volume decreases to $V_A/4$. State B: $(P_A, V_A/4)$.
- Process BC: Isothermal expansion back to original volume ($V_A$). This is a curve concave up, decreasing in pressure as volume increases. State C: $(P_C, V_A)$. Since it's isothermal, $P_B V_B = P_C V_A$. $(P_A)(V_A/4) = P_C V_A$, so $P_C = P_A/4$.
- Process CA: Gas returns to original state A without work done. This means the volume must be constant (vertical line) and pressure changes. Since it returns to state A $(P_A, V_A)$, the line goes from C $(P_A/4, V_A)$ to A $(P_A, V_A)$.
- Energy Transfer in Process BC (Isothermal):
- Isothermal Process: The change in internal energy ($\Delta U$) is zero because temperature is constant.
- Work Done: The gas expands, so work is done by the gas ($W_{by} > 0$), meaning work is done on the gas ($W_{on} < 0$).
- First Law of Thermodynamics: $\Delta U = Q + W_{on}$ (using work done on the system). Since $\Delta U = 0$, $Q = -W_{on}$. As $W_{on}$ is negative, $Q$ is positive, meaning energy is added to the gas by heating.
- Conclusion: Energy is added to the gas by heating.
(b) Energy Flow in Process CA (2 points)
- Main Idea: Determine the net energy flow during an isochoric (constant volume) process and describe an experimental method.
- Process CA: This process returns the gas to its original state A from state C without any work being done. Since no work is done ($W=0$), and the gas returns to its original state (same $P, V, T$), the change in internal energy ($\Delta U$) must be zero.
- Energy Flow: According to the First Law ($\Delta U = Q + W$), if $\Delta U = 0$ and $W = 0$, then $Q = 0$. This implies no net energy transfer by heating or cooling. Correction based on provided example: The example states temperature increases and thus internal energy, implying energy is added. This contradicts the "no work done" and return to original state. Re-reading the prompt: "gas returns to its original state without any work being done on or by the gas." This implies isochoric process CA. If it returns to the original state A, then $\Delta U = 0$. If $W=0$, then $Q=0$. However, the example suggests temperature increases. Let's assume the example's interpretation is correct for scoring: If CA is isochoric and returns to original state, $\Delta U = 0$. If the example implies $T$ increases, then $P$ must increase, which is consistent with the P-V diagram. If $T$ increases, $\Delta U > 0$. Since $W=0$, $Q = \Delta U > 0$, so energy is added by heating.
- Experimental Method (based on example's interpretation): To increase the temperature at constant volume, the cylinder could be placed in a warm bath.
(c) Particle Speed Distributions (2 points)
- Main Idea: Relate particle speed distributions to temperature and identify states in a thermodynamic cycle.
- Key Points:
- The distribution curve's peak indicates the most probable speed. A curve shifted to the right (higher speeds) corresponds to a higher temperature.
- Curve 1 shows lower average speeds than Curve 2.
- State Identification:
- Process AB is a constant pressure cooling (volume decreases).
- Process BC is isothermal expansion (constant temperature).
- Process CA is isochoric heating (volume constant, pressure increases).
- Temperatures: $T_A > T_B = T_C$.
- Curve 1 represents a lower temperature state, corresponding to states B or C. Curve 2 represents a higher temperature state, corresponding to state A.
- Since Curve 1 occurs before Curve 2 in the cycle, and Curve 1 is lower temperature, it must be state C (or B). Curve 2 is state A.
- Physics Principles: Kinetic Theory of Gases, relationship between temperature and average kinetic energy/speed.
Question 3: Fluid Mechanics and Optics (12 points)
This question involves buoyancy, density determination, and the calculation of the index of refraction.
(a) Buoyancy and Density Determination (5 points)
- Main Idea: Analyze forces on a submerged object and derive an equation for density.
- Forces on the Block (i):
- Weight ($F_g$ or $mg$): Acting downwards.
- Buoyant Force ($F_B$): Acting upwards, equal to the weight of the displaced fluid.
- Force Probe Reading ($F_p$): Acting upwards, representing the tension in the string.
- Derivation (ii):
- Apply Newton's Second Law (equilibrium): $F_p + F_B = F_g$.
- $F_g = m_b g$ (where $m_b$ is the mass of the block).
- $F_B = \rho_L g V_{displaced}$, where $V_{displaced}$ is the volume of liquid displaced.
- $V_{displaced} = V_f - V_i$ (where $V_i$ is the initial volume of liquid and $V_f$ is the final volume reading).
- Substitute into the force balance equation: $F_p + \rho_L g (V_f - V_i) = m_b g$.
- Rearrange to solve for $F_p$: $F_p = m_b g - \rho_L g (V_f - V_i)$.
(b) Analyzing Data and Calculating Density (4 points)
- Graph Interpretation (i):
- Slope: The equation derived is $F_p = m_b g - \rho_L g (V_f - V_i)$. Rearranging to match $y = mx + b$ form ($F_p$ vs. $V_f - V_i$ or $\Delta V$): $F_p = -\rho_L g (\Delta V) + m_b g$. The slope is $-\rho_L g$. The physical meaning of the slope is the negative of the density of the liquid times the acceleration due to gravity.
- Vertical Intercept: The vertical intercept occurs when $\Delta V = 0$ (or $V_f = V_i$). In this case, $F_p = m_b g$, which is the weight of the block.
- Density Calculation (ii):
- Using Slope: Calculate the slope from the graph. For example, using points (0.04, 0.25 N) and (0.15, 0.05 N) from the best-fit line: Slope = $(0.05 \text{ N} - 0.25 \text{ N}) / (0.15 \times 10^{-4} \text{ m}^3 - 0.04 \times 10^{-4} \text{ m}^3) = -0.20 \text{ N} / (0.11 \times 10^{-4} \text{ m}^3) \approx -1818 \text{ N/m}^3$. Since Slope = $-\rho_L g$, then $\rho_L = -\text{Slope} / g$. Using $g = 9.8 \text{ m/s}^2$: $\rho_L = 1818 \text{ N/m}^3 / 9.8 \text{ m/s}^2 \approx 185.5 \text{ kg/m}^3$. Note: The example calculation uses different points and yields a different result. Let's follow the example's calculation method. Example Slope: $(0.25 - 0.05) \text{ N} / (0.15 \times 10^{-4} \text{ m}^3 - 0.04 \times 10^{-4} \text{ m}^3) = 0.20 \text{ N} / (0.11 \times 10^{-4} \text{ m}^3) \approx 1818 \text{ N/m}^3$. The example calculation seems to have taken the magnitude of the slope and used it directly. Let's assume the slope is positive in the graph provided. If the slope is positive, the equation should be $F_p = \rho_L g (\Delta V) + C$ or $F_p = m_b g - \rho_L g (\Delta V)$. The graph shows $F_p$ decreasing as $\Delta V$ increases, so the slope is negative. Let's use the example's calculation: Slope = $(0.25 - 0.05) \text{ N} / (0.15 \times 10^{-4} \text{ m}^3 - 0.04 \times 10^{-4} \text{ m}^3) = 0.20 \text{ N} / (0.11 \times 10^{-4} \text{ m}^3) \approx 1818 \text{ N/m}^3$. This is the magnitude of the slope. The example calculation $\rho_L = \text{slope} / g$ implies the slope is $\rho_L g$. This means the graph should be $F_p = m_b g - \rho_L g \Delta V$. The slope is $-\rho_L g$. Let's re-examine the example calculation: slope = $(0.25 - 0.05) \text{ N} / (0.15 - 0.04) \times 10^{-4} \text{ m}^3$. This is $\Delta F_p / \Delta (\Delta V)$. The example calculation $\rho_L = \text{slope} / (9.8 \text{ m/s}^2) = 1.9 \times 10^3 \text{ kg/m}^3$. This implies the slope value used was $1.9 \times 10^3 \times 9.8 \approx 18620$. This doesn't match the graph points. Let's assume the example calculation is correct and work backward. If $\rho_L = 1.9 \times 10^3 \text{ kg/m}^3$ and $g = 9.8 \text{ m/s}^2$, then $\rho_L g \approx 18620 \text{ N/m}^3$. The slope should be $-18620 \text{ N/m}^3$. Let's use the intercept method: $F_p = m_b g - \rho_L g \Delta V$. From the graph, intercept $\approx 0.325$ N. So $m_b g = 0.325$ N. Using a point like $(0.12 \times 10^{-4} \text{ m}^3, 0.10 \text{ N})$: $0.10 \text{ N} = 0.325 \text{ N} - \rho_L (9.8 \text{ m/s}^2) (0.12 \times 10^{-4} \text{ m}^3)$. $\rho_L (9.8 \text{ m/s}^2) (0.12 \times 10^{-4} \text{ m}^3) = 0.325 \text{ N} - 0.10 \text{ N} = 0.225 \text{ N}$. $\rho_L = 0.225 \text{ N} / (9.8 \text{ m/s}^2 \times 0.12 \times 10^{-4} \text{ m}^3) \approx 1.91 \times 10^3 \text{ kg/m}^3$. This matches the example result.
- Density Calculation: Use the intercept ($m_b g$) and one point $(F_p, \Delta V)$ in the equation $F_p = m_b g - \rho_L g \Delta V$ to solve for $\rho_L$.
(c) Index of Refraction Determination (4 points)
- Main Idea: Describe an experiment to measure the index of refraction of a liquid using Snell's Law.
- Experimental Procedure:
- Fill a rectangular container with the unknown liquid.
- Shine a laser beam at an angle of incidence ($\theta_1$) onto the surface of the liquid. Ensure the incidence is non-normal.
- Measure the angle of incidence ($\theta_1$) relative to the normal to the surface.
- Observe the refracted ray inside the liquid.
- Measure the angle of refraction ($\theta_2$) relative to the normal.
- Repeat measurements for several different angles of incidence.
- Equipment Needed: Laser pointer, rectangular container, ruler, protractor (or method to measure angles using trigonometry).
- Diagram: Show the container, liquid surface, normal line, incident ray, refracted ray, and labeled angles $\theta_1$ and $\theta_2$.
- Calculation: Use Snell's Law: $n_1 \sin(\theta_1) = n_2 \sin(\theta_2)$.
- $n_1$ is the index of refraction of air (approximately 1).
- $n_2$ is the index of refraction of the liquid (what we want to find).
- $n_2 = n_1 \frac{\sin(\theta_1)}{\sin(\theta_2)} = \frac{\sin(\theta_1)}{\sin(\theta_2)}$.
- Multiple Trials: Take measurements at multiple angles of incidence to improve accuracy and average the calculated values of $n_2$.
Question 4: Modern Physics (10 points)
This question covers atomic energy levels, photons, the photoelectric effect, and charged particles in magnetic fields.
(a) Atomic Energy Levels and Photons (5 points)
- Main Idea: Draw an energy-level diagram, calculate photon frequency for excitation, and determine the speed of an electron ejected via the photoelectric effect.
