Worked example 1: solution

One conversion factor is needed.

A common first attempt

No unit cancels, and the answer is not in mmHg.

Worked example 1: solution

Both factors come from 1 atm = 760 mmHg, and both equal 1. Only one cancels atm:

Worked example 1: solution

Worked example 1: the route on the map

Practice 1 answer: C

A inverted the factor: 88.2 × 14.7 = 1.30 × 10³, and no unit cancels; the units come out psi²/atm, not atm. B used mercury's 760 in place of psi's 14.7: 88.2 / 760 = 0.116. D used the kPa number 101.325: 88.2 / 101.325 = 0.870.

Worked example 2: solution

Two conversion factors are needed, one per arrow of the plan.

Step 2 · Pick the factor that cancels its unit

The first arrow removes kPa. Its equality gives the factor, with kPa in the denominator:

Worked example 2: solution

Worked example 2: the route on the map

Practice 2 answer: B

A stopped before subtracting: 1.60 atm is the final reading, not the rise. C subtracted in kPa and skipped the last hop: 162.0 − 101.06 = 60.9, still kPa. D subtracted backward: 0.9974 − 1.5988 = −0.601, a fall, though the pressure rose.

Extra practice 1 answer: D

A stopped mid-plan: 11.8 ÷ 14.7 = 0.803 is still atm. B used the kPa number on the second arrow: 11.8 ÷ 14.7 × 101.325 = 81.3, a count in kPa. C flipped the second factor: 11.8 ÷ 14.7 ÷ 760 = 1.06 × 10⁻³.

Extra practice 1 answer: D

Extra practice 2 answer: A

B stopped before subtracting: 0.112 atm is the final reading, not the fall. C subtracted backward: 0.1118 − 0.9864 = −0.875, a rise, though the pump lowered the pressure. D read the mmHg value with the kPa number: 85.0 ÷ 101.325 = 0.8389 atm, and 0.9864 − 0.8389 = 0.148.

Extra practice 2 answer: A

Extra practice 3 answer: C

A stopped the rise at atm: 86.2 ÷ 101.325 = 0.851, and 38.0 − 0.851 = 37.1 mixes atm with psi. B added the rise: 38.0 + 12.5 = 50.5, a reading after a second pump. D used psi's 14.7 on the kPa value: 86.2 ÷ 14.7 = 5.86, and 38.0 − 5.86 = 32.1.

Extra practice 3 answer: C

Worked example 1: solution

One conversion factor is needed.

Step 2 · Map the route

moles → liters. One arrow, one factor: the molar volume.

Worked example 1: solution

The given is moles, so moles go in the denominator: 22.4 L over 1 mol.

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

One conversion factor is needed.

A common first attempt

No unit cancels, and no gas volume carries L²/mol. The factor is upside down.

Worked example 2: solution

liters → moles. One arrow, one factor: the molar volume.

Step 3 · Orient each factor

The given is liters, so liters go in the denominator. Flip it: 1 mol over 22.4 L.

Worked example 2: solution

Worked example 2: solution

Worked example 2: the route on the map

Practice 1 answer: D

A reached for the molar mass, 32.00 g/mol, instead of the molar volume: 3.36 / 32.00 = 0.105, and grams never entered the problem. C left the molar volume upside down: 3.36 × 22.4 = 75.3, with units L²/mol. B used the molar mass upside down: 3.36 × 32.00 = 108. Only 22.4 L/mol, with liters underneath, cancels liters.

Worked example 3: solution

Two conversion factors are needed.

Step 2 · Map the route

g CO₂ → mol → L. Molar mass covers the first arrow; molar volume covers the second.

Worked example 3: solution

Grams cancel with molar mass underneath. Moles cancel with molar volume's mole underneath.

Worked example 3: solution

One continuous chain; each factor cancels the unit before it:

Worked example 3: solution

Worked example 3: the route on the map

Practice 2 answer: A

B stopped at moles: 6.72 / 22.4 = 0.300 mol, an amount, not a mass. C flipped the molar mass: 0.300 / 64.07 = 4.68 × 10⁻³, grams underneath. D ran the grams-to-liters chain on a volume, both factors upside down: 6.72 / 64.07 × 22.4 = 2.35.

Extra practice 1 answer: B

A used the grams as moles: 25.5 × 22.4 = 571, with no molar mass. C flipped the molar volume: 1.50 / 22.4 = 0.0668, in mol²/L. D stopped at moles: 25.5 / 17.03 = 1.50 mol, an amount, not a volume.

