I85PREP
⌘ K
AP Physics 2 study materials
AP Physics 2 study resources

AP Physics 2 practice questions

One question at a time. Build your AP Physics 2 understanding.

First 50 of 200 questions. Each one lists the choices, the correct answer, and an explanation. The checker at the top of this page saves progress in this browser when it is available.

Continue in the practice tool above to load more questions, 50 at a time.

Question 1Thermodynamics

A gas is compressed while it is thermally isolated from its surroundings. Which statement about the gas is correct?

  1. Its internal energy increases and its temperature rises. Correct answer.
  2. Its temperature stays constant because the process is adiabatic.
  3. Its internal energy decreases and its temperature falls.
  4. Its internal energy stays constant because no heat flows.

Explanation

For an adiabatic process Q = 0, so the first law gives ΔU = W, where W is work done on the gas. Compressing the gas means positive work is done on it, so internal energy and temperature both rise.

Question 2Thermodynamics

Two identical containers hold equal numbers of moles of gas at the same temperature. One holds helium (monatomic) and the other holds nitrogen (diatomic). How do their internal energies compare?

  1. The nitrogen has the greater internal energy. Correct answer.
  2. They have equal internal energy.
  3. The helium has the greater internal energy.
  4. The comparison depends on the gas pressure.

Explanation

Internal energy of an ideal gas is (f/2)nRT, where f is the number of degrees of freedom. A diatomic gas has f = 5 versus f = 3 for monatomic, so with equal moles and temperature the diatomic gas stores more energy.

Question 3Thermodynamics

What is the Kelvin equivalent of a temperature of 27 °C?

  1. 354 K
  2. 27 K
  3. 300 K Correct answer.
  4. 246 K

Explanation

Kelvin temperature is Celsius temperature plus 273 (more precisely 273.15). Adding 27 to 273 gives 300 K.

Question 4Thermodynamics

A fixed mass of ideal gas is held at constant temperature while its volume is halved. What happens to the pressure?

  1. It doubles. Correct answer.
  2. It stays the same.
  3. It halves.
  4. It quadruples.

Explanation

Boyle's law states that at constant temperature the product PV is constant. Halving the volume therefore doubles the pressure.

Question 5Thermodynamics

Which quantity is a measure of the average translational kinetic energy of the molecules of an ideal gas?

  1. Volume
  2. Absolute temperature Correct answer.
  3. Pressure
  4. Number of moles

Explanation

The average translational kinetic energy per molecule equals (3/2)kT, so it depends only on absolute temperature. Pressure and volume relate to energy density and total energy, not the per-molecule average.

Question 6Thermodynamics

In an isothermal process for an ideal gas, which statement is true?

  1. The work done by the gas equals the change in internal energy.
  2. The internal energy does not change, so the heat absorbed equals the work done by the gas. Correct answer.
  3. The pressure remains constant throughout the process.
  4. No heat is exchanged with the surroundings.

Explanation

For an ideal gas, internal energy depends only on temperature, so at constant temperature ΔU = 0. The first law then reduces to Q = W, meaning heat absorbed equals work done by the gas.

Question 7Thermodynamics

Which process transfers thermal energy primarily by the bulk motion of a fluid?

  1. Conduction
  2. Radiation
  3. Evaporation
  4. Convection Correct answer.

Explanation

Convection moves thermal energy by the macroscopic flow of a fluid, carrying warmer material from one place to another. Conduction and radiation transfer energy without bulk fluid motion.

Question 8Thermodynamics

A heat engine takes in 500 J of heat from a hot reservoir and exhausts 300 J to a cold reservoir in one cycle. What is its efficiency?

  1. 0.60
  2. 0.20
  3. 0.40 Correct answer.
  4. 0.80

Explanation

Efficiency is W/Q_H, and the work output is the net heat, 500 J − 300 J = 200 J. Dividing 200 J by 500 J gives 0.40.

Question 9Thermodynamics

An ideal gas expands at constant pressure of 2.0 × 10^5 Pa from 0.010 m^3 to 0.030 m^3. How much work does the gas do?