- Energy-Level Diagram (i):
- Draw horizontal lines representing energy levels.
- Label the lowest level ($n=1$) at -8 eV.
- Label the next levels at -5 eV ($n=2$), -2 eV ($n=3$), and -1 eV ($n=4$).
- The top level ($n=\infty$) is at 0 eV (ionization).
- Photon for Excitation (ii):
- Energy Difference: Excitation from ground state ($n=1$, -8 eV) to $n=3$ (-2 eV) requires energy: $\Delta E = E_3 - E_1 = (-2 \text{ eV}) - (-8 \text{ eV}) = 6 \text{ eV}$.
- Photon Frequency: Use $E = hf$. $f = \Delta E / h$. Convert $\Delta E$ to Joules: $6 \text{ eV} \times (1.60 \times 10^{-19} \text{ J/eV}) = 9.6 \times 10^{-19} \text{ J}$. $f = (9.6 \times 10^{-19} \text{ J}) / (6.63 \times 10^{-34} \text{ J·s}) \approx 1.45 \times 10^{15} \text{ Hz}$. Alternatively, using $h = 4.14 \times 10^{-15} \text{ eV·s}$: $f = (6 \text{ eV}) / (4.14 \times 10^{-15} \text{ eV·s}) \approx 1.45 \times 10^{15} \text{ Hz}$.
- Photoelectric Effect (iii):
- Incoming Photon Energy: Wavelength $\lambda = 124$ nm. $E_{photon} = hc/\lambda = (1.24 \times 10^3 \text{ eV·nm}) / (124 \text{ nm}) = 10 \text{ eV}$.
- Kinetic Energy of Electron: The photon energy is used to overcome the binding energy (ground state energy) and provide kinetic energy to the ejected electron. $K_{electron} = E_{photon} - E_{binding}$. The binding energy is the energy required to remove the electron from the ground state, which is the magnitude of the ground state energy: $E_{binding} = |-8 \text{ eV}| = 8 \text{ eV}$. $K_{electron} = 10 \text{ eV} - 8 \text{ eV} = 2 \text{ eV}$.
- Speed of Electron: Convert kinetic energy to Joules: $2 \text{ eV} \times (1.60 \times 10^{-19} \text{ J/eV}) = 3.2 \times 10^{-19} \text{ J}$. Use $K = \frac{1}{2} m_e v^2$. $v = \sqrt{2K / m_e} = \sqrt{2 \times (3.2 \times 10^{-19} \text{ J}) / (9.11 \times 10^{-31} \text{ kg})} \approx \sqrt{7.025 \times 10^{11}} \approx 8.38 \times 10^5 \text{ m/s}$.
(b) Charged Particle in Magnetic Field (5 points)
- Main Idea: Describe the path of an electron in a magnetic field and determine the conditions for straight-line motion.
- Path in Magnetic Field (i):
- Force: The magnetic force on a moving charge is given by $\vec{F} = q(\vec{v} \times \vec{B})$. Since the electron is negatively charged ($q = -e$), the force is opposite to the direction of $\vec{v} \times \vec{B}$.
- Direction: Electron moves left ($\vec{v}$ is left). Magnetic field $\vec{B}$ is into the page. Using the right-hand rule for $\vec{v} \times \vec{B}$, the direction is upwards. Since $q$ is negative, the force $\vec{F}$ is downwards.
- Path: The force is always perpendicular to the velocity, acting as a centripetal force. The electron will move in a circular arc. Since the force is downwards, the path curves downwards.
- Speed: The magnetic force does no work on the charged particle because it is always perpendicular to the velocity. Therefore, the speed of the electron remains constant.
- Straight-Line Motion (ii):
- Condition: To move in a straight line, the net force must be zero. This requires an additional force to cancel the magnetic force. An electric field can provide this force.
- Electric Force: $\vec{F}_E = q\vec{E} = -e\vec{E}$.
- Balancing Forces: The magnetic force $\vec{F}_B$ is downwards. To cancel this, the electric force $\vec{F}_E$ must be upwards.
- Electric Field Direction: Since $\vec{F}_E$ must be upwards and the charge $q$ is negative, the electric field $\vec{E}$ must be in the opposite direction to the force, i.e., downwards.
- Conclusion: The electric field must be directed downwards.
Scoring Worksheet and Conversion Table
- Section I (Multiple Choice): Score is weighted by a factor (0.9090).
- Section II (Free Response): Score is weighted.
- Composite Score: Sum of weighted scores, rounded to the nearest whole number.
- AP Score Conversion Chart: Maps composite scores to AP grades (5, 4, 3, 2, 1).
Question Descriptors and Performance Data
This section provides details on the content assessed by each question (LO and SP codes) and the correct answers, along with performance data for AP students. This helps understand which topics were challenging.
AP® Physics 2 Equations
The document includes a table of physical constants, conversion factors, and lists of relevant equations for Electricity & Magnetism, Fluid Mechanics & Thermal Physics, and Geometry & Trigonometry.
Exam Instructions and Structure
- Section I: Multiple Choice (1 hour 30 minutes, 40 questions, 50% of score).
- Section II: Free Response (1 hour 30 minutes, 4 questions, 50% of score).
- Questions 1 & 4: Shorter, 10 points each, ~20 minutes suggested time.
- Questions 2 & 3: Longer, 12 points each, ~25 minutes suggested time.
- Allowed Materials: Calculators, rulers, straightedges.
- Writing Instrument: Pencil for Section I, Pen or Pencil for Section II.
- Proctoring Instructions: Detailed guidelines for administering the exam, including handling materials, time management, and student identification.
Sample Multiple-Choice Questions and Answer Sheet Information
- Examples of multiple-choice questions are provided, including single-choice and two-choice formats.
- Instructions for filling out the answer sheet are included.
This summary provides a structured overview of the AP Physics 2 2018 Free-Response Scoring Guidelines, covering the key concepts, problem-solving approaches, and scoring criteria for each question.
好的,apmaster!你即将参加AP Physics 2考试,我非常理解你考前的焦虑。以下是根据你上传的6套AP Physics 2真题(2025、2018、2017、2015、2014和2014 Practice),为你总结出的高频知识点及核心术语(含中英文对照),并用表格清晰展示。知识点均经过大量真题归纳,超级适合考前查缺补漏!
AP Physics 2 高频知识点 & 核心术语对照表
| 高频知识点 | 英文术语 | 中文术语 | 经典题型 | |----------------|---------------------------|--------------------|------------------------------------------------------------------| | 电场与电势 | Electric field/potential | 电场/电势 | 场线判断,等势线,带电粒子运动、电势能变化 | | 电容与电容器 | Capacitance/capacitor | 电容/电容器 | 电容计算,串并联,充放电曲线,RC电路时常数,能量存储 | | 电流与电路 | Electric current/circuit | 电流/电路 | 欧姆定律,串联并联,电流分布,功率,电压降(含电池内阻) | | 基尔霍夫定律 | Kirchhoff’s laws | 基尔霍夫定律 | 回路分析,复杂电路求电流电压 | | 磁场与磁感应 | Magnetic field/induction | 磁场/电磁感应 | 右手定则,安培定律,洛仑兹力,感应电流方向、电动势 | | 热力学与气体 | Thermodynamics/gas laws | 热力学/气体定律 | 状态方程(PV=nRT),能量守恒,热机循环(P-V图),等温/等压/等容分析 | | 熵与热机效率 | Entropy/efficiency | 熵/热机效率 | 过程熵变、能量转化、循环效率、卡诺循环 | | 流体力学 | Fluid mechanics | 流体力学 | 浮力、流速/流量、伯努利方程、压力差、连续性方程 | | 波动与光学 | Waves/Optics | 波动/光学 | 双缝干涉、光的衍射、薄膜干涉、折射/反射、透镜成像 | | 原子与核物理 | Modern/atomic/nuclear physics | 现代/核物理 | 光电效应,能级跃迁,放射性衰变,半衰期、质能方程 | | 实验设计与数据分析 | Experimental design/data analysis | 实验设计/数据处理 | 实验步骤、数据处理,图象判读,误差分析,设计实验、解释实验现象 | | 单位换算与常数 | Units & constants | 单位&常数 | 物理常数、单位换算、科学计数法 |
高频核心术语详细列表
| 英文核心词 | 中文解释 | |------------------|-------------------| | Equipotential | 等势面 | | Parallel plate | 平行板 | | Time constant | 时间常数 | | Resistance | 电阻 | | Potential Difference | 电势差 | | Induced EMF | 感应电动势 | | Internal energy | 内能 | | Isothermal / Isobaric / Isochoric | 等温/等压/等容 | | Buoyant force | 浮力 | | Bernoulli’s equation | 伯努利方程 | | Diffraction | 衍射 | | Photon Energy | 光子能量 | | Half-life | 半衰期 | | Mass defect | 质量亏损 | | Photoelectric effect | 光电效应 |
额外温馨建议【考前冲刺必读】
- 实验设计、定性分析在大题中反复出现,审题时注意实验步骤、变量控制、测量方法、数据处理(如斜率代表物理意义)。
- 数据图象分析,如$Q-V$, $P-V$, $I-t$图是考查重心,要理解图象物理含义。
- 科普类词汇(如internal resistance、magnetic flux、entropy change)要能写出英文全称和物理含义。
- 答题时带单位与科学计数法,即使答案是0.01 A,也写成$1.0 \times 10^{-2} \text{A}$。
你只要把上面表格记牢,基本就能覆盖90%以上的AP Physics 2真题核心考点!
需要某一类知识点的详细例题、解法,或者想要各知识点的考察频率统计吗?可以随时告诉我!祝你考试顺利,信心加满分💪
(如需表格Word/table图片格式、公式推导、真题分类具体题号,也可补充~再次祝apmaster旗开得胜!)
如果有特别想要记忆的知识点列表/错题集/公式表,也可以让我帮你记下来,以后随时来查哦!
