3 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 3
A rigid container holds an unknown gas. A student measures the pressure of the gas and its temperature in kelvins. Which additional quantity must the student know to calculate the number of moles of gas in the container?
Answer and reasoning
AThe molar mass of the gas A student who thinks the identity of a gas changes the gas law picks this. PV = nRT applies to every ideal gas with the same R; molar mass is needed only to convert moles to grams.
BThe diameter of a gas molecule A student who thinks larger molecules mean fewer moles fit in a container picks this. Gas molecules take up a negligible part of the volume, so their size does not enter n = PV/RT.
CThe volume of the containerCorrect Rearranging PV = nRT gives n = PV/RT. With P and T measured and R a constant, the volume of the container is the only missing quantity.
DThe mass of the gas sample A student who reads n in PV = nRT as the mass of gas picks this. The mass would give moles only with the molar mass, which is unknown; with P and T known, the volume gives n directly.
Working n = PV/RT. P and T are measured and R is a constant, so the only additional quantity needed is V, the volume of the container (which, for a gas, is the volume of the gas). The molar mass, molecular size and sample mass do not appear in n = PV/RT.
The diagram represents a mixture of gases in a rigid container. The total pressure of the mixture is 1.20 atm. What is the partial pressure of N₂ in the mixture?
Answer and reasoning
A0.539 atm A student who takes partial pressure as proportional to mass picks this: weighting each particle by its molar mass, the N₂ makes up 140 of 312 mass units (0.449), and 0.449 × 1.20 atm = 0.539 atm. Partial pressure follows mole fraction, 5/12.
B0.706 atm A student who counts each atom as a separate gas particle picks this: 10 N atoms out of 17 atoms. Each N₂ molecule is one particle, so XN₂ = 5/12.
C0.400 atm A student who thinks each gas takes an equal share of the total pressure picks this: 1.20 atm ÷ 3. Each gas contributes in proportion to its moles, and N₂ makes up 5 of the 12 particles.
D0.500 atmCorrect There are 12 gas particles: 5 N₂ molecules, 4 Ar atoms and 3 He atoms. XN₂ = 5/12, so PN₂ = (5/12) × 1.20 atm = 0.500 atm.
Working The diagram shows 5 N₂ molecules, 4 Ar atoms and 3 He atoms: 12 gas particles. XN₂ = 5/12 = 0.417. PN₂ = XN₂ × Ptotal = (5/12) × 1.20 atm = 0.500 atm.
A fixed amount of an ideal gas is kept at constant pressure. Which of the numbered graphs best represents the volume of the gas as a function of its temperature in degrees Celsius?
Answer and reasoning
AGraph 1 A student who treats volume as proportional to Celsius temperature picks this line through 0°C. At 0°C (273 K) a gas still has a substantial volume; zero is reached only at −273°C.
BGraph 2 A student who thinks volume is inversely related to temperature picks this falling curve. At constant pressure a gas expands as it is heated: V rises with T.
CGraph 3 A student who thinks the volume of a gas depends only on its amount picks this flat line. With n and P fixed, V changes in proportion to Kelvin temperature.
DGraph 4Correct At constant n and P, V is proportional to Kelvin temperature, so against Celsius temperature the graph is a straight line that would reach zero volume at −273°C (0 K) and is above zero at 0°C.
Working V = (nR/P)T with T in kelvins, so V is directly proportional to Kelvin temperature. In degrees Celsius, T(K) = T(°C) + 273.15, so V plotted against T(°C) is a straight line that reaches V = 0 at −273°C, not at 0°C.
In preparation: 0 of 3 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
3.4.A.1 Ideal gas Fix
Ideal gas
A model gas whose pressure, volume, temperature and amount are related exactly by PV = nRT. Real gases at moderate temperatures and pressures behave very nearly ideally.
Ideal gas law
PV = nRT, which relates the pressure P, volume V, amount n (in moles) and Kelvin temperature T of an ideal gas; R is the gas constant. The law contains no term for the identity or molar mass of the gas.
Gas constant, R
The constant in PV = nRT. Its value depends on the units used: 0.08206 L·atm/(mol·K) when P is in atm and V in L, or 8.314 J/(mol·K).
Kelvin temperature
Temperature on the absolute scale, T(K) = T(°C) + 273.15. The gas laws require Kelvin temperature, because P and V are proportional to T only when T is measured from absolute zero.
