2 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 2
A sealed box contains helium gas. Which description of the helium atoms explains why the gas exerts a steady outward force on every wall of the box?
Answer and reasoning
AAtoms at rest in fixed places repel one another, so the gas pushes out on each wall like a spring. A student who pictures a gas as atoms held apart by repulsion picks this. Gas atoms are not at rest in fixed places; they move in random directions and exert forces only during collisions. The push on the walls comes from atoms colliding with them.
BAtoms moving in random directions keep hitting each wall, and each collision pushes on it.Correct In the kinetic model the atoms move in random directions and collide with the walls. In each collision the wall exerts a force on the atom and the atom exerts an equal and opposite force on the wall. Very many collisions each second, on every wall, add up to a steady average outward force.
CThe atoms swell until together they fill the box, and so they press hard on each of its walls. A student who gives atoms the properties of the bulk gas picks this. The atoms of a gas do not swell to fill the container; they are tiny compared with the spaces between them. They push on the walls because they move and collide with them.
DHeat stored in the helium pushes outward, so the gas presses on each wall of the box. A student who thinks of heat as a substance inside the gas picks this. The gas does not contain heat that pushes. Its pressure comes from its atoms colliding with the walls; heating the gas makes the atoms move faster, so they collide harder and more often.
A container holds a mixture of 1.0 mol of helium and 3.0 mol of argon, both at the same temperature. An argon atom has about ten times the mass of a helium atom. Which quantity is the same for the helium as for the argon?
Answer and reasoning
AAverage speed of the atoms A student who thinks equal temperatures mean equal speeds picks this. Equal temperatures mean equal average kinetic energies; since K = (1/2)mv², the lighter helium atoms move faster on average, by a factor of about √10 ≈ 3.
BTotal kinetic energy of its atoms A student who thinks temperature measures the total kinetic energy picks this. The average kinetic energy is the same, but there are three times as many argon atoms, so the argon has three times the total kinetic energy.
CKinetic energy of every atom A student who thinks every atom at a given temperature has the same energy picks this. In each gas the atoms have a wide range of kinetic energies, as the Maxwell–Boltzmann distribution shows; only the average is fixed by the temperature.
DAverage kinetic energy per atomCorrect Temperature is characterized by the average kinetic energy of the atoms, Kavg = (3/2)kB T. At the same temperature, helium and argon atoms have the same average kinetic energy, whatever their masses.
In preparation: 0 of 2 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
9.1.A.1 Kinetic model of a gas Fix
Kinetic model of a gas
A model in which a gas is made of a very large number of atoms moving in random directions. The atoms collide with one another and with the walls of the container, and in each collision they exert forces on whatever they strike; the pressure and temperature of the gas are explained by this motion.
Momentum in a collision with a wall
In a collision with a fixed, smooth wall, the wall exerts a force on the atom perpendicular to the wall, so only the perpendicular component of the atom's momentum changes. In an elastic collision that component reverses: an atom striking at speed v at angle θ from the normal has a momentum change of magnitude 2mv cos θ, and by Newton's third law it exerts an impulse of the same magnitude on the wall. SI unit of momentum and impulse: kg·m/s = N·s.
Collisions between atoms
When two atoms collide, the forces they exert on each other are equal in magnitude and opposite in direction, so the total momentum of the pair is conserved. In the ideal-gas model these collisions are elastic, so the total kinetic energy of the pair is also unchanged, although energy is usually transferred from one atom to the other.
Pressure, P
The ratio of the sum of the magnitudes of the perpendicular components of the forces exerted on a surface to the area of the surface: P = F⊥/A. For a gas, the forces come from atoms colliding with the surface, and the pressure is the time-averaged total perpendicular force per unit area. SI unit: pascal (Pa = N/m²).
Pressure in the kinetic model
The pressure of a gas on a wall depends on the momentum each atom delivers per collision and on how often atoms strike the wall. Both increase with the speed of the atoms, so the pressure of a fixed amount of gas in a fixed volume is proportional to vrms², and so to the absolute temperature.
Pressure throughout a gas
Pressure is a property of the gas at every point, not only at its boundary. A small surface placed anywhere in the gas is struck by atoms and experiences the pressure of the gas there; in a container of ordinary size the pressure is essentially the same everywhere.