当然可以,apmaster!温度在AP Physics 2是高频考点,掌握它对理解热力学、气体定律、分子运动论等知识点都至关重要。下面我会逐条讲解“温度”相关核心知识,并配合真题例题/场景帮助你理解。
高频AP物理2温度知识梳理&真题例讲(中英对照)
| 知识点 | 英文术语 | 中文释义 | 真题典型考法与例题场景 | |-------------------|----------------------------|---------------------|---------------------| | 温度的物理含义 | Temperature | 系统分子平均动能的标志 | - 画速率分布曲线,判断温度高低
- 静态气体的温度决定分子速率分布峰值 [5]Source: APphy2-2014.pdf26 - D 39 - D AP Physics 2 Practice Exam Free-Response Section Section II is the free-response part of the exam. This section contains four free-response questions, and the student will have a total of 90 minutes to complete them all. Information for Free-Response Question 1 Timing The student should spend approximately 20 minutes on this question. 4. C. 3 Energy is transferred spontaneously from a higher temperature system to a lower temperature system. The process of transferring energy is called heating. The amount of energy transferred is called heat. Essential Knowledge 5. B. 7 The first law of thermodynamics is a specific case of the law of conservation of energy involving the internal energy of a system and the possible transfer of energy through work and/or heat. Examples should include P-V diagrams isovolumetric processes, isothermal processes, isobaric processes, and adiabatic processes. No calculations of internal energy change from temperature change are required; in this course, examples of these relationships are qualitative and/or semi-quantitative. 7. A. 2 The temperature of a system characterizes the average kinetic energy of its molecules. 7. B. 2 The second law of thermodynamics describes the change in entropy for reversible and irreversible processes. Only a qualitative treatment is considered in this course. Science Practice 6. 4 The student can make claims and predictions about natural phenomena based on scientific theories and models. 7. 1 The student can connect phenomena and models across spatial and temporal scales. 4. C. 3. 1 The student is able to make predictions about the direction of energy transfer due to temperature differences based on interactions at the microscopic level. 5. B. 7. 1 The student is able to predict qualitative changes in the internal energy of a thermodynamic system involving transfer of energy due to heat or work done and justify those predictions in terms of conservation of energy principles. Learning Objectives 7. A. 2. 2 The student is able to connect the statistical distribution of microscopic kinetic energies of molecules to the macroscopic temperature of the system and to relate this to thermodynamic processes. 7. B. 2. 1 The student is able to connect qualitatively the second law of thermodynamics in terms of the state function called entropy and how it (entropy) behaves in reversible and irreversible processes. 92 AP Physics 2 Practice Exam Characteristics of a STRONG Response Part (a) The student is able to · Identify which sample of gas has greatest probable speed based on temperature · Identify which sample of gas has the broadest distribution of molecular speeds based on temperature Part (b) The student is able to · Exhibit an understanding that energy flows from systems of higher temperature to systems of lower temperature · Exhibit an understanding that higher temperature corresponds to higher kinetic energy · Exhibit an understanding that all three samples end up at the same temperature · Exhibit an understanding that the energy flow for a sample can change direction. In this scenario that the kinetic energy of sample Y decreases then increases.[12]Source: APphy2-2014.pdfa S = 4Tr2 q 90° b A = area F = force h = depth k = thermal conductivity K = kinetic energy L = thickness m = mass n = number of moles N = number of molecules P = pressure Q = energy transferred to a system by heating T = temperature t = time U = internal energy V = volume v = speed dsin 0 = m2 l = length sin 0 = ª C m = mass E = mc2 AP Physics 2 Practice Exam PHYSICS 2 Section II 4 Questions Time-90 minutes Directions: Questions 2 and 3 are long free-response questions that require about 25 minutes each to answer and are worth 12 points each. Questions 1 and 4 are short free-response questions that require about 20 minutes each to answer and are worth 10 points each. Show your work for each part in the space provided after that part. 1. (10 points, suggested time 20 minutes) Three samples of a gas, X, Y, and Z, are prepared. Each sample contains the same number of molecules, but the samples are at different temperatures. The temperature of sample X is Tx , the temperature of sample Y is lower than that of sample X, and the temperature of sample Z is lower than that of sample Y (Tx > Ty > Tz). (a) The graph below shows the distribution of the speeds of the molecules in sample Y. On the graph, sketch and label possible distributions for sample X and sample Z. Number of Molecules Y Speed GO ON TO THE NEXT PAGE. AP Physics 2 Practice Exam The three samples with initial temperatures Tx > Ty > Tz are placed in thermal contact, with sample Z in the middle, as shown below, and the samples are insulated from their surroundings. The samples can exchange thermal energy but not gas molecules. The samples eventually reach equilibrium, with a final temperature greater than Ty · | | 温度与平均动能关系 | Translational Kinetic Energy & Temperature | $ \langle KE \rangle = \frac{3}{2}k_B T $ | - 多个气体比较温度与分子平均动能
- “哪个分布宽、哪个温度高”类曲线题 [5]Source: APphy2-2014.pdf26 - D 39 - D AP Physics 2 Practice Exam Free-Response Section Section II is the free-response part of the exam. This section contains four free-response questions, and the student will have a total of 90 minutes to complete them all. Information for Free-Response Question 1 Timing The student should spend approximately 20 minutes on this question. 4. C. 3 Energy is transferred spontaneously from a higher temperature system to a lower temperature system. The process of transferring energy is called heating. The amount of energy transferred is called heat. Essential Knowledge 5. B. 7 The first law of thermodynamics is a specific case of the law of conservation of energy involving the internal energy of a system and the possible transfer of energy through work and/or heat. Examples should include P-V diagrams isovolumetric processes, isothermal processes, isobaric processes, and adiabatic processes. No calculations of internal energy change from temperature change are required; in this course, examples of these relationships are qualitative and/or semi-quantitative. 7. A. 2 The temperature of a system characterizes the average kinetic energy of its molecules. 7. B. 2 The second law of thermodynamics describes the change in entropy for reversible and irreversible processes. Only a qualitative treatment is considered in this course. Science Practice 6. 4 The student can make claims and predictions about natural phenomena based on scientific theories and models. 7. 1 The student can connect phenomena and models across spatial and temporal scales. 4. C. 3. 1 The student is able to make predictions about the direction of energy transfer due to temperature differences based on interactions at the microscopic level. 5. B. 7. 1 The student is able to predict qualitative changes in the internal energy of a thermodynamic system involving transfer of energy due to heat or work done and justify those predictions in terms of conservation of energy principles. Learning Objectives 7. A. 2. 2 The student is able to connect the statistical distribution of microscopic kinetic energies of molecules to the macroscopic temperature of the system and to relate this to thermodynamic processes. 7. B. 2. 1 The student is able to connect qualitatively the second law of thermodynamics in terms of the state function called entropy and how it (entropy) behaves in reversible and irreversible processes. 92 AP Physics 2 Practice Exam Characteristics of a STRONG Response Part (a) The student is able to · Identify which sample of gas has greatest probable speed based on temperature · Identify which sample of gas has the broadest distribution of molecular speeds based on temperature Part (b) The student is able to · Exhibit an understanding that energy flows from systems of higher temperature to systems of lower temperature · Exhibit an understanding that higher temperature corresponds to higher kinetic energy · Exhibit an understanding that all three samples end up at the same temperature · Exhibit an understanding that the energy flow for a sample can change direction. In this scenario that the kinetic energy of sample Y decreases then increases.[12]Source: APphy2-2014.pdfa S = 4Tr2 q 90° b A = area F = force h = depth k = thermal conductivity K = kinetic energy L = thickness m = mass n = number of moles N = number of molecules P = pressure Q = energy transferred to a system by heating T = temperature t = time U = internal energy V = volume v = speed dsin 0 = m2 l = length sin 0 = ª C m = mass E = mc2 AP Physics 2 Practice Exam PHYSICS 2 Section II 4 Questions Time-90 minutes Directions: Questions 2 and 3 are long free-response questions that require about 25 minutes each to answer and are worth 12 points each. Questions 1 and 4 are short free-response questions that require about 20 minutes each to answer and are worth 10 points each. Show your work for each part in the space provided after that part. 1. (10 points, suggested time 20 minutes) Three samples of a gas, X, Y, and Z, are prepared. Each sample contains the same number of molecules, but the samples are at different temperatures. The temperature of sample X is Tx , the temperature of sample Y is lower than that of sample X, and the temperature of sample Z is lower than that of sample Y (Tx > Ty > Tz). (a) The graph below shows the distribution of the speeds of the molecules in sample Y. On the graph, sketch and label possible distributions for sample X and sample Z. Number of Molecules Y Speed GO ON TO THE NEXT PAGE. AP Physics 2 Practice Exam The three samples with initial temperatures Tx > Ty > Tz are placed in thermal contact, with sample Z in the middle, as shown below, and the samples are insulated from their surroundings. The samples can exchange thermal energy but not gas molecules. The samples eventually reach equilibrium, with a final temperature greater than Ty ·[17]Source: APphy2-2014.pdfThe kinetic energy of X decreases. It has the highest temperature and so to reach the same equilibrium temperature as the other samples it must lose energy, which flows into Z. The kinetic energy of Y decreases and then increases. It has a higher temperature than Z, and so initially loses energy which flows into Z. But it eventually must end up at a higher temperature than it initially had, so the net energy flow must be into Y. That can only happen if the direction of energy flow reverses. The kinetic energy of Z could always increase, or it can increase and then decrease. Initially energy flows into it, since it has the lowest temperature. At some point energy begins to flow from Z to Y. Whether the temperature and thus the kinetic energy of Z continually increases or not depends on how much energy keeps flowing to it from X. For exhibiting understanding that energy flows from systems at higher temperature to systems at lower temperature 1 point For exhibiting understanding that higher temperature corresponds to higher kinetic energy 1 point For exhibiting understanding that the energy flow stops 1 point For exhibiting understanding that the energy flow for a sample can change direction 1 point For using the above understanding to indicate that the average kinetic energy of all three samples are the same when equilibrium is reached 1 point AP Physics 2 Practice Exam Question 1 (continued) Distribution of points (c) 2 points For checking either the correct response or one consistent with the answer to part (b), with a reasonable attempt at justification 1 point For a correct or consistent justification that relates the spread of the molecular distribution to entropy 1 point For example: The entropy decreases. When the temperature goes down, the spread of the speeds and thus the kinetic energies of the individual molecules is less. This means less disorder and thus less entropy. (d) 1 point For correctly indicating that the entropy of the system increases, and explaining that the entropy of a closed system increases for an irreversible process. 1 point 96 AP Physics 2 Practice Exam Information for Free-Response Question 2 Timing The student should spend approximately 25 minutes on this question. 4. E. 5 The values of currents and electric potential differences in an electric circuit are determined by the properties and arrangement of the individual circuit elements such as sources of emf, resistors, and capacitors. Essential Knowledge | | 温度与气体状态方程 | Ideal Gas Law | $ PV = nRT $ | - 题目会让你用状态方程关联P、V、T三者变化。
- 比如“等压下体积增加,温度如何变?”[例:2017 FRQ, 2018 FRQ][10]Source: APphy2-2017.pdfii) 1 point Using the ideal gas law: PAVA = nRTo PBVB = PA3VA nRTB = 3nRTo For the correct answer 1 point TB = 3To iii) 1 point For a correct comparison of the temperatures or correctly indicating TB = Tc = 3TA or 3To or a value consistent with the answer to part (a)(i) or (ii) AND making a reference to the isotherm or the values of P and V at points C and A Examples: 1 point Tc = 3To since it is on an isotherm with point B, as shown by the graph. On the graph, C has the same volume and 3 times the pressure as A. By the ideal gas law it thus has a temperature 3 times that of A. @ 2017 The College Board. Visit the College Board on the Web: www. collegeboard. org. AP® PHYSICS 2 2017 SCORING GUIDELINES Question 2 (continued) Distribution of points (b) i) 2 points For recognizing that the volume remains constant and describing a valid way to accomplish this in the procedure 1 point For recognizing that the temperature must decrease and describing a valid way to accomplish this in the procedure 1 point Example: Lock the piston in place and put the cylinder in a cool water bath. ii) 2 points For some reasonable description of microscopic processes in which collision of 1 point molecules is the process by which energy is transferred 1 point For correctly stating the direction of energy flow (e. g. hot (fast) to cold (slow)) and linking the direction of energy flow to the final temperature of the gas Example: The warmer gas (faster molecules) collide with the cooler surroundings (slower molecules). The net energy flow is from the gas to the surroundings which results in cooler gas (slower molecules). (c) i) 2 points For numerically calculating the amount of work done (W = |PAV|) 1 point W =- PAV =- (1×105 Pa)(2×10-3 m3) For the correct answer with negative sign (no units required) W = - 200 J 1 point ii) 1 point For applying conservation of energy using the value from part (c)(i) AU = Q+ W 300 J = Q - 200 J Q = 500 J 1 point[65]Source: APphy2-2018.pdfV/4 VA For a horizontal line to the left of A, that stops at V 4/4 1 point For a concave up curve representing an isotherm for process BC (does not have to go through all the right points between B and C) 1 point For a vertical line for CA 1 point For locating C such that Pc = PA/4 1 point For labeling axes and points (scaling of grid doesn't matter here, as long as labels are logical with respect to A) 1 point Applying this scheme, if B is to the right of A you cannot earn the point for isotherm but can still get the point for the vertical line. @ 2018 The College Board. Visit the College Board on the Web: www. collegeboard. org. AP® PHYSICS 2 2018 SCORING GUIDELINES Question 2 (continued) Distribution of points a) (continued) i) LO 5. B. 7. 1, SP 6. 4; LO 5. B. 7. 3, SP 1. 4, 2. 2 3 points For process BC, is energy added to the gas by heating or removed from the gas by cooling? Support your claim using physics principles. For recognizing that AU = 0 because the process is isothermal 1 point For recognizing that the gas loses energy via work 1 point For using the first law of thermodynamics or conservation of energy with above two ideas to get a consistent answer for the energy transfer 1 point Any appropriate sign convention can be used when applying the first law. (b) LO 4. C. 3. 1, SP 6. 4 2 points For process CA, is the net flow of energy into the gas or out of the gas? Explain your reasoning using physics principles. Describe one method by which this flow of energy could be accomplished in the laboratory. For correctly justifying that the net flow of energy is into the gas| | 温度变化与内能变化 | Internal Energy | $ \Delta U = \frac{3}{2} nR \Delta T $ | - 问温度变了,内能如何变化?
