5 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 5
A copper block rests on a laboratory bench at room temperature. Which of the following makes up the internal energy of the block?
Answer and reasoning
AThe kinetic energy of its vibrating atoms, without any potential energy of their arrangement A student who thinks internal energy is only the random kinetic energy of the particles picks this. That is true of an ideal gas, whose atoms exert no conservative forces on one another, but the atoms of a solid are held in place by such forces, so the block's internal energy also includes the potential energy of their arrangement.
BThe heat stored in the block, which it gives out when it is placed next to a colder object A student who thinks of heat as something an object contains picks this. Heating is energy transferred because of a temperature difference; it is not stored as heat. What the block contains is internal energy, the kinetic and potential energy of its atoms, which heating or cooling can change.
CThe kinetic energy of its vibrating atoms plus the potential energy of their arrangementCorrect Internal energy is the sum of the kinetic energies of the objects that make up the system (here, copper atoms vibrating about their positions) and the potential energy of their configuration, which comes from the forces the atoms exert on one another.
DThe block's gravitational potential energy on the bench plus the kinetic energy of the whole block A student who counts the energy of the block as a whole picks this. The gravitational potential energy belongs to the block–Earth system and depends on the block's position, and the kinetic energy of the whole block depends on the motion of its center of mass; neither is internal energy, which concerns the atoms' motion and arrangement relative to one another.
A sealed container of ideal gas rests on a level table. The container is heated, and the temperature of the gas increases. Which statement describes the gas and the container after the heating?
Answer and reasoning
AThe container starts to move, as its gas has gained kinetic energy. A student who thinks that the kinetic energy gained by the atoms must appear as motion of the whole object picks this. The atoms move in random directions, so their extra kinetic energy is internal energy; with no net external horizontal force, the center of mass stays at rest.
BThe gas atoms keep their average speed, but the gas now holds more heat. A student who thinks heat is a substance stored alongside the atoms picks this. The energy transferred by heating becomes internal energy, which for an ideal gas is the kinetic energy of its atoms, so their average speed must increase as the temperature rises.
CThe gas atoms move faster on average, and the container stays at rest.Correct Heating increases the internal energy of the gas, which for an ideal gas is the kinetic energy of its atoms moving randomly relative to one another: a higher temperature means a greater average kinetic energy. The random motions add to zero net momentum, so the center of mass of the gas and container stays at rest.
DThe container now weighs less, as the gas in it is heated and hot gas rises. A student who thinks heated gas weighs less because hot air rises picks this. Hot air rises through cooler air because it is less dense; the sealed container holds the same mass of gas before and after heating, so its weight is unchanged.
A 2.0 kg copper block at 80°C and a 1.0 kg copper block at 20°C are placed in contact inside an insulated box, and the two blocks form an isolated system. They reach the same final temperature. Which statement about the energy of the blocks is correct?
Answer and reasoning
AThe blocks' total energy has fallen, as some was used up in the transfer. A student who thinks energy is used up when it is transferred picks this. Energy is conserved: in an isolated system none can be lost, so all of the energy lost by the hotter block is gained by the colder block.
BThe total energy of the two blocks is the same afterward as it was beforehand.Correct The two blocks form an isolated system, so their total energy is constant. Energy is transferred from the hotter block to the colder one until their temperatures are equal: the internal energy lost by the 2.0 kg block equals the internal energy gained by the 1.0 kg block.
CThe two blocks end with equal internal energies, as they have equal temperatures. A student who thinks equal temperatures mean equal internal energies picks this. Temperature is related to the average kinetic energy per atom; the 2.0 kg block has twice as many atoms at the same final temperature, so it has more internal energy.
DThe cold block keeps the same energy, and its cold passes into the hot block. A student who thinks of cold as something that flows picks this. Energy is transferred from the hotter block to the colder block; the colder block's temperature rises because its internal energy increases.
The P–V diagram shows a process in which a gas expands from state 1 to state 2 along a straight line. Which expression gives the work W done on the gas during the process?
Answer and reasoning
A−6P₀V₀ A student who takes the work as the area between the line and the P axis picks this: the line spans pressures from P₀ to 4P₀ at volumes from 3V₀ down to V₀, an area of (1/2)(V₀ + 3V₀)(3P₀) = 6P₀V₀. Work is the area between the path and the V axis, (1/2)(4P₀ + P₀)(2V₀) = 5P₀V₀, and it is negative because the gas expands.
B−2P₀V₀ A student who uses W = −PΔV with the final pressure, P₀, picks this: −P₀(2V₀). The pressure falls from 4P₀ during the process, so the work is the whole area under the line.
C−5P₀V₀Correct The magnitude of the work equals the area under the line down to the V axis: a trapezoid of width 3V₀ − V₀ = 2V₀ with parallel sides 4P₀ and P₀, area (1/2)(4P₀ + P₀)(2V₀) = 5P₀V₀. The volume increases, so the work done on the gas is negative: W = −5P₀V₀.
D−3P₀V₀ A student who takes only the triangle above the level of the final pressure picks this: (1/2)(3P₀)(2V₀) = 3P₀V₀. The area under the line also includes the rectangle P₀ × 2V₀ beneath the triangle.
