Chemistry · Thermodynamics / Gases

Ideal gas law

Q: How do you calculate the volume of an ideal gas?

Rearrange $pV=nRT$ for the volume: $V=\frac{nRT}{p}$. Insert the amount of substance n in mol, the universal gas constant $R=8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}$, the absolute temperature T in kelvin and the pressure p in pascals, then the volume comes out in cubic metres. Example: 5 mol of nitrogen at 300 K and $1\,\mathrm{atm}=101325\,\mathrm{Pa}$ give $V=\frac{5\cdot8.314\cdot300}{101325}\,\mathrm{m^3}\approx0.123\,\mathrm{m^3}$, that is 123 litres. Make sure to give the pressure in pascals and the temperature in kelvin; atm, bar or Celsius must be converted first. One cubic metre equals 1000 litres.

Q: What happens to the pressure when you heat a gas at constant volume?

At constant volume and constant amount of substance, $pV=nRT$ implies that the pressure is directly proportional to the absolute temperature: $p\propto T$. Doubling the temperature in kelvin doubles the pressure. This is the law of Amontons or Gay-Lussac. Intuitively, at higher temperature the gas particles collide with the walls faster and more often, which raises the pressure. That is why the pressure in a closed container, such as a spray can, rises when heated and can become dangerous. For the calculation $\frac{p_1}{T_1}=\frac{p_2}{T_2}$ holds, each with temperatures in kelvin.

The ideal gas law relates the pressure, volume, amount of substance and temperature of an ideal gas.

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Formula

Calculate

p \cdot V = n \cdot R \cdot T

p in Pa · V in m³ · n in mol · $R\approx8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}$ · T in K

Variables & units

Variables & units
Symbol	Meaning	Unit
$p$	Pressure	Pa
$V$	Volume	m³ or L
$n$	Amount of substance	mol
$R$	Universal gas constant (8.314)	J/(mol·K)
$T$	Temperature (in kelvin!)	K

LaTeX

p \cdot V = n \cdot R \cdot T

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Diagram: two hyperbolas in the p-V diagram, the highlighted isotherm T and further out the isotherm 2T; a shaded rectangle from the origin to a point on the isotherm T stands for p·V = n·R·T. — At a fixed amount n, all states of equal temperature T lie on a hyperbola, the isotherm: the rectangle p·V always equals n·R·T, and at twice the temperature the isotherm lies further out.

Ideal gas law calculator

Find the pressure, volume, amount or temperature of an ideal gas, compare two states, convert gauge pressure or work out the gas density.

Calculation model

Find

Gas volume V

Volume the gas fills, e.g. 22.4 L for 1 mol at 0 °C and standard pressure.

Amount of substance n

Amount in moles, e.g. 1 mol; from the mass, n = m/M.

Absolute temperature T

Temperature of the gas, e.g. 0 °C; the calculation uses kelvin, and 0 °C is 273.15 K.

Filled with example values. Type your own and the result updates straight away.

Result: Absolute pressure p

p = 101.3 kPa

With V = 22.41 L, n = 1 mol and T = 0 °C, p = 101.3 kPa.

Working · rearrange for p, substitute, evaluate

1. Rearrange for p
$p=\frac{nRT}V$
2. Substitute values (in SI units)
- V = 22.41 L = 0.02241 m³
- T = 0 °C = 273.15 K
$\frac{1\cdot 8.31446\cdot 273.15}{0.02241}$
3. Result
p = 101.3 kPa

Quick tips

Benchmark: 1 mol of ideal gas takes up about 22.4 L at 0 °C and 1013 hPa, and about 24 L at 20 °C.
Always use kelvin and absolute pressure: at fixed volume, going from 20 °C to 40 °C raises the pressure by only about 7 %, not to double.

Assumptions and limits

Classical ideal gas in a suitable equilibrium state. Positive absolute pressure, volume, amount and absolute temperature above zero kelvin. SI gas constant R=N_akB. No automatic real-gas correction, heat, work or time-dependent process calculation.

Derivation & background

A synthesis of the Boyle-Mariotte law ( $pV=\mathrm{const.}$ ), the Gay-Lussac law ( $\frac VT=\mathrm{const.}$ ) and Avogadro's law ( $V\propto n$ ). It holds exactly only for ideal gases (no interactions, point-like particles).

Exam blueprint

Validity range

Ideal equilibrium states using absolute pressure, positive kelvin temperature and consistent SI units. For real gases, adequacy depends on species, density, temperature and required accuracy; check condensation proximity and interactions. State comparisons at fixed n require unchanged amount.

Derivation steps

The ideal equation of state combines three empirical proportionalities, each with specified fixed conditions.

1For the same ideal sample at fixed T, $V\propto\frac1p$ .
2At fixed n and p, $V\propto T$ ; T is absolute kelvin temperature.
3At fixed p and T, $V\propto n$ .
4Combine the three empirical proportionalities: $V\propto\frac{nT}{p}$ .
5Introduce the molar gas constant: $V=R\frac{nT}{p}$ ; multiplying by p gives $pV=nRT$ .
6For fixed n, $\frac{p_1V_1}{T_1}=\frac{p_2V_2}{T_2}$ . Compare absolute pressures and kelvin temperatures; check real-gas deviations separately.

Rearrangements

Volume from state variables

V = \frac{nRT}{p}

Always use temperature in kelvin.

Task variant

What happens to V at constant p and n when T doubles?

V doubles because V is proportional to T.

Common mistakes

Using Celsius instead of kelvin.

Gas laws always require absolute temperature.

Exam context

Often used with density, amount of substance, reaction gases or pressure changes.

