4 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 4
Two systems are in thermal contact. Which statement must be true of them?
Answer and reasoning
AThey are touching, so that their particles can collide at the boundary. A student who thinks thermal contact means touching picks this. Radiation transfers energy between systems that do not touch, such as the Sun and Earth, and those systems are in thermal contact.
BThey are at different temperatures, so energy is flowing between them now. A student who thinks thermal contact exists only while energy flows picks this. Systems at the same temperature can be in thermal contact; they are then in thermal equilibrium, with no net energy transfer.
CEnergy can be transferred between them by conduction, convection or radiation.Correct Two systems are in thermal contact if they can transfer energy by thermal processes, which are conduction, convection and radiation. They need not touch, need not have matter between them and need not be at different temperatures; at equal temperatures they are in thermal contact with no net transfer.
DMatter lies between them, since no energy can ever be transferred across a vacuum. A student who thinks nothing can cross empty space picks this. Radiation transfers energy across a vacuum, as it does from the Sun to Earth, so systems separated by a vacuum can be in thermal contact.
A vacuum flask has double glass walls with a vacuum between them. Which statement explains how the vacuum layer helps keep hot coffee in the flask hot?
Answer and reasoning
AIt stops conduction and convection across the gap, since both processes need matter.Correct Conduction needs particles to collide and convection needs a fluid to move, so neither can carry energy across a vacuum. Radiation can cross the vacuum, which is why the glass is usually silvered to reduce it.
BIt stops all three thermal processes, since energy cannot be transferred across a vacuum. A student who thinks nothing can cross empty space picks this. Radiation needs no material medium and does cross a vacuum, as sunlight does; the vacuum stops conduction and convection only.
CIt stops heat rising out of the coffee, since heat naturally moves upward. A student who thinks heat itself rises picks this. Heat has no preferred direction; the vacuum surrounds the coffee on the sides and bottom and works by stopping conduction and convection in every direction.
DIt stops the cold of the air outside from getting in and cooling the coffee. A student who thinks of cold as something that flows in picks this. The coffee cools because energy leaves it; the vacuum slows that by stopping conduction and convection across the gap.
A small hot metal block is lowered into a large tank of cold water. At the surface of the block, atoms of the metal collide with molecules of the water. Which statement about the energy transferred in these collisions is correct?
Answer and reasoning
AEnergy passes from the metal atom to the water molecule in every single collision, as the metal is hotter. A student who thinks every particle of the hotter system is more energetic than every particle of the colder one picks this. Both systems have wide ranges of particle energies, so some water molecules are more energetic than some metal atoms, and in their collisions energy goes to the metal.
BNo energy passes either way, since in an elastic collision each particle keeps its own kinetic energy. A student who thinks each particle keeps its kinetic energy in an elastic collision picks this. Elastic means the total kinetic energy is conserved, not that each particle keeps its own; energy usually passes from one particle to the other.
CEnergy is more likely to pass from the water molecule to the metal atom, since the water has more internal energy. A student who thinks energy flows from the system with more internal energy picks this. The large tank of water may well have more internal energy than the block, but the direction of net transfer is set by temperature: the metal's atoms have the greater average kinetic energy, so energy is more likely to pass from a metal atom to a water molecule.
DEnergy is more likely to pass from the more energetic particle to the less energetic one, though not in every collision.Correct In a collision between two particles, energy is most likely to be transferred from the particle with more kinetic energy to the one with less. The metal's atoms have the greater average kinetic energy, so over very many collisions the net transfer is into the water, although some individual collisions transfer energy the other way.
A thermometer at room temperature, 20°C, is placed in a cup of water at 70°C. Its reading rises for a while and then stays steady. Why does the steady reading give the temperature of the water?
Answer and reasoning
AAll energy transfer between the thermometer and the water has then stopped completely for good. A student who thinks equilibrium means that all exchange stops picks this. Particles of the water and of the thermometer keep colliding and exchanging energy; what is zero at equilibrium is the net transfer.
