Guided example: solution

One conversion factor is needed.

Step 2 · Pick the factor that cancels its unit

The given unit is mmHg, so mmHg goes in the denominator: 1 atm over 760 mmHg. The wanted unit, atm, goes on top.

Guided example: solution

atm survives after one factor, so the conversion stops here.

Guided example: solution

Guided example: the route on the map

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.

Practice 1 answer: C

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: A

B flipped the second factor: 125 ÷ 14.7 ÷ 101.325 = 0.0839. C used mercury's 760 on the second arrow: 125 ÷ 14.7 × 760 = 6.46 × 10³, a count in mmHg. D stopped mid-plan: 8.50 is still atm.

Practice 2 answer: A

Practice 3 answer: D

A stopped mid-plan: 0.555 is still atm. B read pascals as kilopascals: 5.62 × 10⁴ ÷ 101.325 × 760 = 4.22 × 10⁵. C used the psi number on the second arrow: 5.62 × 10⁴ ÷ 101,325 × 14.7 = 8.15. E flipped the second factor: 5.62 × 10⁴ ÷ 101,325 ÷ 760 = 7.30 × 10⁻⁴.

Practice 3 answer: D

Practice 4 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.

Practice 4: the route on the map

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: C

A stopped mid-plan: 195 ÷ 14.7 = 13.3 is still atm. B used the kilopascal number: 195 ÷ 14.7 × 101.325 = 1.34 × 10³, a count in kPa, not Pa. D flipped the second factor: 195 ÷ 14.7 ÷ 101,325 = 1.31 × 10⁻⁴. E used mercury's 760 on the second arrow: 195 ÷ 14.7 × 760 = 1.01 × 10⁴, a count in mmHg.

Extra practice 2 answer: C

Extra practice 3 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 3 answer: A

Guided example: solution

Step 1 · Confirm STP

STP is 0 °C and 1 atm. The pressure is in mmHg, and 760 mmHg = 1 atm. Both conditions match, so 22.4 L/mol applies.

Guided example: solution

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

Guided example: solution

The given is moles, so moles go underneath: 22.4 L over 1 mol.

Guided example: solution

Guided example: solution

Guided example: the route on the map

Worked example 1: 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 1: 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 1: solution

Worked example 1: solution

Worked example 1: the route on the map

Practice 1 answer: B

A ran the grams chain on a mole amount: 0.715 / 2.016 × 22.4 = 7.94, as if 0.715 were grams. C flipped the molar volume: 0.715 / 22.4 = 0.0319, in mol²/L. D reached for the molar mass: 0.715 × 2.016 = 1.44, a mass in grams.

Practice 1: the route on the map

Practice 2 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.

Practice 2: the route on the map

Worked example 2: 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 2: solution

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

Worked example 2: solution

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

Worked example 2: solution

Worked example 2: the route on the map

Worked example 3: solution

Two conversion factors are needed.

Step 2 · Map the route

L CH₄ → mol → g. Molar volume covers the first arrow; molar mass covers the second.

Worked example 3: solution

Liters cancel with the molar volume's liters underneath. Moles cancel with the molar mass's mole underneath.

Worked example 3: solution

The mass chain starts from the given liters, so no rounded value enters it.

Worked example 3: solution

Worked example 3: the route on the map

Practice 3 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.

Practice 3: the route on the map

Worked example 4: solution

Two conversion factors are needed.

Step 2 · Map the route

One mole sits on both bridges: 1 mol O₂ = 32.00 g, and 1 mol O₂ = 22.4 L.

Worked example 4: solution

Density puts grams on top and liters underneath. The moles cancel between the two factors.

Worked example 4: solution

Worked example 4: solution

Worked example 4: the route on the map

Practice 4 answer: C

A counted one oxygen: NO is 14.01 + 16.00 = 30.01 g/mol, and 30.01 / 22.4 = 1.34. B flipped the density: 22.4 / 46.01 = 0.487, in L/g. D stopped at the molar mass: 46.0 g is the mass of a mole, not of a liter.

