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Combined Gas Law Problems — 12 Worked Examples with Step-by-Step Solutions

Combined Gas Law Problems — 12 Worked Examples with Step-by-Step Solutions

Combined Gas Law Problems — Practice Problems with Step-by-Step Solutions

12 fully worked combined gas law practice problems — Boyle’s law, Charles’s law, Gay-Lussac’s law, the combined gas law, and the ideal gas law PV = nRT — with every temperature conversion shown in full

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▌ COMBINED GAS LAW — QUICK REFERENCE
Boyle’s Law: P₁V₁ = P₂V₂ (T and n constant) Charles’s Law: V₁/T₁ = V₂/T₂ (P and n constant) Gay-Lussac’s Law: P₁/T₁ = P₂/T₂ (V and n constant) Combined Gas Law: P₁V₁/T₁ = P₂V₂/T₂ (n constant) Ideal Gas Law: PV = nRT (full equation)

R = 0.082057 L·atm/(mol·K) ← chemistry R = 8.31446 L·kPa/(mol·K) ← SI metric R = 62.364 L·mmHg/(mol·K) ← barometric
K = °C + 273.15 0°C = 273.15 K 25°C = 298.15 K 100°C = 373.15 K −73°C = 200.15 K

Temperature constant → Boyle’s Law Pressure constant → Charles’s Law Volume constant → Gay-Lussac’s Law None constant, n fixed → Combined Gas Law Need moles → Ideal Gas Law

1 atm = 101.325 kPa = 760 mmHg = 14.696 psi 1 L = 1000 mL = 1000 cm³ P↑ → V↓ (Boyle’s Law) T↑ → V↑ (Charles’s Law)

What Is an Ideal Gas? — Definition and Assumptions

An ideal gas is a theoretical gas that obeys the gas laws — including the ideal gas law PV = nRT — perfectly under all conditions. The ideal gas definition in chemistry is: a gas whose molecules occupy negligible volume and exert no intermolecular forces on each other. Real gases approximate ideal gas behavior at low pressures and high temperatures, where molecules are far apart and kinetic energy greatly exceeds intermolecular attraction.

The ideal gas model underlies every formula on this combined gas law worksheet. Understanding what makes a gas ideal — and when a real gas stops behaving ideally — is essential for knowing when gas law formulas give accurate results and when corrections (such as the Van der Waals equation) are needed.

The Five Assumptions of an Ideal Gas

The assumptions of an ideal gas define the conditions under which the combined gas law and ideal gas law apply exactly. All five assumptions must hold for a gas to behave ideally:

1

Negligible Molecular Volume

Gas molecules are treated as point masses — their physical size is negligible compared to the total volume of the container. This assumption of the ideal gas breaks down at very high pressures, where molecules are packed closely and their own volume becomes a significant fraction of the container volume.

2

No Intermolecular Forces

There are no attractive or repulsive forces between gas molecules. This ideal gas assumption fails for polar molecules (NH₃, H₂O) and molecules with large electron clouds (CO₂) that have significant London dispersion forces. Non-ideal gases deviate most at low temperatures where intermolecular forces dominate.

3

Perfectly Elastic Collisions

All collisions — between molecules and between molecules and container walls — are perfectly elastic, meaning no kinetic energy is lost. This assumption of the ideal gas ensures total kinetic energy remains constant, consistent with constant temperature at the molecular level.

4

Constant Random Motion

Molecules move in constant, random motion in all directions with no preferred orientation. This is why gas fills any container uniformly and exerts equal pressure on all walls — a key ideal gas property that makes pressure isotropic and directly calculable from P = F/A.

5

Kinetic Energy ∝ Temperature (Kelvin)

The average kinetic energy of gas molecules is directly proportional to absolute temperature in Kelvin. This is why temperature must always be in Kelvin in all gas law equations — at 0 K (absolute zero), molecules have zero kinetic energy and zero volume, making Kelvin the only physically meaningful zero for gas calculations.

Ideal Gas Examples and Non-Ideal Gases

Most ideal real gases: Helium (He), Neon (Ne), Hydrogen (H₂), Nitrogen (N₂) at room temperature and low pressure. Noble gases are the most ideal because they are monatomic with minimal intermolecular forces — the definition of ideal gas behavior in practice.

Non-ideal gases: CO₂, water vapor (H₂O), and ammonia (NH₃) deviate significantly from the ideal gas definition because they have strong intermolecular forces. High pressure (molecules close together) and low temperature (near condensation) make any gas non-ideal. The difference between an ideal gas and a non-ideal gas is largest near the gas’s condensation point.

Rule of thumb for gas law problems: Gas law formulas give accurate results when P < 10 atm and T > 200 K for most gases. Beyond these limits, use the Van der Waals equation to correct for non-ideal behavior.

Gas Law Formulas — Which One to Use?