BThe thermometer then has as much internal energy as the water in the cup, so the two readings match. A student who thinks equilibrium means equal internal energies picks this. The small thermometer has far less internal energy than the cup of water; what they share at equilibrium is their temperature.
CThe cold in the thermometer has then all flowed out into the water, so it can read correctly. A student who thinks of cold as something that flows picks this. Nothing called cold leaves the thermometer; energy enters it from the hotter water until there is no net transfer.
DThey are then in thermal equilibrium, so no net energy is transferred between them.Correct A thermometer shows its own temperature. While it is colder than the water there is a net transfer of energy into it and its reading rises. When its reading is steady there is no net transfer: it is in thermal equilibrium with the water, so its temperature equals the water's.
In preparation: 0 of 4 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
9.3.A.1 Thermal contact Fix
Thermal contact
Two systems are in thermal contact if energy can be transferred between them by thermal processes (conduction, convection or radiation). They need not touch: systems that exchange energy only by radiation are in thermal contact. Systems at the same temperature can be in thermal contact, with no net energy transfer.
Heating
The transfer of energy into a system by thermal processes, because of a temperature difference between the system and its surroundings. A rise in temperature caused by work, such as compressing a gas, is not heating.
Cooling
The transfer of energy out of a system by thermal processes. A system that is cooled loses energy; the energy is gained by its surroundings, not destroyed.
Students often think Two systems can exchange energy by thermal processes only if they touch, so systems that are not touching are not in thermal contact. In fact No. Systems are in thermal contact if energy can be transferred between them by any thermal process. Radiation needs no contact and no material in between, so the Sun and Earth are in thermal contact.
Students often think Two systems are in thermal contact only while they are at different temperatures and energy is flowing between them. In fact No. Thermal contact means that energy can be transferred between the systems by thermal processes. Two systems at the same temperature can be in thermal contact; that is the situation of thermal equilibrium, with no net energy transfer.
9.3.A.2 Conduction Fix
Conduction
Transfer of energy through a material, or between materials in contact, by collisions between neighboring particles, without the material as a whole moving.
Convection
Transfer of energy by the bulk movement of a fluid. Fluid warmed by a hot surface expands, becomes less dense than the cooler fluid around it and is pushed upward (a buoyant force), while cooler, denser fluid sinks to take its place.
Radiation (thermal)
Transfer of energy by electromagnetic waves emitted by a system because of its temperature. It needs no material medium, so it is the only thermal process that can transfer energy across a vacuum, as from the Sun to Earth.
Students often think Heat naturally moves upward, so heat rises out of hot objects and carries warm air up with it. In fact No. Heat is not a substance and has no preferred direction; energy is transferred by thermal processes from higher to lower temperature, in any direction. What rises in convection is warmed fluid, pushed up by the cooler, denser fluid around it.
Students often think Warming makes the particles of air, or the air itself, lighter, so warm air weighs less and floats up. In fact No. Each molecule keeps its mass. Warmed air expands, so each cubic meter contains fewer molecules and has less mass; the air is less dense, not made of lighter molecules.
9.3.A.3 Spontaneous direction of energy transfer Fix
Spontaneous direction of energy transfer
Energy is transferred by thermal processes, of its own accord, from the system at the higher temperature to the system at the lower temperature. The direction is set by the temperatures, not by which system has more internal energy.
Energy transfer in collisions between atoms
When particles of two systems collide, energy passes between them. It is more likely to pass from the particle with more kinetic energy to the one with less, but individual collisions can go either way; the net transfer, over very many collisions, is from the system whose particles have the greater average kinetic energy.
Most probable state
After very many collisions between the particles of two systems in thermal contact, the sharing of energy that is overwhelmingly the most likely is the one in which both systems have the same temperature.
Students often think Energy flows from the system with more internal energy to the system with less, whatever their temperatures. In fact No. The direction of spontaneous energy transfer by thermal processes is set by temperature: from higher temperature to lower. A large, cooler object can have more internal energy than a small, hotter one and still gain energy from it.