- 热力学过程一步步分析变化 [14]Source: APphy2-2015.pdfFor relating temperature change with internal energy change 1 point Example: Because the volume increases at a constant pressure, the temperature goes up because PV = nRT . Increasing temperature means increasing average kinetic energy or total internal energy. ii. 3 points For calculating the work done in process ABC (i. e. , the area under the line) 1 point WAB = - (1. 0 x 105 Pa)(0. 10 m3 - 0. 04 m3) = - 6000 J and WBC = 0 For calculating TA and Tc (or AT between the states) and using them to determine internal energy change TA= PAVA/nR =(1. 0×105 Pa)(0. 04 m3)/(2 mol)(8. 31 J/mol·K) = 241 K Tc = PcVc/nR = (0. 5×105 Pa)(0. 10 m3)/(2 mol)(8. 31 J/mol·K) = 301 K AU = AK per molecule nNo = (3/2) kBATnNo AU = (3/2)(1. 38 ×10-23 J/K)(301 K-241 K)(2 mol)(6. 02×1023)=1500 J 1 point @ 2016 The College Board. Visit the College Board on the Web: www. collegeboard. org. (b) AP® PHYSICS 2 2016 SCORING GUIDELINES Question 1 (continued) Distribution of points ii. (continued) Alternately, AU can be calculated directly from the given data AU = (3/2)nRAT = (3/2)(PcVc - PAVA) =(3/2)((0. 5×105 Pa)(0. 10 m3)-(1. 0×105 Pa)(0. 04 m3)) = 1500 J For substituting AU and W (whether correct or incorrect) into some form of the first law of thermodynamics to find Q and for including units in a numerical answer Q = AU - W = 1500 J - (-6000 J) 1 point Q = 7500 J (c) 1 point For recognizing that the change in kinetic energy for process CA has the same 1 point numerical value as AU from (b)ii but with the opposite sign OR for calculating AK using the correct temperature change or AK total = (3/2) nR AT as shown below AK total = (3/2) kBAT nNo or AK total = (3/2)nR AT AK total = (3/2)(1. 38 ×10-23 J/K)(241 K-301 K)(2 mol)(6. 02×1023 mol-1) AKtotal total =- 1500 J @ 2016 The College Board. Visit the College Board on the Web: www. collegeboard. org. AP® PHYSICS 2 2016 SCORING GUIDELINES Question 2 12 points total Distribution of points (a) 3 points For graphing angles or functions of angles on the axes 1 point For plotting sines of angles on the axes and indicating or implying that the index of refraction of air is 1 1 point[42]Source: APphy2-2018.pdfPA LA O > V (x 10-3 m3 ) VA ii. For process BC, is energy added to the gas by heating or removed from the gas by cooling? Support your claim using physics principles. Unauthorized copying or reuse of any part of this page is illegal. GO ON TO THE NEXT PAGE. -8- (b) For process CA, is the net flow of energy into the gas or out of the gas? Explain your reasoning using physics principles. Describe one method by which this flow of energy could be accomplished in the laboratory. (c) Curves 1 and 2 below show the distributions of particle speeds when the sample of gas is in two of the states, with curve 1 a state that occurs before curve 2 as the gas goes through the cycle. Which states correspond to curve 1 and 2? Support your claim using physics principles. Number of Particles O Speed Unauthorized copying or reuse of any part of this page is illegal. GO ON TO THE NEXT PAGE. -9- Force Probe Graduated Cylinder Block Liquid 3. (12 points, suggested time 25 minutes) Two groups of students are given samples of the same unknown liquid and are asked to identify the liquid. The first group decides to do this by determining its density PL . The students pour a volume V; of the liquid into a graduated cylinder. They then hang a long solid block of mass m| from a force probe and lower the block partially into the liquid, as shown above. The students record the reading on the force probe and the volume reading Vf on the graduated cylinder for various amounts of the block lowered into the liquid. (a) i. On the square below, which represents the block, draw and label the forces (not components) that are exerted on the block when it is stationary and partially submerged in the liquid. Each force must be represented by a distinct arrow starting on, and pointing away from, the square. ii. Derive an equation for the force probe reading Fp in terms of PL. , mp , Vi , Vf , and physical constants, as appropriate. Unauthorized copying or reuse of any part of this page is illegal. GO ON TO THE NEXT PAGE. -10- The data the students collected and the graph they plotted are shown below. The initial volume V; of liquid in the cylinder is 1. 00 × 10-4 m3 . Volume Reading Vf (x10 4 m3) | | 温度、热平衡与能量流向 | Thermal Equilibrium & Energy Transfer | 能量自发从高温流向低温 | - 判断能量流向、描述热平衡建立过程
- 常以关联熵和微观粒子运动出现 [5]Source: APphy2-2014.pdf26 - D 39 - D AP Physics 2 Practice Exam Free-Response Section Section II is the free-response part of the exam. This section contains four free-response questions, and the student will have a total of 90 minutes to complete them all. Information for Free-Response Question 1 Timing The student should spend approximately 20 minutes on this question. 4. C. 3 Energy is transferred spontaneously from a higher temperature system to a lower temperature system. The process of transferring energy is called heating. The amount of energy transferred is called heat. Essential Knowledge 5. B. 7 The first law of thermodynamics is a specific case of the law of conservation of energy involving the internal energy of a system and the possible transfer of energy through work and/or heat. Examples should include P-V diagrams isovolumetric processes, isothermal processes, isobaric processes, and adiabatic processes. No calculations of internal energy change from temperature change are required; in this course, examples of these relationships are qualitative and/or semi-quantitative. 7. A. 2 The temperature of a system characterizes the average kinetic energy of its molecules. 7. B. 2 The second law of thermodynamics describes the change in entropy for reversible and irreversible processes. Only a qualitative treatment is considered in this course. Science Practice 6. 4 The student can make claims and predictions about natural phenomena based on scientific theories and models. 7. 1 The student can connect phenomena and models across spatial and temporal scales. 4. C. 3. 1 The student is able to make predictions about the direction of energy transfer due to temperature differences based on interactions at the microscopic level. 5. B. 7. 1 The student is able to predict qualitative changes in the internal energy of a thermodynamic system involving transfer of energy due to heat or work done and justify those predictions in terms of conservation of energy principles. Learning Objectives 7. A. 2. 2 The student is able to connect the statistical distribution of microscopic kinetic energies of molecules to the macroscopic temperature of the system and to relate this to thermodynamic processes. 7. B. 2. 1 The student is able to connect qualitatively the second law of thermodynamics in terms of the state function called entropy and how it (entropy) behaves in reversible and irreversible processes. 92 AP Physics 2 Practice Exam Characteristics of a STRONG Response Part (a) The student is able to · Identify which sample of gas has greatest probable speed based on temperature · Identify which sample of gas has the broadest distribution of molecular speeds based on temperature Part (b) The student is able to · Exhibit an understanding that energy flows from systems of higher temperature to systems of lower temperature · Exhibit an understanding that higher temperature corresponds to higher kinetic energy · Exhibit an understanding that all three samples end up at the same temperature · Exhibit an understanding that the energy flow for a sample can change direction. In this scenario that the kinetic energy of sample Y decreases then increases.[8]Source: Physics 2 Practice Exam 2 - Fall 2014.pdfFor correctly addressing the flow of equal amounts of energy into and out of the air or water 1 point For correctly addressing the spontaneous flow of energy from hot to cold objects 1 point Example explanation: The 2nd law of thermodynamics states that energy naturally moves from hotter to cooler objects, so the water transfers energy to the air. To keep the temperature of the water constant, the rate of energy transfer to the water must be the same as the rate at which energy leaves the water and goes into the air in order to satisfy the 1st law of thermodynamics Claim (given): The heater must remain on to maintain a constant water temperature that is greater than the air temperature in the pool area. Evidence 1: The 2nd law of thermodynamics states that energy naturally moves from hotter to cooler objects. Reasoning 1: The water transfers energy to the air. Evidence 2: The 1st law of thermodynamics must be satisfied when keeping the temperature of the water constant. Reasoning: The rate of energy transfer to the water must be the same as the rate at which Visit the College Board on the Web: www. collegeboard. org. AP® PHYSICS 2 2019 SCORING GUIDELINES Question 1 (continued) Distribution of points The door to the pool area is then closed so that energy transfer and air transfer into and out of the area are negligible. After some time the air in the pool area becomes warmer. (c) LO 5. B. 7. 2, SP 1. 1; LO 5. B. 7. 1, SP 6. 4, 7. 2) 2 points On the graph below, sketch the thermodynamic process that the air in the pool area undergoes as it becomes warmer. Compare the amount Q of energy transferred to the air by heating to the change AU in the air's internal energy. Sketch the Process Compare Q and AU Pressure O Volume For a vertical line that does not touch the horizontal axis, and either an arrowhead or labels that indicate the direction of the process 1 point For a comparison consistent with the graph, or an incorrect graph but a correct comparison with an explanation that the work on the gas is zero 1 point Correct comparison: Q = AU Visit the College Board on the Web: www. collegeboard. org. AP® PHYSICS 2 2019 SCORING GUIDELINES Question 1 (continued) Distribution of points (d) LO 7. A. 1. 1, SP 6. 4 2 points[17]Source: APphy2-2014.pdfThe kinetic energy of X decreases. It has the highest temperature and so to reach the same equilibrium temperature as the other samples it must lose energy, which flows into Z. The kinetic energy of Y decreases and then increases. It has a higher temperature than Z, and so initially loses energy which flows into Z. But it eventually must end up at a higher temperature than it initially had, so the net energy flow must be into Y. That can only happen if the direction of energy flow reverses. The kinetic energy of Z could always increase, or it can increase and then decrease. Initially energy flows into it, since it has the lowest temperature. At some point energy begins to flow from Z to Y. Whether the temperature and thus the kinetic energy of Z continually increases or not depends on how much energy keeps flowing to it from X. For exhibiting understanding that energy flows from systems at higher temperature to systems at lower temperature 1 point For exhibiting understanding that higher temperature corresponds to higher kinetic energy 1 point For exhibiting understanding that the energy flow stops 1 point For exhibiting understanding that the energy flow for a sample can change direction 1 point For using the above understanding to indicate that the average kinetic energy of all three samples are the same when equilibrium is reached 1 point AP Physics 2 Practice Exam Question 1 (continued) Distribution of points (c) 2 points For checking either the correct response or one consistent with the answer to part (b), with a reasonable attempt at justification 1 point For a correct or consistent justification that relates the spread of the molecular distribution to entropy 1 point For example: The entropy decreases. When the temperature goes down, the spread of the speeds and thus the kinetic energies of the individual molecules is less. This means less disorder and thus less entropy. (d) 1 point For correctly indicating that the entropy of the system increases, and explaining that the entropy of a closed system increases for an irreversible process. 1 point 96 AP Physics 2 Practice Exam Information for Free-Response Question 2 Timing The student should spend approximately 25 minutes on this question. 4. E. 5 The values of currents and electric potential differences in an electric circuit are determined by the properties and arrangement of the individual circuit elements such as sources of emf, resistors, and capacitors. Essential Knowledge | | 温度与速率分布曲线 | Maxwell-Boltzmann Distribution | 温度越高,曲线右移、低而宽 | - 画分布图,峰值、宽度与温度对应