Extra practice 2 answer: D

A used the carbon subscript: × 2 instead of × 6 gives 2.11 × 10²³, the C atoms. B stopped at moles: 3.92 / 22.4 = 0.175 mol C₂H₆. C stopped at molecules: 0.175 × 6.022 × 10²³ = 1.05 × 10²³ molecules, not H atoms.

Extra practice 3 answer: A

At 0 °C and 1 atm the tank is at STP. Full, it holds one mole per 22.4 L.

B added the gas already present: 0.350 + 0.120 = 0.470. C flipped the molar volume: 7.84 × 22.4 − 0.120 = 175, a number that is not moles. D stopped at the full tank: 0.350 mol, but 0.120 mol is already inside.

Worked example 1: solution

Step 1 · Identify given and wanted

V₁ = 6.0 L, P₁ = 1.0 atm, P₂ = 3.0 atm. Temperature is held constant. The unknown is V₂.

Worked example 1: solution

Temperature does not change, so T₁ = T₂. The two temperature terms are equal and cancel: no kelvin value is even needed.

Worked example 1: solution

With T cancelled, this is Boyle's case: pressure and volume trade off inversely.

Worked example 1: solution

Worked example 1: the route on the chart

Worked example 2: solution

Step 1 · Identify given and wanted

V₁ = 3.00 L. Pressure is held constant, so P₁ = P₂. The unknown is V₂.

Worked example 2: solution

T₁ = 25.0 + 273.15 = 298.15 K. T₂ = 75.0 + 273.15 = 348.15 K.

Worked example 2: solution

Pressure cancels, leaving Charles's case: volume rises in step with the kelvin temperature.

Worked example 2: solution

Worked example 2: the route on the chart

Practice 1 answer: C

A left the temperatures in Celsius and inverted them: 2.50 × (17/77) = 0.552 L. B inverted the kelvin ratio: 2.50 × (290.15/350.15) = 2.07 L, a shrinking gas though it was heated. D left the temperatures in Celsius: 2.50 × (77/17) = 11.3 L, a ratio that far overstates a 60-degree warming.

Worked example 3: solution

A common first attempt

A volume cannot be negative. The Celsius temperature ratio, not the gas, produced the impossible sign.

Worked example 3: solution

V₁ = 2.00 L, P₁ = 1.00 atm, P₂ = 0.400 atm. The unknown is V₂.

Worked example 3: solution

T₁ = 27 + 273.15 = 300.15 K. T₂ = −23 + 273.15 = 250.15 K.

Worked example 3: the volume at altitude

Step 3 · Rearrange for the unknown

Solving for V₂ puts pressure as old over new (P₁/P₂) and temperature as new over old (T₂/T₁). Pressure fell, so its ratio expands the gas; the gas cooled, so its ratio trims it.

Worked example 3: the volume at altitude

Worked example 3: the route on the chart

Practice 2 answer: B

A inverted the PV ratio: 294.15 × 15.68 / 12.0 = 384 K = 111 °C, a gas that warmed as it rose. C stopped at kelvin: 225 K is the right temperature on the wrong scale. D put Celsius into the ratio: 21.0 × 12.0 / 15.68 = 16.1 °C.

Practice 3: answer D

A inverted the pressure ratio: 1.80 × (0.850/1.45) × (308.15/253.15) = 1.28, a gas that shrank as the pressure fell. B held temperature constant: 1.80 × (1.45/0.850) = 3.07. C left the temperatures in Celsius: 1.80 × (1.45/0.850) × (35.0/−20.0) = −5.37, a negative volume. E inverted the temperature ratio: 1.80 × (1.45/0.850) × (253.15/308.15) = 2.52.

Practice 4: answer E

A stopped in atm: 0.975 × (0.520/0.0310) × (793.15/258.15) = 50.2, the pressure in the wrong unit. B held temperature constant: 0.975 × (0.520/0.0310) × 760 = 1.24 × 10⁴. C left the temperatures in Celsius: 0.975 × (0.520/0.0310) × (520./−15.0) × 760 = −4.31 × 10⁵. D inverted the volume ratio: 0.975 × (0.0310/0.520) × (793.15/258.15) × 760 = 136.

Practice 5: answer A

B inverted the pressure ratio: 3.60 × (2.35/1) × (233.15/273.15) = 7.22, a gas that grew as it was squeezed. C held temperature constant: 3.60 × (1/2.35) = 1.53. D took STP pressure as 760 against 2.35 atm: 3.60 × (760/2.35) × (233.15/273.15) = 994. The two pressures must share one unit. E inverted the temperature ratio: 3.60 × (1/2.35) × (273.15/233.15) = 1.79.