Formula cluster

Thermodynamics and amount of substance

Connects macroscopic state variables with chemical calculations.

Q = m·c·ΔT

Heat Capacity: Q = m·c·ΔT

k = A·e^(−EA/RT)

Arrhenius Equation

η = 1 − T_k/T_w

Carnot Efficiency

Worked example

An ideal gas sample containing $n=5\,\mathrm{mol}$ of nitrogen is at $T=300\,\mathrm K$ and absolute $p=1\,\mathrm{atm}=101325\,\mathrm{Pa}$ . With $R=8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}$ , $V=\frac{nRT}{p}\approx0.123\,\mathrm{m^3}=123\,\mathrm L$ .

Applications

Aviation (pressurised cabins), diving (pressure equalisation), chemical reactors, breathing-gas mixtures

How do you calculate with Ideal Gas Law?

Here is how to work through a typical Ideal Gas Law (pV = nRT) task step by step:

1
Task
What happens to V at constant p and n when T doubles?
Solution path
V doubles because V is proportional to T.

Understand and apply the equation

A bicycle pump, a rigid pressure vessel and a movable piston respond differently to heating. The ideal gas equation explains each once the constraints are identified. Learn to compare states, use measurements consistently and recognize where the model stops being adequate.

Understand the situation before calculating

Picture a gas sample beneath a freely moving piston. Stronger molecular impacts at a higher temperature can raise pressure or move the piston. The apparatus determines which response occurs. A temperature reading alone therefore cannot determine the new pressure.[7]

Three questions lead to the right calculation: does the gas remain in the system, is its container rigid, and what pressure balances a piston? The equation of state $pV=nRT$ applies to each appropriate equilibrium state; it is not an equation for the rate of heating.[1]

You need proportionality, simple algebra and conversion from Celsius to kelvin. Heat and work are not needed for a state calculation alone; their balance becomes relevant when you investigate the energy required for a change.[1]

What does the ideal gas equation describe?

A gas fills its available container rather than retaining a fixed shape and volume. The ideal model treats particles as points with no mutual forces between collisions. Its equation of state describes equilibrium, not the time-resolved trajectory of each particle.[2]

pV=nRT

Pressure times volume equals amount of substance times the gas constant times absolute temperature.[1]

The constraints are decisive. A rigid sealed vessel permits pressure to rise while volume stays fixed. A freely moving frictionless piston, heated slowly under a constant load, can increase volume while equilibrium pressure stays constant. Constant external pressure alone does not guarantee constant gas pressure without these mechanical conditions.[7]

Choose consistent quantities and units

The SI form uses p in Pa, V in m³, n in mol and T in K. Use $R\approx8.314462618\,\mathrm{J\,mol^{-1}\,K^{-1}}$ with these units. The defining value $R=N_Ak_B$ is exact; the decimal shown here is shortened.[5][6]

\mathrm{Pa\,m^3}=\mathrm{J}=\mathrm{mol}\,\frac{\mathrm{J}}{\mathrm{mol\,K}}\,\mathrm{K}

Both sides have the unit joule.

Temperature: $\frac T{\mathrm K}=\frac t{{}^\circ\mathrm C}+273.15$ . Doubling the Celsius reading does not double absolute temperature.
Volume: $1\,\mathrm L=10^{-3}\,\mathrm{m^3}$ . Pressure: $1\,\mathrm{bar}=10^5\,\mathrm{Pa}$ .
Use absolute pressure. A gauge reading of 2 bar above an ambient pressure of 1 bar corresponds to 3 bar absolute.

[5][7]

Establish the equation from the gas laws

Each proportionality holds the other quantities fixed. Together they give $V\propto\frac{nT}{p}$ . This is an empirical synthesis for the ideal model, not proof that every real gas follows it exactly.[1]

V\propto\frac1p\quad(n,T\;\mathrm{const.})

At fixed amount and temperature, volume is inversely proportional to pressure.

V\propto T\quad(n,p\;\mathrm{const.})

At fixed amount and pressure, volume is proportional to kelvin temperature.

V\propto n\quad(p,T\;\mathrm{const.})

At fixed pressure and temperature, volume is proportional to amount of substance.

\begin{aligned}V&=R\frac{nT}{p}\\pV&=nRT\end{aligned}

Introduce the common proportionality constant R and multiply by p.

For the same gas sample, n stays fixed. Dividing by T therefore gives a direct comparison of two states. If n changes through a leak or reaction, this simplified comparison no longer applies.

\frac{p_1V_1}{T_1}=\frac{p_2V_2}{T_2}\quad(n\;\mathrm{const.})

Pressure times volume divided by kelvin temperature stays constant for the same ideal gas sample.

How does particle motion explain the relationship?

For the particle explanation, consider random motion in the rest frame of the gas sample. Subtract any common flow of the whole gas. Otherwise, directed kinetic energy would incorrectly be counted as temperature or thermal pressure.[3]

Collisions transfer momentum to the wall and produce pressure. Isotropic motion assigns one third of the mean squared speed to each spatial direction. Kinetic theory gives the first relationship; combining it with the classical mean translational energy yields $pV=Nk_BT$ .[3]

\begin{aligned}&pV=\frac13Nm\langle v^2\rangle\\&\frac12m\langle v^2\rangle=\frac32kT\end{aligned}

Pressure–volume product and mean translational kinetic energy in a classical ideal gas.

pV=NkT=nN_{\!A}kT=nRT

With particle count N equal to n times the Avogadro constant and R equal to N A times k, the molar form follows.