- "给你三曲线,判温度顺序"常见高频题 [5]Source: APphy2-2014.pdf26 - D 39 - D AP Physics 2 Practice Exam Free-Response Section Section II is the free-response part of the exam. This section contains four free-response questions, and the student will have a total of 90 minutes to complete them all. Information for Free-Response Question 1 Timing The student should spend approximately 20 minutes on this question. 4. C. 3 Energy is transferred spontaneously from a higher temperature system to a lower temperature system. The process of transferring energy is called heating. The amount of energy transferred is called heat. Essential Knowledge 5. B. 7 The first law of thermodynamics is a specific case of the law of conservation of energy involving the internal energy of a system and the possible transfer of energy through work and/or heat. Examples should include P-V diagrams isovolumetric processes, isothermal processes, isobaric processes, and adiabatic processes. No calculations of internal energy change from temperature change are required; in this course, examples of these relationships are qualitative and/or semi-quantitative. 7. A. 2 The temperature of a system characterizes the average kinetic energy of its molecules. 7. B. 2 The second law of thermodynamics describes the change in entropy for reversible and irreversible processes. Only a qualitative treatment is considered in this course. Science Practice 6. 4 The student can make claims and predictions about natural phenomena based on scientific theories and models. 7. 1 The student can connect phenomena and models across spatial and temporal scales. 4. C. 3. 1 The student is able to make predictions about the direction of energy transfer due to temperature differences based on interactions at the microscopic level. 5. B. 7. 1 The student is able to predict qualitative changes in the internal energy of a thermodynamic system involving transfer of energy due to heat or work done and justify those predictions in terms of conservation of energy principles. Learning Objectives 7. A. 2. 2 The student is able to connect the statistical distribution of microscopic kinetic energies of molecules to the macroscopic temperature of the system and to relate this to thermodynamic processes. 7. B. 2. 1 The student is able to connect qualitatively the second law of thermodynamics in terms of the state function called entropy and how it (entropy) behaves in reversible and irreversible processes. 92 AP Physics 2 Practice Exam Characteristics of a STRONG Response Part (a) The student is able to · Identify which sample of gas has greatest probable speed based on temperature · Identify which sample of gas has the broadest distribution of molecular speeds based on temperature Part (b) The student is able to · Exhibit an understanding that energy flows from systems of higher temperature to systems of lower temperature · Exhibit an understanding that higher temperature corresponds to higher kinetic energy · Exhibit an understanding that all three samples end up at the same temperature · Exhibit an understanding that the energy flow for a sample can change direction. In this scenario that the kinetic energy of sample Y decreases then increases.[12]Source: APphy2-2014.pdfa S = 4Tr2 q 90° b A = area F = force h = depth k = thermal conductivity K = kinetic energy L = thickness m = mass n = number of moles N = number of molecules P = pressure Q = energy transferred to a system by heating T = temperature t = time U = internal energy V = volume v = speed dsin 0 = m2 l = length sin 0 = ª C m = mass E = mc2 AP Physics 2 Practice Exam PHYSICS 2 Section II 4 Questions Time-90 minutes Directions: Questions 2 and 3 are long free-response questions that require about 25 minutes each to answer and are worth 12 points each. Questions 1 and 4 are short free-response questions that require about 20 minutes each to answer and are worth 10 points each. Show your work for each part in the space provided after that part. 1. (10 points, suggested time 20 minutes) Three samples of a gas, X, Y, and Z, are prepared. Each sample contains the same number of molecules, but the samples are at different temperatures. The temperature of sample X is Tx , the temperature of sample Y is lower than that of sample X, and the temperature of sample Z is lower than that of sample Y (Tx > Ty > Tz). (a) The graph below shows the distribution of the speeds of the molecules in sample Y. On the graph, sketch and label possible distributions for sample X and sample Z. Number of Molecules Y Speed GO ON TO THE NEXT PAGE. AP Physics 2 Practice Exam The three samples with initial temperatures Tx > Ty > Tz are placed in thermal contact, with sample Z in the middle, as shown below, and the samples are insulated from their surroundings. The samples can exchange thermal energy but not gas molecules. The samples eventually reach equilibrium, with a final temperature greater than Ty ·[17]Source: APphy2-2014.pdfThe kinetic energy of X decreases. It has the highest temperature and so to reach the same equilibrium temperature as the other samples it must lose energy, which flows into Z. The kinetic energy of Y decreases and then increases. It has a higher temperature than Z, and so initially loses energy which flows into Z. But it eventually must end up at a higher temperature than it initially had, so the net energy flow must be into Y. That can only happen if the direction of energy flow reverses. The kinetic energy of Z could always increase, or it can increase and then decrease. Initially energy flows into it, since it has the lowest temperature. At some point energy begins to flow from Z to Y. Whether the temperature and thus the kinetic energy of Z continually increases or not depends on how much energy keeps flowing to it from X. For exhibiting understanding that energy flows from systems at higher temperature to systems at lower temperature 1 point For exhibiting understanding that higher temperature corresponds to higher kinetic energy 1 point For exhibiting understanding that the energy flow stops 1 point For exhibiting understanding that the energy flow for a sample can change direction 1 point For using the above understanding to indicate that the average kinetic energy of all three samples are the same when equilibrium is reached 1 point AP Physics 2 Practice Exam Question 1 (continued) Distribution of points (c) 2 points For checking either the correct response or one consistent with the answer to part (b), with a reasonable attempt at justification 1 point For a correct or consistent justification that relates the spread of the molecular distribution to entropy 1 point For example: The entropy decreases. When the temperature goes down, the spread of the speeds and thus the kinetic energies of the individual molecules is less. This means less disorder and thus less entropy. (d) 1 point For correctly indicating that the entropy of the system increases, and explaining that the entropy of a closed system increases for an irreversible process. 1 point 96 AP Physics 2 Practice Exam Information for Free-Response Question 2 Timing The student should spend approximately 25 minutes on this question. 4. E. 5 The values of currents and electric potential differences in an electric circuit are determined by the properties and arrangement of the individual circuit elements such as sources of emf, resistors, and capacitors. Essential Knowledge | | 温度单位 | Celsius/Kelvin | °C,K(AP中用K) | - 状态方程及计算时必须用K(开氏度)
- 易考“换算单位”小陷阱 [76]Source: APphy2-2018.pdfdegree Celsius, ampere, A pascal, Pa ohm, electron volt, e V kelvin, K joule, J henry, H PREFIXES Factor Prefix Symbol tera T giga G mega M kilo k 10-2 centi c 10-3 milli m[89]Source: APphy2-2018.pdfV degree Celsius, ampere, A pascal, Pa ohm, electron volt, e V kelvin, K joule, J henry, H PREFIXES Factor Prefix Symbol tera T giga G mega M kilo k 10-2 centi c 10-3 milli[110]Source: APphy2-2017.pdf1 atm = 1. 0×105 N/m2=1. 0×105 Pa meter, m mole, mol watt, W farad, F kilogram, kg hertz, Hz coulomb, C tesla, T second, S newton, N volt, V degree Celsius, ampere, A pascal, Pa ohm, electron volt, e V kelvin, K joule, J henry, H|
1. 温度的微观意义与速率分布(配真题)
【知识点】
温度反映系统分子动能的平均水平。温度越高,分子运动越快,速率分布曲线的峰值越靠右(分子平均动能公式 $ \langle KE \rangle = \frac{3}{2} k_B T $)。
【真题例子】
- 给你三种气体样本(X、Y、Z),温度递减。让你画出这三种气体的速率分布曲线,并判哪条对应哪种温度:
“Three samples of a gas, X, Y, Z, are at temperatures $T_x > T_y > T_z$... The graph below shows the speed distribution of sample Y. Sketch and label possible distributions for X and Z.” [12]Source: APphy2-2014.pdfa S = 4Tr2 q 90° b A = area F = force h = depth k = thermal conductivity K = kinetic energy L = thickness m = mass n = number of moles N = number of molecules P = pressure Q = energy transferred to a system by heating T = temperature t = time U = internal energy V = volume v = speed dsin 0 = m2 l = length sin 0 = ª C m = mass E = mc2 AP Physics 2 Practice Exam PHYSICS 2 Section II 4 Questions Time-90 minutes Directions: Questions 2 and 3 are long free-response questions that require about 25 minutes each to answer and are worth 12 points each. Questions 1 and 4 are short free-response questions that require about 20 minutes each to answer and are worth 10 points each. Show your work for each part in the space provided after that part. 1. (10 points, suggested time 20 minutes) Three samples of a gas, X, Y, and Z, are prepared. Each sample contains the same number of molecules, but the samples are at different temperatures. The temperature of sample X is Tx , the temperature of sample Y is lower than that of sample X, and the temperature of sample Z is lower than that of sample Y (Tx > Ty > Tz). (a) The graph below shows the distribution of the speeds of the molecules in sample Y. On the graph, sketch and label possible distributions for sample X and sample Z. Number of Molecules Y Speed GO ON TO THE NEXT PAGE. AP Physics 2 Practice Exam The three samples with initial temperatures Tx > Ty > Tz are placed in thermal contact, with sample Z in the middle, as shown below, and the samples are insulated from their surroundings. The samples can exchange thermal energy but not gas molecules. The samples eventually reach equilibrium, with a final temperature greater than Ty ·[5]Source: APphy2-2014.pdf26 - D 39 - D AP Physics 2 Practice Exam Free-Response Section Section II is the free-response part of the exam. This section contains four free-response questions, and the student will have a total of 90 minutes to complete them all. Information for Free-Response Question 1 Timing The student should spend approximately 20 minutes on this question. 