Molar volume of a gas
The volume occupied by one mole of gas, V/n = RT/P. It is the same for every ideal gas at the same temperature and pressure; it equals 22.4 L/mol at STP (273.15 K and 1.0 atm) and generally has a different value at other temperatures and pressures.
Moles from mass
The amount of a gas in moles is its mass divided by its molar mass, n = m/𝐌, where the molar mass of a diatomic element such as O₂ or N₂ is twice the atomic mass.
Students often think Any temperature scale can be used in gas-law calculations, so a gas's pressure or volume is proportional to its temperature in degrees Celsius. In fact No. P and V are proportional to the Kelvin temperature, measured from absolute zero. A temperature in °C must be converted (T = °C + 273.15) before it is used in PV = nRT or in a ratio such as P₂/P₁ = T₂/T₁.
Students often think Gaseous elements such as oxygen and nitrogen consist of single atoms, so the molar mass of oxygen gas is 16.00 g/mol and that of nitrogen gas is 14.01 g/mol. In fact No. Oxygen gas consists of O₂ molecules, so its molar mass is 32.00 g/mol. H₂, N₂, O₂, F₂, Cl₂, Br₂ and I₂ are diatomic as elements.
3.4.A.2 Partial pressure Fix
Partial pressure
The pressure that one component of a gas mixture exerts. Each component's partial pressure is independent of the other components: it is the pressure that component would exert if it occupied the container alone at the same temperature.
Mole fraction
XA = moles of A / total moles of gas. The partial pressure of A is PA = Ptotal × XA, so partial pressures are proportional to mole fractions, not to mass fractions.
Dalton's law of partial pressures
The total pressure of a mixture of ideal gases is the sum of the partial pressures of its components: Ptotal = PA + PB + PC + …
Students often think The partial pressure of each gas in a mixture is proportional to its fraction of the total mass, so heavier gases contribute more pressure. In fact No. Partial pressure is proportional to mole fraction: PA = Ptotal × (moles A / total moles). A gas with a large share of the mass can have a small share of the pressure if its molar mass is large.
Students often think Each gas in a mixture exerts an equal share of the total pressure, so with three gases each partial pressure is one-third of the total. In fact No. Each gas contributes in proportion to its number of moles: a gas that makes up half of the moles in the mixture exerts half of the total pressure, whatever the number of other gases present.
3.4.A.3 Proportional relationships among P, V, T and n Fix
Proportional relationships among P, V, T and n
From PV = nRT: at constant n and T, P is inversely proportional to V; at constant n and V, P is directly proportional to Kelvin T; at constant n and P, V is directly proportional to Kelvin T; at constant T and V, P is directly proportional to n. When two variables change, their factors multiply.
Graphs of gas behavior
A graph of V against Kelvin T (constant n and P), or of P against Kelvin T (constant n and V), is a straight line through the origin; plotted against T in °C, the same line meets zero at −273°C. A graph of P against V at constant n and T is a curve, while P against 1/V is a straight line through the origin.
Slope of a V–T or P–T graph
For V against Kelvin T at constant P, the slope is nR/P; for P against Kelvin T at constant V, the slope is nR/V. A steeper line therefore means more moles of gas (or a lower fixed pressure or volume).
Students often think The pressure or volume of a gas is inversely proportional to its temperature, in the same way that pressure is inversely proportional to volume. In fact No. At constant n and V, P is directly proportional to Kelvin temperature, and at constant n and P, V is directly proportional to Kelvin temperature. The inverse relationship in the gas laws is between P and V.
Students often think The pressure (or volume) of a gas depends only on how much gas is present, so changing the temperature of a sealed sample does not change its pressure or volume. In fact No. With n fixed, P is proportional to T in a rigid container, and V is proportional to T at constant pressure, so heating a gas raises its pressure or its volume even though no gas is added.
11 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 11
A rigid 2.50 L flask contains O₂(g) at a pressure of 1.20 atm and a temperature of 27°C. What is the mass of O₂ in the flask?
Answer and reasoning
A43.3 g A student who uses the temperature in degrees Celsius picks this: n = (1.20)(2.50)/[(0.08206)(27)] = 1.35 mol, 43.3 g. The ideal gas law needs the Kelvin temperature, 300.15 K.
B1.95 g A student who treats oxygen gas as single O atoms picks this, using 16.00 g/mol. The moles are right (0.1218 mol), but O₂ has a molar mass of 32.00 g/mol.
C3.57 g A student who assumes a mole of any gas occupies 22.4 L picks this: 2.50/22.4 = 0.112 mol, 3.57 g. The gas is not at STP; at 1.20 atm and 300.15 K, n = PV/RT = 0.1218 mol.