Pressure from the kinetic theory of gases explains the microscopic origin of pressure more precisely. Thermal RMS speed is measured relative to mean gas flow. A directed particle beam or anisotropic velocity distribution does not automatically have the same isotropic equilibrium pressure.[3]

Rearrange, compare states or calculate density?

Identify the unknown and list all given values with their units.
Decide whether you need one equation of state or a comparison between states.
Rearrange symbolically first, then convert units and substitute numbers.
Check the result against the proportionalities and its units.

\begin{aligned}&p=\frac{nRT}{V}\\&V=\frac{nRT}{p}\\&n=\frac{pV}{RT}\\&T=\frac{pV}{nR}\end{aligned}

Rearrangements for pressure, volume, amount and temperature; denominators must be nonzero.

If mass m is given instead of n, use $n=\frac mM$ with molar mass M. Combining this with $\rho=\frac mV$ gives the density relation. Use M in kg/mol when finding ρ in kg/m³.

\rho=\frac{pM}{RT}

Density of an ideal gas from pressure, molar mass and temperature.

Two temperatures, the same amount of gas

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Each curve contains equilibrium states of the same amount of gas at one fixed temperature. At 5 L they show the pressures from Example 3. Moving between curves is not an isothermal process, and this plot does not determine heat demand.[1]

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At fixed amount of substance, two suitable equilibrium states satisfy $\frac{p_1V_1}{T_1}=\frac{p_2V_2}{T_2}$ . Use absolute pressure and positive absolute temperatures. Mechanical constraints determine whether pressure or volume stays constant; gas loss requires accounting for the change in amount.[1]

Three examples: states and constraints

Volume at a known pressure

Given: $n=0.500\,\mathrm{mol}$ , $T=300\,\mathrm K$ , $p=100\,\mathrm{kPa}$ absolute. Ideal gas at equilibrium.
Find: Find V in litres.

Approach: Use the ideal gas equation $pV=nRT$ , rearranged as $V=\frac{nRT}{p}$ . Convert 100 kPa to 100 000 Pa. Use R in SI units.
Divide by p to isolate V.
$V=\frac{nRT}{p}$
Substitute: SI units satisfy $\mathrm{Pa\,m^3}=\mathrm J$ , giving the calculated volume in cubic metres.
$V=\frac{0.500\cdot8.314462618\cdot300}{100000}\,\mathrm{m^3}\approx0.0124717\,\mathrm{m^3}$
Multiply cubic metres by 1 000 to convert to litres.

Result: V ≈ 12.5 L with appropriate rounding.

Check: One mole would occupy twice this volume under the same conditions, providing a useful plausibility check.

Heating a sealed container

Given: Initial absolute pressure $p_1=150\,\mathrm{kPa}$ . Heating from 20 °C to 80 °C at unchanged amount and volume.
Find: Find the new absolute pressure p₂.

Approach: at fixed n and V, $\frac{p_2}{p_1}=\frac{T_2}{T_1}$ . First convert both temperatures to kelvin: $T_1=293.15\,\mathrm K$ , $T_2=353.15\,\mathrm K$ .
At fixed n and V, these quantities cancel from the state comparison.
$\frac{p_1}{T_1}=\frac{p_2}{T_2}$
Multiply by T₂ and substitute.
$p_2=p_1\frac{T_2}{T_1}=150\,\mathrm{kPa}\cdot\frac{353.15}{293.15}\approx180.70\,\mathrm{kPa}$

Result: p₂ ≈ 181 kPa absolute. Pressure rises by about 20.5 percent.

Check: A ratio of Celsius readings $\frac{80}{20}=4$ would be wrong: kelvin temperature increases by a factor of only 1.205. Both pressures may stay in kPa because this is a ratio.

The following comparison uses only the ideal equation of state. The mechanical constraint determines whether pressure or volume changes.[1][7]

The same heating, two different responses

Given: A closed ideal gas sample: $n=0.200\,\mathrm{mol}$ , $T_1=300\,\mathrm K$ , $V_1=5.00\,\mathrm L$ , heated to $T_2=450\,\mathrm K$ . A: rigid vessel. B: frictionless piston, slow heating at constant equilibrium pressure.
Find: Find initial pressure, final pressure A and final volume B.

Approach: calculate the common initial state first. Then hold volume fixed in A and pressure fixed in B.
$\begin{aligned}p_1&=\frac{nRT_1}{V_1}\\p_{2,A}&=p_1\frac{T_2}{T_1}\\V_{2,B}&=V_1\frac{T_2}{T_1}\end{aligned}$
Substitute: convert litres to cubic metres; the temperature ratio is dimensionless.
$\begin{aligned}p_1&=\frac{(0.200)(8.314462618)(300)}{0.00500}\,\mathrm{Pa}\\&\approx99.774\,\mathrm{kPa}\\p_{2,A}&=\frac{(0.200)(8.314462618)(450)}{0.00500}\,\mathrm{Pa}\\&\approx149.660\,\mathrm{kPa}\\V_{2,B}&=5.00\cdot\frac{450}{300}\,\mathrm L=7.50\,\mathrm L\end{aligned}$

Result: A reaches about 150 kPa at 5.00 L. B reaches 7.50 L at unchanged pressure about 99.8 kPa.

Check: Check: both final states satisfy $p_2V_2=nRT_2\approx748.302\,\mathrm J$ . Equal amount and temperature give equal products, not equal individual values. Heat demand still requires a caloric gas model.