Working |W| = area under the line from V₀ to 3V₀ = (1/2)(4P₀ + P₀)(2V₀) = 5P₀V₀. Expansion, so W = −5P₀V₀. (Final pressure × ΔV = 2P₀V₀; triangle alone (1/2)(3P₀)(2V₀) = 3P₀V₀; area toward the P axis (1/2)(V₀ + 3V₀)(3P₀) = 6P₀V₀.)
An ideal gas in a cylinder is compressed slowly, so that its temperature stays constant. The work done on the gas during the compression is W, which is positive. Which statement about the energy transferred by heating or cooling is correct?
Answer and reasoning
AEnergy equal to W is transferred into the gas by heating. A student who uses the work done by the gas, −W, in ΔU = Q + W picks this, getting Q = +W. With W the work done on the gas, Q = ΔU − W = −W: energy leaves the gas.
BNo energy is transferred, since the temperature stays fixed. A student who thinks constant temperature means no energy transfer by heating or cooling picks this. Heating and cooling are transfers, not temperature changes: here energy W must leave by cooling, or the work done on the gas would raise its temperature.
CLess than W leaves by cooling; the rest is stored in the gas. A student who thinks compressing an ideal gas stores energy in the spacing of its atoms picks this. Ideal-gas atoms exert no conservative forces on one another, so there is no such potential energy; with ΔU = 0, all of W leaves by cooling.
DEnergy equal to W is transferred out of the gas by cooling.Correct For an ideal gas at constant temperature, U = (3/2)nRT is constant, so ΔU = 0. From ΔU = Q + W, Q = −W: the energy added as work leaves the gas by cooling, which is what keeps its temperature constant.
In preparation: 0 of 5 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
9.4.A.1 Internal energy, U Fix
Internal energy, U
The sum of the kinetic energies of the objects (atoms or molecules) that make up a system and the potential energy of their configuration, which comes from the forces they exert on one another. It does not include the kinetic energy of the system's center of mass or the gravitational potential energy of the system with Earth. SI unit: joule (J).
Potential energy of configuration
Energy associated with the arrangement of the objects within a system, due to the conservative forces they exert on one another. In a solid or liquid it changes when the molecules are rearranged, as in melting or boiling, even when the temperature does not change. SI unit: joule (J).
Ideal gas and internal potential energy
In the ideal-gas model the atoms exert no conservative forces on one another (they interact only in brief collisions) and their internal structure is not considered, so an ideal gas has no internal potential energy. Changing the separation of the atoms, by compressing or expanding the gas, therefore changes no potential energy.
Internal energy of an ideal monatomic gas
The sum of the kinetic energies of its atoms: U = (3/2)nRT = (3/2)NkB T, where n is the number of moles, N the number of atoms and T the absolute temperature. For a fixed amount of gas U changes only when T changes; since PV = nRT, it can also be written U = (3/2)PV.
Absolute temperature, T
Temperature on the Kelvin scale, T(K) = T(°C) + 273, which must be used in U = (3/2)nRT and PV = nRT. A ratio of Celsius temperatures is not a ratio of energies. SI unit: kelvin (K).
Students often think The internal energy of an object is only the kinetic energy of its particles, the quantity its temperature measures. In fact No. Internal energy is the kinetic energy of the particles plus the potential energy of their configuration. In the ideal-gas model the atoms exert no conservative forces on one another, so that potential energy is absent; in solids and liquids the molecules do exert such forces, and it is part of the internal energy.
Students often think Heat is a substance that an object contains and stores: heating adds more of it, and it can be squeezed, concentrated or released. In fact No. Heating (Q) is energy transferred between systems because of a temperature difference. What an object contains is internal energy; heating is one way of changing it, and work is another.
9.4.A.2 Internal energy and center-of-mass motion Fix
Internal energy and center-of-mass motion
Internal energy concerns the motion and arrangement of a system's parts relative to one another. Changing it, for example by heating, changes the system's internal structure or behavior, such as the average speed of its atoms, without changing the velocity of its center of mass; moving or lifting the whole system changes other forms of energy, not its internal energy.
Students often think If the particles of an object gain kinetic energy, the object as a whole must gain kinetic energy and start to move. In fact No. The atoms move in random directions, so their extra kinetic energy is internal energy, motion relative to one another. Their momenta still add to zero, and the center of mass stays at rest.
Students often think Heating a gas makes it weigh less, since hot gas rises, so a sealed container of gas becomes lighter when it is heated. In fact No. The mass of the gas, and so the weight of the container, is unchanged. Hot air rises through cooler air because it is less dense, not because it has lost weight.
9.4.B.1 First law of thermodynamics Fix
First law of thermodynamics
Conservation of energy for a thermodynamic system, counting all the energy transferred into or out of the system by work and by heating or cooling. It is not a separate principle but the energy principle applied to systems whose internal energy changes.
Heating and cooling, Q
Q is the energy transferred into a system by thermal processes because of a temperature difference between the system and its surroundings: Q > 0 for heating (energy in), Q < 0 for cooling (energy out). Q describes a transfer during a process, not energy that a system contains. SI unit: joule (J).