Students often think Energy is transferred between two systems until their internal energies are equal, so systems in thermal equilibrium have equal internal energies. In fact No. They have equal temperatures. Their internal energies also depend on how many particles each has and of what kind, so they are usually different.
9.3.A.4 Thermal equilibrium Fix
Thermal equilibrium
The state of two systems in thermal contact when no net energy is transferred between them by thermal processes; they are then at the same temperature. Particles of the two systems keep colliding and exchanging energy, but the transfers in the two directions balance.
Students often think The system that started at the higher temperature keeps transferring net energy to the other, even after the temperatures have become equal. In fact No. The net transfer depends on the present temperatures, not on the starting ones. Once the temperatures are equal, there is no net energy transfer by thermal processes.
Students often think At thermal equilibrium no energy at all passes between the particles of the two systems; all transfer stops. In fact No. The particles keep colliding and energy keeps passing in both directions. At equilibrium the transfers in the two directions balance, so the NET transfer is zero.
10 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 10
A gas is sealed in a cylinder whose walls and piston are insulated, so that no energy can pass through them by thermal processes. The piston is pushed in quickly, and the temperature of the gas rises. A student claims, "The gas was heated, because its temperature went up." Which evaluation of the claim is correct?
Answer and reasoning
ACorrect: a rise in a system's temperature is itself the evidence that the system was heated. A student who treats heating and a rise in temperature as the same thing picks this. A temperature can rise because work is done on a system, as here, where the insulation rules out heating.
BCorrect: pushing the piston in squeezed the heat already in the gas into a smaller space. A student who thinks of heat as a substance the gas contains picks this. The gas contains no heat to be squeezed; the energy that raised its temperature came in as work done by the moving piston.
CIncorrect: the walls are insulated, so no energy of any kind entered the gas at all. A student who thinks insulation blocks every kind of energy transfer picks this. Insulation blocks thermal processes only. The moving piston did work on the gas, transferring energy to it, which is why its temperature rose.
DIncorrect: the energy entered as work done by the piston, not by a thermal process.Correct Heating means energy is transferred into a system by thermal processes. The insulation prevents that, so the gas was not heated. The piston exerted a force on the gas while moving inward, doing work on it; that transfer of energy raised the gas's temperature.
A hot apple pie is set on a counter in a cooler kitchen. Which statement about the pie, with its reasoning, is correct?
Answer and reasoning
AThe pie is cooled, since cold is transferred into it from the counter and the air. A student who thinks of cold as something that flows picks this. Nothing called cold enters the pie; the pie cools because energy leaves it for the cooler counter and air.
BThe pie is cooled, since energy is transferred out of it by thermal processes.Correct Cooling means energy is transferred out of a system by thermal processes. The pie is at a higher temperature than the counter and the air, so energy leaves it by conduction, convection and radiation; the pie is cooled and its surroundings are heated.
CThe pie is cooled, since the energy it loses is used up and ceases to exist. A student who thinks energy is used up picks this. The energy the pie loses is not destroyed: the counter and the air gain exactly as much, and warm slightly.
DThe pie is heated, since it is the pie that gives out energy to its surroundings. A student who attaches 'heating' to the object that gives out energy picks this. Heating means energy transferred into a system. Energy leaves the pie, so the pie is cooled; the counter and air are heated.
The diagram shows two copper blocks, P and Q, placed in contact inside an insulated box, with the mass and temperature of each. Which statement about the energy transfer between the blocks is correct?
Answer and reasoning
AFrom P to Q, since P has more internal energy. A student who thinks energy flows from the system with more energy picks this. P does have more internal energy, with four times as many atoms at a similar temperature, but the direction of transfer is set by temperature, and Q is hotter.
BFrom Q to P, since Q is at the higher temperature.Correct Energy is transferred spontaneously by thermal processes from the system at the higher temperature to the one at the lower temperature. Q is at 60°C and P at 40°C, so the net transfer is from Q to P, even though P, with four times the mass, has more internal energy.