Worked example 1: solution

Step 1 · List the pieces

P = 2.00 atm. V = 5.00 L. T = 25.0 + 273.15 = 298.15 K. The unknown is n.

Worked example 1: solution

Divide both sides by R T to isolate n.

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

Step 1 · List the pieces

V = 8.00 L. T = 27.0 + 273.15 = 300.15 K. The unknown is n.

Worked example 2: solution

Isolate n exactly as before: n = PV/RT.

Worked example 2: solution

Worked example 2: the route on the map

Practice 1 answer: B

A flipped the rearrangement: (0.08206 × 308.15) ÷ (1.25 × 4.00) = 5.06. C left the temperature in Celsius: 1.25 × 4.00 ÷ (0.08206 × 35.0) = 1.74 mol. D left the pressure in mmHg: 950. × 4.00 ÷ (0.08206 × 308.15) = 150. mol.

Worked example 3: solution

A common first attempt

No real gas is lighter than helium at 4.00 g/mol. The temperature went in as Celsius.

Worked example 3: solution

m = 3.60 g. V = 2.00 L. P = 1.00 atm. T = 25.0 + 273.15 = 298.15 K. The unknown is M.

Worked example 3: solution

Worked example 3: the molar mass

Step 3 · Substitute R and cancel units

Only CO₂ matches 44.0 g/mol. The gas is carbon dioxide.

Worked example 3: the molar mass

Step 4 · Check the units and the size

Worked example 3: the route on the map

Practice 2 answer: D

A stopped before dividing by P: 1.97 × 0.08206 × 295.15 = 47.7, as if the pressure were 1 atm. B left the temperature in Celsius: 1.97 × 0.08206 × 22.0 ÷ 0.822 = 4.32. C flipped the mmHg factor: 760/625 = 1.216 atm, and 47.7 ÷ 1.216 = 39.2.

Practice 3 answer: C

A skipped the molar mass: 9.60 × 0.08206 × 288.15 ÷ 2.50 = 90.8. B left the temperature in Celsius: 0.300 × 0.08206 × 15.0 ÷ 2.50 = 0.148. D used 16.00 g/mol, one O atom: 0.600 × 0.08206 × 288.15 ÷ 2.50 = 5.67.

Practice 4 answer: A

B stopped in kelvins: 318. C flipped the torr factor: 760/684 = 1.111 atm gives 393 K, or 120. °C. D added 273.15 instead of subtracting: 318.4 + 273.15 = 592. E left the volume in mL: 0.900 × 180. ÷ (0.00620 × 0.08206) − 273.15 = 3.18 × 10⁵.

Practice 5 answer: B

A inverted the formula, RT/(PM): 24.88 ÷ 16.54 = 1.50 L/g. C left the pressure in kPa: 104.5 × 16.04 ÷ (0.08206 × 303.15) = 67.4. D used the STP molar volume: 16.04 ÷ 22.4 = 0.716, true only at 0 °C and 1 atm. E left the temperature in Celsius: 1.031 × 16.04 ÷ (0.08206 × 30.0) = 6.72.

Worked example 1: solution

Step 1 · List every gas

Three gases share the lamp: helium, neon, and argon. Each presses on the walls on its own.

Worked example 1: solution

The three partial pressures add to the total: 1.20 atm = 0.30 + 0.35 + P(Ar).

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

Step 1 · List every gas

Nitrogen and oxygen fill the cylinder. Only these two share the pressure.

Worked example 2: solution

The two partial pressures add to 3.0 atm, and each gas's share follows its share of the molecules.

Worked example 2: solution

Worked example 2: solution

Worked example 2: the route on the map

Practice 1 answer: A

B removed only nitrogen: 1.30 − 0.50 = 0.80 atm, leaving oxygen's share inside. C removed only oxygen: 1.30 − 0.20 = 1.10 atm, leaving nitrogen's share inside. D added the two known pressures: 0.50 + 0.20 = 0.70 atm, the combined push of nitrogen and oxygen.

Worked example 3: solution

Step 1 · List every gas

Only the products remain: nitrogen and hydrogen. The ammonia is gone.

Worked example 3: solution

P(N₂) + P(H₂) = 866.0 mmHg. The coefficients fix the amounts: of every 4 mol of product, 1 mol is N₂ and 3 mol are H₂.