Every combined gas law problem begins with identifying which of the four gas law formulas applies. The choice depends entirely on which variable is held constant. Use this table as your first step in any gas law practice problem:

Situation Gas Law Formula What’s Constant
Pressure changes, temperature constant P₁V₁ = P₂V₂ (Boyle’s law) T, n
Volume changes, pressure constant V₁/T₁ = V₂/T₂ (Charles’s law) P, n
Pressure changes, volume constant P₁/T₁ = P₂/T₂ (Gay-Lussac’s law) V, n
All three variables change P₁V₁/T₁ = P₂V₂/T₂ (Combined gas law) n only
Need moles or mass PV = nRT (Ideal gas law)

The combined gas law contains all three individual laws:

Set T₁ = T₂ in the combined gas law → Boyle’s law (P₁V₁ = P₂V₂) emerges automatically.

Set P₁ = P₂ in the combined gas law → Charles’s law (V₁/T₁ = V₂/T₂) emerges.

Set V₁ = V₂ in the combined gas law → Gay-Lussac’s law (P₁/T₁ = P₂/T₂) emerges.

This is why the combined gas law P₁V₁/T₁ = P₂V₂/T₂ is the master formula — when in doubt, use it and cancel the constant variable. The gas chemical equation PV = nRT (the ideal gas law) is the further generalization that includes the number of moles.

Boyle’s Law Problems — Pressure and Volume at Constant Temperature

Boyle’s law states that for a fixed amount of gas at constant temperature, pressure and volume are inversely proportional: P₁V₁ = P₂V₂. When pressure increases, volume decreases proportionally, and vice versa. Boyle’s law problems are the only gas law problems where temperature conversion to Kelvin is not required (temperature doesn’t appear in the formula) — but temperature must still be verified as constant.

1
Basic Boyle’s Law — Finding New Volume Boyle’s Law Easy
A gas occupies 4.0 L at a pressure of 2.0 atm. If the pressure is increased to 5.0 atm at constant temperature, what is the new volume?
Identify — Temperature constant → Boyle’s law: P₁V₁ = P₂V₂
Given: P₁ = 2.0 atm V₁ = 4.0 L P₂ = 5.0 atm V₂ = ? (unknown)
Step 1 — Rearrange Boyle’s law for V₂
V₂ = P₁V₁ / P₂
Step 2 — Substitute values
V₂ = (2.0 atm)(4.0 L) / (5.0 atm) V₂ = 8.0 / 5.0
Step 3 — Calculate
V₂ = 1.6 L
Step 4 — Check reasonableness
Pressure increased (2.0 → 5.0 atm) → Volume should DECREASE ✓ (4.0 → 1.6 L) Verify: P₁V₁ = 2.0 × 4.0 = 8.0 P₂V₂ = 5.0 × 1.6 = 8.0 ✓
✓ V₂ = 1.6 L
2
Scuba Tank Air Supply — Boyle’s Law Real-World Application Boyle’s Law Medium
A scuba tank contains air at 200 atm and occupies 12 L. A diver breathes this air at a depth where the pressure is 3.0 atm. Assuming constant temperature, what volume of air does the tank provide at this depth?
Identify — Temperature constant → Boyle’s law: P₁V₁ = P₂V₂
Given: P₁ = 200 atm (tank pressure) V₁ = 12 L (tank volume) P₂ = 3.0 atm (pressure at depth) V₂ = ?
Step 1 — Rearrange Boyle’s law for V₂
V₂ = P₁V₁ / P₂
Step 2 — Substitute and calculate
V₂ = (200)(12) / 3.0 V₂ = 2400 / 3.0 V₂ = 800 L of air at 3.0 atm
Step 3 — Real-world context
At 3.0 atm (≈ 20 m depth), the 12 L tank provides 800 L of breathable air. Diver breathing ~15 L/min → 800 / 15 ≈ 53 minutes of air supply.
✓ V₂ = 800 L at 3.0 atm
Why scuba tanks “run out” faster at depth: Boyle’s law shows that at higher pressure (greater depth), each breath drawn from the tank removes more mass of gas per litre exhaled. A diver at 30 m (4 atm) uses air four times faster than at the surface. The combined gas law and Boyle’s law are therefore central to dive planning and tank capacity calculations.

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Charles’s Law Problems — Volume and Temperature at Constant Pressure

Charles’s law states that for a fixed amount of gas at constant pressure, volume is directly proportional to absolute temperature: V₁/T₁ = V₂/T₂. Temperature must always be in Kelvin — using Celsius produces answers that can be wrong by a factor of 3 or more, as Problem 3 demonstrates explicitly. This is the most critical rule in all Charles’s law problems.