Students often think A gas presses on its container because its atoms repel one another and push outward, like compressed springs, whether or not they are moving. In fact No. In the kinetic model the atoms move in random directions and exert forces only when they collide. A gas pushes on a wall because atoms keep striking it; each collision exerts a brief force on the wall, and the very many collisions add up to a steady average force.
Students often think The atoms of a gas swell when it is heated, and it is their larger size that makes the gas expand or press harder on its container. In fact No. The atoms keep the same size. Heating increases their average kinetic energy, so they move faster and strike the walls harder and more often; when a gas expands, the spaces between the atoms grow, not the atoms.
9.1.B.1 Temperature, T Fix
Temperature, T
A property of a system characterized by the average kinetic energy of its atoms. In the ideal-gas equations T is the absolute temperature, T(K) = T(°C) + 273, which is proportional to the average kinetic energy of the atoms. SI unit: kelvin (K).
Average kinetic energy, Kavg
The mean of the kinetic energies of all the atoms of a system. For an ideal gas, Kavg = (3/2)kB T, the same for every gas at the same temperature whatever the mass of its atoms. SI unit: joule (J).
Maxwell–Boltzmann distribution
A graph of the number of atoms in a gas against their speed (or kinetic energy) at a given temperature. It rises from zero, peaks at the most probable speed and has a long tail toward high speeds. At a higher temperature the distribution shifts to higher speeds and spreads out, so for the same number of atoms its peak is lower; the area under the curve represents the number of atoms.
Boltzmann constant, kB
The constant that links the average kinetic energy of an atom to the absolute temperature: kB = 1.38 × 10⁻²³ J/K. It is the gas constant per atom; R = N0 kB is the gas constant per mole.
Root-mean-square speed, vrms
The speed of an atom whose kinetic energy equals the average kinetic energy of the atoms of the gas: (1/2)mvrms² = (3/2)kB T, so vrms = √(3kB T/m), where m is the mass of one atom. It is proportional to √T and to 1/√m. SI unit: m/s.
Students often think At the same temperature, the atoms of every gas move with the same average speed. In fact No. At the same temperature they have the same average kinetic energy, (3/2)kB T. Since K = (1/2)mv², atoms of smaller mass move faster: vrms is proportional to 1/√m.
Students often think Temperature measures the total kinetic energy of a system's atoms, so two samples at the same temperature have the same total kinetic energy. In fact No. Temperature is characterized by the average kinetic energy of the atoms. The total kinetic energy also depends on how many atoms there are.
8 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 8
The diagram shows, from above, a helium atom of mass 6.64 × 10⁻²⁷ kg that strikes a fixed, smooth wall of its container with speed 1.20 × 10³ m/s and rebounds elastically with the same speed. The angles are marked on the diagram. What is the magnitude of the change in the atom's momentum during the collision?
Answer and reasoning
A3.98 × 10⁻²⁴ kg·m/s A student who counts only the stopping part of the collision picks this: mv cos 60° = 3.98 × 10⁻²⁴. The wall first brings the perpendicular momentum to zero and then pushes the atom back out with the same perpendicular momentum in the opposite direction, so the change is twice as large.
B1.59 × 10⁻²³ kg·m/s A student who reverses the whole velocity picks this: 2mv = 1.59 × 10⁻²³, the result for a head-on collision. The atom strikes at an angle, and a smooth wall exerts no force parallel to itself, so only the perpendicular component, v cos 60°, reverses.
C7.97 × 10⁻²⁴ kg·m/sCorrect The smooth wall exerts a force only perpendicular to itself, so the component of velocity parallel to the wall is unchanged. The angle is measured from the normal, so the perpendicular component is v cos 60°; it reverses, from toward the wall to away from it. |Δp| = 2mv cos 60° = 2(6.64 × 10⁻²⁷ kg)(1.20 × 10³ m/s)(0.500) = 7.97 × 10⁻²⁴ kg·m/s.
D1.38 × 10⁻²³ kg·m/s A student who uses the sine picks this: 2mv sin 60° = 1.38 × 10⁻²³. The 60° angle is measured from the normal, so the component perpendicular to the wall is v cos 60°; v sin 60° is the component parallel to the wall, which does not change.