CFrom Q to P, in every collision between their atoms. A student who thinks every atom of the hotter block has more energy than every atom of the cooler one picks this. The net transfer is from Q to P, but both blocks have wide ranges of atomic energies, so in some collisions energy passes from an atom of P to an atom of Q.
DCold moves from P to Q, and no energy moves from Q into P. A student who thinks of cold as something that flows picks this. Only energy is transferred, from the hotter block Q to the cooler block P; Q cools because it loses energy.
A hot metal block is placed in cold water in an insulated container. Some time later the block and the water are at the same temperature. Which statement best explains why this final state is reached?
Answer and reasoning
AEnergy flows from the block into the water until the block and the water have equal internal energies. A student who thinks transfer stops when internal energies are equal picks this. The transfer stops when the temperatures are equal; the block and the water, with different numbers and kinds of particles, generally have different internal energies then.
BThe block's store of heat runs out, so that it has no more heat left over to pass on to the water. A student who thinks of heat as a stored substance picks this. The block does not contain heat that runs out; it still has plenty of internal energy at the end. The net transfer stops because the temperatures are equal.
COf all the ways the energy can be shared, those giving both the same temperature are by far the most likely.Correct Collisions between particles of the block and of the water keep passing energy back and forth. Of all the possible ways of sharing the energy, the ones in which the two systems have the same temperature (the same average kinetic energy of their particles) are overwhelmingly the most probable, so after many collisions that is the state found.
DThe collisions keep on sharing out energy until every particle has exactly the same energy as the rest. A student who thinks equilibrium means equal energies for all particles picks this. At the common temperature the particles still have a wide range of energies; what is equal is the average kinetic energy of the particles of each system.
A metal block X is placed in water Y inside an insulated container. The graph shows the temperatures of X and Y as functions of time t. Which claim about the energy transfer between X and Y at t = 8 min, with its reasoning, is correct?
Answer and reasoning
AThere is no net transfer, because X and Y are now at the same temperature.Correct The graph shows the two curves meeting at about 35°C by t ≈ 6 min and staying together. At t = 8 min X and Y are at the same temperature, so they are in thermal equilibrium: no net energy is transferred between them by thermal processes.
BThere is still a net transfer from X to Y, because X started out hotter. A student who keeps X labeled as 'the hot one' picks this. The net transfer depends on the temperatures at the moment in question; by t = 8 min the graph shows X and Y at the same temperature, so there is no net transfer.
CThere is no net transfer, because X and Y now have equal internal energies. A student who thinks equilibrium means equal internal energies picks this. There is no net transfer, but because the temperatures are equal; the internal energies of the block and the water need not be equal.
DNo energy passes between the particles of X and Y, as their temperatures match. A student who thinks all energy exchange stops at equilibrium picks this. The particles of X and Y keep colliding and energy keeps passing both ways; at equal temperatures the two directions balance, so only the net transfer is zero.
Working Read the graph: X falls from 80°C and Y rises from 20°C; the curves meet at about 35°C near t = 6 min and are level and equal after that. At t = 8 min, TX = TY ⇒ thermal equilibrium ⇒ no net energy transfer (energy still passes both ways between particles).
Air directly above a hot radiator rises toward the ceiling, and cooler air moves in to take its place. Which statement correctly explains why the warmed air rises?
Answer and reasoning
AWarmed air expands, so it is less dense than the cooler air around it, which sinks and pushes it up.Correct The warmed air expands, so each cubic meter holds fewer molecules and less mass: it is less dense than the surrounding cooler air. The denser air sinks and pushes it upward (the buoyant force exceeds its weight). This bulk movement of the fluid is convection.
BHeat itself naturally rises, and as it moves upward it carries the warmed air along with it. A student who thinks heat itself rises picks this. Heat has no preferred direction; it is the warmed air that rises, because it is less dense than the air around it.
