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General Chemistry · Gases

Gas laws practice problems, with worked answers

Most gas problems aren't hard math. The hard part is the first ten seconds: figuring out which equation you need and getting your units ready for it. So this page starts there, then works through the ideal gas law, the combined gas law, and partial pressures.

Short on time? Jump to the 9 practice problems or the worked examples.

Which gas law do I use?

Ask what the problem is describing. These four setups cover the pressure, volume, and reaction problems practiced here. Assume ideal gas behavior throughout this guide, and use absolute pressures (not gauge pressures).

Match the problem to the law
What the problem looks likeUse
One gas, one set of conditions, and you need P, V, n, T, or massPV = nRT
One sample changes from a "before" to an "after," with moles unchangedP1V1 ÷ T1 = P2V2 ÷ T2
Two or more gases share a container, or a gas is collected over waterPtotal = PA + PB + … and PA = XAPtotal
A reaction makes or uses a gasStoichiometry for moles, then PV = nRT

Try it before doing any math. For each one, pick the most direct setup. The combined gas law and partial-pressure equations can also be derived from PV = nRT.

Pick the law 1A sealed syringe of air at 1.00 atm is squeezed from 30.0 mL to 12.0 mL at constant temperature. What is the new pressure?

Show solution for pick the law 1

One sample, two states, same moles. With T constant, it shrinks to P1V1 = P2V2.

Pick the law 2How many moles of argon are in a 5.0 L tank at 2.0 atm and 300 K?

Show solution for pick the law 2

One state, and you want n. That is exactly what PV = nRT is for.

Pick the law 3A tank holds 0.30 mol O2 and 0.90 mol N2 at 4.0 atm total. What is the pressure of the O2 alone?

Show solution for pick the law 3

Two gases and one total: use the mole fraction, X(O2) = 0.25, so the O2 pressure is 1.0 atm.

Pick the law 4How many liters of CO2 at 1.00 atm and 25 °C form when 10.0 g of propane burns completely in excess oxygen?

Show solution for pick the law 4

Grams of propane to moles, mole ratio to CO2, then PV = nRT turns those moles into liters.

Three checks before you calculate

  1. Is the temperature in kelvin? For these gas-law equations, convert Celsius to kelvin by adding 273.15.
  2. Do your units match R? Either convert the pressure and volume, or pick the R that fits them.
  3. Is the pressure really your gas? If water vapor or another gas is in the mix, the total isn't the pressure of the gas you care about.
Pressure units and the R that goes with them, with volume in liters
Pressure unitEqual to 1 atmR to use
atm1 atm0.08206 L·atm/(mol·K)
kPa101.325 kPa8.314 L·kPa/(mol·K)
torr or mmHg760 torr62.36 L·torr/(mol·K)

Volume in mL? Divide by 1,000 to get liters. Every example below uses R = 0.08206 unless it says otherwise. Keep unrounded values in your calculator; the displayed intermediate values are rounded. Treat torr and mmHg as equivalent at this precision.

Ideal gas law examples

PV = nRT

This connects pressure, volume, moles, and temperature for one gas at one moment. Know three, solve for the fourth. Since n = mass ÷ molar mass, it also lets you work with grams.

Example 1: Solve for pressure

A 10.0 L container holds 0.500 mol of N2 at 25 °C. What is the pressure?

  1. T = 25 + 273.15 = 298.15 K.
  2. Rearrange: P = nRT ÷ V.
  3. P = (0.500 mol)(0.08206 L·atm/(mol·K))(298.15 K) ÷ 10.0 L.

Answer: 1.22 atm

The easy mistake: plugging in 25 for T gives 0.103 atm, which is about 12 times too low.

Example 2: Start from grams, in kPa

What volume does 9.00 g of CO2 (44.01 g/mol) take up at 30.0 °C and 95.0 kPa?

  1. n = 9.00 g ÷ 44.01 g/mol = 0.2045 mol.
  2. T = 303.15 K. The pressure is in kPa, so use R = 8.314 L·kPa/(mol·K).
  3. V = nRT ÷ P = (0.2045)(8.314)(303.15) ÷ 95.0.