Internal energy of a monatomic ideal gas answers a different question from the equation of state: it assigns energy to temperature. Required heat additionally depends on work and process constraints. The same temperature increase can require different heat in a rigid vessel and under a moving piston.[7][8]

Try it: five exercises with solutions

Calculate · Foundation

1. An ideal gas sample contains 1.00 mol at 300 K and 100 kPa. Find its volume.

Show hint

Use $V=\frac{nRT}{p}$ ; convert kPa to Pa first.

Learning goal: Find gas volume from amount, absolute temperature and pressure.

Show solution and calculation

Approach: rearrange the equation of state for volume and express pressure in pascals.
Volume is about 24.9 L.

Check: only amount is doubled compared with Example 1, so volume must double.

\begin{aligned}V&=\frac{nRT}{p}\\&=\frac{(1.00)(8.314462618)(300)}{100000}\,\mathrm{m^3}\\&\approx0.0249434\,\mathrm{m^3}=24.9434\,\mathrm L\end{aligned}

Calculate · Application

2. At fixed amount and temperature, pressure rises from 100 kPa to 250 kPa. Initial volume: 10.0 L. Find V₂.

Show hint

Use $p_1V_1=p_2V_2$ . Pressure units cancel in the ratio; volume can stay in litres.

Learning goal: Calculate isothermal volume change at fixed amount of gas.

Show solution and calculation

Approach: pressure times volume stays constant for the same amount at equal temperature.
Solve for final volume and use the dimensionless pressure ratio.

Check: pressure increases by 2.5, so volume must be divided by 2.5.

\begin{aligned}p_1V_1&=p_2V_2\\V_2&=\frac{p_1}{p_2}V_1\\&=\frac{100\,\mathrm{kPa}}{250\,\mathrm{kPa}}(10.0\,\mathrm L)=4.00\,\mathrm L\end{aligned}

Error analysis · Application

3. A sealed container is heated at fixed volume from 27 °C to 127 °C. Does pressure increase by a factor of $\frac{127}{27}$ ?

Show hint

The temperature ratio must use kelvin.

Learning goal: Identify erroneous pressure scaling using Celsius readings.

Show solution and calculation

Approach: at fixed amount and volume, the pressure ratio equals the ratio of absolute temperatures.
Convert Celsius to kelvin first.
No: pressure increases by about one third.

Check: equal additive Celsius intervals do not give equal temperature ratios because the scales have different zeros.

\begin{aligned}T_1&=300.15\,\mathrm K\\T_2&=400.15\,\mathrm K\\\frac{p_2}{p_1}&=\frac{T_2}{T_1}=\frac{400.15}{300.15}\approx1.33317\end{aligned}

Interpret · Application

A rigid ideal-gas vessel initially has absolute pressure 200 kPa at 300 K, later 240 kPa at 400 K. Volume stays constant. A sketch shows an open valve. What change in amount of gas do the data require? Can the vessel be treated as closed with fixed n throughout the change?

Show hint

At constant volume, n is proportional to $\frac pT$ , not to pressure alone.

Learning goal: Recognize changing amount of gas from state data rather than assuming a closed process.

Show solution and calculation

Approach: divide the ideal-gas laws for both states; volume and gas constant cancel.
Final amount is 90% of the initial amount, a 10% decrease.
A closed model with unchanged n does not fit these data.

Check: fixed n would require final pressure about 266.7 kPa. The states do not determine the detailed time-dependent valve flow.

\begin{aligned}\frac{n_2}{n_1}&=\frac{p_2T_1}{p_1T_2}=\frac{240\cdot300}{200\cdot400}=0.90\\p_{2,n=\mathrm{const}}&=p_1\frac{T_2}{T_1}\approx266.7\,\mathrm{kPa}\end{aligned}

[1]

Error analysis · Application

An ideal gas occupies 12.0 L at 300 K. A gauge reads 150 kPa above ambient, with ambient pressure 100 kPa. Someone inserts 150 kPa into $n=\frac{pV}{RT}$ . Using $R=8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}$ , find the correct amount and the factor by which this calculation is too small.

Show hint

The equation of state requires absolute pressure. Add ambient pressure and convert litres to cubic metres.

Learning goal: Convert gauge pressure to absolute pressure before using the equation of state.

Show solution and calculation

Approach: establish the correct pressure reference, then solve the equation of state for n.
Absolute pressure 250 kPa gives about 1.203 mol, rather than 0.722 mol.
The correct value is 1.667 times the incorrect one.

Check: pressure times volume has units of joules; positive ambient pressure must increase the inferred amount.

\begin{aligned}p_{\mathrm{abs}}&=150\,\mathrm{kPa}+100\,\mathrm{kPa}=250\,\mathrm{kPa}\\n&=\frac{p_{\mathrm{abs}}V}{RT}=\frac{(250\,000\,\mathrm{Pa})(0.0120\,\mathrm{m^3})}{(8.314\,\mathrm{J/(mol\,K)})(300\,\mathrm K)}\approx1.203\,\mathrm{mol}\\\frac{n_{\mathrm{correct}}}{n_{\mathrm{wrong}}}&=\frac{250}{150}=\frac53\end{aligned}

[1][7]

Limits: when is the ideal model insufficient?

At high density, particle volume and interactions matter; deviations may be substantial near condensation. There is no universal pressure limit valid for every gas. Suitability depends on the gas, temperature and required accuracy.[4]

Z=\frac{pV}{nRT}

The compressibility factor Z measures deviation; Z equals one for an ideal gas.[4]

A real-gas calculation may use measured Z values or a suitable real equation of state. The van der Waals equation approximately accounts for attraction and particle volume; it too has limits. A temperature change alone does not specify heat or work.[4]

Key points

$pV=nRT$ describes an equilibrium state of an ideal gas.
Use absolute temperature and absolute pressure; match the units to R.
Identify fixed quantities before comparing two states.
$n=\frac mM$ gives amount of substance; $\rho=\frac{pM}{RT}$ gives ideal-gas density.
Assess real deviations using Z or an appropriate equation of state.