  1. 6.0 × 10^3 J
  2. 8.0 × 10^3 J
  3. 2.0 × 10^3 J
  4. 4.0 × 10^3 J Correct answer.

Explanation

For a constant-pressure process the work done by the gas is PΔV. The volume change is 0.020 m^3, and multiplying by 2.0 × 10^5 Pa gives 4.0 × 10^3 J.

Question 10Thermodynamics

A rigid sealed tank contains an ideal gas. Heat is added to the gas. Which combination of quantities changes?

  1. Temperature rises and volume rises; pressure stays fixed.
  2. Pressure rises and volume rises; temperature stays fixed.
  3. Temperature and pressure rise; volume stays fixed. Correct answer.
  4. Only the number of moles changes.

Explanation

A rigid tank fixes the volume, so at constant V the ideal gas law gives P proportional to T. Adding heat raises the temperature, which raises the pressure.

Question 11Thermodynamics

An ideal gas is taken through a complete cycle and returns to its initial state. Which statement about the cycle is correct?

  1. The net heat absorbed is zero, but the net change in internal energy need not be zero.
  2. Both the net work done and the net heat absorbed are necessarily zero.
  3. Both the net heat absorbed and the net change in internal energy are zero.
  4. The net change in internal energy is zero, but the net heat absorbed need not be zero. Correct answer.

Explanation

Internal energy is a state function, so over a closed cycle ΔU = 0. Heat and work are path-dependent, so the net heat absorbed equals the net work done, which need not be zero.

Question 12Thermodynamics

Two moles of an ideal monatomic gas are heated from 300 K to 400 K at constant volume. What is the change in internal energy? (R = 8.31 J/(mol·K))

  1. 8.3 × 10^3 J
  2. 2.5 × 10^3 J Correct answer.
  3. 4.2 × 10^3 J
  4. 1.7 × 10^3 J

Explanation

For a monatomic ideal gas ΔU = (3/2)nRΔT. Substituting gives 1.5 × 2 × 8.31 × 100, which is about 2.5 × 10^3 J.

Question 13Thermodynamics

A gas is compressed adiabatically and quasistatically. Compared with an isothermal compression between the same two volumes, the adiabatic compression requires:

  1. Less work, because the pressure stays lower throughout.
  2. The same work, because work depends only on the volumes.
  3. Less work, because no heat is removed.
  4. More work, because the temperature rises during compression. Correct answer.

Explanation

In adiabatic compression the temperature rises, so the pressure stays higher than in an isothermal compression over the same volume range. Higher pressure over the same displacement means more work is done on the gas.

Question 14Thermodynamics

A Carnot engine operates between reservoirs at 600 K and 300 K. What is its maximum efficiency?

  1. 0.50 Correct answer.
  2. 1.00
  3. 0.25
  4. 0.75

Explanation

The Carnot efficiency is 1 − T_C/T_H using absolute temperatures. Here 1 − 300/600 = 0.50.

Question 15Thermodynamics

A sample of gas undergoes an isochoric (constant-volume) heating. Which of the following is true?

  1. The gas does work, and its internal energy is unchanged.
  2. The gas does no work, and all heat added raises internal energy. Correct answer.
  3. No heat is exchanged, and the pressure rises.
  4. The gas does work equal to the heat added.

Explanation

Work done by a gas is PΔV, which is zero when volume is fixed. The first law then gives Q = ΔU, so all the added heat increases internal energy.

Question 16Thermodynamics

The pressure of an ideal gas is doubled while its absolute temperature is held constant. What happens to the average distance between molecules?

  1. It stays the same.
  2. It increases by a factor of 2.
  3. It decreases by a factor of 2.
  4. It decreases by a factor of about 1.26. Correct answer.

Explanation

At constant temperature, doubling pressure halves the volume, so the number density doubles. The spacing between molecules scales as the cube root of the volume per molecule, so it shrinks by 2^(1/3) ≈ 1.26.

Question 17Thermodynamics

In one cycle, a refrigerator removes 400 J of heat from a cold region while 100 J of work is done on it. How much heat is delivered to the hot region?