Answer: 5.43 L

The easy mistake: mixing kPa with R = 0.08206. Your R has to speak the same units as your pressure.

Example 3: Find a gas's molar mass

A 0.500 L flask holds 1.27 g of an unknown gas at 27 °C and 742 torr. What is its molar mass?

  1. P = 742 ÷ 760 = 0.9763 atm. T = 300.15 K.
  2. n = PV ÷ RT = (0.9763)(0.500) ÷ [(0.08206)(300.15)] = 0.01982 mol.
  3. Molar mass = grams ÷ moles = 1.27 g ÷ 0.01982 mol.

Answer: 64.1 g/mol, consistent with SO2 (64.06 g/mol). Molar mass alone does not prove the gas's identity.

The easy mistake: stopping at moles. The question asks for molar mass, so divide grams by moles.

Combined gas law examples

P1V1 ÷ T1 = P2V2 ÷ T2

This is PV = nRT with n and R canceled, so it applies to an ideal gas when moles stay fixed. You don't have to memorize three smaller laws. Cross out whatever stays constant and they appear on their own.

What's left when something stays constant
Held constantWhat's leftNameIn words
TP1V1 = P2V2Boyle's lawSqueeze it, pressure goes up
PV1 ÷ T1 = V2 ÷ T2Charles's lawHeat it, it expands
VP1 ÷ T1 = P2 ÷ T2Gay-Lussac's lawHeat it in a rigid can, pressure goes up

Example 4: A rising weather balloon

A balloon holds 2.50 L of helium at 745 torr and 22 °C. It rises to where the pressure is 525 torr and the temperature is −8 °C. What is its new volume?

  1. T1 = 295.15 K and T2 = 265.15 K.
  2. Solve for V2: V2 = V1 × (P1 ÷ P2) × (T2 ÷ T1).
  3. V2 = 2.50 L × (745 ÷ 525) × (265.15 ÷ 295.15).

Answer: 3.19 L

The easy mistake: leaving −8 in the formula. That gives a negative volume, which can't happen. The pressures are different: you don't need to convert them, because torr cancels in the ratio.

Partial pressure examples

Ptotal = PA + PB + PC + …

PA = XA × Ptotal, where XA = nA ÷ ntotal

In an ideal mixture, each gas contributes the pressure it would exert alone at the same temperature and volume. This is Dalton's law. Pressure splits up by moles, not by mass, so convert grams to moles before you find a mole fraction. Mole fractions always add up to 1, which makes a quick check.

Example 5: A gas mixture from grams

A 12.0 L tank at 15 °C holds 4.00 g of He (4.003 g/mol) and 14.0 g of N2 (28.01 g/mol). Find the total pressure and each partial pressure.

  1. n(He) = 4.00 ÷ 4.003 = 0.9993 mol. n(N2) = 14.0 ÷ 28.01 = 0.4998 mol. Total = 1.499 mol.
  2. Ptotal = nRT ÷ V = (1.499)(0.08206)(288.15) ÷ 12.0 = 2.954 atm.
  3. X(He) = 0.9993 ÷ 1.4991 = 0.6666, so P(He) = 0.6666 × 2.954 atm = 1.97 atm.
  4. X(N2) = 0.4998 ÷ 1.4991 = 0.3334, so P(N2) = 0.3334 × 2.954 atm = 0.985 atm.

Answer: 2.95 atm total, 1.97 atm He, 0.985 atm N2

The easy mistake: splitting by mass. Nitrogen is most of the mass here, but helium is most of the moles, so helium has most of the pressure.

Gas collected over water: an upside-down tube stands in a water bath, with gas trapped above the water inside the tube. In the tube:your gas pluswater vapor Level waterinside and out gas in
Original schematic by GenChemDrill. Gas bubbles up through water and pushes the water down the tube. The trapped gas is wet, so its total pressure is your gas plus water vapor. When the water levels match, that total equals the room's pressure.

Example 6: Gas collected over water

Zinc reacts with HCl, and you collect 0.240 L of H2 over water at 25 °C. The total pressure is 755 torr. The vapor pressure of water at 25 °C is 23.8 torr. How many moles of H2 did you collect?