3
Balloon Heating — Charles’s Law with Kelvin Conversion Charles’s Law Easy
A balloon has a volume of 3.0 L at 27°C. What volume will it occupy at 127°C if pressure remains constant?
Identify — Pressure constant → Charles’s law: V₁/T₁ = V₂/T₂
⚠ CRITICAL: CONVERT TEMPERATURE TO KELVIN FIRST — NEVER USE CELSIUS IN GAS LAW EQUATIONST₁ = 27°C + 273.15 = 300.15 K ≈ 300 K T₂ = 127°C + 273.15 = 400.15 K ≈ 400 K
Given values (after Kelvin conversion)
V₁ = 3.0 L T₁ = 300 K ← converted from 27°C T₂ = 400 K ← converted from 127°C V₂ = ?
Step 1 — Rearrange Charles’s law for V₂
V₂ = V₁ × T₂/T₁
Step 2 — Substitute (using Kelvin temperatures)
V₂ = 3.0 × (400 K / 300 K) V₂ = 3.0 × 1.333
Step 3 — Calculate
V₂ = 4.0 L Check: Temperature ratio = 400/300 = 4/3 Volume ratio = 4.0/3.0 = 4/3 ✓ (directly proportional as Charles’s law predicts)
✓ V₂ = 4.0 L
❌ Most Common Mistake in Charles’s Law Problems — Using Celsius Instead of Kelvin
❌ WRONG — Using CelsiusV₂ = 3.0 × (127/27) V₂ = 3.0 × 4.704 V₂ = 14.1 L ← WRONG (3.5× too large!)
✓ CORRECT — Using KelvinV₂ = 3.0 × (400/300) V₂ = 3.0 × 1.333 V₂ = 4.0 L ← CORRECT Temperature must be in Kelvin.

The Celsius error produces a number (14.1 L) that looks plausible — it has units, it’s positive — but it is 3.5× wrong. There is no calculation error to spot. The only protection is the habit of converting to Kelvin before every single substitution, every single time.

4
Finding Final Temperature — Charles’s Law Rearranged Charles’s Law Medium
A gas occupies 500 mL at −73°C. At what Celsius temperature will it occupy 750 mL at the same pressure?
Identify — Pressure constant → Charles’s law: V₁/T₁ = V₂/T₂
⚠ CRITICAL: CONVERT TEMPERATURE TO KELVIN FIRST — NEVER USE CELSIUS IN GAS LAW EQUATIONST₁ = −73°C + 273.15 = 200.15 K ≈ 200 K Note: Negative Celsius temperatures are common in Charles’s law problems. A negative Celsius temperature is still a positive Kelvin value — never use a negative number directly in a Charles’s law or combined gas law equation.
Given values (after Kelvin conversion)
V₁ = 500 mL T₁ = 200 K ← converted from −73°C V₂ = 750 mL T₂ = ? (solve in K, then convert back to °C)
Step 1 — Rearrange Charles’s law for T₂
T₂ = T₁ × V₂/V₁
Step 2 — Substitute
T₂ = 200 K × (750 mL / 500 mL) T₂ = 200 × 1.5
Step 3 — Calculate
T₂ = 300 K
Step 4 — Convert back to Celsius
T₂ = 300 K − 273.15 = 26.85°C ≈ 27°C
✓ T₂ = 300 K = 27°C
Check: Volume increased (500 → 750 mL) → Temperature must increase ✓ (200 K → 300 K). The volume ratio (750/500 = 1.5) equals the temperature ratio (300/200 = 1.5), confirming Charles’s law: volume is directly proportional to absolute temperature.

Gay-Lussac’s Law Problems — Pressure and Temperature at Constant Volume

Gay-Lussac’s law states that for a fixed amount of gas in a rigid container (constant volume), pressure is directly proportional to absolute temperature: P₁/T₁ = P₂/T₂. Gay-Lussac’s law applies whenever the problem mentions a “rigid container,” “sealed vessel,” or “constant volume.” Temperature must always be in Kelvin — the same rule applies to Gay-Lussac’s law as to every other gas law formula.