Worked example 3: solution

Worked example 3: solution

Worked example 3: the route on the map

Practice 2 answer: C

A used the mass fraction and skipped the molar masses: 36.0 ÷ 42.0 × 1.50 = 1.29 atm. Pressure counts particles, not grams. B flipped the mole fraction: 2.400 ÷ 0.901 × 1.50 = 4.00 atm, more than the whole mixture. D stopped at moles: 0.901 mol of argon is an amount, not a pressure.

Practice 3 answer: D

A split the total three ways: 884 ÷ 3 = 295 torr, as if the three gases formed in equal amounts. B found carbon dioxide's share: 1 ÷ 4 × 884 = 221 torr. C counted the solid as a gas: 2 ÷ 5 × 884 = 354 torr, but the ammonium carbonate is gone.

Practice 3 answer: D

Practice 4 answer: A

B split the total three ways: 10.4 ÷ 3 = 3.47 atm, but the three amounts differ. C used the mass fraction: 24.0 ÷ 113.0 × 10.4 = 2.21 atm. Pressure counts particles, not grams. D left helium out of the total: 0.7500 ÷ 3.748 × 10.4 = 2.08 atm.

Practice 4 answer: A

Practice 5 answer: B

Two moves are needed. Move 1: oxygen's partial pressure, from PV = nRT.

Practice 5 answer: B

Practice 5 answer: B

Practice 5 answer: B

Worked example 1: solution

Step 1 · Level and read

The levels match, so the wet gas is at atmospheric pressure: P_total = 755.0 torr.

Worked example 1: solution

At 25 °C the table gives 23.8 torr. The temperature alone sets it.

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: the dry pressure

Step 1 · Level and read

The levels are matched, so the wet gas is at atmospheric pressure: P_total = 762 torr.

Worked example 2: the dry pressure

From the table, water contributes 22.4 torr at 24 °C, set only by the temperature.

Worked example 2: the dry pressure

Worked example 2: the moles

Step 4 · Use the dry pressure

Worked example 2: the moles

Step 4 · Use the dry pressure

Worked example 2: the route on the map

Practice 1 answer: A

B subtracted the vapor out of habit: 727.4 − 23.8 = 703.6 torr, but the dry pressure is already the part; the whole is larger. C read the 20 °C row: 727.4 + 17.5 = 744.9 torr, the wrong temperature. D skipped the vapor: 727.4 torr is the methane alone, not the whole wet mixture that balances the atmosphere.

Practice 2 answer: D

A used the wet total: 748 ÷ 760 = 0.984 atm in the denominator gives 0.367 L, with the vapor's share counted as oxygen. B left the pressure in torr: dividing by 730.5 gives 4.94 × 10⁻⁴ L. C left the temperature in Celsius: 0.08206 × 20.0 on top gives 0.0256 L.

Practice 3 answer: C

A converted the barometer but skipped the vapor: 753.1 torr is propane and water vapor together. B added the vapor: 753.1 + 26.7 = 779.8 torr, a part larger than the whole. D subtracted across units: 100.4 − 26.7 = 73.7 takes torr away from kPa.

Practice 4 answer: B

A flipped the pressure ratio: 24.5 × 760/732.3 × 273.15/294.15 = 23.6 mL. C flipped the temperature ratio: 24.5 × 732.3/760 × 294.15/273.15 = 25.4 mL. D used the wet 751 torr: 24.5 × 751/760 × 273.15/294.15 = 22.5 mL, counting the vapor as oxygen.

Practice 5 solution: moles of H₂

Four moves: dry pressure, moles of H₂ from PV = nRT, the mole ratio, then grams.

Practice 5 solution: moles of H₂

Practice 5 answer: A

B skipped the mole ratio: 0.01248 × 6.94 = 0.0866 g. C used 23 °C for T: 2.23 g. D stopped at 0.0250 mol Li, a count, not a mass. E used the wet 753 torr: 0.178 g, counting the vapor as hydrogen.

Worked example 1: solution

Three conversion factors are needed.

Step 1 · To moles

At STP the molar volume converts the given volume to moles; the liters cancel:

Worked example 1: solution

The mole ratio, written NH₃ over H₂ (2 : 3), crosses substances; mol H₂ cancels.

Worked example 1: solution

Convert out with NH₃'s molar mass, in one continuous setup:

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

Three conversion factors are needed. The gas volume is the answer now, so the molar volume comes last.

Step 1 · To moles

NaN₃'s molar mass converts the given mass to moles; grams cancel.