Working Perpendicular (normal) component of velocity: v cos 60° = (1.20 × 10³ m/s)(0.500) = 600 m/s, reversed by the collision; the parallel component v sin 60° is unchanged (smooth wall). |Δp| = 2mv cos 60° = 2(6.64 × 10⁻²⁷ kg)(600 m/s) = 7.968 × 10⁻²⁴ ≈ 7.97 × 10⁻²⁴ kg·m/s. Errors: stop only, mv cos 60° = 3.98 × 10⁻²⁴; whole velocity reversed, 2mv = 1.59 × 10⁻²³; sine for cosine, 2mv sin 60° = 1.38 × 10⁻²³.
In a computer simulation of a gas, the atoms that strike one flat wall of the container during a time interval of 0.020 s deliver a total impulse of 2.0 × 10⁻² N·s to the wall, directed perpendicular to it. The wall has an area of 4.0 cm². What is the average pressure the gas exerts on the wall during this interval?
Answer and reasoning
A2.5 × 10³ PaCorrect The average total perpendicular force is the impulse divided by the time interval: F = (2.0 × 10⁻² N·s)/(0.020 s) = 1.0 N. The area is 4.0 cm² = 4.0 × 10⁻⁴ m². P = F⊥/A = 1.0 N/(4.0 × 10⁻⁴ m²) = 2.5 × 10³ Pa.
B5.0 × 10¹ Pa A student who uses the impulse as if it were the force picks this: (2.0 × 10⁻² N·s)/(4.0 × 10⁻⁴ m²) = 5.0 × 10¹. Impulse is force multiplied by time, so the average force is the impulse divided by 0.020 s, which is 1.0 N.
C2.5 × 10¹ Pa A student who converts the area as if it were a length picks this, using 4.0 cm² = 4.0 × 10⁻² m²: 1.0 N/(4.0 × 10⁻² m²) = 2.5 × 10¹ Pa. Since 1 cm = 10⁻² m, 1 cm² = 10⁻⁴ m², so the area is 4.0 × 10⁻⁴ m².
D1.0 × 10⁰ Pa A student who takes the total force on the wall as its pressure picks this: the average force is 1.0 N. Pressure is force per unit area, so the 1.0 N must be divided by the wall's area, 4.0 × 10⁻⁴ m².
Working F⊥,avg = J/Δt = (2.0 × 10⁻² N·s)/(0.020 s) = 1.0 N. A = 4.0 cm² = 4.0 × (10⁻² m)² = 4.0 × 10⁻⁴ m². P = F⊥/A = 1.0 N/(4.0 × 10⁻⁴ m²) = 2.5 × 10³ Pa. Errors: impulse used as force, 2.0 × 10⁻²/4.0 × 10⁻⁴ = 5.0 × 10¹ Pa; area converted with 10⁻², 1.0/(4.0 × 10⁻²) = 2.5 × 10¹ Pa; force taken as pressure, 1.0 × 10⁰.
A rigid, sealed container holds an ideal gas. The absolute temperature of the gas is doubled. Using the kinetic model, in which the pressure comes from atoms colliding with the walls, by what factor is the pressure of the gas multiplied?
Answer and reasoning
A1.4 A student who counts only the harder collisions picks this: each collision delivers √2 ≈ 1.4 times as much momentum. The faster atoms also cross the container in less time, so they strike the walls √2 times as often; the pressure is multiplied by (√2)² = 2.0.
B4.0 A student who thinks the atoms' speed is proportional to the absolute temperature picks this: speed ×2, so momentum per collision ×2 and collisions per second ×2, giving ×4. The average kinetic energy, not the speed, is proportional to T, so the speed increases by √2 and the pressure by 2.0.
C1.0 A student who thinks pressure depends on how crowded the atoms are picks this: the container is rigid and sealed, so the number of atoms per unit volume is unchanged. The atoms move faster, however, so they strike the walls harder and more often, and the pressure doubles.
D2.0Correct Doubling T doubles Kavg, so vrms is multiplied by √2. Each atom then delivers √2 times as much momentum per collision, and, moving √2 times as fast in the same container, strikes the walls √2 times as often. The pressure is multiplied by √2 × √2 = 2.0.
