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

Practice 1 — answer: C

A inverted the factor: 88.2 × 14.7 = 1.30 × 10³ psi²/atm, and no unit cancels. 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

Practice 2 — answer: D

A stopped mid-plan: 570 / 760 = 0.750 atm, not kPa. B inverted the kPa factor: 0.750 / 101.325 = 7.40 × 10⁻³. C used psi's 14.7 in place of kPa's 101.325: 0.750 × 14.7 = 11.0.

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 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

Practice 1 — answer: B

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. D 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

Practice 2 — answer: C

A put grams straight onto the molar volume, skipping molar mass: 34.0 × 22.4 = 762 L. B left the molar volume upside down: 34.0 / 17.03 / 22.4 = 0.0891. D stopped at moles: 34.0 / 17.03 = 2.00 mol, a count, not a volume.

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 2 — solution

Step 1 · Identify given and wanted

V₁ = 3.0 L. Pressure is held constant, so P₁ = P₂. The unknown is V₂.

Worked example 2 — solution

T₁ = 25 + 273.15 = 298.15 K. T₂ = 75 + 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

Practice 1 — answer: C

A left the temperatures in Celsius and inverted them: 4.0 × (27/127) = 0.850 L. B inverted the kelvin ratio: 4.0 × (300.15/400.15) = 3.00 L, a shrinking gas though it was heated. D left the temperatures in Celsius: 4.0 × (127/27) = 18.8 L, a ratio that far overstates a 100-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 = 2.0 L, P₁ = 1.0 atm, P₂ = 0.40 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

Practice 2 — answer: B

A inverted the pressure ratio: 3.0 × (0.80/1.2) × (350.15/293.15) = 2.39 L, a shrinking gas though the pressure dropped. C inverted the temperature ratio: 3.0 × (1.2/0.80) × (293.15/350.15) = 3.77 L. D left the temperatures in Celsius: 3.0 × (1.2/0.80) × (77/20) = 17.3 L.

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 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

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

Practice 2 — answer: B

32.1 g/mol matches O₂. A left the temperature in Celsius: 4.00 × 0.08206 × 20.0 ÷ 3.00 = 2.19 g/mol. C dropped the volume: 4.00 × 0.08206 × 293.15 ÷ 1.00 = 96.2 g/mol. D stopped at the moles: n = PV/RT = 0.125 mol, before dividing the mass by it.

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 2 — solution

Step 1 · List every gas

Nitrogen and oxygen fill the cylinder. No water is present, so 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

Practice 1 — answer: B

A split the total three ways: 1.25 ÷ 3 = 0.42 atm, which assumes equal amounts. C added the two known pressures: 0.50 + 0.20 = 0.70 atm, the combined push of nitrogen and oxygen. D removed only nitrogen: 1.25 − 0.50 = 0.75 atm, leaving oxygen's share inside.

Worked example 3 — solution

A common first attempt

No single gas can push harder than the whole mixture. The vapor was added when it should be removed.

Worked example 3 — solution

The cylinder holds oxygen and water vapor. Together they make the 761 torr the barometer reads.

Worked example 3 — solution

The wet total splits into the dry gas and the vapor: 761 torr = P(O₂) + 24 torr.

Worked example 3 — the dry oxygen

Step 3 · Find the unknown

Worked example 3 — the dry oxygen

Practice 2 — answer: B

A added the vapor instead of subtracting: 745 + 20 = 765 torr, a part larger than the whole. C used the standard 760 torr instead of the measured 745: 760 − 20 = 740 torr. D reported the barometer reading itself, 745 torr, forgetting the gas came out wet.

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 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

Practice 1 — answer: B

A skipped the mole ratio: 8.10/26.98 × 22.4 = 6.72. C stopped at moles: 8.10/26.98 × 3/2 = 0.450 mol, not liters. D put grams into the ratio: 8.10 × 3/2 × 22.4 = 272.

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

The molar volume is 22.4 L/mol only at STP. At 27 °C each mole spreads out farther, so the shortcut undercounts.

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):

Practice 2 — answer: B

A used 22.4 L/mol off STP: 0.198 × 22.4 = 4.43. C skipped the mole ratio: 0.396 mol KNO₃ into PV = nRT gives 13.0 L. D forgot to convert to kelvin: (0.198)(0.08206)(127) = 2.06.