  1. 100 J
  2. 400 J
  3. 500 J Correct answer.
  4. 300 J

Explanation

For a refrigerator, energy conservation requires Q_H = Q_C + W. Adding 400 J removed from the cold region to 100 J of work input gives 500 J rejected to the hot region.

Question 18Thermodynamics

A wall separates two rooms at different temperatures. The rate of heat conduction through the wall is doubled. Which change alone could produce this?

  1. Halving the wall area
  2. Halving the thermal conductivity
  3. Doubling the wall thickness
  4. Doubling the temperature difference across the wall Correct answer.

Explanation

Conduction rate is kAΔT/L, so it is proportional to the temperature difference. Doubling ΔT doubles the rate, whereas doubling L or halving A or k would reduce it.

Question 19Thermodynamics

A gas absorbs 300 J of heat while doing 120 J of work on its surroundings. What is the change in its internal energy?

  1. −180 J
  2. +420 J
  3. +180 J Correct answer.
  4. −420 J

Explanation

Using ΔU = Q − W, with Q = +300 J absorbed and W = +120 J done by the gas, ΔU = 300 − 120 = +180 J.

Question 20Thermodynamics

A container holds a mixture of 2 mol of helium and 3 mol of argon at temperature T. Both are monatomic. What is the total internal energy?

  1. (3/2)(3)R T
  2. (5/2)(5)R T
  3. (3/2)(5)R T Correct answer.
  4. (3/2)(2)R T

Explanation

Each monatomic gas has internal energy (3/2)nRT. The total is (3/2)(2 + 3)RT = (3/2)(5)RT since both gases are monatomic and share the same temperature.

Question 21Thermodynamics

An ideal gas is taken from state A to state B by two different paths. Which quantity must be the same for both paths?

  1. The change in internal energy Correct answer.
  2. The heat absorbed by the gas
  3. The work done by the gas
  4. The sum of heat absorbed and work done by the gas

Explanation

Internal energy is a state function, so its change depends only on the initial and final states, not the path. Heat and work are path-dependent, though their difference Q − W is fixed.

Question 22Thermodynamics

An ideal gas undergoes an adiabatic free expansion into a vacuum inside an insulated rigid container. Which statement is correct?

  1. The temperature of the gas decreases because it does work.
  2. The temperature of the gas does not change. Correct answer.
  3. The pressure of the gas does not change.
  4. The temperature of the gas increases because no heat escapes.

Explanation

In free expansion into a vacuum the gas does no work (no external pressure to push against) and exchanges no heat (insulated), so ΔU = 0. For an ideal gas, unchanged internal energy means unchanged temperature.

Question 23Thermodynamics

A real heat engine operates between 500 K and 300 K and has an efficiency of 0.30. How does this compare with the maximum possible efficiency?

  1. It exceeds the Carnot efficiency of 0.25, so it is impossible.
  2. It equals the Carnot efficiency, so it is reversible.
  3. It exceeds the Carnot efficiency of 0.40, so it is impossible.
  4. It is less than the Carnot efficiency of 0.40, so it is allowed. Correct answer.

Explanation

The Carnot efficiency between these reservoirs is 1 − 300/500 = 0.40. A real engine at 0.30 is below this upper bound, so it is physically permitted.

Question 24Thermodynamics

A gas follows the path shown by its description: it is heated at constant volume from 1.0 × 10^5 Pa to 3.0 × 10^5 Pa, then expands at constant pressure to triple its original volume. If the initial volume is 0.020 m^3, what is the net work done by the gas over the whole process?

  1. 0 J
  2. 8.0 × 10^3 J Correct answer.
  3. 1.2 × 10^4 J
  4. 6.0 × 10^3 J

Explanation

No work is done during the isochoric heating. During the constant-pressure expansion the volume goes from 0.020 m^3 to 0.060 m^3, a change of 0.040 m^3, and multiplying by 3.0 × 10^5 Pa gives 1.2 × 10^4 J.

Question 25Thermodynamics

Two identical samples of the same ideal gas are each heated through the same temperature change, one at constant volume and one at constant pressure. Which requires more heat, and why?