4. C. 3 Energy is transferred spontaneously from a higher temperature system to a lower temperature system. The process of transferring energy is called heating. The amount of energy transferred is called heat. Essential Knowledge 5. B. 7 The first law of thermodynamics is a specific case of the law of conservation of energy involving the internal energy of a system and the possible transfer of energy through work and/or heat. Examples should include P-V diagrams isovolumetric processes, isothermal processes, isobaric processes, and adiabatic processes. No calculations of internal energy change from temperature change are required; in this course, examples of these relationships are qualitative and/or semi-quantitative. 7. A. 2 The temperature of a system characterizes the average kinetic energy of its molecules. 7. B. 2 The second law of thermodynamics describes the change in entropy for reversible and irreversible processes. Only a qualitative treatment is considered in this course. Science Practice 6. 4 The student can make claims and predictions about natural phenomena based on scientific theories and models. 7. 1 The student can connect phenomena and models across spatial and temporal scales. 4. C. 3. 1 The student is able to make predictions about the direction of energy transfer due to temperature differences based on interactions at the microscopic level. 5. B. 7. 1 The student is able to predict qualitative changes in the internal energy of a thermodynamic system involving transfer of energy due to heat or work done and justify those predictions in terms of conservation of energy principles. Learning Objectives 7. A. 2. 2 The student is able to connect the statistical distribution of microscopic kinetic energies of molecules to the macroscopic temperature of the system and to relate this to thermodynamic processes. 7. B. 2. 1 The student is able to connect qualitatively the second law of thermodynamics in terms of the state function called entropy and how it (entropy) behaves in reversible and irreversible processes. 92 AP Physics 2 Practice Exam Characteristics of a STRONG Response Part (a) The student is able to · Identify which sample of gas has greatest probable speed based on temperature · Identify which sample of gas has the broadest distribution of molecular speeds based on temperature Part (b) The student is able to · Exhibit an understanding that energy flows from systems of higher temperature to systems of lower temperature · Exhibit an understanding that higher temperature corresponds to higher kinetic energy · Exhibit an understanding that all three samples end up at the same temperature · Exhibit an understanding that the energy flow for a sample can change direction. In this scenario that the kinetic energy of sample Y decreases then increases.
解读:$T_x$最高,对应“矮而宽、右移”的曲线;$T_z$最低,对应“高而窄、左移”的曲线。
2. 温度与气体状态方程($PV=nRT$,常高频)
【知识点】
理想气体状态方程 $ PV = nRT $,任何一项改变都能影响温度。
- 等压(P定)膨胀,$T ↑$;
- 等温(T定),$PV = constant$。
【真题例】
- “等压下,体积增加一倍,温度变为多少?”
解:$T_{new} = T_{old} \times (V_{new}/V_{old})$ [10]Source: APphy2-2017.pdfii) 1 point Using the ideal gas law: PAVA = nRTo PBVB = PA3VA nRTB = 3nRTo For the correct answer 1 point TB = 3To iii) 1 point For a correct comparison of the temperatures or correctly indicating TB = Tc = 3TA or 3To or a value consistent with the answer to part (a)(i) or (ii) AND making a reference to the isotherm or the values of P and V at points C and A Examples: 1 point Tc = 3To since it is on an isotherm with point B, as shown by the graph. On the graph, C has the same volume and 3 times the pressure as A. By the ideal gas law it thus has a temperature 3 times that of A. @ 2017 The College Board. Visit the College Board on the Web: www. collegeboard. org. AP® PHYSICS 2 2017 SCORING GUIDELINES Question 2 (continued) Distribution of points (b) i) 2 points For recognizing that the volume remains constant and describing a valid way to accomplish this in the procedure 1 point For recognizing that the temperature must decrease and describing a valid way to accomplish this in the procedure 1 point Example: Lock the piston in place and put the cylinder in a cool water bath. ii) 2 points For some reasonable description of microscopic processes in which collision of 1 point molecules is the process by which energy is transferred 1 point For correctly stating the direction of energy flow (e. g. hot (fast) to cold (slow)) and linking the direction of energy flow to the final temperature of the gas Example: The warmer gas (faster molecules) collide with the cooler surroundings (slower molecules). The net energy flow is from the gas to the surroundings which results in cooler gas (slower molecules). (c) i) 2 points For numerically calculating the amount of work done (W = |PAV|) 1 point W =- PAV =- (1×105 Pa)(2×10-3 m3) For the correct answer with negative sign (no units required) W = - 200 J 1 point ii) 1 point For applying conservation of energy using the value from part (c)(i) AU = Q+ W 300 J = Q - 200 J Q = 500 J 1 point[65]Source: APphy2-2018.pdfV/4 VA For a horizontal line to the left of A, that stops at V 4/4 1 point For a concave up curve representing an isotherm for process BC (does not have to go through all the right points between B and C) 1 point For a vertical line for CA 1 point For locating C such that Pc = PA/4 1 point For labeling axes and points (scaling of grid doesn't matter here, as long as labels are logical with respect to A) 1 point Applying this scheme, if B is to the right of A you cannot earn the point for isotherm but can still get the point for the vertical line. @ 2018 The College Board. Visit the College Board on the Web: www. collegeboard. org. AP® PHYSICS 2 2018 SCORING GUIDELINES Question 2 (continued) Distribution of points a) (continued) i) LO 5. B. 7. 1, SP 6. 4; LO 5. B. 7. 3, SP 1. 4, 2. 2 3 points For process BC, is energy added to the gas by heating or removed from the gas by cooling? Support your claim using physics principles. For recognizing that AU = 0 because the process is isothermal 1 point For recognizing that the gas loses energy via work 1 point For using the first law of thermodynamics or conservation of energy with above two ideas to get a consistent answer for the energy transfer 1 point Any appropriate sign convention can be used when applying the first law. (b) LO 4. C. 3. 1, SP 6. 4 2 points For process CA, is the net flow of energy into the gas or out of the gas? Explain your reasoning using physics principles. Describe one method by which this flow of energy could be accomplished in the laboratory. For correctly justifying that the net flow of energy is into the gas
3. 温度变化与内能($ \Delta U $)
【知识点】
理想气体内能变化只取决于温度变化:$ \Delta U = \frac{3}{2} nR \Delta T $
- $ \Delta T > 0 $,内能增加
- $ \Delta T < 0 $,内能减少
【真题例】
- “对于等压膨胀,温度升高,内能增加…”[2016 FRQ, 2017 FRQ][14]Source: APphy2-2015.pdfFor relating temperature change with internal energy change 1 point Example: Because the volume increases at a constant pressure, the temperature goes up because PV = nRT . Increasing temperature means increasing average kinetic energy or total internal energy. ii. 3 points For calculating the work done in process ABC (i. e. , the area under the line) 1 point WAB = - (1. 0 x 105 Pa)(0. 10 m3 - 0. 04 m3) = - 6000 J and WBC = 0 For calculating TA and Tc (or AT between the states) and using them to determine internal energy change TA= PAVA/nR =(1. 0×105 Pa)(0. 04 m3)/(2 mol)(8. 31 J/mol·K) = 241 K Tc = PcVc/nR = (0. 5×105 Pa)(0. 10 m3)/(2 mol)(8. 31 J/mol·K) = 301 K AU = AK per molecule nNo = (3/2) kBATnNo AU = (3/2)(1. 38 ×10-23 J/K)(301 K-241 K)(2 mol)(6. 02×1023)=1500 J 1 point @ 2016 The College Board. Visit the College Board on the Web: www. collegeboard. org. (b) AP® PHYSICS 2 2016 SCORING GUIDELINES Question 1 (continued) Distribution of points ii. (continued) Alternately, AU can be calculated directly from the given data AU = (3/2)nRAT = (3/2)(PcVc - PAVA) =(3/2)((0. 5×105 Pa)(0. 10 m3)-(1. 0×105 Pa)(0. 04 m3)) = 1500 J For substituting AU and W (whether correct or incorrect) into some form of the first law of thermodynamics to find Q and for including units in a numerical answer Q = AU - W = 1500 J - (-6000 J) 1 point Q = 7500 J (c) 1 point For recognizing that the change in kinetic energy for process CA has the same 1 point numerical value as AU from (b)ii but with the opposite sign OR for calculating AK using the correct temperature change or AK total = (3/2) nR AT as shown below AK total = (3/2) kBAT nNo or AK total = (3/2)nR AT AK total = (3/2)(1. 38 ×10-23 J/K)(241 K-301 K)(2 mol)(6. 02×1023 mol-1) AKtotal total =- 1500 J @ 2016 The College Board. Visit the College Board on the Web: www. collegeboard. org. AP® PHYSICS 2 2016 SCORING GUIDELINES Question 2 12 points total Distribution of points (a) 3 points For graphing angles or functions of angles on the axes 1 point For plotting sines of angles on the axes and indicating or implying that the index of refraction of air is 1 1 point[42]Source: APphy2-2018.pdfPA LA O > V (x 10-3 m3 ) VA ii. For process BC, is energy added to the gas by heating or removed from the gas by cooling? Support your claim using physics principles. Unauthorized copying or reuse of any part of this page is illegal. GO ON TO THE NEXT PAGE. -8- (b) For process CA, is the net flow of energy into the gas or out of the gas? Explain your reasoning using physics principles. Describe one method by which this flow of energy could be accomplished in the laboratory. (c) Curves 1 and 2 below show the distributions of particle speeds when the sample of gas is in two of the states, with curve 1 a state that occurs before curve 2 as the gas goes through the cycle. Which states correspond to curve 1 and 2? Support your claim using physics principles. Number of Particles O Speed Unauthorized copying or reuse of any part of this page is illegal. GO ON TO THE NEXT PAGE. -9- Force Probe Graduated Cylinder Block Liquid 3. (12 points, suggested time 25 minutes) Two groups of students are given samples of the same unknown liquid and are asked to identify the liquid. The first group decides to do this by determining its density PL . The students pour a volume V; of the liquid into a graduated cylinder. They then hang a long solid block of mass m| from a force probe and lower the block partially into the liquid, as shown above. The students record the reading on the force probe and the volume reading Vf on the graduated cylinder for various amounts of the block lowered into the liquid. (a) i. On the square below, which represents the block, draw and label the forces (not components) that are exerted on the block when it is stationary and partially submerged in the liquid. Each force must be represented by a distinct arrow starting on, and pointing away from, the square. ii. Derive an equation for the force probe reading Fp in terms of PL. , mp , Vi , Vf , and physical constants, as appropriate. Unauthorized copying or reuse of any part of this page is illegal. GO ON TO THE NEXT PAGE. -10- The data the students collected and the graph they plotted are shown below. The initial volume V; of liquid in the cylinder is 1. 00 × 10-4 m3 . Volume Reading Vf (x10 4 m3)