  1. P(H2) = 755 − 23.8 = 731.2 torr.
  2. Convert: 731.2 ÷ 760 = 0.9621 atm. T = 298.15 K.
  3. n = PV ÷ RT = (0.9621)(0.240) ÷ [(0.08206)(298.15)].

Answer: 9.44 × 10−3 mol H2

The easy mistake: using 755 torr in PV = nRT. That counts the water vapor as hydrogen. Water's vapor pressure depends on temperature, so problems give it or give you a table. It's 17.5 torr at 20 °C and 31.8 torr at 30 °C.

Gases in reactions

Balanced equations count moles, so a gas volume is just the last step of a normal problem. If the mole-ratio part feels shaky, the stoichiometry practice guide covers it.

Example 7: Grams of solid to liters of gas

Sodium azide breaks down by 2 NaN3 → 2 Na + 3 N2. After 50.0 g of NaN3 (65.01 g/mol) decomposes completely, what volume does the N2 occupy when measured at 25 °C and 1.00 atm?

  1. Moles of NaN3 = 50.0 ÷ 65.01 = 0.7691 mol.
  2. Mole ratio: 0.7691 mol × (3 mol N2 ÷ 2 mol NaN3) = 1.154 mol N2.
  3. V = nRT ÷ P = (1.154)(0.08206)(298.15) ÷ 1.00.

Answer: 28.2 L

The easy mistake: using 22.4 L/mol here. That shortcut only works at 0 °C and 1 atm, and this gas is at 25 °C.

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Practice problems

Type a number without units, such as 3.37, 1.33e-2, or 1.33 × 10^-2. Answers within about 1% count, so rounding differences and using 273 instead of 273.15 won't hurt you. If you get it wrong, the feedback will often name the exact slip.

Problem 1

A 15.0 L tank holds 2.00 mol of O2 at 35 °C. What is the pressure in atm?

Show solution for problem 1
  1. T = 35 + 273.15 = 308.15 K.
  2. P = nRT ÷ V = (2.00 mol)(0.08206 L·atm/(mol·K))(308.15 K) ÷ 15.0 L.
  3. P = 3.37 atm.

Problem 2

A 5.00 L flask of neon is at 850 torr and 22 °C. How many moles of neon are there?

Show solution for problem 2
  1. P = 850 torr ÷ 760 torr/atm = 1.118 atm. T = 295.15 K.
  2. n = PV ÷ RT = (1.118 atm)(5.00 L) ÷ [(0.08206)(295.15 K)].
  3. n = 0.231 mol.

Problem 3

How many grams of methane, CH4 (16.04 g/mol), fill a 12.0 L tank at 1.50 atm and 18 °C?

Show solution for problem 3
  1. T = 291.15 K.
  2. n = PV ÷ RT = (1.50)(12.0) ÷ [(0.08206)(291.15)] = 0.753 mol.
  3. Mass = 0.753 mol × 16.04 g/mol = 12.1 g.

Problem 4

A fixed amount of gas takes up 6.00 L at 1.20 atm and 25 °C. It moves to 0.800 atm and 50 °C. What is the new volume?

Show solution for problem 4
  1. T1 = 298.15 K and T2 = 323.15 K.
  2. V2 = V1 × (P1 ÷ P2) × (T2 ÷ T1).
  3. V2 = 6.00 L × (1.20 ÷ 0.800) × (323.15 ÷ 298.15) = 9.75 L.
  4. Sense check: lower pressure and higher temperature both make it bigger.

Problem 5

A sealed rigid can is at 2.20 atm and 20 °C. It sits in the sun and reaches 60 °C. What is the new pressure?

Show solution for problem 5
  1. Rigid means V is constant, and sealed means n is constant.
  2. P2 = P1 × (T2 ÷ T1) = 2.20 atm × (333.15 ÷ 293.15).
  3. P2 = 2.50 atm.

Problem 6

A mixture has 0.40 mol He, 1.20 mol Ar, and 0.40 mol Ne at 6.00 atm total. What is the partial pressure of the argon?

Show solution for problem 6
  1. Total moles = 0.40 + 1.20 + 0.40 = 2.00 mol.
  2. X(Ar) = 1.20 ÷ 2.00 = 0.600.
  3. P(Ar) = 0.600 × 6.00 atm = 3.60 atm.