D3.90 gCorrect T = 300.15 K, so n = PV/RT = (1.20)(2.50)/[(0.08206)(300.15)] = 0.1218 mol, and the mass is 0.1218 mol × 32.00 g/mol = 3.90 g.
Working T = 27 + 273.15 = 300.15 K. n = PV/RT = (1.20 atm)(2.50 L) / [(0.08206 L·atm/(mol·K))(300.15 K)] = 0.1218 mol O₂. Mass = 0.1218 mol × 32.00 g/mol = 3.90 g.
A rigid steel tank contains a gas at a pressure of 2.40 atm and a temperature of 27°C. The tank is heated to 327°C, and no gas escapes. What is the pressure in the tank at 327°C?
Answer and reasoning
A29.1 atm A student who treats pressure as proportional to the Celsius temperature picks this: 2.40 × 327/27 = 29.1 atm. The ratio must use kelvins: 600.15/300.15 = 2.00.
B1.20 atm A student who thinks pressure is inversely proportional to temperature picks this. In a rigid container P = (nR/V)T, so raising T raises P.
C4.80 atmCorrect With n and V fixed, P is proportional to Kelvin temperature. T rises from 300.15 K to 600.15 K, a factor of 2.00, so P = 2.40 atm × 2.00 = 4.80 atm.
D2.40 atm A student who thinks the pressure of a sealed gas depends only on the amount of gas picks this. With n and V fixed, P still rises in proportion to the Kelvin temperature.
Working n and V are constant, so P₂/P₁ = T₂/T₁ in kelvins. T₁ = 300.15 K, T₂ = 600.15 K. P₂ = 2.40 atm × 600.15/300.15 = 4.80 atm.
Two rigid containers, X and Y, have the same volume. Container Y holds three times as many moles of gas as container X does, and the Kelvin temperature of the gas in Y is twice that of the gas in X. The pressure in container Y is how many times the pressure in container X?
Answer and reasoning
A5.0 A student who adds the two factors instead of multiplying them picks this. P is proportional to the product nT, so the factors combine as 3 × 2 = 6.
B6.0Correct P = nRT/V, and V is the same, so the pressure ratio is the product of the mole ratio and the temperature ratio: 3 × 2 = 6.0.
C3.0 A student who thinks pressure depends only on the amount of gas picks this, ignoring the temperature. Doubling the Kelvin temperature at constant n and V doubles P, so the ratio is 3 × 2 = 6.
D1.5 A student who thinks pressure is inversely proportional to temperature picks this: 3 ÷ 2. P = nRT/V, so P rises with T: 3 × 2 = 6.
Working P = nRT/V. With V the same, PY/PX = (nY/nX)(TY/TX) = (3)(2) = 6.0.
The diagram shows two sealed, rigid flasks, one containing He(g) and the other containing Ar(g), with the volume, temperature and pressure of each. Which statement about the contents of the two flasks is correct?
Answer and reasoning
AThe flasks contain equal numbers of atoms, as the pressure, volume and temperature all match.Correct Both flasks have V = 1.00 L, T = 25°C and P = 1.50 atm, so n = PV/RT is the same, 0.0613 mol, in each. The ideal gas law does not depend on which gas is present, so the flasks hold equal numbers of atoms.
BThe Ar flask contains fewer atoms, because each Ar atom takes up more room in the flask. A student who thinks larger gas particles fill the space picks this. Gas atoms are far apart, so their size has a negligible effect on how many are present; both flasks contain 0.0613 mol.
CThe He flask contains fewer atoms, as faster He atoms each strike the walls more often. A student who thinks lighter gas particles exert more pressure picks this. The ideal gas law contains no term for the identity of the gas, so the same P, V and T mean the same number of moles of He and of Ar.
DThe flasks contain equal masses of gas, as equal volumes of gases have equal masses. A student who thinks equal gas volumes contain equal masses picks this. The flasks contain equal moles, 0.0613 mol, so the Ar sample (2.45 g) has about ten times the mass of the He sample (0.245 g).
A rigid container holds the mixture of gases shown in the table. The total pressure of the mixture is 4.20 atm. What is the partial pressure of O₂ in the mixture?
B1.29 atm A student who takes partial pressure as proportional to mass picks this: 16.00 g of 52.04 g is 0.307 of the mass, 0.307 × 4.20 atm. Partial pressure follows the mole fraction, 0.250.