CThe molecules of the air become lighter as they warm, so the warmed air weighs much less. A student who thinks warming makes particles lighter picks this. Each molecule keeps its mass; the warmed air is less dense because its molecules are spread farther apart.
DThe molecules of the air swell as they warm, and these larger molecules then float upward. A student who thinks particles swell when heated picks this. The molecules keep their size; the air expands because the faster-moving molecules spread farther apart, which lowers its density.
Two rigid, sealed containers are placed in contact inside an insulated box, and they exchange energy only with each other, by thermal processes through their walls. Container 1 has volume V and holds N atoms of helium. Container 2 has volume 4V and holds 3N atoms of argon. Treat both gases as ideal, and take the mass of an argon atom to be 10 times the mass of a helium atom. After a long time the two gases are in thermal equilibrium, and the pressure of the helium is P₀. What is the pressure of the argon?
Answer and reasoning
A1.00P₀ A student who thinks gases at the same temperature must be at the same pressure picks this. Thermal equilibrium makes the temperatures equal, but the pressure also depends on the number of atoms per unit volume, 3N/(4V) for the argon against N/V for the helium, so the argon's pressure is 0.75P₀.
B0.75P₀Correct In thermal equilibrium the two gases are at the same temperature T. For the helium, P₀ = NkBT/V. The argon has three times as many atoms in four times the volume at the same temperature, so its pressure is (3N)kBT/(4V) = (3/4)P₀ = 0.75P₀. The masses of the atoms do not affect the pressure of an ideal gas at a given temperature and number of atoms per unit volume.
C0.25P₀ A student who thinks the gases stop exchanging energy when their total energies are equal picks this: N(3/2)kBTHe = 3N(3/2)kBTAr gives TAr = THe/3, and the argon's pressure is (3N)kB(THe/3)/(4V) = 0.25P₀. Equilibrium means equal temperatures, not equal energies, so the argon's pressure is (3N)kBT/(4V) = 0.75P₀.
D7.50P₀ A student who thinks the atoms of the two gases end up with the same average speed picks this: an argon atom has 10 times the mass, so equal speeds would mean 10 times the average kinetic energy, TAr = 10THe, and a pressure of 7.50P₀. At equilibrium the atoms have equal average kinetic energies, (3/2)kBT, so the argon atoms move more slowly, and the pressure is 0.75P₀.
Working Thermal equilibrium: both gases at the same temperature T. Helium: P₀ = NkBT/V, so kBT = P₀V/N. Argon: P = (3N)kBT/(4V) = (3/4)P₀ = 0.75P₀. The atomic masses do not enter. Errors: equal pressures → 1.00P₀; equal total kinetic energies, N(3/2)kBTHe = 3N(3/2)kBTAr, so TAr = THe/3 → 0.25P₀; equal average speeds, argon atoms 10 times as massive, so TAr = 10THe → 7.50P₀.
An insulated, rigid box is divided into two compartments, A and B, by a fixed wall that conducts energy by thermal processes. Compartment A holds N atoms of an ideal monatomic gas at absolute temperature 3T₀, and compartment B holds N atoms of the same gas at T₀. The gases are left until they reach thermal equilibrium, and during this time energy E₁ is transferred out of the gas in A by thermal processes. The experiment is then repeated with the gas in A exactly as before, but with 2N atoms of the gas, at T₀, in compartment B. The box and the dividing wall take in negligible energy. How much energy is transferred out of the gas in A in the second experiment?
Answer and reasoning
A1.00E₁ A student who takes the equilibrium temperature to be halfway between the starting temperatures, 2T₀, in both experiments picks this, since A then cools by T₀ each time. With twice as many atoms in B, the common temperature is closer to B's starting temperature: N(3T₀ − Tf) = 2N(Tf − T₀) gives (5/3)T₀, so A cools further and loses 1.33E₁.
