4 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 4
Which of the following is one of the assumptions of the classical model of an ideal gas?
Answer and reasoning
AThe atoms exert appreciable forces on one another only while they collide.Correct The ideal-gas model assumes that the only appreciable forces on the atoms are those that occur during collisions. Between collisions each atom moves in a straight line at constant speed. (The model also assumes random velocities, negligible atomic volumes and elastic collisions.)
BThe atoms lose a little kinetic energy in every collision with one another. A student who thinks the motion of gas atoms runs down picks this. The model assumes that the collisions are elastic: kinetic energy passes from one atom to another, but the total is unchanged.
CThe atoms are packed so closely that they fill most of the gas's volume. A student who pictures gas atoms as closely packed picks this. The model assumes the opposite: the volumes of the atoms are negligible compared with the volume the gas occupies, which is mostly empty space.
DAll the atoms move at one common speed, and in random directions. A student who thinks every atom has the same speed picks this. The model assumes random velocities, meaning a range of speeds as well as random directions; collisions keep changing the speed of each atom.
A rigid container of volume 10.0 L holds 0.500 mol of an ideal gas at 27°C. What is the pressure of the gas? Use R = 8.31 J/(mol·K) and 1 L = 1.00 × 10⁻³ m³.
Answer and reasoning
A1.12 × 10⁴ Pa A student who uses the Celsius temperature picks this: (0.500)(8.31)(27)/(1.00 × 10⁻²) = 1.12 × 10⁴. The gas law needs the absolute temperature, 300 K.
B1.25 × 10² Pa A student who substitutes the volume in liters picks this: (0.500)(8.31)(300)/10.0 = 1.25 × 10². With R in J/(mol·K), V must be in cubic meters: 10.0 L = 1.00 × 10⁻² m³.
C1.87 × 10⁵ Pa A student who sets PV equal to the total kinetic energy of the atoms, (3/2)nRT, picks this: (3/2)(0.500)(8.31)(300)/(1.00 × 10⁻²) = 1.87 × 10⁵. The ideal gas law is PV = nRT, without the factor 3/2.
D1.25 × 10⁵ PaCorrect P = nRT/V, with T = 27 + 273 = 300 K and V = 10.0 L = 1.00 × 10⁻² m³: P = (0.500 mol)(8.31 J/(mol·K))(300 K)/(1.00 × 10⁻² m³) = 1.25 × 10⁵ Pa, about 1.2 atm.
The P–V diagram shows three states, A, B and C, of a fixed amount of ideal gas. Which ranks the temperatures of the gas in the three states, from highest to lowest?
Answer and reasoning
AA > C > B A student who judges temperature by pressure alone picks this: 3P₀ > 2P₀ > P₀. State A has the highest pressure but the smallest volume; the temperature depends on the product PV, which is largest for C.
BB > C > A A student who judges temperature by volume alone picks this: 3V₀ > 2V₀ > V₀. State B has the largest volume but the lowest pressure; the temperature depends on the product PV, which is largest for C.
CC > A = BCorrect For a fixed amount of ideal gas, T = PV/(nR), so T is proportional to PV. A: (3P₀)(V₀) = 3P₀V₀; B: (P₀)(3V₀) = 3P₀V₀; C: (2P₀)(2V₀) = 4P₀V₀. So C is hottest, and A and B are at the same, lower temperature.
DA = B = C A student who thinks states on one straight line of the diagram are at one temperature picks this: the three points do lie on a straight line. Equal temperatures need equal PV, which traces a curve; PV is 3P₀V₀ at A and B but 4P₀V₀ at C.
Working T ∝ PV for fixed n. A: 3P₀ × V₀ = 3P₀V₀. B: P₀ × 3V₀ = 3P₀V₀. C: 2P₀ × 2V₀ = 4P₀V₀. TC > TA = TB. (A, C and B are collinear, on the line P + (P₀/V₀)V = 4P₀, which is not an isotherm.)
