Ideal gas law: practice problems with step-by-step solutions

Also called: PV = nRT

Free Ideal gas law practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: P from n, V, T, from mass, state change or density.

Free practice problemEasy
n = 1.5 molV = 20 LT = 35 °CP = ?

A rigid tank holds an ideal gas. Find its pressure P.

n = 1.5 molV = 20 LT = 35 °CR = 8.314 J/(mol·K)
kPa
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What you practise

  1. EasyP from n, V, T
  2. Mediumfrom mass
  3. Hardstate change or density

Key formulas

Ideal gas law
pV = nRT
Gas constant
R = 8.314 J/(mol·K)
Combined gas law
p1V1/T1 = p2V2/T2
Moles
n = m/M
Temperature
T[K] = T[°C] + 273.15

Learn Ideal gas law from scratch

1What the ideal gas law is

A gas is a huge number of molecules flying around and hitting the walls of their container. Those hits push on the walls: that push per square metre is the pressure.

Squeeze the gas into a smaller volume, and the molecules hit the walls more often: the pressure rises. Heat it, and they move faster: the pressure rises too.

For most gases at ordinary pressures, one equation links all of this:

P · V = n · R · T
  • P: pressure in Pa (1 kPa = 1000 Pa).
  • V: volume in m³ (1 L = 0.001 m³).
  • n: amount of gas in mol.
  • T: absolute temperature in kelvin.
  • R = 8.314 J/(mol·K), the gas constant.
Self-check Which temperature do you put into PV = nRT for a gas at 20 °C?
  1. 20 K
  2. 20 °C
  3. 293.15 K
  4. −253.15 K

Answer: 293.15 K. The gas law needs absolute temperature: T = 20 + 273.15 = 293.15 K. At 0 °C a gas still has pressure, so a scale that starts at 0 °C cannot work.

2Units that match R

R = 8.314 J/(mol·K) works with SI units. A joule is a Pa·m³, so put P in Pa and V in m³. Then everything fits.

  • Temperature: T[K] = T[°C] + 273.15.
  • Volume: 1 L = 0.001 m³, so 20 L = 0.020 m³.
  • Pressure: the result comes out in Pa. Divide by 1000 for kPa.
  • Mass: if you know the mass m in grams, find the moles with n = m/M, where M is the molar mass in g/mol (N₂: 28.01, CO₂: 44.01).

A useful shortcut: with V in litres, n·R·T/V gives P directly in kPa, because the factor 1000 for litres and the factor 1000 for kPa cancel.

Self-check What is 5 L in cubic metres?
  1. 0.005 m³
  2. 0.05 m³
  3. 0.5 m³
  4. 5000 m³

Answer: 0.005 m³. One cubic metre holds 1000 litres, so 5 L = 5/1000 = 0.005 m³.

3Changing the state of a gas

If a fixed amount of gas goes from one state to another, n and R stay the same. So PV/T stays the same:

P₁V₁T₁P₂V₂T₂
P1·V1 / T1 = P2·V2 / T2

This is the combined gas law. Any pressure and volume units work, as long as they are the same on both sides. The temperatures must still be in kelvin.

The density of a gas follows from n = m/M as well:

ρ = P · M / (R · T)

With P in kPa and M in g/mol, ρ comes out in kg/m³.

4The method step by step

  1. List the given values and the unknown.
  2. Convert the temperature to kelvin.
  3. If a mass is given, find the moles: n = m/M.
  4. Convert the volume to m³ (or keep litres and use the kPa shortcut).
  5. Rearrange PV = nRT for the unknown, then fill in the numbers.
  6. For a change of state, use P1V1/T1 = P2V2/T2 instead.

5A worked example

Follow every step. Try to predict the next one before you open it.

Worked example

A rigid tank of 20 L contains 88 g of CO₂ (M = 44.01 g/mol) at 25 °C. Treat it as an ideal gas and find the pressure P.

1

List the values

m = 88 gV = 20 LT = 25 °CP = ?

We know the mass, the volume and the temperature. First turn the mass into moles.

2

Moles

n = m/M = 88/44.01 = 2.000 mol
3

Convert the units

T = 25 + 273.15 = 298.15 K
V = 20 L = 0.020 m³
4

Ideal gas law

P = nRT/V = 2.000 · 8.314 · 298.15 / 0.020 ≈ 247 800 Pa

That is about 247.8 kPa.

5

Check the answer

One mole of gas at 25 °C and 1 atm (101.3 kPa) takes about 24.5 L. Here 2 mol are squeezed into 20 L, about 2.5 times less room per mole, so about 2.5 atm. 247.8 kPa fits.