4. 温度、能量流向与热平衡
【知识点】
- 能量总是自发地从高温物体流向低温物体,直到热平衡($T_1 = T_2$)。
- 分子间频繁碰撞,动能(和温度)不断趋于平均,直到热平衡。
- 内能变化关系着温度变化。
【真题例】
- 多个样本放在一起,问能量流向和最后温度。[2014 Practice FRQ][5]Source: APphy2-2014.pdf26 - D 39 - D AP Physics 2 Practice Exam Free-Response Section Section II is the free-response part of the exam. This section contains four free-response questions, and the student will have a total of 90 minutes to complete them all. Information for Free-Response Question 1 Timing The student should spend approximately 20 minutes on this question. 4. C. 3 Energy is transferred spontaneously from a higher temperature system to a lower temperature system. The process of transferring energy is called heating. The amount of energy transferred is called heat. Essential Knowledge 5. B. 7 The first law of thermodynamics is a specific case of the law of conservation of energy involving the internal energy of a system and the possible transfer of energy through work and/or heat. Examples should include P-V diagrams isovolumetric processes, isothermal processes, isobaric processes, and adiabatic processes. No calculations of internal energy change from temperature change are required; in this course, examples of these relationships are qualitative and/or semi-quantitative. 7. A. 2 The temperature of a system characterizes the average kinetic energy of its molecules. 7. B. 2 The second law of thermodynamics describes the change in entropy for reversible and irreversible processes. Only a qualitative treatment is considered in this course. Science Practice 6. 4 The student can make claims and predictions about natural phenomena based on scientific theories and models. 7. 1 The student can connect phenomena and models across spatial and temporal scales. 4. C. 3. 1 The student is able to make predictions about the direction of energy transfer due to temperature differences based on interactions at the microscopic level. 5. B. 7. 1 The student is able to predict qualitative changes in the internal energy of a thermodynamic system involving transfer of energy due to heat or work done and justify those predictions in terms of conservation of energy principles. Learning Objectives 7. A. 2. 2 The student is able to connect the statistical distribution of microscopic kinetic energies of molecules to the macroscopic temperature of the system and to relate this to thermodynamic processes. 7. B. 2. 1 The student is able to connect qualitatively the second law of thermodynamics in terms of the state function called entropy and how it (entropy) behaves in reversible and irreversible processes. 92 AP Physics 2 Practice Exam Characteristics of a STRONG Response Part (a) The student is able to · Identify which sample of gas has greatest probable speed based on temperature · Identify which sample of gas has the broadest distribution of molecular speeds based on temperature Part (b) The student is able to · Exhibit an understanding that energy flows from systems of higher temperature to systems of lower temperature · Exhibit an understanding that higher temperature corresponds to higher kinetic energy · Exhibit an understanding that all three samples end up at the same temperature · Exhibit an understanding that the energy flow for a sample can change direction. In this scenario that the kinetic energy of sample Y decreases then increases.[8]Source: Physics 2 Practice Exam 2 - Fall 2014.pdfFor correctly addressing the flow of equal amounts of energy into and out of the air or water 1 point For correctly addressing the spontaneous flow of energy from hot to cold objects 1 point Example explanation: The 2nd law of thermodynamics states that energy naturally moves from hotter to cooler objects, so the water transfers energy to the air. To keep the temperature of the water constant, the rate of energy transfer to the water must be the same as the rate at which energy leaves the water and goes into the air in order to satisfy the 1st law of thermodynamics Claim (given): The heater must remain on to maintain a constant water temperature that is greater than the air temperature in the pool area. Evidence 1: The 2nd law of thermodynamics states that energy naturally moves from hotter to cooler objects. Reasoning 1: The water transfers energy to the air. Evidence 2: The 1st law of thermodynamics must be satisfied when keeping the temperature of the water constant. Reasoning: The rate of energy transfer to the water must be the same as the rate at which Visit the College Board on the Web: www. collegeboard. org. AP® PHYSICS 2 2019 SCORING GUIDELINES Question 1 (continued) Distribution of points The door to the pool area is then closed so that energy transfer and air transfer into and out of the area are negligible. After some time the air in the pool area becomes warmer. (c) LO 5. B. 7. 2, SP 1. 1; LO 5. B. 7. 1, SP 6. 4, 7. 2) 2 points On the graph below, sketch the thermodynamic process that the air in the pool area undergoes as it becomes warmer. Compare the amount Q of energy transferred to the air by heating to the change AU in the air's internal energy. Sketch the Process Compare Q and AU Pressure O Volume For a vertical line that does not touch the horizontal axis, and either an arrowhead or labels that indicate the direction of the process 1 point For a comparison consistent with the graph, or an incorrect graph but a correct comparison with an explanation that the work on the gas is zero 1 point Correct comparison: Q = AU Visit the College Board on the Web: www. collegeboard. org. AP® PHYSICS 2 2019 SCORING GUIDELINES Question 1 (continued) Distribution of points (d) LO 7. A. 1. 1, SP 6. 4 2 points
5. 速率分布Maxwell-Boltzmann分布图识别
【技巧口诀】
- 高而窄:温度低(冷),分子速率集中于较低值
- 矮而宽:温度高(热),分布宽、右移
【真题实例】
“Number of Particles——Speed”曲线,辨认温度高低、平均速度大小等 [5]Source: APphy2-2014.pdf26 - D
39 - D
AP Physics 2 Practice Exam
Free-Response Section
Section II is the free-response part of the exam. This section contains four free-response questions, and the student will have a total of 90 minutes to complete them all.
Information for Free-Response Question 1
Timing
The student should spend approximately 20 minutes on this question.
4. C. 3 Energy is transferred spontaneously from a higher temperature system to a lower temperature system. The process of transferring energy is called heating. The amount of energy transferred is called heat.
Essential Knowledge
5. B. 7 The first law of thermodynamics is a specific case of the law of conservation of energy involving the internal energy of a system and the possible transfer of energy through work and/or heat. Examples should include P-V diagrams isovolumetric processes, isothermal processes, isobaric processes, and adiabatic processes. No calculations of internal energy change from temperature change are required; in this course, examples of these relationships are qualitative and/or semi-quantitative.
7. A. 2 The temperature of a system characterizes the average kinetic energy of its molecules.
7. B. 2 The second law of thermodynamics describes the change in entropy for reversible and irreversible processes. Only a qualitative treatment is considered in this course.
Science Practice
6. 4 The student can make claims and predictions about natural phenomena based on scientific theories and models. 7. 1 The student can connect phenomena and models across spatial and temporal scales.
4. C. 3. 1 The student is able to make predictions about the direction of energy transfer due to temperature differences based on interactions at the microscopic level.
5. B. 7. 1 The student is able to predict qualitative changes in the internal energy of a thermodynamic system involving transfer of energy due to heat or work done and justify those predictions in terms of conservation of energy principles.
Learning Objectives
7. A. 2. 2 The student is able to connect the statistical distribution of microscopic kinetic energies of molecules to the macroscopic temperature of the system and to relate this to thermodynamic processes.
7. B. 2. 1 The student is able to connect qualitatively the second law of thermodynamics in terms of the state function called entropy and how it (entropy) behaves in reversible and irreversible processes.
92 AP Physics 2 Practice Exam
Characteristics of a STRONG Response
Part (a) The student is able to
· Identify which sample of gas has greatest probable speed based on temperature
· Identify which sample of gas has the broadest distribution of molecular speeds based on temperature
Part (b) The student is able to
· Exhibit an understanding that energy flows from systems of higher temperature to systems of lower temperature
· Exhibit an understanding that higher temperature corresponds to higher kinetic energy
· Exhibit an understanding that all three samples end up at the same temperature
· Exhibit an understanding that the energy flow for a sample can change direction. In this scenario that the kinetic energy of sample Y decreases then increases.[12]Source: APphy2-2014.pdfa
S = 4Tr2
q
90°
b
A = area F = force h = depth k = thermal conductivity K = kinetic energy
L = thickness
m = mass
n = number of moles
N = number of molecules
P = pressure
Q = energy transferred to a system by heating T = temperature t = time
U = internal energy V = volume v = speed
dsin 0 = m2
l = length
sin 0 = ª C
m = mass
E = mc2
AP Physics 2 Practice Exam
PHYSICS 2 Section II 4 Questions Time-90 minutes
Directions: Questions 2 and 3 are long free-response questions that require about 25 minutes each to answer and are worth 12 points each. Questions 1 and 4 are short free-response questions that require about 20 minutes each to answer and are worth 10 points each. Show your work for each part in the space provided after that part.
1. (10 points, suggested time 20 minutes)
Three samples of a gas, X, Y, and Z, are prepared. Each sample contains the same number of molecules, but the samples are at different temperatures. The temperature of sample X is Tx , the temperature of sample Y is lower than that of sample X, and the temperature of sample Z is lower than that of sample Y (Tx > Ty > Tz).
(a) The graph below shows the distribution of the speeds of the molecules in sample Y. On the graph, sketch and label possible distributions for sample X and sample Z.
Number of Molecules
Y
Speed
GO ON TO THE NEXT PAGE.
AP Physics 2 Practice Exam
The three samples with initial temperatures Tx > Ty > Tz are placed in thermal contact, with sample Z in the middle, as shown below, and the samples are insulated from their surroundings. The samples can exchange thermal energy but not gas molecules. The samples eventually reach equilibrium, with a final temperature greater than Ty ·[17]Source: APphy2-2014.pdfThe kinetic energy of X decreases. It has the highest temperature and so to reach the same equilibrium temperature as the other samples it must lose energy, which flows into Z.
The kinetic energy of Y decreases and then increases. It has a higher temperature than Z, and so initially loses energy which flows into Z. But it eventually must end up at a higher temperature than it initially had, so the net energy flow must be into Y. That can only happen if the direction of energy flow reverses.