Working Kavg = (3/2)kB T, so T ×2 gives Kavg ×2 and vrms ×√2. Pressure ∝ (momentum per collision ∝ v) × (collisions per second ∝ v) ∝ v² ∝ T. Factor = (√2)(√2) = 2.0. Errors: impulse factor alone √2 ≈ 1.4; speed ∝ T gives 2 × 2 = 4.0; crowding only gives 1.0.
A small, thin pressure sensor hangs on a thread at the center of a sealed box of gas, far from every wall. An identical sensor is mounted on one of the side walls at the same height. Which claim about the two readings, with its reasoning, is correct?
Answer and reasoning
AThey are equal, since atoms at rest in contact pass the walls' push on to the center. A student who pictures a gas as atoms at rest, packed in contact and passing on a push, picks this. The readings are equal, but not for that reason: the atoms are far apart and moving, and the pressure at the center comes from atoms colliding with the sensor there.
BThe central one reads zero, since a gas exerts its pressure where it pushes on its walls. A student who thinks a gas has pressure only at its container's walls picks this. Pressure exists throughout the gas: any surface in the gas, including the central sensor, is struck by moving atoms.
CThey are equal, since atoms strike the central sensor just as they strike the wall sensor.Correct Pressure exists throughout the gas, not only at its boundary. Atoms moving in random directions strike the central sensor as often and as hard as they strike the sensor on the wall, so the two sensors read the same pressure.
DThe central one reads less, since it feels only the weight of the gas above it. A student who explains gas pressure by the weight of the gas above picks this. In a box of gas the weight of the gas is tiny compared with the forces from collisions, so the pressure is essentially the same everywhere, and the central sensor reads the same as the wall sensor.
The graph shows the distributions of atomic speeds for two samples, 1 and 2, of the same ideal gas. The two samples contain the same number of atoms. Which claim about the temperatures of the samples, with its reasoning, is correct?
Answer and reasoning
ASample 1 is hotter, since the peak of its curve is the taller of the two. A student who reads the taller curve as the hotter gas picks this. The height of the peak shows how many atoms have speeds near the most probable speed. Curve 1 is tall and narrow because its atoms are crowded into a range of lower speeds; it is the cooler sample.
BSample 2 is hotter, since its distribution lies at higher speeds.Correct The temperature is characterized by the average kinetic energy of the atoms. Curve 2 peaks at a higher speed and has more atoms at high speeds, so its atoms have the greater average kinetic energy and sample 2 is at the higher temperature. Its peak is lower because the same number of atoms is spread over a wider range of speeds.
CThey are equally hot, since the areas under the two curves are equal. A student who thinks the area under a speed distribution shows temperature picks this. The area represents the number of atoms, which the stem says is the same for both samples. Temperature shows in how far along the speed axis the distribution lies, and curve 2 lies farther.
DNeither has one temperature, since its atoms move at a range of speeds. A student who thinks all the atoms at one temperature move at the same speed picks this. A gas at a single temperature always has a distribution of speeds; the temperature fixes the average kinetic energy, which is greater for sample 2.
Working Same gas, same number of atoms. Curve 2 peaks at a higher speed (by a factor of about √2 on this sketch) and extends farther to high speeds, so its average kinetic energy, and so its absolute temperature, is greater. Equal areas reflect equal numbers of atoms; the lower, broader peak of curve 2 follows from spreading the same number of atoms over a wider speed range.
A sample of helium gas is at 27°C. The mass of a helium atom is 6.64 × 10⁻²⁷ kg. What is the root-mean-square speed of the helium atoms? Use kB = 1.38 × 10⁻²³ J/K.
B4.10 × 10² m/s A student who uses the Celsius temperature picks this: √(3(1.38 × 10⁻²³)(27)/(6.64 × 10⁻²⁷)) = 4.10 × 10². The equation needs the absolute temperature, 300 K.
C9.67 × 10² m/s A student who writes the kinetic energy as mv², without the ½, picks this: mv² = (3/2)kB T gives √(3kB T/(2m)) = 9.67 × 10². With K = (1/2)mv², vrms = √(3kB T/m) = 1.37 × 10³ m/s.