Isolated system
A system that exchanges neither energy nor matter with its surroundings. Its total energy is constant, although energy can change form or be transferred between objects inside it.
Closed system
A system that exchanges no matter with its surroundings but can exchange energy with them by work and by heating or cooling. For a closed system, ΔU = Q + W.
Work done on a system, W
In ΔU = Q + W, W is the work done ON the system by its surroundings: positive when a gas is compressed (energy transferred in), negative when it expands (energy transferred out), zero when its volume does not change. SI unit: joule (J).
Work at constant pressure
W = −PΔV, where P is the constant (or average) external pressure and ΔV = Vf − Vi. For a piston of area A that moves a distance d, |ΔV| = Ad. Units: 1 Pa·m³ = 1 J.
Students often think A gas's temperature follows its pressure: a higher pressure in itself makes a gas hotter, and a lower pressure makes it colder. In fact No. For an ideal gas PV = nRT, so the temperature depends on the product of pressure and volume. A gas can be compressed at constant temperature, with its pressure rising, or can reach a lower pressure at the same temperature by expanding.
Students often think Some energy is used up or lost whenever energy is transferred, so the total energy of an isolated system decreases. In fact No. Energy is conserved. In an isolated system the energy lost by one object is gained by another, and the total stays constant.
9.4.B.2 Pressure–volume (PV) diagram Fix
Pressure–volume (PV) diagram
A graph of the pressure of a gas against its volume. Each point is a state of the gas, and a line or curve with an arrow is a process. A closed loop is a cycle, which returns the gas to its initial state, so ΔU = 0 for one complete cycle.
Isotherm
A curve on a PV diagram joining states of equal temperature. For a fixed amount of ideal gas PV = nRT is constant along it, so P is inversely proportional to V; isotherms farther from the origin are at higher temperatures.
Work as area on a PV diagram
The magnitude of the work done on a gas equals the area between the process curve and the volume axis. W is negative when the volume increases, positive when it decreases, and zero along a vertical (constant-volume) line. For a cycle, |W| equals the area enclosed by the loop, and W < 0 when the loop is traversed clockwise (V to the right, P upward).
Students often think A gas's temperature can be judged from its volume alone: a larger volume means a higher temperature. In fact No. For an ideal gas the temperature depends on PV: a larger volume at a proportionally lower pressure, as along an isotherm, means the same temperature.
Students often think The work for a straight-line process on a PV diagram is the area of the triangle between the line and the level of the lower pressure. In fact No. The area under the line down to the V axis is a trapezoid: the triangle above the lower pressure plus the rectangle beneath it. |W| = (1/2)(P₁ + P₂)|ΔV|.
9.4.B.3 Isovolumetric (isochoric) process Fix
Isovolumetric (isochoric) process
A process at constant volume. No work is done (W = 0), so ΔU = Q.
Isothermal process
A process at constant temperature. For an ideal gas ΔU = 0, so Q = −W: energy added as work leaves by cooling, or energy added by heating leaves as work.
Isobaric process
A process at constant pressure. W = −PΔV; for an ideal monatomic gas ΔU = (3/2)nRΔT = (3/2)PΔV, so Q = ΔU − W = (5/2)PΔV.
Adiabatic process
A process in which no energy is transferred to or from the system by heating or cooling: Q = 0, so ΔU = W. An adiabatic expansion of an ideal gas lowers its temperature and an adiabatic compression raises it; on a PV diagram an adiabatic expansion curve falls more steeply than the isotherm through the same starting state.
Students often think No energy transfer by heating means no change in temperature, and no change in temperature means no energy transfer by heating, so adiabatic and isothermal processes are the same. In fact No. Adiabatic means Q = 0; the temperature of an ideal gas then changes whenever work is done, since ΔU = W. Isothermal means ΔT = 0; energy is then transferred by heating or cooling so that Q = −W.
Students often think The energy that must be transferred by heating to raise a gas's temperature by a given amount is the same whatever the process. In fact No. The heating needed depends on the process. At constant volume Q = ΔU; at constant pressure the gas also expands and does work on its surroundings, so more heating is needed for the same ΔT.
16 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 16
Equal masses of liquid water and water vapor are both at 100°C. Which claim about their internal energies, with its reasoning, is correct?
Answer and reasoning
AThe vapor's is greater, as its molecules move faster and so they have more kinetic energy. A student who thinks the molecules of a gas move faster than those of the liquid at the same temperature picks this. The average kinetic energy of the molecules is set by the temperature, which is the same for both; the difference is in the potential energy of their configuration.
BThe vapor's is greater, as its molecules have more potential energy of configuration.Correct At the same temperature the molecules have the same average kinetic energy. Separating the molecules against the attractive forces between them, as happens in boiling, increases the potential energy of their configuration, so the vapor has the greater internal energy. That is why energy must be transferred to water to boil it, although its temperature stays at 100°C.
CThey are equal, since the two samples have equal masses and are at the same temperature. A student who treats temperature as a measure of internal energy picks this. Equal temperatures mean equal average kinetic energies of the molecules, but internal energy also includes the potential energy of their configuration, which is greater in the vapor.