5
Rigid Container Heating — Basic Gay-Lussac’s Law Gay-Lussac’s Law Easy
A gas in a rigid container has a pressure of 1.50 atm at 25°C. What pressure will the gas exert at 75°C?
Identify — Volume constant (rigid container) → Gay-Lussac’s law: P₁/T₁ = P₂/T₂
⚠ CRITICAL: CONVERT TEMPERATURE TO KELVIN FIRST — NEVER USE CELSIUS IN GAS LAW EQUATIONST₁ = 25°C + 273.15 = 298.15 K T₂ = 75°C + 273.15 = 348.15 K
Given values
P₁ = 1.50 atm T₁ = 298.15 K ← converted from 25°C T₂ = 348.15 K ← converted from 75°C P₂ = ?
Step 1 — Rearrange Gay-Lussac’s law for P₂
P₂ = P₁ × T₂/T₁
Step 2 — Substitute and calculate
P₂ = 1.50 × (348.15 / 298.15) P₂ = 1.50 × 1.1677 P₂ = 1.752 atm ≈ 1.75 atm
✓ P₂ = 1.75 atm
Real-world application (Gay-Lussac’s law): This explains why tyre pressure increases when a car warms up after driving — the rigid tyre maintains constant volume while temperature rises, so Gay-Lussac’s law predicts an increased pressure. Always check tyre pressure when cold; never release air from hot tyres to compensate.
6
Aerosol Can Safety — Finding Critical Temperature with Gay-Lussac’s Law Gay-Lussac’s Law Medium
An aerosol can has a pressure of 1.20 atm at 20°C. At what temperature (°C) will the pressure reach 1.80 atm? (Safety concern: aerosol cans warn never to incinerate — this problem shows why.)
Identify — Volume constant → Gay-Lussac’s law: P₁/T₁ = P₂/T₂
⚠ CRITICAL: CONVERT TEMPERATURE TO KELVIN FIRST — NEVER USE CELSIUS IN GAS LAW EQUATIONST₁ = 20°C + 273.15 = 293.15 K T₂ = unknown — solve in Kelvin, then convert to °C at the end
Given values
P₁ = 1.20 atm T₁ = 293.15 K ← converted from 20°C P₂ = 1.80 atm T₂ = ?
Step 1 — Rearrange Gay-Lussac’s law for T₂
T₂ = T₁ × P₂/P₁
Step 2 — Substitute
T₂ = 293.15 × (1.80 / 1.20) T₂ = 293.15 × 1.500 T₂ = 439.7 K
Step 3 — Convert back to Celsius
T₂ = 439.7 − 273.15 = 166.6°C ≈ 167°C
✓ T₂ = 439.7 K = 167°C
Safety context: Gay-Lussac’s law reveals why aerosol cans are dangerous near heat sources. A 50% pressure increase occurs at just 167°C — well within the range of a hot car dashboard or near a campfire. Most aerosol cans are rated to fail at around twice their working pressure. Fires can reach 300–500°C, producing pressures far beyond can tolerances.

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Combined Gas Law Problems — All Three Variables Change

The combined gas law P₁V₁/T₁ = P₂V₂/T₂ applies when pressure, volume, and temperature all change simultaneously, but the amount of gas (n) remains constant. These combined gas law practice problems require careful algebraic rearrangement and temperature must always be in Kelvin for both T₁ and T₂. The combined gas law is the most versatile formula on this worksheet — when in doubt, start here.

7
All Three Variables — Standard Combined Gas Law Problem Combined Gas Law Medium
A gas sample has a pressure of 1.50 atm, volume of 4.00 L, and temperature of 300 K. The conditions change to P₂ = 2.00 atm and T₂ = 350 K. Find the new volume.
Identify — All three variables change, n constant → Combined gas law: P₁V₁/T₁ = P₂V₂/T₂
⚠ CRITICAL: TEMPERATURE MUST BE IN KELVIN — CHECK BOTH T₁ AND T₂T₁ = 300 K ✓ (already in Kelvin — no conversion needed) T₂ = 350 K ✓ (already in Kelvin — no conversion needed)
Given values
P₁ = 1.50 atm P₂ = 2.00 atm V₁ = 4.00 L V₂ = ? T₁ = 300 K T₂ = 350 K
Step 1 — Rearrange the combined gas law for V₂
P₁V₁/T₁ = P₂V₂/T₂ V₂ = P₁V₁T₂ / (T₁P₂) Note: T₂ is in the NUMERATOR and T₁ is in the DENOMINATOR. Swapping these gives 4.67 L — the most common combined gas law error.
Step 2 — Substitute values
V₂ = (1.50)(4.00)(350) / (300)(2.00) V₂ = 2100 / 600
Step 3 — Calculate
V₂ = 3.50 L
Step 4 — Verify using the combined gas law
P₁V₁/T₁ = (1.50)(4.00)/300 = 6.00/300 = 0.0200 P₂V₂/T₂ = (2.00)(3.50)/350 = 7.00/350 = 0.0200 ✓
✓ V₂ = 3.50 L
Reasonableness check: Pressure increase (1.5 → 2.0 atm) tends to decrease volume. Temperature increase (300 → 350 K) tends to increase volume. The pressure effect is slightly larger here, so the net result is a small volume decrease (4.00 → 3.50 L). Both opposing effects are visible in the combined gas law calculation.
8
Weather Balloon at Altitude — Combined Gas Law Combined Gas Law Medium
A weather balloon has a volume of 1,000 L at sea level where T = 20°C and P = 1.00 atm. At high altitude, the pressure drops to 0.25 atm and temperature falls to −40°C. What is the new volume?
Identify — All three variables change → Combined gas law: P₁V₁/T₁ = P₂V₂/T₂
⚠ CRITICAL: CONVERT BOTH TEMPERATURES TO KELVIN BEFORE ANY CALCULATIONT₁ = 20°C + 273.15 = 293.15 K T₂ = −40°C + 273.15 = 233.15 K Both temperatures are positive Kelvin values even though T₂ is −40°C.
Given values (after Kelvin conversion)
P₁ = 1.00 atm P₂ = 0.25 atm V₁ = 1000 L V₂ = ? T₁ = 293.15 K T₂ = 233.15 K
Step 1 — Rearrange combined gas law for V₂
V₂ = P₁V₁T₂ / (T₁P₂)
Step 2 — Substitute and calculate
V₂ = (1.00)(1000)(233.15) / (293.15)(0.25) V₂ = 233,150 / 73.29 V₂ = 3,181 L ≈ 3,180 L
Step 3 — Effect analysis
Pressure effect: 1.00/0.25 = 4.00× volume increase Temperature effect: 233.15/293.15 = 0.795× volume decrease Net combined effect: 4.00 × 0.795 = 3.18× volume increase ✓ (3,180 / 1,000 = 3.18 ✓)
✓ V₂ ≈ 3,180 L — the balloon expands to 3.18× its original volume
Why weather balloons are launched partially inflated: The combined gas law predicts a 3.18× volume expansion as the balloon rises. A fully inflated balloon would burst before reaching its target altitude. Meteorologists account for this using the combined gas law to calculate the launch volume needed for the balloon to reach the correct size at the target altitude.
9
Compressed STP Gas — Finding Final Pressure with Combined Gas Law Combined Gas Law Hard
A 2.50 L sample of gas at STP (0°C, 1.00 atm) is compressed to 0.800 L while the temperature is raised to 150°C. What is the final pressure?
Identify — All three change → Combined gas law: P₁V₁/T₁ = P₂V₂/T₂
⚠ CRITICAL: CONVERT BOTH TEMPERATURES TO KELVIN — STP IS 0°C = 273.15 K, NOT ZERO KELVINT₁ = 0°C + 273.15 = 273.15 K ← STP temperature T₂ = 150°C + 273.15 = 423.15 K
Given values (after Kelvin conversion)
P₁ = 1.00 atm P₂ = ? V₁ = 2.50 L V₂ = 0.800 L T₁ = 273.15 K T₂ = 423.15 K
Step 1 — Rearrange combined gas law for P₂
P₂ = P₁V₁T₂ / (V₂T₁)
Step 2 — Substitute and calculate
P₂ = (1.00)(2.50)(423.15) / (0.800)(273.15) P₂ = 1,057.9 / 218.52 P₂ = 4.84 atm
Step 3 — Convert to other pressure units
P₂ = 4.84 atm = 490.2 kPa = 3,638 mmHg = 71.2 psi
Step 4 — Effect analysis (verify)
Volume compression: V₁/V₂ = 2.50/0.800 = 3.125× pressure increase Temperature increase: T₂/T₁ = 423.15/273.15 = 1.550× pressure increase Net: 1.00 × 3.125 × 1.550 = 4.84 atm ✓
✓ P₂ = 4.84 atm = 490.2 kPa