D1.87 × 10⁶ m/s A student who stops at vrms² picks this: 3kB T/m = 1.87 × 10⁶, which is in m²/s². The rms speed is its square root, 1.37 × 10³ m/s.
The atoms of an ideal gas each have mass m, and their root-mean-square speed is vrms. Which expression gives the absolute temperature T of the gas?
Answer and reasoning
A2mvrms²/(3kB) A student who writes the kinetic energy as mv², without the ½, picks this: mvrms² = (3/2)kB T gives T = 2mvrms²/(3kB), twice the correct value.
Bmvrms/(3kB) A student who takes the temperature to be proportional to the speed itself picks this. The temperature is proportional to the average kinetic energy, which depends on the square of the speed: T = mvrms²/(3kB).
Cmvrms²/(3kB)Correct The average kinetic energy is Kavg = (1/2)mvrms² = (3/2)kB T. Multiplying both sides by 2/(3kB) gives T = mvrms²/(3kB).
Dmvrms²/(3R) A student who uses the gas constant R picks this. R is the constant per mole; the equation concerns one atom of mass m, so the Boltzmann constant kB, the constant per atom, is needed.
Working (1/2)mvrms² = (3/2)kB T ⇒ T = (2/(3kB))(1/2)mvrms² = mvrms²/(3kB). Errors: K = mv² gives 2mvrms²/(3kB); dropping the square gives mvrms/(3kB); R for kB gives mvrms²/(3R).
A single atom of mass m moves inside a closed rectangular box, and no other atoms are present. Two opposite walls of the box, W and W′, each have area A and are a distance L apart, and the atom moves in a plane perpendicular to W. The walls are smooth and every collision with them is elastic, so the atom's speed is always v, and its velocity always makes an angle θ with the line perpendicular to W. Ignore gravity. Averaged over a long time, which expression gives the pressure P that the atom exerts on wall W?
Answer and reasoning
AP = mv² cosθ/(AL) A student who thinks the atom's whole velocity is reversed at the wall picks this: Δp = 2mv at each collision, and 2mv/(2L/(v cos θ)) = mv² cos θ/L of force. A smooth wall pushes only perpendicular to itself, so only the component v cos θ is reversed and Δp = 2mv cos θ, which gives P = mv² cos²θ/(AL).
BP = mv² cos²θ/(AL)Correct A smooth wall pushes only perpendicular to itself, so at W only the component v cos θ is reversed and each collision changes the atom's momentum by 2mv cos θ. Collisions with the other walls do not change this component, so between hits on W the atom covers 2L perpendicular to W (to W′ and back) at v cos θ, taking 2L/(v cos θ). The average force is (2mv cos θ)/(2L/(v cos θ)) = mv² cos²θ/L, and dividing by A gives P = mv² cos²θ/(AL).
CP = 2mv² cos²θ/(AL) A student who takes the time between collisions with W as L/(v cos θ) picks this, doubling the collision rate. After striking W the atom must cross to W′ and come back, 2L perpendicular to W, before it strikes W again, so the time is 2L/(v cos θ) and P = mv² cos²θ/(AL).
DP = (2mv cosθ)/A A student who uses the impulse of one collision, 2mv cos θ, as the force on the wall picks this; its units, N·s/m², are not those of pressure. The force is the momentum delivered per unit time: dividing 2mv cos θ by the time between hits, 2L/(v cos θ), gives mv² cos²θ/L, and P = mv² cos²θ/(AL).
Working Component perpendicular to W: v cos θ; collisions with the other walls do not change it. Each collision with W: Δp⊥ = 2mv cos θ. Time between hits on W (across to W′ and back, 2L at v cos θ): Δt = 2L/(v cos θ). Average force F = (2mv cos θ)/(2L/(v cos θ)) = mv² cos²θ/L. P = F/A = mv² cos²θ/(AL). Errors: whole velocity reversed, Δp = 2mv → mv² cosθ/(AL); return trip taken as L → 2mv² cos²θ/(AL); impulse of one collision used as the force → (2mv cosθ)/A.
Compiled from the AP Physics 2 Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