DThe liquid's is greater, as its molecules are held together by bonds that store energy. A student who thinks attractions between molecules store energy that is released when they are pulled apart picks this. Pulling molecules apart against attractive forces requires energy, so the more tightly held molecules of the liquid have less potential energy, not more.
A sample of ideal gas in a cylinder expands slowly, pushing a piston outward, while its temperature is kept constant. How does the internal energy Uf of the gas at the end of the expansion compare with its internal energy Ui at the start?
Answer and reasoning
AUf > Ui, since with its atoms farther apart the gas stores more potential energy A student who thinks an ideal gas stores potential energy in the spacing of its atoms picks this. Ideal-gas atoms exert no conservative forces on one another, so moving them apart changes no potential energy; with T constant, U is unchanged.
BUf < Ui, since the gas does work on the piston and loses that much internal energy A student who counts only the work picks this. The gas does transfer energy out as work, but an equal amount is transferred in by heating, which is what keeps its temperature constant: ΔU = Q + W = 0.
CThey cannot be compared without first knowing how much energy was added to the gas by heating A student who thinks internal energy depends on the process rather than on the state picks this. For a fixed amount of ideal gas, U = (3/2)nRT depends only on the temperature, which is the same at the start and at the end, so no value of Q is needed.
DUf = Ui, since spacing stores no energy and the atoms' average kinetic energy is unchangedCorrect The atoms of an ideal gas exert no conservative forces on one another, so the gas has no internal potential energy and the change in their spacing does not matter. Its internal energy is the kinetic energy of its atoms, which for a fixed amount of gas depends only on the temperature; T is constant, so Uf = Ui. (The energy transferred in by heating equals the energy transferred out as work.)
A sealed, rigid steel container holds an ideal monatomic gas at 27°C, and the internal energy of the gas is U₀. The gas is heated until its temperature is 327°C. What is the internal energy of the gas now?
Answer and reasoning
A2.00U₀Correct For a fixed amount of ideal monatomic gas, U = (3/2)nRT is proportional to the absolute temperature. The temperature rises from 27 + 273 = 300 K to 327 + 273 = 600 K, so the internal energy doubles.
B12.1U₀ A student who uses the Celsius temperatures picks this: 327/27 ≈ 12.1. U = (3/2)nRT needs absolute temperatures, and 300 K to 600 K is a factor of 2.
C1.41U₀ A student who uses the factor for the rms speed picks this: vrms is proportional to √T, so it increases by √2 ≈ 1.41. The internal energy is the sum of the atoms' kinetic energies, which is proportional to T itself, so it doubles.
D1.00U₀ A student who thinks internal energy changes only when work is done picks this, since no work is done in a rigid container. Energy is transferred to the gas by heating, so ΔU = Q + W = Q > 0, and the internal energy doubles.
Working U = (3/2)nRT with T in kelvins; n is fixed. T₁ = 27 + 273 = 300 K, T₂ = 327 + 273 = 600 K. U/U₀ = 600 K/300 K = 2.00, so U = 2.00U₀. (Celsius ratio 327/27 ≈ 12.1; √2 ≈ 1.41 is the factor for vrms; 1.00U₀ ignores the heating.)
A cylinder holds 0.40 mol of an ideal monatomic gas at 77°C. What is the internal energy of the gas? Use R = 8.31 J/(mol·K).
Answer and reasoning
A1.2 × 10³ J A student who uses nRT as the internal energy picks this: (0.40)(8.31)(350) = 1.2 × 10³ J. nRT equals PV; the internal energy of a monatomic ideal gas is (3/2)nRT.
B4.4 × 10³ J A student who thinks U depends only on the temperature, not on the amount of gas, picks this: (3/2)(8.31)(350) = 4.4 × 10³ J, the internal energy of one mole. The cylinder holds 0.40 mol, so U is 0.40 times as great.
C3.8 × 10² J A student who uses the Celsius temperature picks this: (3/2)(0.40)(8.31)(77) = 3.8 × 10² J. The equation needs the absolute temperature, 350 K.
D1.7 × 10³ JCorrect The internal energy of an ideal monatomic gas is U = (3/2)nRT, with T in kelvins: T = 77 + 273 = 350 K. U = (3/2)(0.40 mol)(8.31 J/(mol·K))(350 K) = 1.7 × 10³ J.
Working T = 77 + 273 = 350 K. U = (3/2)nRT = (3/2)(0.40 mol)(8.31 J/(mol·K))(350 K) = 1745 J ≈ 1.7 × 10³ J.
The outlet of a bicycle pump is blocked, and the frictionless piston is pushed in quickly, compressing the air inside. The piston, the cylinder and the air start at the same temperature. Model the air as an ideal gas. The temperature of the air rises. Which explanation is correct?
Answer and reasoning
AThe piston does work on the air, transferring energy that raises its internal energy.Correct The first law is conservation of energy: ΔU = Q + W. In the quick compression the piston does positive work on the air, W > 0, and there is almost no time for energy to leave by cooling, so the internal energy of the air, and with it the temperature, increases.