Combined Gas Law — Finding Temperature Problems

Some combined gas law practice problems ask you to find the final temperature. The method is the same — rearrange the combined gas law for T₂ — but you must remember to convert your answer back to Celsius if the question asks for °C. Always solve for T₂ in Kelvin first, then convert.

10
Finding Required Temperature — Combined Gas Law for T₂ Combined Gas Law Medium
A gas at 2.00 atm and 15°C occupies 3.50 L. What temperature (°C) is needed to increase the volume to 5.00 L at a pressure of 1.50 atm?
Identify — Combined gas law, solving for T₂
⚠ CRITICAL: CONVERT T₁ TO KELVIN FIRST — SOLVE FOR T₂ IN KELVIN, THEN CONVERT ANSWER TO °CT₁ = 15°C + 273.15 = 288.15 K T₂ = unknown → solve in K → convert to °C at the final step
Given values
P₁ = 2.00 atm P₂ = 1.50 atm V₁ = 3.50 L V₂ = 5.00 L T₁ = 288.15 K T₂ = ?
Step 1 — Rearrange the combined gas law for T₂
T₂ = T₁ × P₂V₂ / (P₁V₁)
Step 2 — Substitute
T₂ = 288.15 × (1.50 × 5.00) / (2.00 × 3.50) T₂ = 288.15 × 7.50 / 7.00 T₂ = 288.15 × 1.0714
Step 3 — Calculate
T₂ = 308.7 K
Step 4 — Convert back to Celsius
T₂ = 308.7 − 273.15 = 35.5°C
✓ T₂ = 308.7 K = 35.5°C

Ideal Gas Law Problems — Using PV = nRT

The ideal gas law PV = nRT is used when you need to find the number of moles (n) or mass of gas. Unlike the combined gas law (which compares two states), the ideal gas law relates all four variables — pressure, volume, moles, and temperature — in a single state. Temperature must be in Kelvin, and the gas constant R must match your pressure units: R = 0.082057 L·atm/(mol·K) when P is in atm; R = 8.31446 L·kPa/(mol·K) when P is in kPa.