Answer

P ≈ 247.8 kPa.

6Common mistakes

  • Using °C instead of kelvin. This is the most common mistake and can make the answer many times too small or too large.
  • Putting litres into the formula with R = 8.314 and expecting Pa.
  • Forgetting to turn grams into moles.
  • Turning the pressure ratio upside down in the combined gas law.
Self-check A gas at 100 kPa, 10 L and 27 °C goes to 200 kPa and 327 °C. What is the new volume?
  1. 10.0 L
  2. 40.0 L
  3. 5.0 L
  4. 60.6 L

Answer: 10.0 L. In kelvin: V₂ = 10 · (100/200) · (600.15/300.15) = 10.0 L. Using °C gives 10 · 0.5 · 327/27 = 60.6 L. Ignoring the temperature gives 5 L, and turning the pressure ratio around gives 40 L.

7How exam questions look

  • Easy: find P from n, V and T in a rigid tank.
  • Medium: the amount is given as a mass. Find n = m/M first, then P or V, as in the example.
  • Hard: a change of state (find V2 with the combined gas law) or the density of a gas.

Density example: N₂ (M = 28.01 g/mol) at 200 kPa and 27 °C has ρ = 200 · 28.01/(8.314 · 300.15) = 2.24 kg/m³.

Learn it step by step in the appLearn it step by step in the app

Worked examples

Example 1Easy
n = 5 molV = 15 LT = 30 °CP = ?

A rigid tank holds an ideal gas. Find its pressure P.

n = 5 molV = 15 LT = 30 °CR = 8.314 J/(mol·K)

Find P (kPa)

Show hints
  1. Use PV = nRT with absolute temperature (K).
  2. Units: with V in m³ and R = 8.314, P comes out in Pa. 1 L = 0.001 m³.
  3. T = 303.1 K.
Show the answer and the step-by-step solution

Ideal gas law: PV = nRT, with absolute temperature (K) and units that match R.

Convert units

T = 30 + 273.15 = 303.1 K
V = 15 L = 0.015 m³

Ideal gas law

P = nRT/V = 5·8.314·303.1/0.015 = 840130 Pa = 840.1 kPa

Answer

P = 840.1 kPa
Example 2Medium
500 g of O₂V = 10 LT = 50 °CP = ?

A tank of 10 L contains 500 g of O₂ at 50 °C. Treat it as an ideal gas and find the pressure P.

m = 500 gM = 32 g/molV = 10 LT = 50 °CR = 8.314 J/(mol·K)

Find P (kPa)

Show hints
  1. Convert mass to moles first: n = m/M.
  2. Then PV = nRT with T in K.
  3. n = 15.63 mol.
Show the answer and the step-by-step solution

Ideal gas law: PV = nRT, with n = m/M.

Moles

n = m/M = 500/32 = 15.63 mol
T = 50 + 273.15 = 323.1 K

Pressure

P = nRT/V = 15.63·8.314·323.1/0.01 = 4198 kPa

Answer

P = 4198 kPa
Example 3Hard
He, M = 4.003 g/molP = 850 kPaT = 70 °Cρ = ?

Find the density ρ of He at 850 kPa and 70 °C, treating it as an ideal gas.

M = 4.003 g/molP = 850 kPaT = 70 °CR = 8.314 J/(mol·K)

Find ρ (kg/m³)

Show hints
  1. Write n = m/M in PV = nRT and rearrange for m/V.
  2. ρ = PM/(RT). With P in kPa and M in g/mol, ρ comes out in kg/m³.
  3. T = 343.1 K.
Show the answer and the step-by-step solution

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 70 + 273.15 = 343.1 K

Density

ρ = 850·4.003/(8.314·343.1) = 1.193 kg/m³

Answer

ρ = 1.193 kg/m³

Questions and answers

What is Ideal gas law?

A gas is a huge number of molecules flying around and hitting the walls of their container. Those hits push on the walls: that push per square metre is the pressure.

Squeeze the gas into a smaller volume, and the molecules hit the walls more often: the pressure rises. Heat it, and they move faster: the pressure rises too.

How do I solve Ideal gas law problems step by step?

Ideal gas law: PV = nRT, with absolute temperature (K) and units that match R.

What do the Ideal gas law exercises cover?
  • Easy: P from n, V, T
  • Medium: from mass
  • Hard: state change or density

Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.

How practising works

  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Your progress, mistakes and saved problems are kept, so you can review before the exam.

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Printable worksheet

Ideal gas law worksheet with answers
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