B2.00E₁ A student who reasons that twice as many atoms in B take in twice as much energy picks this. That would be true only if B's temperature still rose by T₀; the larger B warms less, only to (5/3)T₀, and the energy A loses is 1.33E₁.
C1.33E₁Correct The energy of each ideal monatomic gas is the total kinetic energy of its atoms, N(3/2)kBT, and the energy A loses equals the energy B gains. In the first experiment N(3T₀ − Tf) = N(Tf − T₀) gives Tf = 2T₀, so A cools by T₀ and E₁ = (3/2)NkBT₀. In the second, N(3T₀ − Tf) = 2N(Tf − T₀) gives Tf = (5/3)T₀, so A cools by (4/3)T₀ and loses (4/3)E₁ ≈ 1.33E₁.
D0.50E₁ A student who thinks the gases stop exchanging energy when their total energies are equal picks this. The total is (3/2)kB(3NT₀ + 2NT₀); half of it is (3/2)NkB(2.5T₀), so A would cool only to 2.5T₀ and lose 0.50E₁. Equilibrium is reached when the temperatures are equal, at (5/3)T₀, and A loses 1.33E₁.
Working Energy of an ideal monatomic gas = total kinetic energy of its atoms = N(3/2)kBT. Energy lost by A = energy gained by B. Experiment 1: N(3T₀ − Tf) = N(Tf − T₀) → Tf = 2T₀; E₁ = (3/2)NkB(3T₀ − 2T₀) = (3/2)NkBT₀. Experiment 2: N(3T₀ − Tf) = 2N(Tf − T₀) → Tf = (5/3)T₀; E₂ = (3/2)NkB(3T₀ − 5T₀/3) = (3/2)NkB(4/3)T₀. E₂/E₁ = 4/3 ≈ 1.33. Errors: halfway temperature 2T₀ both times → 1.00E₁; twice the atoms take twice the energy → 2.00E₁; equal total energies at equilibrium, A keeps half of (3/2)kB(3N + 2N)T₀ = (3/2)NkB(2.5T₀) → E₂ = (3/2)NkB(0.5T₀) = 0.50E₁.
A sealed, rigid container holding 8.0 g of helium (molar mass 4.0 g/mol) at 420 K is placed in contact with a sealed, rigid container holding 40.0 g of argon (molar mass 40 g/mol) at 270 K. The two containers are inside an insulated box and exchange energy only with each other, by thermal processes through their walls, and the containers themselves take in negligible energy. Treat both gases as ideal monatomic gases. Use R = 8.31 J/(mol·K). How much energy is transferred out of the helium by cooling before the gases reach thermal equilibrium?
Answer and reasoning
A1.9 × 10³ J A student who takes the common temperature to be halfway between the starting temperatures, 345 K, picks this: (3/2)(2.0)(8.31)(75) = 1.9 × 10³ J. The helium has twice as many atoms as the argon, so it changes temperature half as much, and the common temperature, 370 K, is closer to the helium's starting temperature.
B1.2 × 10³ JCorrect The helium is 2.0 mol and the argon 1.0 mol. The energy of each gas is the total kinetic energy of its atoms, (3/2)nRT, and the energy the helium loses equals the energy the argon gains: 2.0(420 K − Tf) = 1.0(Tf − 270 K), so Tf = 370 K. The helium cools by 50 K and loses (3/2)(2.0 mol)(8.31 J/(mol·K))(50 K) = 1.2 × 10³ J.
C3.1 × 10³ J A student who weights the temperatures by the masses of the samples picks this: (8.0 × 420 + 40.0 × 270)/48.0 = 295 K, so the helium would lose (3/2)(2.0)(8.31)(125) = 3.1 × 10³ J. The energy of each gas depends on its number of atoms, not its mass: the 8.0 g of helium is 2.0 mol, twice the 1.0 mol of argon, so Tf = 370 K and Q = 1.2 × 10³ J.