Worked example 2: solution

The mole ratio, written N₂ over NaN₃ (3 : 2), crosses substances.

Worked example 2: solution

At STP the molar volume converts moles of N₂ to liters. The complete chain:

Worked example 2: solution

Worked example 2: the route on the map

Practice 1 answer: D

A skipped the mole ratio: 8.10/26.98 × 22.4 = 6.72. B stopped at moles: 8.10/26.98 × 3/2 = 0.450 mol, not liters. C flipped the mole ratio: 8.10/26.98 × 2/3 × 22.4 = 4.48, as if fewer moles of H₂ formed than Al reacted.

Worked example 3: solution

Grams cross to moles of CO₂ first; the volume then comes from PV = nRT.

Step 1 · To moles Step 2 · Moles → moles

Molar mass, then the mole ratio (1 CO₂ : 2 NaHCO₃), gives moles of CO₂:

Worked example 3: solution

22.4 L/mol holds only at STP. At 27 °C a mole takes more room, so 5.60 L is too small.

Worked example 3: solution

Worked example 3: solving PV = nRT

Step 3 · To the wanted unit

Solve PV = nRT for volume, with T in kelvin (27 + 273.15 = 300.15 K):

Worked example 3: solving PV = nRT

Step 3 · To the wanted unit

Solve PV = nRT for volume, with T in kelvin (27 + 273.15 = 300.15 K):

Worked example 3: the route on the map

Practice 2 answer: A

B used 22.4 L/mol off STP: 0.198 × 22.4 = 4.43. C stopped at moles: 0.198 mol O₂ is the amount, not the volume it fills. D forgot to convert to kelvin: (0.198)(0.08206)(127) = 2.06.

Extra practice 1 answer: C

0.900 g is 0.00500 mol of glucose. A skipped the mole ratio: 0.00500 × 22.4 × 1000 = 112. B stopped at liters: 0.224 L. D inverted the ratio: 0.00500 × (1/2) × 22.4 × 1000 = 56.0. E stopped at moles: 0.00999 mol CO₂.

Extra practice 2 answer: B

A used °C: (0.265)(0.08206)(35.0) ÷ 0.971 = 0.784. C kept mmHg: (0.265)(0.08206)(308.15) ÷ 738 = 0.00908. D used 22.4 L/mol off STP: 0.265 × 22.4 = 5.94. E stopped at moles: 0.265 mol CO₂.

Extra practice 3 answer: C

A skipped the mole ratio: 0.0170 × 22.99 = 0.391. B skipped mL → L: 381 ÷ 22.4 × 2 × 22.99 = 782. D stopped at moles: 0.0170 × 2 = 0.0340 mol Na. E inverted the ratio: 0.0170 × (1/2) × 22.99 = 0.196.

Worked example 1: solution

Step 1 · Convert T to kelvin

T = 27 + 273.15 = 300.15 K.

Worked example 1: solution

About 3.74 kJ per mole. Helium or xenon at the same temperature gives the identical number.

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the strip

Worked example 2: solution

Step 1 · Convert T to kelvin

T = 25 + 273.15 = 298.15 K for all three flasks.

Worked example 2: solution

Same T, same average KE. (3/2)RT has no mass term, so CO₂ carries no more energy than He. No gas wins.

Worked example 2: solution

Worked example 2: solution

Worked example 2: the route on the strip

Practice 1 answer: B

Average kinetic energy depends only on Kelvin temperature, so all three are equal. A confuses speed with energy: He is the fastest, but that is how it carries the same KE with less mass. C assumes mass sets the energy; mass sets the speed. D adds a variable that never appears: (3/2)kT has no pressure term.

Practice 2 answer: D

A applied the same-temperature rule, but these flasks sit at four different temperatures. B compared the raw numbers, setting 300 K against Celsius readings; convert every T to kelvin first. C ranked by speed, lightest fastest, but speed is not energy: mass never enters (3/2)RT.

Practice 3 answer: C

A picked the heaviest gas, which is the slowest. B picked the single atom, but Ar (39.95 g/mol) is heavier than the two light molecules. D is close, but 17.03 > 16.04, so ammonia trails methane by 3%. E is true of the average kinetic energy, not the speed.

Practice 4 answer: B

A took the molar-mass ratio without the root: 83.80 ÷ 20.18 = 4.15. C inverted the fraction: √(20.18 ÷ 83.80) = 0.491, which would make the lighter gas slower. D inverted it and skipped the root: 20.18 ÷ 83.80 = 0.241.