Students measure the pressure of three samples of gas, each sealed in a rigid container, at several temperatures between 0°C and 100°C. The samples are of different gases and different amounts, and each behaves as an ideal gas. For each sample they plot P against T in °C and get a straight line. They then extend each best-fit line to P = 0. What do they find?
Answer and reasoning
AAll three lines reach P = 0 at the same temperature, 0°C. A student who takes 0°C to be the zero of temperature picks this. At 0°C the atoms are still moving and each gas still has a pressure; the lines reach P = 0 at about −273°C, which is 0 K.
BAll three lines reach P = 0 at the same temperature, about −273°C.Correct For an ideal gas at constant volume, P = (nR/V)(TC + 273), which is zero at TC = −273°C whatever n, V or the kind of gas. The lines have different slopes, but all extrapolate to the same point on the temperature axis, absolute zero.
CThe steepest line reaches P = 0 at the highest temperature. A student who judges where a line meets the axis by its slope alone picks this. The steeper lines also start at higher pressures, and every line reaches P = 0 at the same temperature, about −273°C.
DEach line reaches P = 0 at a temperature set by its kind of gas. A student who thinks the zero-pressure temperature is a property of each substance, like a boiling point, picks this. For every ideal gas the extrapolated temperature is the same, about −273°C, which is why it defines the zero of the Kelvin scale.
Working Ideal gas at constant volume: P = (nR/V)T with T = TC + 273, so P = (nR/V)(TC + 273), a straight line in TC whose slope nR/V differs from sample to sample but which is zero at TC = −273°C for every sample.
In preparation: 0 of 4 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
9.2.A.1 Ideal gas (classical model) Fix
Ideal gas (classical model)
A model gas whose atoms have random instantaneous velocities, have volumes negligible compared with the volume the gas occupies, collide elastically, and exert appreciable forces only during collisions. Real gases behave nearly ideally when their atoms are far apart, as at low pressure.
Elastic collision (in the gas model)
A collision in which the total kinetic energy of the colliding objects is the same afterward as before. Because the collisions of an ideal gas are elastic, collisions alone do not change the total kinetic energy of the atoms of an isolated sample, although they constantly change the speeds of individual atoms.
Students often think Gas atoms lose a little kinetic energy in each collision, so the atoms of a gas left alone gradually slow down. In fact No, not in the ideal-gas model: collisions between atoms are elastic, so the total kinetic energy of the colliding atoms is unchanged, although energy usually passes from one atom to the other. A gas left in an insulated, rigid container keeps its temperature indefinitely.
Students often think The atoms of a gas are packed closely together and fill most of the volume the gas occupies. In fact No. In the ideal-gas model the volumes of the atoms are negligible compared with the volume of the gas: the gas is mostly empty space, which is why it can be compressed so easily.
9.2.A.2 Ideal gas law Fix
Ideal gas law
PV = nRT = NkB T, relating the pressure P, the volume V, the amount of gas (n moles or N atoms) and the absolute temperature T of an ideal gas. Consistent SI units must be used: P in Pa, V in m³, T in K.
Amount of gas: n and N
n is the number of moles (SI unit: mol) and N the number of atoms (a pure number); N = nN0, where N0 = 6.02 × 10²³ mol⁻¹. n goes with the gas constant R and N with the Boltzmann constant kB.
Gas constant, R
R = 8.31 J/(mol·K), the constant in PV = nRT when the amount of gas is counted in moles. R = N0 kB.
Absolute (Kelvin) temperature
Temperature measured from absolute zero: T(K) = T(°C) + 273. Only absolute temperatures may be used in PV = nRT or in ratios of temperatures.
Students often think Temperatures in degrees Celsius can be substituted directly into PV = nRT. In fact No. The ideal gas law uses the absolute temperature in kelvins, T(K) = T(°C) + 273.
Students often think A volume given in liters can be substituted into PV = nRT with R = 8.31 J/(mol·K) without converting it. In fact No. With R in J/(mol·K), P must be in Pa and V in m³. 1 L = 1.00 × 10⁻³ m³, so 10.0 L = 1.00 × 10⁻² m³.