  1. The constant-volume heating, because all heat goes into internal energy.
  2. The constant-pressure heating, because its internal energy change is larger.
  3. Both require the same heat, because the temperature change is the same.
  4. The constant-pressure heating, because the gas also does work as it expands. Correct answer.

Explanation

At constant pressure the gas absorbs heat both to raise its internal energy and to do expansion work, so Q = nC_PΔT exceeds Q = nC_VΔT. The internal energy change is the same for both since ΔT is the same.

Question 26Thermodynamics

A gas is taken from state A to state B. Along path 1, it absorbs 500 J of heat and does 200 J of work. Along path 2, it absorbs 300 J of heat. What is the work done by the gas along path 2?

  1. 500 J
  2. 100 J Correct answer.
  3. 0 J
  4. 200 J

Explanation

The change in internal energy is path-independent, so ΔU = Q1 - W1 = 500 J - 200 J = 300 J for both paths. For path 2, ΔU = Q2 - W2 gives 300 J = 300 J - W2, so W2 = 0 J.

Question 27Thermodynamics

A monatomic ideal gas is taken through a process in which its pressure is directly proportional to its volume (P = kV). How does the heat capacity of the gas in this process compare with its heat capacity at constant volume, C_V?

  1. It is equal to C_P.
  2. It is less than C_V but greater than zero.
  3. It is equal to C_V.
  4. It is greater than C_V but less than C_P. Correct answer.

Explanation

For a process with P proportional to V, the molar heat capacity is C = C_V + R/2. For a monatomic gas, C_V = (3/2)R, so C = 2R, which lies between C_V = 1.5R and C_P = 2.5R.

Question 28Thermodynamics

Which statement best describes the second law of thermodynamics?

  1. The total energy of an isolated system is conserved.
  2. The internal energy of an ideal gas depends only on its temperature.
  3. The entropy of an isolated system never decreases. Correct answer.
  4. Heat spontaneously flows from a colder object to a hotter object.

Explanation

The second law states that the entropy of an isolated system cannot decrease; it increases for irreversible processes and remains constant for reversible ones. This principle also implies that heat cannot spontaneously flow from cold to hot without external work.

Question 29Thermodynamics

A sample of gas is taken through a cyclic process consisting of two isothermal and two adiabatic steps. Which statement about this cycle is correct?

  1. The efficiency of the cycle is 100%.
  2. The change in entropy of the gas over one complete cycle is zero. Correct answer.
  3. The net heat absorbed by the gas is negative.
  4. The net work done by the gas is zero.

Explanation

Entropy is a state function, so over any complete cycle the change in entropy of the gas is zero. The net work done equals the net heat absorbed and is generally nonzero, and the efficiency is less than 100% unless the cold reservoir is at absolute zero.

Question 30Electric Force, Field, and Potential

Two point charges are separated by a distance r. If the distance between them is doubled while both charges are unchanged, the magnitude of the electric force between them becomes

  1. twice the original force
  2. one half of the original force
  3. four times the original force
  4. one quarter of the original force Correct answer.

Explanation

Coulomb's law states that the force is inversely proportional to the square of the separation, F = kq1q2/r². Doubling r multiplies the denominator by four, so the force falls to one quarter.

Question 31Electric Force, Field, and Potential

The SI unit of electric field strength is equivalent to which of the following?

  1. newton per coulomb Correct answer.
  2. coulomb per newton
  3. newton meter squared per coulomb
  4. joule per coulomb

Explanation

Electric field is defined as force per unit charge, E = F/q, so its unit is the newton per coulomb. The joule per coulomb is the volt, which measures potential rather than field.

Question 32Electric Force, Field, and Potential

A positive point charge is placed at the center of an isolated, uncharged conducting spherical shell. Which statement about the induced charge on the shell is correct?

  1. The inner surface acquires negative charge and the outer surface positive charge Correct answer.
  2. No charge appears on either surface
  3. The inner surface acquires positive charge and the outer surface negative charge
  4. Both surfaces acquire positive charge

Explanation

The field of the central positive charge attracts electrons to the inner surface, giving it negative charge. Charge conservation forces an equal positive charge onto the outer surface.