11
Finding Moles with PV = nRT — Ideal Gas Law Ideal Gas Law Medium
How many moles of gas are present in a 5.00 L container at 2.50 atm and 27°C?
Identify — Need moles → Ideal gas law: PV = nRT
⚠ CRITICAL: CONVERT TEMPERATURE TO KELVIN BEFORE SUBSTITUTING INTO PV = nRTT = 27°C + 273.15 = 300.15 K ≈ 300 K
Given values
P = 2.50 atm V = 5.00 L T = 300 K ← converted from 27°C R = 0.082057 L·atm/(mol·K) ← use this R because P is in atm n = ?
Step 1 — Rearrange PV = nRT for n
n = PV / (RT)
Step 2 — Calculate numerator (PV)
PV = 2.50 atm × 5.00 L = 12.50 L·atm
Step 3 — Calculate denominator (RT)
RT = 0.082057 × 300 = 24.617 L·atm/mol
Step 4 — Divide
n = 12.50 / 24.617 n = 0.5078 mol ≈ 0.508 mol Unit check: [L·atm] / [L·atm/mol] = mol ✓
Step 5 — Number of molecules
N = n × Nₐ = 0.508 × 6.022×10²³ = 3.06×10²³ molecules
✓ n = 0.508 mol ≈ 3.06×10²³ molecules
⚠ Common Mistake — Wrong R value Using R = 8.314 L·kPa/(mol·K) when pressure is given in atm gives: n = 12.50/(8.314 × 300) = 0.00501 mol — 101.325× too small. Always match R to your pressure units. For P in atm, use R = 0.082057 L·atm/(mol·K). For P in kPa, use R = 8.31446 L·kPa/(mol·K).

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Challenge Problem — Multi-Step Gas Law

This challenge combined gas law problem requires applying three different gas law formulas sequentially. It represents the highest level of combined gas law practice problems — multi-step problems where the output of one step becomes the input of the next. Identify each step independently before calculating.

12
Three-Step Gas Law — Cooling, Escape, Heating Combined Gas Law Hard
A 10.0 L cylinder at 25°C contains gas at 5.00 atm. The cylinder is cooled to −25°C, then a valve opens allowing gas to escape until pressure is 2.00 atm. Finally, the remaining gas is heated back to 25°C in a flexible container. What is the final volume?
Strategy: This multi-step problem applies three different gas laws sequentially. Identify each sub-problem independently: Step A = constant volume cooling (Gay-Lussac’s law); Step B = gas escape at constant T and V; Step C = constant pressure heating (Charles’s law).
⚠ CRITICAL: CONVERT ALL TEMPERATURES TO KELVIN BEFORE BEGINNING ANY STEPT₁ = 25°C + 273.15 = 298 K (initial and final temperature) T_A = −25°C + 273.15 = 248 K (cold temperature after cooling)
Step A — Cool from 25°C to −25°C in rigid cylinder (constant volume → Gay-Lussac’s law)
P₁/T₁ = P_A/T_A P_A = P₁ × T_A/T₁ = 5.00 × (248/298) = 5.00 × 0.8322 P_A = 4.161 atm After Step A: P = 4.161 atm, V = 10.0 L, T = 248 K
Step B — Valve opens, gas escapes until P = 2.00 atm (T and V unchanged at 248 K, 10.0 L)
Gas escapes — this is NOT a gas law rearrangement. The remaining gas occupies V = 10.0 L at P = 2.00 atm, T = 248 K. Fraction of gas remaining = P_final / P_A = 2.00 / 4.161 = 0.4806 (≈ 48% of original gas remains) After Step B: P = 2.00 atm, V = 10.0 L, T = 248 K
Step C — Heat remaining gas from −25°C to 25°C in flexible container (constant pressure → Charles’s law)
V_B/T_B = V_f/T_f V_B = 10.0 L, T_B = 248 K, T_f = 298 K V_f = V_B × T_f/T_B = 10.0 × (298/248) = 10.0 × 1.2016 V_f = 12.02 L ≈ 12.0 L
Summary of all three steps
Step A (Gay-Lussac): P: 5.00 → 4.16 atm (cooling at const. volume) Step B (Gas escape): P: 4.16 → 2.00 atm (48% of gas remains) Step C (Charles’s): V: 10.0 → 12.0 L (heating at const. pressure)
✓ Final volume ≈ 12.0 L (48% of the original gas, expanded by heating)

Gas Law Assumptions and When They Break Down

The combined gas law and ideal gas law give accurate results only when the assumptions of an ideal gas approximately hold. Understanding when these gas law assumptions fail — and why — tells you when to distrust your calculated answer and apply corrections.

High pressure (P > 10 atm): Gas molecules are close together. Their actual volume becomes a significant fraction of the container, and intermolecular attractive forces matter. Both ideal gas model assumptions (negligible volume, no intermolecular forces) break down. The Van der Waals equation adds correction terms for both effects.

Low temperature (near condensation point): Kinetic energy decreases, and intermolecular attractive forces become dominant. The gas begins to behave like a liquid, and the ideal gas definition no longer applies. Gas law formulas give increasingly inaccurate results as temperature approaches the condensation point.