9.2.A.3 Graphs of gas properties Fix
Graphs of gas properties
Graphs of P, V and T for an ideal gas show its behavior. For a fixed amount of gas, a P–V graph at constant temperature is a curve on which PV is constant (an isotherm); P against T at constant volume, and V against T at constant pressure, are straight lines through the origin when T is in kelvins. The temperature of a state on a P–V diagram is proportional to the product PV.
Students often think A gas's temperature follows its pressure: a higher pressure in itself means a hotter gas with faster atoms. In fact No. For an ideal gas PV = nRT, so for a fixed amount of gas the temperature depends on the product PV. A gas can be at high pressure and low temperature if its volume is small.
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 means the same temperature.
9.2.A.4 Absolute zero Fix
Absolute zero
The temperature, about −273°C (0 K), at which the pressure of an ideal gas would be zero. It is found by extrapolating a graph of pressure against temperature for a gas at constant volume to P = 0; every ideal-gas sample extrapolates to the same temperature, whatever the gas or the amount.
Extrapolation
Extending a graph beyond the range of the measured data, following the trend of the data. Extending a straight-line P–T graph to P = 0 gives the zero-pressure temperature, although a real gas would condense before reaching it.
Students often think The zero of the Celsius scale is the zero of temperature, so a gas would have zero pressure at 0°C. In fact No. At 0°C (273 K) the atoms are still moving and the gas still exerts pressure. Extrapolating P against T to P = 0 gives about −273°C, which is 0 K.
Students often think When lines are extended toward an axis, the steepest one reaches the axis first, so a sample whose pressure falls fastest as it cools reaches zero pressure at the highest temperature. In fact No. Where a line meets an axis depends on both its slope and its starting value. For ideal gases at constant volume, lines of P against T with different slopes also start at different pressures, and they all reach P = 0 at the same temperature, about −273°C.
7 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 7
Careful measurements show that at very high pressures a real gas no longer obeys PV = nRT closely. Which reasoning, based on the assumptions of the ideal-gas model, best supports this observation?
Answer and reasoning
AThe atoms are squeezed smaller, so they no longer behave like the atoms of the model. A student who thinks compression squeezes the atoms themselves picks this. Compressing a gas reduces the space between the atoms, not the atoms' size; the departure arises because the atoms' unchanged volume becomes significant.
BThe high pressure makes the atoms move faster, so the gas is hotter than PV = nRT predicts. A student who thinks pressure in itself makes a gas hotter picks this. The average kinetic energy of the atoms is set by the temperature, not by the pressure; a gas at high pressure and low temperature has slow atoms.
CThe atoms are crowded so close that their own volume is no longer negligible.Correct The ideal-gas model assumes that the volumes of the atoms are negligible compared with the volume the gas occupies. At very high pressure the atoms are pushed close together, their own volume becomes a significant fraction of the gas's volume, and that assumption, and so PV = nRT, no longer holds well.
DThe atoms collide more often, and each collision takes away some of their energy. A student who thinks collisions between atoms lose energy picks this. Collisions between gas atoms conserve kinetic energy, however often they happen, so more frequent collisions do not explain the departure from PV = nRT.
In the ideal-gas model, a sample of gas is sealed in a rigid container whose walls let no energy in or out. The gas is left alone for a long time. Which claim about the average kinetic energy of its atoms, with its reasoning, is correct?
Answer and reasoning
AIt decreases, since the atoms lose some energy in each collision and slow down. A student who thinks the motion of gas atoms runs down picks this. The collisions in the model are elastic, so the atoms do not lose kinetic energy in them; the gas keeps its temperature.
BIt stays the same, since every collision, with an atom or a wall, is elastic.Correct In the ideal-gas model every collision is elastic and the atoms exert no other appreciable forces, so collisions only pass kinetic energy from atom to atom. With no energy entering or leaving through the walls, the total kinetic energy, and so the average, stays the same indefinitely.