Question 33Electric Force, Field, and Potential

The electric potential at a point in space is defined as

  1. the work done in moving any charge around a closed loop
  2. the electric potential energy per unit positive charge at that point Correct answer.
  3. the total electric potential energy stored in the region
  4. the electric force per unit positive charge at that point

Explanation

Electric potential is potential energy divided by charge, V = U/q, and is a scalar property of the location. Field, by contrast, is force per unit charge.

Question 34Electric Force, Field, and Potential

A uniform electric field of magnitude 500 N/C points in the +x direction. What is the change in electric potential when moving 0.20 m in the +x direction?

  1. +100 V
  2. -100 V Correct answer.
  3. +2500 V
  4. -2500 V

Explanation

For a uniform field, ΔV = -E·Δx along the field direction. With E = 500 N/C and Δx = 0.20 m, the potential drops by 100 V.

Question 35Electric Force, Field, and Potential

An electron is released from rest in a uniform electric field. Which description of its motion is correct?

  1. It accelerates opposite the field direction, losing kinetic energy
  2. It accelerates opposite the field direction, gaining kinetic energy Correct answer.
  3. It moves at constant velocity along the field direction
  4. It accelerates along the field direction, losing kinetic energy

Explanation

The force on a charge is qE; for a negative charge this points opposite the field. The electron therefore accelerates opposite the field and the field does positive work, increasing its kinetic energy.

Question 36Electric Force, Field, and Potential

Two identical positive point charges are fixed a distance apart. At the midpoint between them, the electric field and the electric potential are respectively

  1. nonzero and zero
  2. both zero
  3. zero and nonzero Correct answer.
  4. both nonzero

Explanation

At the midpoint the two field vectors are equal and opposite, so they cancel to zero. The potentials are scalars and add, giving a positive nonzero value.

Question 37Electric Force, Field, and Potential

The electric field inside a solid conductor in electrostatic equilibrium is

  1. equal to the applied external field
  2. zero everywhere inside the material Correct answer.
  3. uniform and equal to the surface field
  4. directed radially outward from the center

Explanation

In electrostatic equilibrium, free charges rearrange until the internal field vanishes. Any residual internal field would drive further charge motion.

Question 38Electric Force, Field, and Potential

A parallel-plate capacitor has plate separation d and potential difference V. If V is doubled while d is unchanged, the magnitude of the electric field between the plates

  1. is halved
  2. is quadrupled
  3. stays the same
  4. is doubled Correct answer.

Explanation

For closely spaced plates the field is approximately uniform, with E = V/d. Doubling V at fixed d therefore doubles E.

Question 39Electric Force, Field, and Potential

A charge of +2.0 μC is moved from a point at potential 50 V to a point at potential 150 V. What is the change in its electric potential energy?

  1. +1.0×10⁻⁴ J
  2. +2.0×10⁻⁴ J Correct answer.
  3. +3.0×10⁻⁴ J
  4. -2.0×10⁻⁴ J

Explanation

The change in potential energy is ΔU = qΔV. Here q = 2.0×10⁻⁶ C and ΔV = 100 V, giving ΔU = +2.0×10⁻⁴ J.

Question 40Electric Force, Field, and Potential

Two point charges, +q and -q, are separated by a distance. Which statement about the electric field on the perpendicular bisector of the line joining them is correct?

  1. It points perpendicular to the line joining the charges
  2. It points parallel to the line joining the charges, toward the negative charge Correct answer.
  3. It points parallel to the line joining the charges, toward the positive charge
  4. It is zero everywhere on the bisector

Explanation

On the perpendicular bisector, the components of the two fields perpendicular to the axis cancel, while the components along the axis add. The net field points from the positive toward the negative charge.

Question 41Electric Force, Field, and Potential

The electric potential at a distance r from an isolated point charge Q is given by

  1. kQ/r²
  2. kQr
  3. kQ/r Correct answer.
  4. kQ²/r

Explanation

Potential from a point charge falls off as 1/r, V = kQ/r, and is a scalar. The field, by contrast, falls off as 1/r².