Large or polar molecules: CO₂, NH₃, and H₂O vapor deviate significantly from ideal gas behavior because they have strong London dispersion forces (CO₂) or permanent dipoles and hydrogen bonding (NH₃, H₂O). Helium and hydrogen are the most ideal real gases because they are small and nonpolar — closest to the ideal gas definition.

Rule of thumb for gas law practice problems: Combined gas law and ideal gas law formulas give results accurate to within ~1% when P < 10 atm and T > 200 K for most common gases. Beyond these limits, the assumptions of an ideal gas become increasingly unreliable.

How to Identify Which Gas Law to Use — Decision Flowchart

Use this decision tree at the start of every combined gas law problem or gas law practice problem. Correctly identifying which gas law formula to apply before calculating is the single most important skill for combined gas law worksheets.

START — Read the gas law problem carefully
Do you need to find moles (n), mass, or number of molecules?
YES → Use PV = nRT (Ideal Gas Law) — the only gas law formula that includes n
NO → Are you comparing two states of the same gas (before/after)?
What is held constant between the two states?
Temperature (T) constant → Boyle’s Law: P₁V₁ = P₂V₂
Pressure (P) constant → Charles’s Law: V₁/T₁ = V₂/T₂  ⚠ Kelvin required
Volume (V) constant (rigid container) → Gay-Lussac’s Law: P₁/T₁ = P₂/T₂  ⚠ Kelvin required
Nothing constant (all three change) → Combined Gas Law: P₁V₁/T₁ = P₂V₂/T₂  ⚠ Kelvin required
⚠ Remember: Temperature must always be in Kelvin for Charles’s law, Gay-Lussac’s law, the combined gas law, and the ideal gas law PV = nRT. Only Boyle’s law (P₁V₁ = P₂V₂) does not involve temperature — but verify it is actually constant.

Common Mistakes in Gas Law Problems

These five mistakes account for the majority of wrong answers on combined gas law worksheets and gas law practice problem sets. Each example shows the wrong approach and the correct approach side by side.

❌ Mistake 1 — Using Celsius Instead of Kelvin (Most Common Error in All Gas Law Problems)

Temperature must always be in Kelvin in every gas law formula except Boyle’s Law. Using Celsius produces a plausible-looking number that is completely wrong.

❌ WRONG — CelsiusP₁=1 atm, V₁=2 L T₁=27°C, T₂=54°C V₂ = 2 × (54/27) V₂ = 2 × 2.0 = 4.0 L ← WRONG (ratio 2.0 is meaningless in Celsius)
✓ CORRECT — KelvinT₁ = 27+273 = 300 K T₂ = 54+273 = 327 K V₂ = 2 × (327/300) V₂ = 2 × 1.09 = 2.18 L ✓ ← Only Kelvin gives correct ratio
❌ Mistake 2 — Applying the Combined Gas Law When a Simpler Law Applies

If temperature is constant, use Boyle’s law directly. If you apply the combined gas law and cancel T₁ = T₂, you get the right answer — but the extra algebra increases the risk of arithmetic errors. Always identify the constant variable first.

❌ INEFFICIENT (error-prone)Applying combined law when T is constant: P₁V₁/T₁ = P₂V₂/T₂ → extra T division, extra chance of error
✓ CORRECT — Identify constant firstT constant → Boyle’s Law P₁V₁ = P₂V₂ → one step, no T needed, faster and less error-prone
❌ Mistake 3 — Wrong Gas Constant R for the Ideal Gas Law PV = nRT

The value of R depends on your pressure units. Mixing R values with incompatible units gives answers off by a factor of 101.325.

❌ WRONG — R mismatchP = 2.50 atm, V = 5.00 L T = 300 K R = 8.314 ← kPa value, wrong for atm n = PV/RT = 12.50/(8.314×300) n = 0.00501 mol ← 101× too small
✓ CORRECT — R matched to P unitsP in atm → use R = 0.082057 n = 12.50/(0.082057×300) n = 12.50/24.617 n = 0.508 mol ✓
❌ Mistake 4 — Mixed Pressure Units on Each Side of the Combined Gas Law

P₁ and P₂ must be in the same units. V₁ and V₂ must be in the same units. Mixing atm on one side with kPa on the other produces a completely wrong answer without any obvious error to spot.

❌ WRONG — Mixed unitsP₁ = 1.00 atm P₂ = 202.65 kPa ← not converted P₁V₁/T₁ = P₂V₂/T₂ → 1.00 vs 202.65 — wrong ratio → answer off by 100:1
✓ CORRECT — Convert firstP₂ = 202.65 kPa = 202.65/101.325 = 2.00 atm Now both sides use atm: P₁=1.00 atm, P₂=2.00 atm ✓
❌ Mistake 5 — T₁ and T₂ Swapped in the Combined Gas Law Rearrangement for V₂

When rearranging the combined gas law for V₂, T₂ goes in the numerator and T₁ in the denominator. Swapping them gives a wrong answer that is off by the square of the temperature ratio.