D3.6 × 10³ J A student who thinks energy is transferred until the two gases have equal energies picks this: the total is (3/2)R(2.0 mol × 420 K + 1.0 mol × 270 K), and if each gas ended with half of it, the helium would lose (3/2)(8.31 J/(mol·K))(840 − 555) mol·K = 3.6 × 10³ J, leaving the argon at 555 K, hotter than the helium ever was. Transfer stops when the temperatures are equal, at 370 K, so the helium loses 1.2 × 10³ J.
Working Amounts: nHe = 8.0 g/(4.0 g/mol) = 2.0 mol; nAr = 40.0 g/(40 g/mol) = 1.0 mol. Energy of each gas = total kinetic energy of its atoms = N(3/2)kBT = (3/2)nRT (NkB = nR). Energy lost by helium = energy gained by argon: (3/2)R(2.0)(420 − Tf) = (3/2)R(1.0)(Tf − 270), so Tf = 370 K. Q = (3/2)(2.0 mol)(8.31 J/(mol·K))(420 K − 370 K) = 1246 J ≈ 1.2 × 10³ J. Errors: halfway, Tf = 345 K → 1870 J ≈ 1.9 × 10³ J; weighting by mass, Tf = (8.0 × 420 + 40.0 × 270)/48.0 = 295 K → 3116 J ≈ 3.1 × 10³ J; equal energies at the end, each gas has (3/2)R(1110 mol·K)/2, helium keeps (3/2)R(555 mol·K) → Q = (3/2)(8.31)(840 − 555) = 3553 J ≈ 3.6 × 10³ J.
A small, sealed, rigid container holds helium gas at absolute temperature 4T₀. It is lowered into a large tank of argon gas at absolute temperature T₀; the tank holds about 10⁴ times as many atoms as the container. It is left until no net energy is transferred between the helium and the argon. An argon atom has 10 times the mass of a helium atom. Treat both gases as ideal monatomic gases. By what factor does the rms speed of the helium atoms change?
Answer and reasoning
A×0.25 A student who takes the rms speed to be proportional to the absolute temperature picks this. It is the average kinetic energy, (1/2)mvrms², that is proportional to T, so the speed changes by √(1/4) = 0.50 when the temperature falls from 4T₀ to T₀.
B×0.50Correct When no net energy is transferred, the helium and the argon are at the same temperature; because the argon has about 10⁴ times as many atoms, its temperature rises by only about 0.0003T₀, so the common temperature is T₀ to within 0.03%. The average kinetic energy of the helium atoms, (1/2)mvrms² = (3/2)kBT, falls to one-fourth, so their rms speed changes by √(1/4) = 0.50.
C×0.79 A student who takes the common temperature to be halfway between 4T₀ and T₀, 2.5T₀, picks this: √(2.5/4) = 0.79. The argon has about 10⁴ times as many atoms, so its temperature hardly changes; the common temperature is very nearly T₀, and the helium's rms speed changes by √(1/4) = 0.50.
D×0.16 A student who thinks equilibrium means the helium atoms end with the same average speed as the argon atoms picks this: argon atoms at very nearly T₀ have rms speed √(1/4) × √(1/10) = 0.16 times the helium's starting value. At equilibrium the atoms have the same average kinetic energy, not the same speed, so the light helium atoms stay faster and their rms speed changes by 0.50.
Working No net transfer → thermal equilibrium → common temperature. Energy lost by the helium = energy gained by the argon (each atom's average energy (3/2)kBT): N(4T₀ − Tf) = 10⁴N(Tf − T₀), so Tf = 1.0003T₀ ≈ T₀. (1/2)mvrms² = (3/2)kBT, so vrms ∝ √T: factor = √(T₀/4T₀) = 1/2 → ×0.50. The argon atoms' mass is not needed. Errors: vrms ∝ T → ×0.25; common temperature halfway, 2.5T₀ → √(2.5/4) = 0.79; helium atoms end with the argon atoms' average speed: vAr = vHe,i √(T₀/4T₀) √(1/10) → ×0.16.
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