C1.40 atm A student who thinks each gas takes an equal share of the total pressure picks this: 4.20 atm ÷ 3. The gases are present in different numbers of moles, so their shares differ.
D1.20 atm A student who treats oxygen and nitrogen gases as single atoms picks this, using 16.00 g/mol for O₂ and 14.01 g/mol for N₂: 1.000 mol of 3.500 mol, and (1.000/3.500) × 4.20 atm. O₂ and N₂ are diatomic, with molar masses 32.00 and 28.02 g/mol.
Working Moles: CH₄ 8.02 g ÷ 16.04 g/mol = 0.500 mol; O₂ 16.00 g ÷ 32.00 g/mol = 0.500 mol; N₂ 28.02 g ÷ 28.02 g/mol = 1.000 mol. Total = 2.000 mol. XO₂ = 0.500/2.000 = 0.250. PO₂ = 0.250 × 4.20 atm = 1.05 atm.
A rigid flask contains N₂(g). Some Ar(g) is then added to the flask while the temperature is held constant. Which statement about the partial pressure of N₂ after the Ar is added is correct?
Answer and reasoning
AIt decreases, because the mole fraction of N₂ falls while the total pressure stays the same. A student who treats the total pressure as fixed picks this. The mole fraction of N₂ does fall, but the total pressure rises by just enough that XN₂ × Ptotal is unchanged.
BIt stays the same, because the amount of N₂, the volume and the temperature are unchanged.Correct The partial pressure of each gas in a mixture is independent of the other gases: PN₂ = nN₂RT/V, and none of nN₂, V or T has changed. The total pressure rises by the partial pressure of the added Ar.
CIt increases, because the Ar atoms collide with the N₂ molecules and drive them into the walls. A student who thinks the added gas pushes the first gas into the walls picks this. Each gas's partial pressure depends only on its own moles, V and T; the rise in total pressure is the Ar's own partial pressure.
DIt becomes equal to the partial pressure of Ar, because the two gases share the total equally. A student who thinks each gas takes an equal share of the total pressure picks this. Each partial pressure is set by that gas's own moles; nothing makes the N₂ and Ar partial pressures equal.
A rigid container holds 5.60 g of N₂(g). The graph shows the pressure of the gas as a function of its temperature. What is the volume of the container?
Answer and reasoning
A1.64 L A student who uses the temperature in degrees Celsius picks this: (0.1999)(0.08206)(327)/3.28. The graph's line reaches zero pressure at −273°C, showing that P is proportional to Kelvin temperature, 600.15 K.
B4.48 L A student who assumes a mole of gas always occupies 22.4 L picks this: 0.1999 mol × 22.4 L/mol. That molar volume is the value at STP; this gas is at 327°C and 3.28 atm, so the volume must come from V = nRT/P with the graph's values.
C3.00 LCorrect The marked point gives P = 3.28 atm at 327°C (600.15 K). n = 5.60/28.02 = 0.1999 mol, so V = nRT/P = (0.1999)(0.08206)(600.15)/3.28 = 3.00 L.
D6.00 L A student who treats nitrogen gas as single N atoms picks this, using 14.01 g/mol to get 0.400 mol. N₂ has a molar mass of 28.02 g/mol.
Working From the graph, P = 3.28 atm at 327°C = 600.15 K. n = 5.60 g ÷ 28.02 g/mol = 0.1999 mol. V = nRT/P = (0.1999 mol)(0.08206 L·atm/(mol·K))(600.15 K)/(3.28 atm) = 3.00 L.
The graph shows the volume of two gas samples, A and B, as a function of temperature. Both samples are kept at the same constant pressure. Which statement accounts for the difference between the two lines?
Answer and reasoning
ASample A contains more moles of gas than Sample B does.Correct Both lines pass through the origin, so V = (nR/P)T with slope nR/P. At the same pressure the steeper line, A, must have the larger n; its slope is twice B's, so A contains twice as many moles.
BThe molecules in Sample A are larger than those in Sample B. A student who thinks larger gas molecules take up more of the volume picks this. Gas molecules occupy a negligible part of a gas's volume, so the slope nR/P depends only on the number of moles.
CThe molecules in Sample A have a greater mass than those in B. A student who thinks heavier gas molecules occupy more volume picks this. V = nRT/P contains no molar mass, so at the same T and P the volume depends only on the number of moles.
DSample A was heated more quickly than Sample B was. A student who reads the temperature axis as time picks this. The graph shows the volume at each temperature; how quickly the samples were heated cannot be read from it.