BThe heat already in the air is squeezed into a smaller volume, so the air is hotter. A student who thinks of heat as a substance that can be concentrated picks this. The air does not contain heat; its internal energy increases because work is done on it, so the temperature rise comes from new energy, not the same energy in less space.
CEnergy is transferred to the air by heating, as heating is what raises temperature. A student who thinks only heating can raise a gas's internal energy or temperature picks this. Nothing in contact with the air is hotter than it, so there is no heating; the energy comes from the work done by the piston.
DThe air's pressure rises, and a higher pressure in itself makes a gas hotter. A student who thinks temperature follows pressure alone picks this. A slow compression at constant temperature also raises the pressure; what raises the temperature here is the work done on the air with no time for cooling.
A block of mass m slides across a rough, level floor, and its speed decreases from v to v/2. Consider the system of the block and the floor, and assume that no energy is transferred between this system and its surroundings, so that it is isolated. By how much does the internal energy of the system increase while the speed decreases?
Answer and reasoning
A(3/4)mv² A student who drops the ½ from the kinetic energy picks this: mv² − m(v/2)² = (3/4)mv². With K = (1/2)mv², the decrease is (1/2)mv² − (1/8)mv² = (3/8)mv².
B(1/8)mv² A student who squares the change in speed picks this: (1/2)m(v − v/2)² = (1/8)mv². Kinetic energy depends on the square of the speed, so each kinetic energy must be found before subtracting.
C(3/8)mv²Correct The system is isolated, so its total energy is constant. The floor is level, so only kinetic energy and internal energy change: the kinetic energy falls from (1/2)mv² to (1/2)m(v/2)² = (1/8)mv², a decrease of (3/8)mv², and the internal energy of the block and floor (the motion and arrangement of their atoms at the rubbing surfaces) increases by that amount.
D(1/2)mv² A student who takes the whole initial kinetic energy as the increase in internal energy picks this. The block is still moving at v/2, so it keeps (1/8)mv² of kinetic energy; only the decrease, (3/8)mv², becomes internal energy.
Working Isolated system, level floor: ΔK + ΔUint = 0. ΔK = (1/2)m(v/2)² − (1/2)mv² = (1/8)mv² − (4/8)mv² = −(3/8)mv², so the internal energy increases by (3/8)mv². (Errors: dropping ½ gives (3/4)mv²; (1/2)m(Δv)² gives (1/8)mv²; the initial kinetic energy is (1/2)mv².)
A gas in a cylinder is compressed by a piston. During the compression, 250 J of work is done on the gas, and 400 J of energy is transferred from the gas to its surroundings by cooling. What is the change in the internal energy of the gas?
Answer and reasoning
A−150 JCorrect ΔU = Q + W. The work done on the gas is W = +250 J, and energy leaving by cooling makes Q = −400 J. ΔU = −400 J + 250 J = −150 J: the gas loses more energy by cooling than it gains as work.
B−650 J A student who takes W as the work done by the gas picks this: W = −250 J gives ΔU = −400 J − 250 J = −650 J. In ΔU = Q + W, W is the work done on the gas, +250 J for a compression.
C+650 J A student who adds the two energy transfers without signs picks this: 400 J + 250 J = 650 J. Energy leaving the gas by cooling makes Q negative, −400 J.
D−400 J A student who counts only the energy transferred by cooling picks this. The 250 J of work done on the gas is also a transfer of energy into it, which makes up part of the 400 J lost.
Working ΔU = Q + W. W = +250 J (work done on the gas); Q = −400 J (energy leaves by cooling). ΔU = −400 J + 250 J = −150 J.
The P–V diagram shows a process in which a sample of ideal monatomic gas goes from state i to state f. During the process, 1.0 × 10³ J of energy is transferred to the gas by heating. What is the change in the internal energy of the gas?
Answer and reasoning
A+1.4 × 10³ J A student who takes W as the work done by the gas, +4.0 × 10² J, picks this. In ΔU = Q + W, W is the work done on the gas, which is negative in an expansion.
B+6.0 × 10² JCorrect The gas expands at constant pressure, so the work done on it is W = −PΔV = −(2.0 × 10⁵ Pa)(2.0 × 10⁻³ m³) = −4.0 × 10² J: the gas transfers energy to its surroundings as it pushes outward. ΔU = Q + W = 1.0 × 10³ J − 4.0 × 10² J = +6.0 × 10² J.
C+1.0 × 10³ J A student who thinks all of the energy transferred by heating stays in the gas picks this. As it expands, the gas transfers 4.0 × 10² J to its surroundings as work, so its internal energy increases by only 6.0 × 10² J.
D−4.0 × 10² J A student who counts only the work picks this, leaving out the 1.0 × 10³ J transferred in by heating. Both transfers change the internal energy: ΔU = Q + W.
Working Graph: horizontal line at P = 2.0 × 10⁵ Pa from V = 1.0 × 10⁻³ m³ to 3.0 × 10⁻³ m³ (constant-pressure expansion). W = −PΔV = −(2.0 × 10⁵ Pa)(2.0 × 10⁻³ m³) = −4.0 × 10² J. ΔU = Q + W = 1.0 × 10³ J − 4.0 × 10² J = +6.0 × 10² J. (Check: for a monatomic ideal gas at constant pressure ΔU = (3/2)PΔV = 6.0 × 10² J and Q = (5/2)PΔV = 1.0 × 10³ J, consistent with the stem.)