Before comparing states, decide which quantities remain fixed. Absolute pressure and kelvin readings are essential. The gas equation connects states; the actual path additionally determines heat and work.[1]

Sources for this explanation

Quanta · Updated October 2, 2026

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This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 2, 2026Added a state comparison with absolute pressure and absolute temperature, distinctions from kinetic theory and internal energy, and a source on heat capacities of ideal gases; preserved existing derivations, exercises and original figures.
October 1, 2026Expanded explanations and learning transitions; added a further fully worked example and an original quantitative diagram; clarified assumptions and calculation steps.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.

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Quanta. Ideal Gas Law. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/ideales-gasgesetz

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Frequently asked questions about Ideal Gas Law

How do you calculate the volume of an ideal gas?+

Rearrange $pV=nRT$ for the volume: $V=\frac{nRT}{p}$ . Insert the amount of substance n in mol, the universal gas constant $R=8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}$ , the absolute temperature T in kelvin and the pressure p in pascals, then the volume comes out in cubic metres. Example: 5 mol of nitrogen at 300 K and $1\,\mathrm{atm}=101325\,\mathrm{Pa}$ give $V=\frac{5\cdot8.314\cdot300}{101325}\,\mathrm{m^3}\approx0.123\,\mathrm{m^3}$ , that is 123 litres. Make sure to give the pressure in pascals and the temperature in kelvin; atm, bar or Celsius must be converted first. One cubic metre equals 1000 litres.

Why must the temperature in the gas law be in kelvin?+

Gas laws compare absolute temperatures because classical molecular motion depends on that scale. At fixed amount and constant pressure, $V\propto T$ ; at fixed volume, instead $p\propto T$ . Both quantities cannot be treated as independently proportional to temperature at the same time. Heating from 10 to 20 °C is therefore not a doubling: the absolute temperatures are 283.15 and 293.15 K. Convert to kelvin before taking a temperature ratio. For temperature differences, however, the constant scale offset cancels. This classical model does not mean that real gas remains ideal down to 0 K; condensation and quantum effects can become important first.

What happens to the pressure when you heat a gas at constant volume?+

At constant volume and constant amount of substance, $pV=nRT$ implies that the pressure is directly proportional to the absolute temperature: $p\propto T$ . Doubling the temperature in kelvin doubles the pressure. This is the law of Amontons or Gay-Lussac. Intuitively, at higher temperature the gas particles collide with the walls faster and more often, which raises the pressure. That is why the pressure in a closed container, such as a spray can, rises when heated and can become dangerous. For the calculation $\frac{p_1}{T_1}=\frac{p_2}{T_2}$ holds, each with temperatures in kelvin.

When does the ideal gas law fail?+

The ideal model neglects particle volume and interactions between collisions. Whether this approximation is adequate depends on species, temperature, density and required accuracy. High density and proximity to condensation are common warning signs, but no pressure threshold applies to every gas. Check the compressibility factor $Z=\frac{pV}{nRT}$ against measurements or an appropriate real equation of state. If it is close to one, the ideal approximation may meet the requested accuracy. A van der Waals equation is one possible extended model, though it is not universally accurate either. Appropriate state data and acceptable deviation determine which description to use.

What is the difference between the amount of substance n and the number of particles?+

The amount of substance n is given in moles and counts portions of particles. One mole contains exactly the Avogadro number of particles, that is about $6.022\times{}10^{23}$ . The number of particles N is the actual count of individual atoms or molecules and is linked to the amount of substance through $N=nN_A$ . In the ideal gas law $pV=nRT$ you work with the amount of substance and the gas constant R. Alternatively there is the form $pV=Nk_BT$ with the number of particles and the Boltzmann constant $k_B=\frac{R}{N_A}$ . Both forms are equivalent; you choose, depending on the task, whether moles or the particle number is given.

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax Chemistry 2e, §9.2 Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law

Related formulas

\left(p+a\frac{n^2}{V^2}\right)(V-nb)=nRT

van der Waals Equation

K_p/K_c=(RTc^\circ/p^\circ)^{\Delta n_{gas}}

Relationship between Kp and Kc

More formulas: Gases and energetics

\Delta G = \Delta H - T \cdot \Delta S

Gibbs Energy (Free Enthalpy)

\Delta H_R^0 = \sum \Delta H_f^0(\text{Produkte}) - \sum \Delta H_f^0(\text{Edukte})

Reaction Enthalpy (Hess's Law)

\Delta H_{\text{Gitter}} = \Delta H_f^0 - \Delta H_{\text{Sub}} - I - \tfrac{1}{2} D - E_{\text{EA}}

Lattice Energy (Born-Haber Cycle)

C_g = k_H \cdot p_g

Henry's Law (Concentration Convention)

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Ideal gas law

The ideal gas law relates the pressure, volume, amount of substance and temperature of an ideal gas.

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Symbol

Meaning

Unit

p

Pressure

V

Volume

m³ or L

n

Amount of substance

mol

R

Universal gas constant (8.314)

J/(mol·K)

T

Temperature (in kelvin!)

Exam blueprint

Validity range

Derivation steps

The ideal equation of state combines three empirical proportionalities, each with specified fixed conditions.