A student will measure the pressure of a fixed amount of gas in a rigid container at several temperatures to test the claim that the pressure is directly proportional to the absolute temperature. Which graph of the data, if it is a straight line through the origin, would support the claim?
Answer and reasoning
AA graph of pressure versus the temperature in °C A student who thinks pressure is proportional to Celsius temperature picks this. A gas still has pressure at 0°C; plotted against °C, the line meets zero pressure at −273°C, not at the origin.
BA graph of pressure versus the temperature in kelvinsCorrect Direct proportionality, P = kT, gives a straight line through the origin only when T is measured from absolute zero, that is, in kelvins.
CA graph of pressure versus 1/(kelvin temperature) A student who thinks pressure is inversely related to temperature picks this. A straight line through the origin on this graph would show P inversely proportional to T, which is not the claim.
DA graph of pressure versus time since heating began A student who treats a temperature graph as a record over time picks this. Time is not the variable in the claim; the pressure depends on how hot the gas is, not on how long it has been heated.
A rigid 2.50 L container holds a mixture of 0.200 mol of He(g) and 0.300 mol of Ne(g) at 127°C. What is the partial pressure of Ne in the container?
Answer and reasoning
A1.25 atm A student who uses the temperature in degrees Celsius picks this: (0.300)(0.08206)(127)/2.50. The ideal gas law needs the Kelvin temperature, 400.15 K.
B3.28 atm A student who thinks each gas takes an equal share of the total pressure picks this, half of 6.57 atm. Ne makes up 0.300 of the 0.500 mol of gas, so it exerts 0.600 of the total pressure.
C3.94 atmCorrect Each gas in a mixture of ideal gases exerts the pressure it would exert if it were alone in the container: PNe = nNe RT/V = (0.300)(0.08206)(400.15)/2.50 = 3.94 atm. This equals the mole fraction of Ne, 0.600, times the total pressure, 6.57 atm.
D5.80 atm A student who takes partial pressure as proportional to mass picks this: Ne is 6.05 g of the 6.85 g of gas (0.883), and 0.883 × 6.57 atm = 5.80 atm. Partial pressure follows the mole fraction, 0.600.
Working The partial pressure of each gas is independent of the other gas, so PNe = nNe RT/V. T = 127 + 273.15 = 400.15 K. PNe = (0.300 mol)(0.08206 L·atm/(mol·K))(400.15 K)/(2.50 L) = 3.94 atm. (Check: Ptotal = (0.500)(0.08206)(400.15)/2.50 = 6.57 atm and XNe = 0.600, so PNe = 0.600 × 6.57 atm = 3.94 atm.)
A rigid flask contains N₂(g). Ar(g) is added to the flask in portions at constant temperature, and the total pressure is measured after each addition. The graph shows the results. What is the mole fraction of Ar in the flask after 0.30 mol of Ar has been added?
Answer and reasoning
A0.60Correct The intercept, 1.0 atm, is the partial pressure of N₂, which stays the same as Ar is added. At 0.30 mol of Ar the total pressure is 2.5 atm, so the partial pressure of Ar is 1.5 atm and XAr = 1.5 atm/2.5 atm = 0.60.
B0.50 A student who thinks the two gases share the total pressure equally picks this. The graph shows that N₂ exerts 1.0 atm of the 2.5 atm and Ar the remaining 1.5 atm, so the shares are 0.40 and 0.60.
C0.43 A student who counts each atom as a separate gas particle picks this. The graph shows 0.5 atm for each 0.10 mol of gas, so the flask holds 0.20 mol of N₂; counting it as 0.40 mol of N atoms gives 0.30/0.70. Each N₂ molecule is one particle, so XAr = 0.30/0.50.
D0.68 A student who takes a gas's share of the pressure to be its share of the mass picks this: 12.0 g of Ar in 17.6 g of gas (the flask holds 0.20 mol, 5.6 g, of N₂). The share of the pressure is the mole fraction, 1.5 atm/2.5 atm.
Working Before any Ar is added the total pressure is the pressure of N₂ alone, 1.0 atm, and the partial pressure of N₂ does not change when Ar is added (n, V and T of N₂ are constant). After 0.30 mol of Ar, Ptotal = 2.5 atm, so PAr = 2.5 − 1.0 = 1.5 atm. XAr = PAr/Ptotal = 1.5/2.5 = 0.60.
Compiled from the AP Chemistry Course and Exam Description (effective Fall 2024) and our question bank · Specialist review in progress. How these pages are made · Free, no account