CIt decreases, since pushing on the walls uses up some of the atoms' energy. A student who thinks exerting a force uses up energy picks this. The walls are rigid and do not move, so the atoms' forces on them do no work and transfer no energy.
DIt increases, since collisions between the atoms produce heat, as friction does. A student who thinks collisions generate heat picks this. The kinetic energy of the atoms is itself the thermal energy of the gas; elastic collisions pass it around without adding any.
The diagram shows three sealed containers, X, Y and Z, each holding a sample of ideal gas, with the pressure, volume and absolute temperature of each sample. Which ranks the samples by the number of atoms they contain, from greatest to least? The containers are not drawn to scale.
Answer and reasoning
AZ > X = Y A student who compares the products PV and leaves out the temperature picks this: PV is 300 for X and Y and 600 for Z (kPa·L). Z is also twice as hot, so N = PV/(kB T) is the same for all three.
BY > Z > X A student who takes the pressure as a measure of the amount of gas picks this: 300 kPa > 200 kPa > 100 kPa. Y's high pressure comes from its small volume and Z's from its high temperature; N ∝ PV/T is equal for all three.
CX = Z > Y A student who takes the volume as a measure of the amount of gas picks this: 3.0 L, 3.0 L, 1.0 L. A gas fills its container, so volume alone does not measure the amount; N ∝ PV/T is equal for all three.
DX = Y = ZCorrect From PV = NkB T, N = PV/(kB T), so N is proportional to PV/T. X: (100)(3.0)/300 = 1.0; Y: (300)(1.0)/300 = 1.0; Z: (200)(3.0)/600 = 1.0 (in kPa·L/K). All three samples contain the same number of atoms.
Working N = PV/(kB T) ∝ PV/T. X: 100 × 3.0/300 = 1.0; Y: 300 × 1.0/300 = 1.0; Z: 200 × 3.0/600 = 1.0 (kPa·L/K). Equal. Errors: PV only gives Z (600) > X = Y (300); P only gives Y > Z > X; V only gives X = Z > Y.
A rigid steel tank contains an ideal gas at 27°C and a pressure of 2.0 × 10⁵ Pa. A valve is opened until half of the gas has escaped, and then the gas remaining in the tank is heated to 327°C. What is the pressure of the gas now?
Answer and reasoning
A2.0 × 10⁵ PaCorrect At constant volume P = NkB T/V is proportional to N T. Half of the atoms remain, and the absolute temperature rises from 27 + 273 = 300 K to 327 + 273 = 600 K, a factor of 2. P = (2.0 × 10⁵ Pa)(1/2)(2) = 2.0 × 10⁵ Pa: the two changes cancel.
B1.2 × 10⁶ Pa A student who uses the Celsius temperatures picks this: (2.0 × 10⁵)(1/2)(327/27) = 1.2 × 10⁶. The gas law needs absolute temperatures, and 300 K to 600 K is a factor of 2.
C4.0 × 10⁵ Pa A student who thinks the pressure in a rigid tank depends on the temperature alone picks this, doubling the pressure for the doubled absolute temperature. Half the atoms have escaped, which halves the pressure, so the net factor is 1.
D1.4 × 10⁵ Pa A student who takes the pressure to be proportional to the atoms' speed picks this: vrms increases by √2, giving (2.0 × 10⁵)(1/2)(√2) = 1.4 × 10⁵. Faster atoms also strike the walls more often, so the pressure is proportional to T itself, and the factor is (1/2)(2) = 1.
Working Rigid tank: V constant, so P ∝ NT. N → N/2; T: 300 K → 600 K (×2). P = 2.0 × 10⁵ Pa × (1/2) × 2 = 2.0 × 10⁵ Pa. Errors: Celsius ratio 327/27 gives 1.2 × 10⁶ Pa; ignoring the escaped gas gives 4.0 × 10⁵ Pa; P ∝ √T gives 1.4 × 10⁵ Pa.