Question 42Electric Force, Field, and Potential

A small sphere carrying charge q hangs from an insulating thread in a uniform horizontal electric field E. In equilibrium the thread makes a small angle with the vertical. If q is doubled while E and the mass are unchanged, the tangent of the angle

  1. is unchanged
  2. is doubled Correct answer.
  3. is halved
  4. is quadrupled

Explanation

At equilibrium the horizontal electric force qE balances the horizontal component of tension, so tan θ = qE/(mg). Doubling q doubles tan θ.

Question 43Electric Force, Field, and Potential

A solid conducting sphere of radius R carries net positive charge Q. Which statement about the electric field is correct?

  1. It equals kQ/r² both inside and outside the sphere
  2. It is zero inside and equals kQ/r² outside at distance r from the center Correct answer.
  3. It increases linearly with r inside the sphere
  4. It is zero outside and nonzero inside the sphere

Explanation

Charge resides on the outer surface of a conductor, so the internal field is zero. Outside, the sphere behaves like a point charge at its center, giving E = kQ/r².

Question 44Electric Force, Field, and Potential

The work done by the electric field when a charge moves between two points depends on

  1. the speed of the charge
  2. the path taken between the points
  3. only the initial and final potentials Correct answer.
  4. the mass of the charge

Explanation

The electrostatic force is conservative, so the work depends only on the endpoints, W = -qΔV. The path, speed, and mass do not affect the work done by the field.

Question 45Electric Force, Field, and Potential

Two parallel plates separated by 4.0 mm have a potential difference of 200 V. What is the approximate magnitude of the electric field between them?

  1. 8.0×10⁻¹ V/m
  2. 8.0×10⁵ V/m
  3. 5.0×10⁴ V/m Correct answer.
  4. 5.0×10⁻⁴ V/m

Explanation

For parallel plates E = V/d. With V = 200 V and d = 4.0×10⁻³ m, E = 5.0×10⁴ V/m.

Question 46Electric Force, Field, and Potential

A positive charge is released from rest near a fixed positive charge. Which graph shape best describes how the electric potential energy of the system varies with separation r?

  1. A curve decreasing toward zero as r increases Correct answer.
  2. A horizontal line independent of r
  3. A curve increasing toward a constant as r increases
  4. A straight line decreasing with r

Explanation

The potential energy of two like charges is U = kq1q2/r, positive and falling toward zero as r grows. The 1/r dependence makes the graph a decreasing curve, not a line.

Question 47Electric Force, Field, and Potential

An electron and a proton are each accelerated through the same potential difference from rest. Which particle gains the greater kinetic energy?

  1. Neither gains kinetic energy
  2. They gain equal kinetic energy Correct answer.
  3. The proton gains more
  4. The electron gains more

Explanation

The work done by the field is qΔV, and both particles have the same charge magnitude, so they gain equal kinetic energy. Their masses differ, so their speeds differ, but not their energies.

Question 48Electric Force, Field, and Potential

The electric field at a point is best described as

  1. the total charge enclosed by a surface around that point
  2. the force that acts on any charge placed at that point
  3. the potential energy per unit charge at that point
  4. the force per unit positive charge that would act at that point Correct answer.

Explanation

Field is defined through the force on a small positive test charge, E = F/q, so it is force per unit charge. It is a property of the location independent of the test charge.

Question 49Electric Force, Field, and Potential

A point charge is placed at the center of a hollow conducting sphere that carries no net charge. Which statement about the field in the conductor material is correct?

  1. It is zero because the induced inner-surface charge cancels the field of the center charge Correct answer.
  2. It equals the field of the outer-surface charge alone
  3. It equals the field of the center charge alone
  4. It points radially outward through the conductor

Explanation

The center charge induces an equal and opposite charge on the inner surface, and together these produce zero field within the conductor material. The outer surface charge then produces no field inside itself.

Question 50Electric Force, Field, and Potential

Four point charges are arranged so that the net electric field at a certain point is zero. What can be concluded about the electric potential at that same point?

  1. It must be negative
  2. It is generally not zero, since potential is a scalar sum Correct answer.
  3. It must be positive
  4. It must also be zero

Explanation

Field is a vector that can cancel by symmetry, but potential is a scalar that adds algebraically. A zero net field therefore does not require zero potential.