SCIENCE · INVESTIGATE AND EXPLAIN

Relate ideal-gas pressure, volume and temperature

Compare equilibrium states while keeping amount, units and controlled variables explicit.

Learning outcome: Apply PV = nRT with absolute pressure and kelvin temperature, compare constrained state changes and identify limits of the ideal model.

Starting recap

A mole measures amount of substance. Absolute pressure is measured relative to vacuum; temperature ratios require kelvin.

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Try

A sealed gas sample doubles its volume. Is its final pressure necessarily half its initial pressure?

Open the experiment

Predict

Compare 300 K at 20 L with 300 K at 40 L, then compare 300 K at 20 L with 600 K at 40 L. Predict which final state restores the original pressure.

Your prediction and reason

Experiment

Capture the reference and both changed states. Calculate PV/T for each. Explain which variable was controlled and why selecting endpoints does not specify heat or work.

Change an ideal gas state while retaining one mole

Worked teaching model · synthetic values · One mole of nonreacting ideal gas at equilibrium. Pressure is absolute, not gauge pressure. R is rounded to 8.314 L kPa/(mol K). States are imposed; the model does not calculate transient heating, piston dynamics, leaks, condensation or real-gas corrections.

Absolute gas temperature: 300 K Gas volume: 20 L

Reset keeps your prediction, captures and practice history.

The moving boundary represents selected chamber volume on a schematic scale. The twelve marks remind you that amount stays fixed; they are not twelve molecules or measured particle positions. The diagram does not simulate collisions or piston motion. No time evolution is implied by changing a control.

Calculated values on the same fixed scale
Row	Quantity	Value (kPa)
1	Current absolute pressure	124.71
2	Reference: 1 mol at 300 K and 20 L	124.71

P = 1 × 8.314 × 300 / 20 = 124.710 kPa absolute.

Current PV = 2494.200 L kPa. Temperature 300 K equals 26.85 °C. Compare states at fixed temperature to test inverse volume scaling, or at fixed volume to test absolute-temperature scaling. This is an equilibrium relation, not a predicted process path.

Observation to attach to a capture

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Keep the state variables and units together

The investigation retains one mole of ideal gas while you select its temperature and volume. At 300 K and 20 L, pressure is 124.710 kPa absolute. Predict the result of doubling volume before moving the control. Your prediction needs a condition: what happens to temperature?

The ideal-gas relation is PV = nRT. P is absolute pressure, V volume, n amount in moles, T absolute temperature in kelvin and R the molar gas constant. The widget uses rounded R = 8.314 L kPa/(mol K), so its pressure is in kPa when volume is in litres. One L kPa equals one joule because 0.001 m³ × 1000 Pa = 1 J. Thus the same numerical R is 8.314 J/(mol K) when using cubic metres and pascals consistently.

One mole contains 6.02214076 × 10²³ elementary entities. Specify what those entities are: molecules for a molecular gas, atoms for a monatomic gas. A mole is not a fixed mass across species. The diagram’s twelve marks are schematic reminders of a fixed amount, not twelve molecules, a collision simulation or measured positions.

Compare controlled states

At fixed n and T, PV is constant. At 300 K, changing volume from 20 to 40 L reduces pressure from 124.710 to 62.355 kPa. Halving volume to 10 L raises it to 249.420 kPa. The plotted chamber width uses one fixed volume scale so the geometry agrees with the selected amount of space.

At fixed n and V, P/T is constant. Changing from 300 to 600 K at 20 L doubles pressure to 249.420 kPa. That temperature change is from 26.85 to 326.85 °C. Celsius readings did not double: ratios require kelvin.

Change both temperature and volume from 300 K, 20 L to 600 K, 40 L. The ratio T/V remains 15 K/L, so pressure returns to 124.710 kPa. A pressure reading alone does not determine temperature and volume separately.

For a fixed amount, the general comparison is P₂/P₁ = (T₂/T₁)(V₁/V₂). Subscripts label the two states, not time steps of a simulated process. At constant pressure the relation instead requires V/T constant.

Use absolute pressure

The gas law uses pressure relative to vacuum. A gauge reading is relative to a reference pressure, often local atmospheric pressure: Pabsolute = Pgauge + Preference. If the reference is 100 kPa, a gauge reading of 50 kPa corresponds to 150 kPa absolute. Doubling that absolute pressure gives 300 kPa absolute, or 200 kPa gauge at the same reference. Doubling the gauge reading would be incorrect.

A negative gauge pressure can still be positive absolute pressure. The current gas controls always yield positive absolute pressure and never represent a gas at zero kelvin or zero volume.

An equation of state is not a process simulation

The OpenStax ideal-gas reference describes the equilibrium relationship and its molecular assumptions. It does not say that every rapid compression stays at constant temperature. Keeping T fixed during a real slow compression generally requires heat exchange with surroundings. An insulated compression can change T.

The controls select equilibrium endpoints. They do not calculate piston forces, elapsed time, heat supplied, work done or the path between states. Two different processes can share endpoints but exchange different amounts of heat and work. These distinctions become essential in the first-law lessons.

The ideal approximation neglects finite particle volume and intermolecular interactions except collisions. A real gas may deviate at high density or near condensation; the widget does not determine a species-specific validity boundary. Its 200–600 K range is a teaching domain, not a promise that every substance remains an ideal gas there.

Transfer to different amount and units

A sealed ideal gas begins at 150 kPa absolute, 12 L and 300 K. It ends at 450 K and 18 L with the same amount. Find final absolute pressure and amount in moles. Then express the initial PV in joules.

Check the gas-state transfer

P₂ = 150 × (450/300) × (12/18) = 150 kPa. The amount is n = 150 × 12/(8.314 × 300), approximately 0.722 mol. Initial PV = 1800 L kPa = 1800 J. The amount is not the widget's fixed one mole, so this is an independent transfer calculation. No heat or work value follows from these endpoints alone.

Apply

Public practice checks a specific claim. It does not establish independent mastery.

Explain the conditions behind inverse pressure-volume scaling, distinguish absolute from gauge pressure and state why the gas law alone cannot predict a process.

Your explanation and limitations

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

Skipping retains experiments and awards no mastery.

Previous: Combine convection and radiation at a surface Next: Connect molecular motion to gas pressure

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