Two rigid tanks are joined by a thin tube that is closed by a valve. Tank X has volume V and holds an ideal gas at pressure 3P₀. Tank Y has volume 2V and holds the same gas at pressure P₀. The gas in both tanks is at the same absolute temperature T. The valve is opened, and when the gas has come to equilibrium its temperature is again T. The volume of the tube is negligible. What is the final pressure of the gas?
Answer and reasoning
A1.67P₀Correct From PV = nRT, tank X holds nX = 3P₀V/(RT) and tank Y holds nY = P₀(2V)/(RT) = 2P₀V/(RT), so there is 5P₀V/(RT) of gas in all. After the valve opens, this gas fills the volume 3V at temperature T, so P = nRT/(3V) = (5/3)P₀ ≈ 1.67P₀.
B2.00P₀ A student who averages the two pressures picks this: (3P₀ + P₀)/2 = 2.00P₀. The average ignores the volumes: the gas at P₀ fills twice the volume, so the amounts of gas are in the ratio 3 : 2, and P = (3P₀V + 2P₀V)/(3V) = 1.67P₀.
C4.00P₀ A student who adds the pressures of the two samples picks this: 3P₀ + P₀ = 4.00P₀. After the valve opens, each sample spreads through the combined volume 3V, so each contributes less than its original pressure: P = (3P₀V + 2P₀V)/(3V) = 1.67P₀.
D1.00P₀ A student who lets the gas from X spread into Y as though Y were empty picks this: 3P₀V/(3V) = 1.00P₀. Y already holds 2P₀V/(RT) of gas, which also fills the combined volume, so the final pressure is (3P₀V + 2P₀V)/(3V) = 1.67P₀.
Working PV = nRT for each tank: nX = 3P₀V/(RT), nY = P₀(2V)/(RT) = 2P₀V/(RT). Total n = 5P₀V/(RT), in volume V + 2V = 3V at temperature T. P = nRT/(3V) = (5/3)P₀ ≈ 1.67P₀. Errors: average of the pressures (3P₀ + P₀)/2 = 2.00P₀; pressures added 3P₀ + P₀ = 4.00P₀; gas in Y left out, X's gas spread over 3V: 3P₀V/(3V) = 1.00P₀.
Two identical glass bulbs, each of volume V, are joined by a thin tube of negligible volume and together contain a fixed amount of an ideal gas. At first both bulbs are at absolute temperature T₀, and the pressure of the gas is P₀. One bulb is then placed in a bath that holds it at 2T₀, while the other bulb is held at T₀. The volumes of the bulbs do not change. When the gas has come to equilibrium, what is the pressure of the gas in the heated bulb?
Answer and reasoning
A1.50P₀ A student who treats all of the gas as if it were at the average temperature, 1.5T₀, picks this: P = nR(1.5T₀)/(2V) = 1.5P₀. Averaging the temperatures counts the two bulbs as holding equal amounts of gas, but at the common pressure the cold bulb holds twice as much gas as the hot one; adding the amounts, PV/(RT₀) + PV/(2RT₀) = 2P₀V/(RT₀), gives 1.33P₀.
B2.00P₀ A student who thinks a gas's pressure is set by its temperature alone picks this: the heated bulb's temperature doubles, so its pressure is taken to double, and the gas that flows out into the other bulb is taken not to matter. With less gas left in it, the heated bulb's pressure rises only to 1.33P₀, the same as in the cold bulb.
C1.00P₀ A student who thinks the amount of gas is fixed by its pressure and volume alone picks this: the amount of gas and the total volume have not changed, so the pressure is taken to be unchanged. Temperature matters: n = PV/(RT) for each bulb, and with one bulb at 2T₀ the fixed amount of gas needs a pressure of 1.33P₀.