The kinetic energy of Z could always increase, or it can increase and then decrease. Initially energy flows into it, since it has the lowest temperature. At some point energy begins to flow from Z to Y. Whether the temperature and thus the kinetic energy of Z continually increases or not depends on how much energy keeps flowing to it from X.
For exhibiting understanding that energy flows from systems at higher temperature to systems at lower temperature
1 point
For exhibiting understanding that higher temperature corresponds to higher kinetic energy
1 point
For exhibiting understanding that the energy flow stops 1 point
For exhibiting understanding that the energy flow for a sample can change direction
1 point
For using the above understanding to indicate that the average kinetic energy of all three samples are the same when equilibrium is reached
1 point
AP Physics 2 Practice Exam
Question 1 (continued)
Distribution of points
(c) 2 points
For checking either the correct response or one consistent with the answer to part (b), with a reasonable attempt at justification
1 point
For a correct or consistent justification that relates the spread of the molecular distribution to entropy
1 point
For example: The entropy decreases. When the temperature goes down, the spread of the speeds and thus the kinetic energies of the individual molecules is less. This means less disorder and thus less entropy.
(d) 1 point
For correctly indicating that the entropy of the system increases, and explaining that the entropy of a closed system increases for an irreversible process.
1 point
96 AP Physics 2 Practice Exam
Information for Free-Response Question 2
Timing
The student should spend approximately 25 minutes on this question.
4. E. 5 The values of currents and electric potential differences in an electric circuit are determined by the properties and arrangement of the individual circuit elements such as sources of emf, resistors, and capacitors.
Essential Knowledge
6. 温度单位换算“陷阱题”
【知识点】
AP中所有温度一定要用开尔文$K$
- $T(K) = t(^{\circ}C) + 273.15$
- 状态方程不能用°C
【真题提醒】
状态题和表格题中“温度单位”极容易考查——要特别注意!!!
汇总 Q&A 真题串讲(答题技巧)
Q1:
题目:三种气体,T1>T2>T3,画分布曲线并判断能量流向。
- 画:T最大,分布图右移且平矮;T最小,最左且最高
- 能量流向:总是从高温流到低温,直到所有温度相等 [5]Source: APphy2-2014.pdf26 - D 39 - D AP Physics 2 Practice Exam Free-Response Section Section II is the free-response part of the exam. This section contains four free-response questions, and the student will have a total of 90 minutes to complete them all. Information for Free-Response Question 1 Timing The student should spend approximately 20 minutes on this question. 4. C. 3 Energy is transferred spontaneously from a higher temperature system to a lower temperature system. The process of transferring energy is called heating. The amount of energy transferred is called heat. Essential Knowledge 5. B. 7 The first law of thermodynamics is a specific case of the law of conservation of energy involving the internal energy of a system and the possible transfer of energy through work and/or heat. Examples should include P-V diagrams isovolumetric processes, isothermal processes, isobaric processes, and adiabatic processes. No calculations of internal energy change from temperature change are required; in this course, examples of these relationships are qualitative and/or semi-quantitative. 7. A. 2 The temperature of a system characterizes the average kinetic energy of its molecules. 7. B. 2 The second law of thermodynamics describes the change in entropy for reversible and irreversible processes. Only a qualitative treatment is considered in this course. Science Practice 6. 4 The student can make claims and predictions about natural phenomena based on scientific theories and models. 7. 1 The student can connect phenomena and models across spatial and temporal scales. 4. C. 3. 1 The student is able to make predictions about the direction of energy transfer due to temperature differences based on interactions at the microscopic level. 5. B. 7. 1 The student is able to predict qualitative changes in the internal energy of a thermodynamic system involving transfer of energy due to heat or work done and justify those predictions in terms of conservation of energy principles. Learning Objectives 7. A. 2. 2 The student is able to connect the statistical distribution of microscopic kinetic energies of molecules to the macroscopic temperature of the system and to relate this to thermodynamic processes. 7. B. 2. 1 The student is able to connect qualitatively the second law of thermodynamics in terms of the state function called entropy and how it (entropy) behaves in reversible and irreversible processes. 92 AP Physics 2 Practice Exam Characteristics of a STRONG Response Part (a) The student is able to · Identify which sample of gas has greatest probable speed based on temperature · Identify which sample of gas has the broadest distribution of molecular speeds based on temperature Part (b) The student is able to · Exhibit an understanding that energy flows from systems of higher temperature to systems of lower temperature · Exhibit an understanding that higher temperature corresponds to higher kinetic energy · Exhibit an understanding that all three samples end up at the same temperature · Exhibit an understanding that the energy flow for a sample can change direction. In this scenario that the kinetic energy of sample Y decreases then increases.[17]Source: APphy2-2014.pdfThe kinetic energy of X decreases. It has the highest temperature and so to reach the same equilibrium temperature as the other samples it must lose energy, which flows into Z. The kinetic energy of Y decreases and then increases. It has a higher temperature than Z, and so initially loses energy which flows into Z. But it eventually must end up at a higher temperature than it initially had, so the net energy flow must be into Y. That can only happen if the direction of energy flow reverses. The kinetic energy of Z could always increase, or it can increase and then decrease. Initially energy flows into it, since it has the lowest temperature. At some point energy begins to flow from Z to Y. Whether the temperature and thus the kinetic energy of Z continually increases or not depends on how much energy keeps flowing to it from X. For exhibiting understanding that energy flows from systems at higher temperature to systems at lower temperature 1 point For exhibiting understanding that higher temperature corresponds to higher kinetic energy 1 point For exhibiting understanding that the energy flow stops 1 point For exhibiting understanding that the energy flow for a sample can change direction 1 point For using the above understanding to indicate that the average kinetic energy of all three samples are the same when equilibrium is reached 1 point AP Physics 2 Practice Exam Question 1 (continued) Distribution of points (c) 2 points For checking either the correct response or one consistent with the answer to part (b), with a reasonable attempt at justification 1 point For a correct or consistent justification that relates the spread of the molecular distribution to entropy 1 point For example: The entropy decreases. When the temperature goes down, the spread of the speeds and thus the kinetic energies of the individual molecules is less. This means less disorder and thus less entropy. (d) 1 point For correctly indicating that the entropy of the system increases, and explaining that the entropy of a closed system increases for an irreversible process. 1 point 96 AP Physics 2 Practice Exam Information for Free-Response Question 2 Timing The student should spend approximately 25 minutes on this question. 4. E. 5 The values of currents and electric potential differences in an electric circuit are determined by the properties and arrangement of the individual circuit elements such as sources of emf, resistors, and capacitors. Essential Knowledge
Q2:
题目:气体状态变化,判断温度与P、V的关系。
- 记住$PV/nR = T$,任何一步都可以解出温度变化 [10]Source: APphy2-2017.pdfii) 1 point Using the ideal gas law: PAVA = nRTo PBVB = PA3VA nRTB = 3nRTo For the correct answer 1 point TB = 3To iii) 1 point For a correct comparison of the temperatures or correctly indicating TB = Tc = 3TA or 3To or a value consistent with the answer to part (a)(i) or (ii) AND making a reference to the isotherm or the values of P and V at points C and A Examples: 1 point Tc = 3To since it is on an isotherm with point B, as shown by the graph. On the graph, C has the same volume and 3 times the pressure as A. By the ideal gas law it thus has a temperature 3 times that of A. @ 2017 The College Board. Visit the College Board on the Web: www. collegeboard. org. AP® PHYSICS 2 2017 SCORING GUIDELINES Question 2 (continued) Distribution of points (b) i) 2 points For recognizing that the volume remains constant and describing a valid way to accomplish this in the procedure 1 point For recognizing that the temperature must decrease and describing a valid way to accomplish this in the procedure 1 point Example: Lock the piston in place and put the cylinder in a cool water bath. ii) 2 points For some reasonable description of microscopic processes in which collision of 1 point molecules is the process by which energy is transferred 1 point For correctly stating the direction of energy flow (e. g. hot (fast) to cold (slow)) and linking the direction of energy flow to the final temperature of the gas Example: The warmer gas (faster molecules) collide with the cooler surroundings (slower molecules). The net energy flow is from the gas to the surroundings which results in cooler gas (slower molecules). (c) i) 2 points For numerically calculating the amount of work done (W = |PAV|) 1 point W =- PAV =- (1×105 Pa)(2×10-3 m3) For the correct answer with negative sign (no units required) W = - 200 J 1 point ii) 1 point For applying conservation of energy using the value from part (c)(i) AU = Q+ W 300 J = Q - 200 J Q = 500 J 1 point[65]Source: APphy2-2018.pdfV/4 VA For a horizontal line to the left of A, that stops at V 4/4 1 point For a concave up curve representing an isotherm for process BC (does not have to go through all the right points between B and C) 1 point For a vertical line for CA 1 point For locating C such that Pc = PA/4 1 point For labeling axes and points (scaling of grid doesn't matter here, as long as labels are logical with respect to A) 1 point Applying this scheme, if B is to the right of A you cannot earn the point for isotherm but can still get the point for the vertical line. @ 2018 The College Board. Visit the College Board on the Web: www. collegeboard. org. AP® PHYSICS 2 2018 SCORING GUIDELINES Question 2 (continued) Distribution of points a) (continued) i) LO 5. B. 7. 1, SP 6. 4; LO 5. B. 7. 3, SP 1. 4, 2. 2 3 points For process BC, is energy added to the gas by heating or removed from the gas by cooling? Support your claim using physics principles. For recognizing that AU = 0 because the process is isothermal 1 point For recognizing that the gas loses energy via work 1 point For using the first law of thermodynamics or conservation of energy with above two ideas to get a consistent answer for the energy transfer 1 point Any appropriate sign convention can be used when applying the first law. (b) LO 4. C. 3. 1, SP 6. 4 2 points For process CA, is the net flow of energy into the gas or out of the gas? Explain your reasoning using physics principles. Describe one method by which this flow of energy could be accomplished in the laboratory. For correctly justifying that the net flow of energy is into the gas
超高频真题情境总结
- 分布曲线判温度
- 用$PV=nRT$或者分子动能公式对温度进行定量/定性判断
- 能量流动、热平衡、微观动能与温度等效分析
✔️ 温度-英文高频专用术语表
| 英文术语 | 中文释义 | |---------------------------|------------------| | Temperature | 温度 | | Thermal equilibrium | 热平衡 | | Translational kinetic energy | 平动动能 | | Maxwell-Boltzmann distribution | 麦克斯韦-玻尔兹曼速率分布 | | Internal energy ($U$) | 内能 | | Average kinetic energy | 平均动能 | | Heat transfer | 热传递 | | Isothermal process | 等温过程 | | Ideal gas law ($PV=nRT$) | 理想气体定律 | | Kelvin | 开尔文(K) |
总结口诀记忆
- 温度高分布右移更平坦,温度低分布左移高且窄。
- $ PV = nRT $用K,无论多复杂,公式全靠它。
- 能量总从高温向低温,为热平衡打基础。
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