Practice 4: the route on the map

Practice 5 answer: E

A stopped at the density: 1.108 / 0.350 = 3.17 g/L. B used the filled bulb, glass and all: 36.570 / 0.01563 = 2.34 × 10³. C skipped the volume, as if the bulb held 1 L: 1.108 × 22.4 = 24.8. D flipped the density: 0.350 / 1.108 × 22.4 = 7.08.

Practice 5: the route on the map

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 divided the molar volume by the density: 22.4 / 1.78 = 12.6, units that never reduce to g/mol. B flipped the molar volume: 1.78 / 22.4 = 0.0795, in g·mol/L². C took the density for the molar mass: 1.78 g is the mass of one liter, not of one mole.

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.

Guided example: pick the law

Step 1 · Identify given and wanted

Guided example: pick the law

Guided example: pick the law

Guided example: solve and check

Step 3 · Rearrange for the unknown

Move 3: multiply both sides by T₂. The new temperature lands on top.

Guided example: solve and check

Guided example: solve and check

Guided example: the route on the chart

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

Practice 1: answer B

A inverted the volume ratio: 745 × 60.0/36.0 = 1.24 × 10³, a pressure that rose as the gas expanded. C multiplied P₁ by V₁ and stopped: 745 × 36.0 = 2.68 × 10⁴. D swapped P₁ and V₂ when rearranging, P₂ = V₁V₂/P₁: 36.0 × 60.0/745 = 2.90, in mL²/mmHg, not a pressure.

Practice 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 2 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.

Practice 2: the route on the chart

Practice 3: answer D

A stopped at kelvin: 257 K is the right temperature on the wrong scale. B put Celsius into the ratio: 23.0 × 2.95/3.40 = 20.0 °C. C inverted the volume ratio: 296.15 × 3.40/2.95 = 341 K = 68 °C, a gas that warmed as it shrank.

Practice 3: the route on the chart

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 4: answer A

B left the temperatures in Celsius: 2.80 × 0.400 × (37.0/10.0) = 4.14. C inverted the volume ratio: 2.80 × (1.50/0.600) × (310.15/283.15) = 7.67. D inverted the temperature ratio: 2.80 × 0.400 × (283.15/310.15) = 1.02. E held temperature constant: 2.80 × 0.400 = 1.12.

Practice 4: the route on the chart

Practice 5 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 5: the route on the chart

Practice 6: 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 7: 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 8: 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.

Guided example: solution

Step 1 · List the pieces

n = 0.250 mol and P = 1.00 atm are already in R's units. Only the temperature converts: T = 20.0 + 273.15 = 293.15 K. The unknown is V.

Guided example: solution

V is multiplied by P. Divide both sides by P to leave V alone.

Guided example: solution

Guided example: the route on the map

Practice 1 answer: C

A reported the temperature in Celsius: 682 − 273.15 = 409. B flipped the rearrangement: (0.0150 × 0.08206) ÷ (2.40 × 0.350) = 1.47 × 10⁻³. D left n out: 2.40 × 0.350 ÷ 0.08206 = 10.2.

Practice 1: the route on the map

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 2 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.

Practice 2: the route on the map

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 3 answer: A

B stopped at moles: 0.607. C divided by the molar mass: 0.607 ÷ 20.18 = 0.0301. D left the temperature in Celsius: 3.20 × 4.50 ÷ (0.08206 × 16.0) = 10.97 mol, × 20.18 = 221.

Practice 3: the route on the map

Practice 4 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 4: the route on the map

Practice 5 answer: B

A skipped the molar mass: 8.50 × 0.08206 × 297.15 ÷ 0.962 = 215. C left the pressure in kPa: 0.499 × 0.08206 × 297.15 ÷ 97.5 = 0.125. D used 22.4 L/mol, which holds only at STP: 0.499 × 22.4 = 11.2. E left the temperature in Celsius: 0.499 × 0.08206 × 24.0 ÷ 0.962 = 1.02.