D1.33P₀Correct The bulbs are connected, so at equilibrium the gas in both is at one pressure P. The total amount of gas is fixed: 2P₀V/(RT₀) = PV/(RT₀) + PV/(R·2T₀) = (3/2)PV/(RT₀). So P = (4/3)P₀ ≈ 1.33P₀. Some gas moves from the hot bulb into the cold one, which is why the pressure rises by less than the factor of 2 by which the hot bulb's temperature rose.
Working The tube connects the bulbs, so at equilibrium the pressure P is the same in both. Total amount (fixed): n = 2P₀V/(RT₀). At equilibrium, n = PV/(RT₀) + PV/(R·2T₀) = (3/2)PV/(RT₀). So (3/2)P = 2P₀ and P = (4/3)P₀ ≈ 1.33P₀; gas has moved from the hot bulb into the cold one. Errors: whole gas at the average temperature 1.5T₀ in volume 2V, P = nR(1.5T₀)/(2V) = 1.5P₀; heated bulb's pressure doubles with its temperature, no matter that gas leaves it (m13) → 2.00P₀; amount fixed by P and V alone, temperature ignored (m10) → 1.00P₀.
An air bubble of volume 1.0 cm³ is released at the bottom of a lake, 12 m below the surface, where the water temperature is 8.0°C. It rises to just below the surface, where the water temperature is 23.0°C. The atmospheric pressure at the surface is 1.0 × 10⁵ Pa, and the density of the water is 1.0 × 10³ kg/m³. Treat the air in the bubble as a fixed amount of ideal gas that is always at the temperature and pressure of the water around it. Use g = 9.8 m/s². What is the volume of the bubble just below the surface?
Answer and reasoning
A1.2 cm³ A student who takes the pressure at the bottom to be ρgh alone, 1.18 × 10⁵ Pa, picks this: V₂ = (1.0 cm³)(1.18)(296/281) = 1.2 cm³. The atmosphere also presses on the lake's surface, so the absolute pressure at the bottom is P₀ + ρgh = 2.18 × 10⁵ Pa, and the bubble grows to 2.3 cm³.
B2.3 cm³Correct At the bottom the pressure is P₀ + ρgh = 1.0 × 10⁵ Pa + (1.0 × 10³ kg/m³)(9.8 m/s²)(12 m) = 2.18 × 10⁵ Pa; near the surface it is 1.0 × 10⁵ Pa. The temperatures are 281 K and 296 K. For a fixed amount of gas, P₁V₁/T₁ = P₂V₂/T₂, so V₂ = (1.0 cm³)(2.18)(296 K/281 K) = 2.3 cm³.
C6.3 cm³ A student who uses the Celsius temperatures picks this: V₂ = (1.0 cm³)(2.18)(23.0/8.0) = 6.3 cm³. The gas law needs absolute temperatures, 296 K and 281 K, whose ratio is only 1.05, so V₂ = 2.3 cm³.
D2.2 cm³ A student who uses PV = constant, leaving out the rise in temperature, picks this: V₂ = (1.0 cm³)(2.18) = 2.2 cm³. The air also warms from 281 K to 296 K as it rises, which makes it expand a little more: V₂ = (1.0 cm³)(2.18)(296/281) = 2.3 cm³.
Working Pressure at the bottom: P₁ = P₀ + ρgh = 1.0 × 10⁵ Pa + (1.0 × 10³ kg/m³)(9.8 m/s²)(12 m) = 2.176 × 10⁵ Pa. Near the surface P₂ = 1.0 × 10⁵ Pa. T₁ = 281 K, T₂ = 296 K. Fixed amount of gas: P₁V₁/T₁ = P₂V₂/T₂, so V₂ = V₁(P₁/P₂)(T₂/T₁) = (1.0 cm³)(2.176)(296/281) = 2.29 cm³ ≈ 2.3 cm³. Errors: pressure at the bottom taken as ρgh = 1.176 × 10⁵ Pa → 1.24 ≈ 1.2 cm³; Celsius temperatures, 23.0/8.0 → 6.26 ≈ 6.3 cm³; temperature change ignored (PV constant) → 2.18 ≈ 2.2 cm³.
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
