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The ideal gas equation of state

Apply the ideal-gas equation of state with consistent units, checking its assumptions and limits.

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  • University Physics Year 1
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An equation of state connects quantities that describe an equilibrium state. For a dilute classical ideal gas, PV = nRT = NkBT. It tells you the pressure of a specified amount at a temperature and volume. It does not say how the gas reached that state or how much heat was supplied: those require an energy balance and a process.

Use n for amount in moles, N for particle count, R = NAkB for the molar gas constant and kB for the single-particle constant. Build on the mole concept, pressure and volume units.

Make the pressure and temperature absolute

P is absolute pressure, measured relative to vacuum. A gauge reports the difference from the local ambient pressure, so P_abs = P_gauge + P_ambient. A zero gauge reading does not mean no molecular pressure. Use kelvin for T; the conversion is T/K = θ/°C + 273.15.

In SI units, Pa times m³ gives J, matching mol times J/(mol K) times K. Convert litres to m³ before substitution. For particle count use kB; for moles use R. Changing the units does not change the molecules.

Understand what the ideal model omits

The gas occupies an equilibrium volume with negligible molecular volume compared with that volume. Intermolecular interactions make negligible corrections to its state relation. The classical treatment also excludes quantum-degenerate regimes. Random molecular motion transfers momentum to the walls, producing the thermodynamic pressure; temperature sets the mean translational kinetic-energy scale.

The law is not limited to monatomic gases. Rotational and vibrational modes can change heat capacity without changing the ideal state relation. Conversely, PV is an energy-sized quantity but is not the gas’s whole internal energy: even a classical monatomic ideal gas has thermal internal energy 3nRT/2, rather than nRT.

The compressibility factor Z = PV/nRT compares a measured state with the ideal prediction. Z close to one supports that prediction for the specified conditions and required accuracy; it does not prove that every thermodynamic property is ideal. Dense gases, strong attractions near condensation and very low-temperature quantum regimes can require another model. A density expansion such as Z = 1 + B₂(T)(N/V) + … makes the first correction explicit; in this number-density convention B₂ has units of volume per particle. Do not use a molar-density coefficient without converting its convention.

Use the state law to connect changes

For fixed amount, comparing equilibrium endpoints gives PfVf/Tf = PᵢVᵢ/Tᵢ. At fixed volume, pressure follows absolute temperature; at fixed temperature, pressure is inversely proportional to volume. These are state constraints. A piston path, heat input or leak is extra information, and a fixed-amount ratio is invalid if gas enters or escapes.

Guided practice 1

Predict a cylinder gauge reading

About 8 min

Problem

A sealed rigid ideal-gas cylinder has gauge pressure 150 kPa at 27 °C. Local atmospheric pressure stays at 100 kPa. It warms to 87 °C without leaking. Using the exact Celsius-to-kelvin offset 273.15, find the initial absolute pressure, final absolute pressure and final gauge pressure. Explain why scaling the gauge value directly would fail.

Work through the model, then compare your reasoning

Unit: kPa
Unit: kPa
Unit: kPa

Hints

Hint 1

Check the model

Add the ambient pressure before taking the temperature ratio; subtract it only after finding the new absolute pressure.
Show solution step by step

  1. Construct the state variables

    Method

    Use absolute pressure and kelvin temperature.

    Reason

    Gauge zero and Celsius zero are offsets, so neither belongs in a proportionality.

    Working

    Answer

    Pᵢ = 150 + 100 = 250 kPa, Tᵢ = 300.15 K and Tf = 360.15 K.

  2. Apply the fixed-volume relation

    Method

    Calculate Pf = PᵢTf/Tᵢ.

    Reason

    The sealed rigid cylinder fixes both amount and volume.

    Working

    Answer

    Pf = 250(360.15/300.15) = 299.975… kPa. The final gauge reading is Pf-100 = 199.975… kPa, about 200 kPa. Scaling 150 kPa directly would predict about 180 kPa and omit the ambient part of the initial molecular pressure. No heat input has been calculated: the equation of state alone does not provide it.

Pressure of helium in a cylinder

Worked example 2

Helium pressure

Problem

A cylinder contains 0.50 mol of helium at 300 K in 6.0 × 10⁻³ m³. Treat it as ideal and find its pressure using R = 8.314 J mol⁻¹K⁻¹.
Show full solution

  1. Choose the molar gas law

    Method

    Rearrange PV = nRT to P = nRT/V.

    Reason

    The amount is in moles, so use R, not kB. The supplied volume and temperature are already in SI units.

    Working

    Answer

    P = nRT/V = (0.50)(8.314)(300)/(6.0 × 10⁻³) = 2.08 × 10⁵ Pa.

  2. State the pressure

    Working

    Answer

    P ≈ 2.1 × 10⁵ Pa to two significant figures, about twice atmospheric pressure. The units are J/m³ = Pa.

Judge the model against a measured state

Check your understanding 1

A gas has n = 0.50 mol, T = 300 K and V = 6.0 × 10⁻³ m³. Its measured absolute pressure is 230 kPa. Use R = 8.314 J/(mol K) to find the ideal prediction and Z. An application permits a pressure prediction error of 3% relative to the measured pressure. Does the ideal prediction meet that requirement? Does the direction of the discrepancy identify its cause uniquely?

Show answer

P_ideal = nRT/V = 207850 Pa and Z = 230000/207850 ≈ 1.11. The ideal prediction is about 9.6% below the measurement, so it fails the stated 3% requirement. The sign alone does not identify a microscopic cause; interactions, model conditions and measurement or input uncertainty need examination. State the observed mismatch before choosing a better model.

Practice set

Check your understanding 2

An ideal gas has P = 1.2 × 10⁵ Pa, V = 0.020 m³ and T = 400 K. Find its amount in moles using R = 8.314 J mol⁻¹K⁻¹.

Show hint
Make n the subject of PV = nRT.
Show answer
n = PV/(RT) = (1.2 × 10⁵)(0.020)/(8.314 × 400) ≈ 0.72 mol. Pressure times volume has units J, so dividing by RT gives mol.

Check your understanding 3

An ideal gas’s volume triples at fixed amount and temperature. What happens to pressure?

Show hint
The product PV stays constant.

Show answer

Pressure becomes one-third of its initial value: Pf(3Vᵢ) = PᵢVᵢ, so Pf = Pᵢ/3.

Check your understanding 4

An ideal gas has 3.0 × 10²² molecules at 350 K in 1.0 × 10⁻³ m³. Find P using kB = 1.38 × 10⁻²³ J K⁻¹.

Show hint
Use the particle-count form, not the molar form.
Show answer
P = NkBT/V = (3.0 × 10²²)(1.38 × 10⁻²³)(350)/(1.0 × 10⁻³) = 1.449 × 10⁵ Pa ≈ 1.4 × 10⁵ Pa to two significant figures. Using R with a molecule count would give the wrong scale.

What to keep

Use the ideal gas law with absolute pressure, kelvin temperature and a consistent particle-count or mole convention. Its state relation does not supply heat, work or internal energy. Check model accuracy against conditions or data before relying on an ideal prediction.

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Syllabus and review details

No official syllabus alignment is listed for this lesson.

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