The P–V diagram shows two ways, path 1 and path 2, of taking a sample of ideal gas from state i to state f. Q₁ and Q₂ are the energies transferred to the gas by heating along paths 1 and 2, W₁ and W₂ are the work done on the gas, and ΔU₁ and ΔU₂ are the changes in its internal energy. Which comparison is correct?
Answer and reasoning
AQ₂ > Q₁, since ΔU is the same and W₁ is greater than W₂ A student who takes W as the work done by the gas picks this. With that meaning W₁ is indeed greater, but in ΔU = Q + W, W is the work done on the gas, which is more negative along path 1, so Q₁ is the greater.
BQ₁ = Q₂, since both paths start and end at the same states A student who treats heat as a property of a state picks this. Q describes a transfer during a process: the paths have different areas under them, so different work, and since ΔU is the same, different Q.
CΔU₁ > ΔU₂, since path 1 goes through higher temperatures A student who thinks the change in internal energy depends on the path picks this. U depends only on the state of the gas; both paths start at i and end at f, so ΔU₁ = ΔU₂, even though path 1 passes through higher temperatures on the way.
DQ₁ > Q₂, since ΔU is the same and W₁ is more negative than W₂Correct Internal energy depends only on the state, so ΔU is the same for both paths. Both paths include an expansion, so W is negative on each; path 1 expands at the higher pressure, so the area under it is greater and W₁ is more negative. From Q = ΔU − W, Q₁ > Q₂: along path 1 more energy must be supplied by heating to make up for the greater energy transferred out as work.
Working Same end states: ΔU₁ = ΔU₂ = ΔU. Path 1: expansion at the higher pressure, then constant volume; path 2: constant volume, then expansion at the lower pressure. Both expansions have the same ΔV, so |W₁| > |W₂| and W₁ < W₂ < 0. Q = ΔU − W, so Q₁ > Q₂.
A vertical cylinder of gas is closed at the top by a piston of mass M and area A that slides without friction. The atmosphere above the piston exerts a pressure P₀. The gas is heated slowly, and the piston rises a distance d. Which expression gives the work W done on the gas?
Answer and reasoning
A−(P₀A + Mg)dCorrect The piston rises slowly, so the net force on it is zero: the gas pushes up with PA, and the atmosphere and Earth push down with P₀A + Mg. The gas pressure is therefore constant at P = P₀ + Mg/A. The volume increases by Ad, so W = −PΔV = −(P₀ + Mg/A)Ad = −(P₀A + Mg)d, negative because the gas expands.
B−(P₀) × (Ad) A student who takes the gas pressure to be atmospheric pressure picks this: W = −P₀(Ad). The gas must hold up the piston as well as push against the atmosphere, so P = P₀ + Mg/A and W = −(P₀A + Mg)d.
C−(P₀A − Mg)d A student who puts the piston's weight on the same side as the gas's upward push in the force balance picks this, getting P = P₀ − Mg/A. The gas must hold up the piston as well as push against the atmosphere, so P = P₀ + Mg/A.
D−(P₀ + Mg)Ad A student who adds the piston's weight directly to the atmospheric pressure picks this. A force cannot be added to a pressure; the weight adds Mg/A to the pressure, which gives W = −(P₀ + Mg/A)Ad = −(P₀A + Mg)d.
Working Piston in equilibrium (rises slowly): PA = P₀A + Mg, so P = P₀ + Mg/A (constant). ΔV = +Ad. W = −PΔV = −(P₀ + Mg/A)(Ad) = −(P₀A + Mg)d. Error: taking the gas pressure as atmospheric, P = P₀, gives W = −P₀(Ad) = −P₀Ad.
The P–V diagram shows a cycle abcda that a sample of ideal gas goes through in the direction shown by the arrows. For one complete cycle, W is the net work done on the gas and Q is the net energy transferred to the gas by heating. Which is correct?
Answer and reasoning
AW > 0 and Q < 0 A student who takes W as the work done by the gas picks this. In this clockwise cycle the gas does positive net work on its surroundings, so the work done on the gas is negative, and Q = −W is positive.
BW = 0 and Q = 0 A student who thinks work and heat, like internal energy, return to their starting values when the gas returns to its initial state picks this. W and Q depend on the path: the net work has the magnitude of the area enclosed by the loop, which is not zero.
CW < 0 and Q = 0 A student who takes Q to be equal to ΔU, as if heating were the only way the internal energy could change, picks this. ΔU = 0 over the cycle, but ΔU = Q + W, so Q = −W, which is positive.
DW < 0 and Q > 0Correct The gas returns to state a, so ΔU = 0 for the cycle. Along ab it expands at the higher pressure, and along cd it is compressed at the lower pressure, so the negative work during the expansion is larger in magnitude than the positive work during the compression: W is minus the enclosed area, W < 0. Then Q = ΔU − W = −W > 0.