Problem 7

You collect 0.350 L of O2 over water at 30 °C. The total pressure is 752 torr, and water's vapor pressure at 30 °C is 31.8 torr. How many moles of O2 did you collect?

Show solution for problem 7
  1. P(O2) = 752 − 31.8 = 720.2 torr = 0.9476 atm.
  2. T = 303.15 K.
  3. n = PV ÷ RT = (0.9476)(0.350) ÷ [(0.08206)(303.15)] = 0.0133 mol.

Problem 8

CaCO3 (100.09 g/mol) breaks down by CaCO3 → CaO + CO2. After heating 5.00 g of CaCO3 until it decomposes completely, what volume does the CO2 occupy when cooled to 25 °C at 1.00 atm?

Show solution for problem 8
  1. Moles of CaCO3 = 5.00 g ÷ 100.09 g/mol = 0.04996 mol.
  2. The ratio is 1:1, so that is 0.04996 mol CO2.
  3. V = nRT ÷ P = (0.04996)(0.08206)(298.15) ÷ 1.00 = 1.22 L.

Problem 9

A gas has a density of 1.96 g/L at 0 °C and 1.00 atm. What is its molar mass in g/mol?

Show solution for problem 9
  1. Start from M = dRT ÷ P.
  2. M = (1.96 g/L)(0.08206 L·atm/(mol·K))(273.15 K) ÷ 1.00 atm.
  3. M = 43.9 g/mol, which fits CO2 (44.01 g/mol).

Answer key

Pick the law

  1. The combined gas law
  2. The ideal gas law, PV = nRT
  3. Partial pressures (Dalton's law)
  4. Stoichiometry, then PV = nRT

Practice problems

  1. 3.37 atm: PV = nRT with T = 308.15 K.
  2. 0.231 mol: convert torr to atm, then n = PV/RT.
  3. 12.1 g: 0.753 mol × 16.04 g/mol.
  4. 9.75 L: 6.00 × (1.20 ÷ 0.800) × (323.15 ÷ 298.15).
  5. 2.50 atm: P rises in proportion to kelvin temperature.
  6. 3.60 atm: X(Ar) = 0.600.
  7. 0.0133 mol: subtract the water vapor first.
  8. 1.22 L: 0.04996 mol of CO2 at 298.15 K.
  9. 43.9 g/mol: M = dRT/P. It fits CO2.

Frequently asked questions

Why do gas laws need kelvin?

For a fixed amount of ideal gas, pressure is proportional to absolute temperature at constant volume; volume is proportional to absolute temperature at constant pressure. Kelvin is the absolute scale used with the R values here. Heating from 20 °C to 40 °C takes you from 293.15 K to 313.15 K, about a 7% rise, not a doubling. Add 273.15 to each Celsius temperature first.

Which value of R should I use?

Use the one whose units match your pressure and volume. With liters and atm, use 0.08206 L·atm/(mol·K). With liters and kPa, use 8.314 L·kPa/(mol·K). With liters and torr, use 62.36 L·torr/(mol·K). The other option is to convert the pressure to atm first.

Is one mole of gas always 22.4 L?

No. About 22.4 L/mol is the usual reference value for an ideal gas at 0 °C and 1 atm. Some sources define STP as 0 °C and 1 bar, which gives about 22.7 L/mol. At any other conditions, use PV = nRT. Check which STP your course uses.

When do I use the combined gas law instead of PV = nRT?

Use the combined gas law when one sample of gas changes from a first state to a second state and the moles stay the same. Use PV = nRT when you have one state and need moles, mass, or one missing variable.

Does it matter which gas it is?

For PV = nRT, ideal gases with the same moles, temperature, and volume have the same pressure. Identity matters when converting grams to moles, balancing a reaction, or accounting for real-gas behavior.

When does the ideal gas law stop working?

The ideal gas law works best at low pressure and temperatures well above condensation. At high pressure, particle volume and attractions matter more; at low temperature, attractions matter more relative to molecular motion. The size of the deviation depends on the gas. The problems on this page assume ideal behavior.

Sources

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