❌ WRONG — T₁ and T₂ swappedV₂ = P₁V₁T₁/(P₂T₂) ↑ T₁ in numerator — WRONG For Problem 7: V₂ = 1.50×4.00×300/(350×2.00) V₂ = 1800/700 = 2.57 L ← WRONG
✓ CORRECT — T₂ in numeratorV₂ = P₁V₁T₂/(T₁P₂) ↑ T₂ in numerator — CORRECT V₂ = 1.50×4.00×350/(300×2.00) V₂ = 2100/600 = 3.50 L ✓

Memory aid: the combined gas law formula is P₁V₁/T₁ = P₂V₂/T₂. When you cross-multiply and rearrange for V₂, T₂ moves to the numerator and T₁ moves to the denominator — opposite sides from where they start.

Frequently Asked Questions — Combined Gas Law Problems

What is the combined gas law and when do you use it?

The combined gas law is P₁V₁/T₁ = P₂V₂/T₂, where P is pressure, V is volume, and T is temperature in Kelvin. It applies when a fixed amount of gas (constant n) undergoes a change in which all three variables — pressure, volume, and temperature — change simultaneously. If any one variable is held constant, a simpler formula (Boyle’s law, Charles’s law, or Gay-Lussac’s law) applies instead. The combined gas law contains all three simpler laws as special cases: set T₁ = T₂ to get Boyle’s law, set P₁ = P₂ to get Charles’s law, set V₁ = V₂ to get Gay-Lussac’s law.

When do you use Boyle’s law vs Charles’s law vs Gay-Lussac’s law?

The choice depends on which variable is constant. Use Boyle’s law (P₁V₁ = P₂V₂) when temperature is constant. Use Charles’s law (V₁/T₁ = V₂/T₂) when pressure is constant. Use Gay-Lussac’s law (P₁/T₁ = P₂/T₂) when volume is constant — typically indicated by a “rigid container” or “sealed vessel” in the problem. Use the combined gas law (P₁V₁/T₁ = P₂V₂/T₂) when all three variables change. Use the ideal gas law (PV = nRT) when you need moles or mass. Temperature must be in Kelvin for all laws except Boyle’s law.

What is the ideal gas law PV = nRT?

The ideal gas law PV = nRT relates pressure (P), volume (V), moles (n), the gas constant (R), and temperature (T) for an ideal gas. Unlike the combined gas law (which compares two states), the ideal gas law describes a single state completely. The gas constant R = 0.082057 L·atm/(mol·K) when pressure is in atm and volume in litres; R = 8.31446 L·kPa/(mol·K) when pressure is in kPa. Temperature must always be in Kelvin. The ideal gas law PV = nRT is the most general gas law formula — the combined gas law is derived from it by holding n constant.

Why must temperature always be in Kelvin in gas law formulas?

Temperature must always be in Kelvin because gas laws describe proportional relationships that require an absolute temperature scale. On the Kelvin scale, 0 K (absolute zero) is the point where gas molecules have zero kinetic energy and zero volume — a true physical zero. Celsius zero (0°C = 273.15 K) is arbitrary. Using Celsius produces completely wrong answers: in Charles’s law (V₁/T₁ = V₂/T₂), a gas at 0°C would appear to have infinite volume relative to a gas at 0.001°C. Using Celsius temperatures like 27°C and 54°C gives a temperature ratio of 2.0, implying volume doubles — but the Kelvin ratio (300 K / 327 K = 1.09) correctly shows only a 9% increase. Always convert: K = °C + 273.15.

What is an ideal gas and what are its assumptions?

An ideal gas is a theoretical gas that obeys PV = nRT exactly under all conditions. The five assumptions of an ideal gas are: (1) molecules have negligible volume (treated as point masses); (2) no intermolecular forces between molecules; (3) all collisions are perfectly elastic; (4) molecules move in constant, random motion; (5) average kinetic energy is directly proportional to absolute temperature in Kelvin. The ideal gas definition in chemistry is a gas obeying these assumptions. Real gases approximate ideal behavior at low pressure and high temperature. Noble gases like helium and neon are the best real-world ideal gas examples. Non-ideal gases like CO₂ and NH₃ deviate because of strong intermolecular forces.

What are gas formulas in chemistry and which is most important?

The main gas formulas in chemistry are: Boyle’s law (P₁V₁ = P₂V₂), Charles’s law (V₁/T₁ = V₂/T₂), Gay-Lussac’s law (P₁/T₁ = P₂/T₂), the combined gas law (P₁V₁/T₁ = P₂V₂/T₂), and the ideal gas law (PV = nRT). The most important and general gas formula is the ideal gas law PV = nRT, from which all others can be derived. The combined gas law is the most versatile for combined gas law practice problems because it handles all scenarios involving two states of a gas. For gas law practice problems on exams, mastering the identification of which formula applies is as important as knowing the formulas themselves.

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