Working ΔU = 0 for a cycle. ab: expansion at higher pressure, Wab = −Phigh ΔV < 0; cd: compression at lower pressure, Wcd = +Plow ΔV > 0; bc, da: constant volume, W = 0. Net W = −(Phigh − Plow)ΔV = −(enclosed area) < 0. Q = −W > 0.
The P–V diagram shows a sample of ideal gas taken from state X to state Y along the path X → Z → Y. The dashed curve is an isotherm. Which claim about the temperatures TX and TY, with its evidence, is correct?
Answer and reasoning
ATY > TX, because the gas occupies a larger volume at Y A student who judges temperature from volume alone picks this. The volume triples from X to Y, but the pressure falls to a third, so PV = nRT, and the temperature, are unchanged.
BTY = TX, because X and Y lie on the same isothermCorrect An isotherm joins states of equal temperature, and X and Y both lie on the dashed isotherm. The values agree: PV = (6.0 × 10⁵ Pa)(1.0 × 10⁻³ m³) = 6.0 × 10² J at X and (2.0 × 10⁵ Pa)(3.0 × 10⁻³ m³) = 6.0 × 10² J at Y, and PV = nRT. The gas is hotter at Z, but its temperature at Y depends only on the state.
CTY < TX, because the gas is at a lower pressure at Y A student who judges temperature from pressure alone picks this. The pressure at Y is a third of that at X, but the volume is three times as great, so PV, and the temperature, are the same.
DTY depends on the path, so the states alone cannot give it A student who thinks the temperature of a state depends on how the gas reached it picks this. Temperature is a property of the state: Y lies on the same isotherm as X, whichever path is followed.
Working X: P = 6.0 × 10⁵ Pa, V = 1.0 × 10⁻³ m³, PV = 6.0 × 10² J. Y: P = 2.0 × 10⁵ Pa, V = 3.0 × 10⁻³ m³, PV = 6.0 × 10² J. Same PV (and on the drawn isotherm), so TY = TX. Z: PV = 1.8 × 10³ J (hotter).
The P–V diagram shows a sample of gas taken from state A to state B and then to state C. What is the total work done on the gas from A to C?
Answer and reasoning
A+7.5 × 10² J A student who thinks work, like internal energy, depends only on the initial and final states picks this, taking the straight path from A to C: the area under that line is (1/2)(1.0 × 10⁵ Pa + 4.0 × 10⁵ Pa)(3.0 × 10⁻³ m³) = 7.5 × 10² J, positive for a compression. Work depends on the path actually followed: under A → B → C the only area is the rectangle under A → B, 3.0 × 10² J.
B+3.0 × 10² JCorrect From A to B the gas is compressed at constant pressure: W = −PΔV = −(1.0 × 10⁵ Pa)(2.0 × 10⁻³ m³ − 5.0 × 10⁻³ m³) = +3.0 × 10² J, positive because the volume decreases. From B to C the volume is constant, so no work is done (there is no area under a vertical line). Total: +3.0 × 10² J.
C+6.0 × 10² J A student who thinks work is done when the pressure changes picks this, taking VΔP = (2.0 × 10⁻³ m³)(3.0 × 10⁵ Pa) for B to C and nothing for A to B. Work is the area under the path down to the V axis: it is done when the volume changes (A to B) and is zero at constant volume (B to C).
D+1.2 × 10³ J A student who uses the final pressure, 4.0 × 10⁵ Pa, with the whole change in volume picks this. During the compression the pressure was 1.0 × 10⁵ Pa; it rose afterward, at constant volume, when no work is done.
Working A → B: constant P = 1.0 × 10⁵ Pa, V from 5.0 × 10⁻³ m³ to 2.0 × 10⁻³ m³: W = −PΔV = −(1.0 × 10⁵ Pa)(−3.0 × 10⁻³ m³) = +3.0 × 10² J. B → C: constant volume, W = 0. Total W = +3.0 × 10² J.
The P–V diagram shows two processes in which a sample of ideal gas expands from the same initial state i to the same final volume Vf. Process 1 is isothermal and process 2 is adiabatic. Which claim is supported by the diagram and the first law of thermodynamics?
Answer and reasoning
AThe gas ends process 2 at a higher temperature than process 1. A student who takes W as the work done by the gas picks this, getting ΔU = +W > 0. The work done on an expanding gas is negative, so with Q = 0 its internal energy and temperature fall, which is why curve 2 ends at the lower pressure.
BThe gas ends both processes at the same temperature as at i. A student who thinks an adiabatic process keeps the temperature constant picks this. Adiabatic means Q = 0, not ΔT = 0: with no heating to replace the energy transferred out as work, the temperature falls, which is why curve 2 ends below curve 1.
CThe gas ends process 2 at a lower temperature than process 1.Correct In process 2, Q = 0 and the gas expands, so W < 0 and ΔU = W < 0: the temperature falls. The diagram agrees: at Vf curve 2 ends at a lower pressure than curve 1, and at the same volume a lower pressure means a lower temperature (PV = nRT). Process 1 ends at the initial temperature.