1For the same ideal sample at fixed T, $V\propto\frac1p$ .
2At fixed n and p, $V\propto T$ ; T is absolute kelvin temperature.
3At fixed p and T, $V\propto n$ .
4Combine the three empirical proportionalities: $V\propto\frac{nT}{p}$ .
5Introduce the molar gas constant: $V=R\frac{nT}{p}$ ; multiplying by p gives $pV=nRT$ .
6For fixed n, $\frac{p_1V_1}{T_1}=\frac{p_2V_2}{T_2}$ . Compare absolute pressures and kelvin temperatures; check real-gas deviations separately.

Understand the situation before calculating

What does the ideal gas equation describe?

pV=nRT

Pressure times volume equals amount of substance times the gas constant times absolute temperature.[1]

Choose consistent quantities and units

\mathrm{Pa\,m^3}=\mathrm{J}=\mathrm{mol}\,\frac{\mathrm{J}}{\mathrm{mol\,K}}\,\mathrm{K}

Both sides have the unit joule.

Temperature: $\frac T{\mathrm K}=\frac t{{}^\circ\mathrm C}+273.15$ . Doubling the Celsius reading does not double absolute temperature.
Volume: $1\,\mathrm L=10^{-3}\,\mathrm{m^3}$ . Pressure: $1\,\mathrm{bar}=10^5\,\mathrm{Pa}$ .
Use absolute pressure. A gauge reading of 2 bar above an ambient pressure of 1 bar corresponds to 3 bar absolute.

[5][7]

Establish the equation from the gas laws

V\propto\frac1p\quad(n,T\;\mathrm{const.})

At fixed amount and temperature, volume is inversely proportional to pressure.

V\propto T\quad(n,p\;\mathrm{const.})

At fixed amount and pressure, volume is proportional to kelvin temperature.

V\propto n\quad(p,T\;\mathrm{const.})

At fixed pressure and temperature, volume is proportional to amount of substance.

\begin{aligned}V&=R\frac{nT}{p}\\pV&=nRT\end{aligned}

Introduce the common proportionality constant R and multiply by p.

For the same gas sample, n stays fixed. Dividing by T therefore gives a direct comparison of two states. If n changes through a leak or reaction, this simplified comparison no longer applies.

\frac{p_1V_1}{T_1}=\frac{p_2V_2}{T_2}\quad(n\;\mathrm{const.})

Pressure times volume divided by kelvin temperature stays constant for the same ideal gas sample.

How does particle motion explain the relationship?

\begin{aligned}&pV=\frac13Nm\langle v^2\rangle\\&\frac12m\langle v^2\rangle=\frac32kT\end{aligned}

Pressure–volume product and mean translational kinetic energy in a classical ideal gas.

pV=NkT=nN_{\!A}kT=nRT

With particle count N equal to n times the Avogadro constant and R equal to N A times k, the molar form follows.

Rearrange, compare states or calculate density?

Identify the unknown and list all given values with their units.
Decide whether you need one equation of state or a comparison between states.
Rearrange symbolically first, then convert units and substitute numbers.
Check the result against the proportionalities and its units.

\begin{aligned}&p=\frac{nRT}{V}\\&V=\frac{nRT}{p}\\&n=\frac{pV}{RT}\\&T=\frac{pV}{nR}\end{aligned}

Rearrangements for pressure, volume, amount and temperature; denominators must be nonzero.

If mass m is given instead of n, use $n=\frac mM$ with molar mass M. Combining this with $\rho=\frac mV$ gives the density relation. Use M in kg/mol when finding ρ in kg/m³.

\rho=\frac{pM}{RT}

Density of an ideal gas from pressure, molar mass and temperature.

Two temperatures, the same amount of gas

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Three examples: states and constraints

Volume at a known pressure

Given: $n=0.500\,\mathrm{mol}$ , $T=300\,\mathrm K$ , $p=100\,\mathrm{kPa}$ absolute. Ideal gas at equilibrium.
Find: Find V in litres.

Approach: Use the ideal gas equation $pV=nRT$ , rearranged as $V=\frac{nRT}{p}$ . Convert 100 kPa to 100 000 Pa. Use R in SI units.
Divide by p to isolate V.
$V=\frac{nRT}{p}$
Substitute: SI units satisfy $\mathrm{Pa\,m^3}=\mathrm J$ , giving the calculated volume in cubic metres.
$V=\frac{0.500\cdot8.314462618\cdot300}{100000}\,\mathrm{m^3}\approx0.0124717\,\mathrm{m^3}$
Multiply cubic metres by 1 000 to convert to litres.

Result: V ≈ 12.5 L with appropriate rounding.

Check: One mole would occupy twice this volume under the same conditions, providing a useful plausibility check.

Heating a sealed container

Given: Initial absolute pressure $p_1=150\,\mathrm{kPa}$ . Heating from 20 °C to 80 °C at unchanged amount and volume.
Find: Find the new absolute pressure p₂.

Approach: at fixed n and V, $\frac{p_2}{p_1}=\frac{T_2}{T_1}$ . First convert both temperatures to kelvin: $T_1=293.15\,\mathrm K$ , $T_2=353.15\,\mathrm K$ .
At fixed n and V, these quantities cancel from the state comparison.
$\frac{p_1}{T_1}=\frac{p_2}{T_2}$
Multiply by T₂ and substitute.
$p_2=p_1\frac{T_2}{T_1}=150\,\mathrm{kPa}\cdot\frac{353.15}{293.15}\approx180.70\,\mathrm{kPa}$

Result: p₂ ≈ 181 kPa absolute. Pressure rises by about 20.5 percent.