Practice 5 answer: A

B ranked by temperature alone, as if all three had one mass. C ranked by mass alone, lightest fastest, as if all three shared one temperature. D put M over T, the fraction under the root upside down, so the order runs backward.

Practice 6 answer: C

A stopped at the energy factor: 673.15 ÷ 293.15 = 2.30 is how the average KE grows, and the speed follows its square root. B divided the Celsius readings with no root: 400. ÷ 20.0 = 20.0. D divided the Celsius readings, then took the root: √(400. ÷ 20.0) = 4.47.

Practice 7 answer: B

A counted only how often the particles hit: the lighter O₂ molecules hit more often, but each hit is softer. C counted only how hard they hit: krypton atoms hit harder, but less often. The two effects cancel. D gave equal energies equal speeds; heavier krypton moves slower to carry the same KE. E gave the faster gas more energy; at one temperature every gas has the same average KE.

Practice 8 answer: D

A rejected statement 3, giving the heavier SF₆ more energy; at one temperature every gas has the same average KE, and SF₆ simply moves slower. B rejected statement 1, as if collisions used up energy; elastic collisions never let the gas run down. C accepted statement 2; the model puts no forces between particles. E accepted statement 2 and rejected statement 1.

Worked example 1: solution

Step 1 · List

Worked example 1: solution

Step 1 · List

At moderate pressure the a term controls the deviation. CO₂ has the largest a, so the strongest attractions, and its Z falls farthest below 1.

Worked example 1: solution

Step 1 · List

Worked example 2: solution

Step 1 · List

Worked example 2: solution

Step 1 · List

a = 3.640 is large: strong attractions will pull the pressure well below ideal.

Worked example 2: solution

Step 1 · List

V − n·b = 0.500 L − 1.00 × 0.04267 L = 0.4573 L.

Worked example 2: solution

Step 1 · List

Worked example 2: solution

Step 1 · List

Worked example 2: the route on the map

Practice 1 answer: B

A picked the heaviest atom: molar mass is not one of the two corrections. C read the b column: CH₄ has the largest b, 0.04278 L/mol, and the b term pushes P up, not down. D picked the smallest a: H₂ is the most nearly ideal of the four.

Practice 2 answer: C

The volume term lifts P by 62.82 − 57.44 = 5.38 atm; the attraction term drops it by 9.13 atm, so the net is 3.76 atm below ideal. A kept only the volume correction: 62.8 atm. B kept only the attraction term: 57.44 − 9.13 = 48.3 atm. D added the a term instead of subtracting it: 62.82 + 9.13 = 71.9 atm.

Guided example: solution

Step 1 · List

Guided example: solution

Step 1 · List

a = 0.0346 is tiny: helium atoms barely attract. At this crowding the molecules' own volume counts more.

Guided example: solution

Step 1 · List

V − n·b = 0.250 L − 2.00 × 0.0238 L = 0.2024 L.

Guided example: solution

Step 1 · List

Guided example: solution

Step 1 · List

Practice 3 answer: E

A left n unsquared: 74.32 − 13.12 = 61.2. B added the a term: 74.32 + 26.28 = 101. C divided by V, not V²: 74.32 − 21.02 = 53.3. D skipped both corrections: 51.78 ÷ 0.800 = 64.7, the ideal pressure.

Extra practice 1 answer: D

High pressure crowds molecules so their volume and mutual attraction both count; low temperature slows them so attractions take hold. A is the opposite extreme, far apart and fast, where the gas is nearly ideal. B and C each meet only one of the two conditions, so their deviations are milder.

Extra practice 2 answer: B

The constant b is the volume excluded per mole, so 3.00 mol exclude 3.00·b, not b. A flipped the sign of the attraction term: attractions lower the measured pressure, so a·n²/V² is subtracted. C dropped the squares: the term is a·n²/V², which has units of atm. D flipped the volume correction: molecules fill space, so they take volume away.

Extra practice 3 answer: A

B left n out of nRT: 35.7 ÷ (0.08206 × 310.15) = 1.40. C used the Celsius temperature: 35.7 ÷ (1.50 × 0.08206 × 37) = 7.84. D flipped the ratio: 38.18 ÷ 35.7 = 1.07, which is nRT/PV.

Extra practice 4 answer: D

A skipped both corrections: 88.62 ÷ 4.00 = 22.2, the ideal pressure. B subtracted the a term but left V uncorrected: 22.16 − 3.65 = 18.5. C added the a term instead of subtracting it: 23.13 + 3.65 = 26.8.