DThe magnitude of the work done is greater in process 2 than in 1. A student who links work to the change in pressure picks this: curve 2 has the larger pressure drop. |W| is the area under a curve down to the V axis, and curve 2 lies below curve 1 throughout, so the magnitude of the work is smaller in process 2.
Working Process 1 (isothermal): ΔU = 0, Tf = Ti. Process 2 (adiabatic): Q = 0, expansion so W < 0, ΔU = W < 0, Tf < Ti. Diagram: at Vf, P₂ < P₁, so T₂ < T₁ (PV = nRT). Area under curve 2 < area under curve 1, so |W₂| < |W₁|.
A sample of ideal monatomic gas is heated at constant volume, and its temperature rises by ΔT when energy Q₀ is transferred to it by heating. The same sample, starting from the same state, is instead heated at constant pressure until its temperature has risen by the same ΔT. How much energy must be transferred to the gas by heating in this second process?
Answer and reasoning
A1.67Q₀Correct At constant volume W = 0, so Q₀ = ΔU = (3/2)nRΔT. At constant pressure ΔU is the same, (3/2)nRΔT = Q₀, because it depends only on the temperature change, but the gas also expands: W = −PΔV = −nRΔT = −(2/3)Q₀. So Q = ΔU − W = Q₀ + (2/3)Q₀ = (5/3)Q₀ ≈ 1.67Q₀.
B0.33Q₀ A student who takes W as the work done by the gas, +nRΔT = +(2/3)Q₀, picks this: Q = ΔU − W = Q₀ − (2/3)Q₀ = (1/3)Q₀. The expanding gas transfers energy out as work, so more heating is needed, not less.
C2.00Q₀ A student who uses U = nRT picks this: then Q₀ = nRΔT, and at constant pressure Q = nRΔT + nRΔT = 2Q₀. For a monatomic ideal gas U = (3/2)nRT, so the work, nRΔT, is (2/3)Q₀, not Q₀.
D1.00Q₀ A student who thinks the energy needed depends only on the temperature change picks this. At constant pressure the gas also does work on its surroundings as it expands, and that energy must be supplied by heating as well.
Working Constant V: W = 0, Q₀ = ΔU = (3/2)nRΔT. Constant P: ΔU = (3/2)nRΔT = Q₀ (same ΔT); PΔV = nRΔT, so W = −nRΔT = −(2/3)Q₀. Q = ΔU − W = Q₀ + (2/3)Q₀ = (5/3)Q₀ ≈ 1.67Q₀.
A sample of ideal monatomic gas is in state A, with pressure P₀ and volume V₀. It is heated at constant volume until its pressure is 2P₀ (state B), and then heated at constant pressure until its volume is 3V₀ (state C). Which expression gives the total energy transferred to the gas by heating as it goes from state A to state C?
Answer and reasoning
A3.50P₀V₀ A student who takes W in ΔU = Q + W to be the work done BY the gas, +4P₀V₀ for the expansion, picks this: Q = 7.5P₀V₀ − 4P₀V₀. In the AP convention W is the work done ON the gas, −4P₀V₀ here, so Q = ΔU − W = 11.5P₀V₀: the gas must be given energy for its internal energy rise and for the work it does.
B11.5P₀V₀Correct For an ideal monatomic gas U = (3/2)nRT = (3/2)PV, so U goes from (3/2)P₀V₀ at A to (3/2)(2P₀)(3V₀) = 9P₀V₀ at C, and ΔU = 7.5P₀V₀. No work is done from A to B; from B to C the work done on the gas is W = −(2P₀)(2V₀) = −4P₀V₀. So Q = ΔU − W = 7.5P₀V₀ + 4P₀V₀ = 11.5P₀V₀.
C7.50P₀V₀ A student who thinks only heating changes a gas's internal energy, so that Q = ΔU, picks this. From B to C the gas expands and work W = −4P₀V₀ is done on it, so energy leaves as work; ΔU = Q + W then needs Q = 7.5P₀V₀ + 4P₀V₀ = 11.5P₀V₀.
D9.00P₀V₀ A student who takes the internal energy to be nRT = PV, without the factor 3/2, picks this: ΔU = 6P₀V₀ − P₀V₀ = 5P₀V₀ and Q = 5P₀V₀ + 4P₀V₀. For an ideal monatomic gas U = (3/2)PV, so ΔU = 7.5P₀V₀ and Q = 11.5P₀V₀.
Working U = (3/2)nRT = (3/2)PV. UA = (3/2)P₀V₀; UC = (3/2)(2P₀)(3V₀) = 9P₀V₀; ΔU = 7.5P₀V₀. Work done on the gas: A→B, ΔV = 0, W = 0; B→C, W = −PΔV = −(2P₀)(2V₀) = −4P₀V₀. ΔU = Q + W → Q = ΔU − W = 7.5P₀V₀ + 4P₀V₀ = 11.5P₀V₀. Errors: W taken as the work done by the gas, +4P₀V₀ (m21) → Q = 7.5 − 4 = 3.50P₀V₀; work ignored, Q = ΔU (m23) → 7.50P₀V₀; U = nRT = PV without 3/2 (m09) → ΔU = 5P₀V₀, Q = 5 + 4 = 9.00P₀V₀.
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