Practice 5: the route on the map

Practice 6 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 7 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 8 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.

Guided example: solution

Step 1 · List every gas

Two gases fill the cylinder: nitrogen and oxygen.

Guided example: solution

P_total = P(N₂) + P(O₂). The law adds pressures, so both must carry one unit.

Guided example: solution

Two moves are needed. Move 1: put nitrogen in atm.

Guided example: solution

Two moves are needed. Move 1: put nitrogen in atm.

Guided example: solution

Two moves are needed. Move 1: put nitrogen in atm.

Guided example: the route on the map

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.

Practice 1 answer: A

Practice 2 answer: B

A split the total into equal shares: 4940 ÷ 2 = 2470 mmHg, as if the two gases were present in equal amounts. C added oxygen instead of removing it: 4940 + 1330 = 6270 mmHg. D stopped at the total: 4940 mmHg is the whole mixture, not helium.

Practice 2 answer: B

Practice 3 answer: D

A split the total three ways: 720 ÷ 3 = 240 mmHg, but the three amounts differ. B found the first-listed gas: 1.20 ÷ 1.80 × 720 = 480 mmHg is nitrogen's share, not oxygen's. C divided by nitrogen's moles, not the total: 0.45 ÷ 1.20 × 720 = 270 mmHg.

Practice 3 answer: D

Worked example 3: solution

Step 1 · List every gas

Only the gases count: ammonia and hydrogen sulfide. The ammonium sulfide is a solid, and it is gone.

Worked example 3: solution

P(NH₃) + P(H₂S) = 867.0 mmHg. The coefficients fix the amounts: of every 3 mol of gas, 2 mol are NH₃ and 1 mol is H₂S.

Worked example 3: solution

Worked example 3: solution

Worked example 3: the route on the map

Practice 4 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 4 answer: C

Worked example 4: solution

Step 1 · List every gas

Hydrogen and nitrogen share the flask, at the same V and T.

Worked example 4: solution

P_total = P(H₂) + P(N₂). Each partial is nRT ÷ V with the same R, T, and V, so the total is n_totalRT ÷ V.

Worked example 4: solution

Three moves are needed: grams to moles, add the moles, then PV = nRT.

Worked example 4: solution

Three moves are needed: grams to moles, add the moles, then PV = nRT.

Worked example 4: solution

Three moves are needed: grams to moles, add the moles, then PV = nRT.

Worked example 4: check

Step 4 · Check

Each gas alone, in the same flask:

Worked example 4: check

Step 4 · Check

Each gas alone, in the same flask:

Worked example 4: the route on the map

Practice 5 answer: B

A stopped at helium: 1.999 × 0.08206 × 291.15 ÷ 12.0 = 3.98 atm is helium's partial pressure alone. C put grams in for moles: 40.0 × 0.08206 × 291.15 ÷ 12.0 = 79.6 atm. D left T in Celsius: 2.999 × 0.08206 × 18.0 ÷ 12.0 = 0.369 atm.

Practice 5 answer: B

Practice 6 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 6 answer: D

Practice 7 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 7 answer: A

Practice 8 answer: C

Two moves are needed. Move 1: total moles from the molar volume.

Practice 8 answer: C

Practice 8 answer: C

Practice 8 answer: C

Guided example: solution

Step 1 · Level and read

Move 1. The levels match, so the bottle pushes exactly as hard as the air: P_total = 745.0 torr. That reading holds O₂ and water vapor together.

Guided example: solution

Move 2. The water's temperature picks the row: 18 °C gives 15.5 torr. The gas and the barometer play no part.

Guided example: solution

Move 3. Take the vapor's share out of the total.

Guided example: the route on the map

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

Your turn: the route on the map

Practice 1 answer: B

A skipped the vapor: 774.0 torr is hydrogen and water vapor together. C subtracted from the standard atmosphere, not today's reading: 760 − 25.2 = 734.8 torr. D added the vapor: 774.0 + 25.2 = 799.2 torr, a part larger than the whole.