Check: A ratio of Celsius readings $\frac{80}{20}=4$ would be wrong: kelvin temperature increases by a factor of only 1.205. Both pressures may stay in kPa because this is a ratio.

The following comparison uses only the ideal equation of state. The mechanical constraint determines whether pressure or volume changes.[1][7]

The same heating, two different responses

Given: A closed ideal gas sample: $n=0.200\,\mathrm{mol}$ , $T_1=300\,\mathrm K$ , $V_1=5.00\,\mathrm L$ , heated to $T_2=450\,\mathrm K$ . A: rigid vessel. B: frictionless piston, slow heating at constant equilibrium pressure.
Find: Find initial pressure, final pressure A and final volume B.

Approach: calculate the common initial state first. Then hold volume fixed in A and pressure fixed in B.
$\begin{aligned}p_1&=\frac{nRT_1}{V_1}\\p_{2,A}&=p_1\frac{T_2}{T_1}\\V_{2,B}&=V_1\frac{T_2}{T_1}\end{aligned}$
Substitute: convert litres to cubic metres; the temperature ratio is dimensionless.
$\begin{aligned}p_1&=\frac{(0.200)(8.314462618)(300)}{0.00500}\,\mathrm{Pa}\\&\approx99.774\,\mathrm{kPa}\\p_{2,A}&=\frac{(0.200)(8.314462618)(450)}{0.00500}\,\mathrm{Pa}\\&\approx149.660\,\mathrm{kPa}\\V_{2,B}&=5.00\cdot\frac{450}{300}\,\mathrm L=7.50\,\mathrm L\end{aligned}$

Result: A reaches about 150 kPa at 5.00 L. B reaches 7.50 L at unchanged pressure about 99.8 kPa.

Try it: five exercises with solutions

Calculate · Foundation

1. An ideal gas sample contains 1.00 mol at 300 K and 100 kPa. Find its volume.

Show hint

Use $V=\frac{nRT}{p}$ ; convert kPa to Pa first.

Learning goal: Find gas volume from amount, absolute temperature and pressure.

Show solution and calculation

Approach: rearrange the equation of state for volume and express pressure in pascals.
Volume is about 24.9 L.

Check: only amount is doubled compared with Example 1, so volume must double.

\begin{aligned}V&=\frac{nRT}{p}\\&=\frac{(1.00)(8.314462618)(300)}{100000}\,\mathrm{m^3}\\&\approx0.0249434\,\mathrm{m^3}=24.9434\,\mathrm L\end{aligned}

Calculate · Application

2. At fixed amount and temperature, pressure rises from 100 kPa to 250 kPa. Initial volume: 10.0 L. Find V₂.

Show hint

Use $p_1V_1=p_2V_2$ . Pressure units cancel in the ratio; volume can stay in litres.

Learning goal: Calculate isothermal volume change at fixed amount of gas.

Show solution and calculation

Approach: pressure times volume stays constant for the same amount at equal temperature.
Solve for final volume and use the dimensionless pressure ratio.

Check: pressure increases by 2.5, so volume must be divided by 2.5.

\begin{aligned}p_1V_1&=p_2V_2\\V_2&=\frac{p_1}{p_2}V_1\\&=\frac{100\,\mathrm{kPa}}{250\,\mathrm{kPa}}(10.0\,\mathrm L)=4.00\,\mathrm L\end{aligned}

Error analysis · Application

3. A sealed container is heated at fixed volume from 27 °C to 127 °C. Does pressure increase by a factor of $\frac{127}{27}$ ?

Show hint

The temperature ratio must use kelvin.

Learning goal: Identify erroneous pressure scaling using Celsius readings.

Show solution and calculation

Approach: at fixed amount and volume, the pressure ratio equals the ratio of absolute temperatures.
Convert Celsius to kelvin first.
No: pressure increases by about one third.

Check: equal additive Celsius intervals do not give equal temperature ratios because the scales have different zeros.

\begin{aligned}T_1&=300.15\,\mathrm K\\T_2&=400.15\,\mathrm K\\\frac{p_2}{p_1}&=\frac{T_2}{T_1}=\frac{400.15}{300.15}\approx1.33317\end{aligned}

Interpret · Application

A rigid ideal-gas vessel initially has absolute pressure 200 kPa at 300 K, later 240 kPa at 400 K. Volume stays constant. A sketch shows an open valve. What change in amount of gas do the data require? Can the vessel be treated as closed with fixed n throughout the change?

Show hint

At constant volume, n is proportional to $\frac pT$ , not to pressure alone.

Learning goal: Recognize changing amount of gas from state data rather than assuming a closed process.

Show solution and calculation

Approach: divide the ideal-gas laws for both states; volume and gas constant cancel.
Final amount is 90% of the initial amount, a 10% decrease.
A closed model with unchanged n does not fit these data.

Check: fixed n would require final pressure about 266.7 kPa. The states do not determine the detailed time-dependent valve flow.

\begin{aligned}\frac{n_2}{n_1}&=\frac{p_2T_1}{p_1T_2}=\frac{240\cdot300}{200\cdot400}=0.90\\p_{2,n=\mathrm{const}}&=p_1\frac{T_2}{T_1}\approx266.7\,\mathrm{kPa}\end{aligned}

[1]

Error analysis · Application

An ideal gas occupies 12.0 L at 300 K. A gauge reads 150 kPa above ambient, with ambient pressure 100 kPa. Someone inserts 150 kPa into $n=\frac{pV}{RT}$ . Using $R=8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}$ , find the correct amount and the factor by which this calculation is too small.

Show hint

The equation of state requires absolute pressure. Add ambient pressure and convert litres to cubic metres.