Practice 1: the route on the map

Practice 2 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: the route on the map

Practice 3 answer: C

A added the vapor: (739 + 28.3) ÷ 760 = 1.01 atm. B converted the vapor instead of the gas: 28.3 ÷ 760 = 0.0372 atm is water's share. D skipped the subtraction: 739 ÷ 760 = 0.972 atm counts the vapor as nitrogen.

Practice 3: the route on the map

Practice 4 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 4: the route on the map

Practice 5 answer: A

Four moves come before the gas law: subtract, then match three units.

Practice 5 answer: A

B kept mmHg: 729.2 × 0.385 ÷ 23.6 = 11.9. C used 15 °C for T: 0.300. D used the wet 742 mmHg, 0.976 atm: 0.0159. E kept 385 mL as 385: 15.6.

Practice 5: the route on the map

Practice 6 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 7 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 8 solution: moles of butane

Four moves: subtract, match units, solve for moles, then convert to grams.

Practice 8 solution: moles of butane

Practice 8 answer: D

A used the wet 744 torr: 1.01 g. E added the vapor, 760.5 torr: 1.03 g. B used 19 °C for T: 15.2 g. C kept torr: 727.5 × 0.425 ÷ 23.97 × 58.12 = 750.

Guided example: solution

Two conversion factors are needed.

Step 1 · To moles

The given is already in moles: 0.400 mol HgO. No molar mass is needed.

Guided example: solution

The mole ratio, written O₂ over HgO (1 : 2), crosses substances; mol HgO cancels:

Guided example: solution

At STP the molar volume converts moles of O₂ to liters. The complete chain:

Guided example: solution

Guided example: the route on the map

Practice 1 answer: C

A inverted the ratio: 4.50 × (2/1) = 9.00. B assumed a 1 : 1 volume ratio: 4.50 L. D divided by 22.4: 2.25 ÷ 22.4 = 0.100 would be moles, but the stem never says STP and the question asks for liters.

Practice 1: the route on the map

Practice 2 answer: B

A read CO₂'s coefficient: 0.250 × (3/1) = 0.750. C inverted the ratio: 0.250 × (1/4) = 0.0625. D skipped the mole ratio: 5.60 ÷ 22.4 = 0.250 mol is the propane, not the water.

Practice 2: the route on the map

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 3 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.

Practice 3: the route on the map

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 4 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.

Practice 4: the route on the map

Practice 5 answer: D

A skipped the mole ratio: 0.0300 × 122.55 = 3.68. B stopped at moles: 0.0300 × (2/3) = 0.0200 mol KClO₃. C inverted the ratio: 0.0300 × (3/2) × 122.55 = 5.51. E skipped mL → L: 672 ÷ 22.4 × (2/3) × 122.55 = 2.45 × 10³.

Practice 5: the route on the map

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: A

B skipped the mole ratio: 2.25 × 23.95 = 53.9. C flipped the molar volume: 50.4 × 22.4 × 2 × 23.95 = 5.41 × 10⁴; liters on top do not cancel. D stopped at moles: 2.25 × 2 = 4.50 mol LiOH. E inverted the ratio: 2.25 × (1/2) × 23.95 = 26.9.

Worked example 1: solution

Step 1 · Convert T to kelvin

Worked example 1: solution

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 1: the route on the strip

Practice 2 answer: A

B treated the balloon as a rigid tank. Its wall is flexible, so it shrinks until the inside hits balance the outside air again. C kept the speed fixed, but KE follows T: colder particles move slower. The particles themselves never shrink. D held everything fixed because no gas left. The particles stay, but they slow down.

Practice 2: the route on the strip

Practice 3 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: the route on the strip

Practice 4 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: the route on the strip

Worked example 3: energy and speed

Step 1 · Convert T to kelvin

Worked example 3: energy and speed

Worked example 3: energy and speed

Worked example 3: energy and speed