Learning goal: Convert gauge pressure to absolute pressure before using the equation of state.

Show solution and calculation

Approach: establish the correct pressure reference, then solve the equation of state for n.
Absolute pressure 250 kPa gives about 1.203 mol, rather than 0.722 mol.
The correct value is 1.667 times the incorrect one.

Check: pressure times volume has units of joules; positive ambient pressure must increase the inferred amount.

\begin{aligned}p_{\mathrm{abs}}&=150\,\mathrm{kPa}+100\,\mathrm{kPa}=250\,\mathrm{kPa}\\n&=\frac{p_{\mathrm{abs}}V}{RT}=\frac{(250\,000\,\mathrm{Pa})(0.0120\,\mathrm{m^3})}{(8.314\,\mathrm{J/(mol\,K)})(300\,\mathrm K)}\approx1.203\,\mathrm{mol}\\\frac{n_{\mathrm{correct}}}{n_{\mathrm{wrong}}}&=\frac{250}{150}=\frac53\end{aligned}

[1][7]

Limits: when is the ideal model insufficient?

Z=\frac{pV}{nRT}

The compressibility factor Z measures deviation; Z equals one for an ideal gas.[4]

Key points

$pV=nRT$ describes an equilibrium state of an ideal gas.
Use absolute temperature and absolute pressure; match the units to R.
Identify fixed quantities before comparing two states.
$n=\frac mM$ gives amount of substance; $\rho=\frac{pM}{RT}$ gives ideal-gas density.
Assess real deviations using Z or an appropriate equation of state.

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

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An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 2, 2026Added a state comparison with absolute pressure and absolute temperature, distinctions from kinetic theory and internal energy, and a source on heat capacities of ideal gases; preserved existing derivations, exercises and original figures.
October 1, 2026Expanded explanations and learning transitions; added a further fully worked example and an original quantitative diagram; clarified assumptions and calculation steps.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.

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Quanta. Ideal Gas Law. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/ideales-gasgesetz

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Frequently asked questions about Ideal Gas Law

How do you calculate the volume of an ideal gas?+

Why must the temperature in the gas law be in kelvin?+

What happens to the pressure when you heat a gas at constant volume?+

When does the ideal gas law fail?+

What is the difference between the amount of substance n and the number of particles?+

Ideal gas law

Variables & units

Ideal gas law calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Thermodynamics and amount of substance

Worked example

Applications

How do you calculate with Ideal Gas Law?

Understand the situation before calculating

What does the ideal gas equation describe?

Choose consistent quantities and units

Establish the equation from the gas laws

How does particle motion explain the relationship?

Rearrange, compare states or calculate density?

Two temperatures, the same amount of gas

Three examples: states and constraints

Volume at a known pressure

Heating a sealed container

The same heating, two different responses

Try it: five exercises with solutions

1. An ideal gas sample contains 1.00 mol at 300 K and 100 kPa. Find its volume.

2. At fixed amount and temperature, pressure rises from 100 kPa to 250 kPa. Initial volume: 10.0 L. Find V2.

3. A sealed container is heated at fixed volume from 27 °C to 127 °C. Does pressure increase by a factor of 12727\frac{127}{27}27127​?

A rigid ideal-gas vessel initially has absolute pressure 200 kPa at 300 K, later 240 kPa at 400 K. Volume stays constant. A sketch shows an open valve. What change in amount of gas do the data require? Can the vessel be treated as closed with fixed n throughout the change?

Limits: when is the ideal model insufficient?

Key points

Sources for this explanation

Frequently asked questions about Ideal Gas Law

Subject references and further reading

Related formulas

More formulas: Gases and energetics

Retain Ideal Gas Law for exams

Ideal gas law

Variables & units

Ideal gas law calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Thermodynamics and amount of substance

Worked example

Applications

How do you calculate with Ideal Gas Law?

Understand the situation before calculating

What does the ideal gas equation describe?

Choose consistent quantities and units

Establish the equation from the gas laws

How does particle motion explain the relationship?

Rearrange, compare states or calculate density?

Two temperatures, the same amount of gas

Three examples: states and constraints

Volume at a known pressure

Heating a sealed container

The same heating, two different responses

Try it: five exercises with solutions

1. An ideal gas sample contains 1.00 mol at 300 K and 100 kPa. Find its volume.

2. At fixed amount and temperature, pressure rises from 100 kPa to 250 kPa. Initial volume: 10.0 L. Find V2.

3. A sealed container is heated at fixed volume from 27 °C to 127 °C. Does pressure increase by a factor of 12727\frac{127}{27}27127​?

A rigid ideal-gas vessel initially has absolute pressure 200 kPa at 300 K, later 240 kPa at 400 K. Volume stays constant. A sketch shows an open valve. What change in amount of gas do the data require? Can the vessel be treated as closed with fixed n throughout the change?

Limits: when is the ideal model insufficient?

Key points

Sources for this explanation

Frequently asked questions about Ideal Gas Law

Subject references and further reading

Related formulas

More formulas: Gases and energetics

Retain Ideal Gas Law for exams

2. At fixed amount and temperature, pressure rises from 100 kPa to 250 kPa. Initial volume: 10.0 L. Find V₂.

3. A sealed container is heated at fixed volume from 27 °C to 127 °C. Does pressure increase by a factor of $\frac{127}{27}$ ?

2. At fixed amount and temperature, pressure rises from 100 kPa to 250 kPa. Initial volume: 10.0 L. Find V₂.

3. A sealed container is heated at fixed volume from 27 °C to 127 °C. Does pressure increase by a factor of $\frac{127}{27}$ ?