SCIENCE · INVESTIGATE AND EXPLAIN

Explain gas pressure with a controlled volume comparison

Keep amount and temperature fixed while testing the inverse pressure-volume relationship.

Learning outcome: Use absolute pressure, units and controlled comparisons to explain ideal-gas compression and identify model limits.

Starting recap

Pressure is force per area. A mole counts amount of substance; litres measure volume. The same amount can occupy different volumes.

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Try

One mole at 300 K occupies 20 L. What happens to the model pressure if its equilibrium volume becomes 10 L while temperature is held fixed?

Open the experiment

Predict

Predict the pressure ratio for halving volume. State what must remain fixed for that prediction to apply.

Your prediction and reason

Experiment

Capture (300,20), (300,10), and (300,40). Compare P and PV. Then change temperature as well and explain why that is no longer a volume-only test.

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

Compress the space without changing the amount

Start with one mole of ideal gas at 300 K in 20 L. The model gives 124.71 kPa absolute pressure. Reduce the equilibrium volume to 10 L while retaining one mole and 300 K: the pressure becomes 249.42 kPa. Expand instead to 40 L and it becomes 62.355 kPa. The amount is unchanged in all three states.

Predict these comparisons before moving the control. The fixed-volume scale and unchanged inventory marks help distinguish a denser sample from a larger amount. The marks are schematic reminders of fixed amount, not simulated molecular trajectories or a literal count of the particles in a mole.

Pressure connects microscopic motion with a macroscopic force

Gas particles move in many directions and transfer momentum when they collide with a wall. The resulting force distributed over wall area is pressure. In an ideal-gas picture, reducing the container volume at the same temperature and amount increases the collision activity per wall area in a way that increases pressure. A gas does not exert pressure only on the bottom of a container as if it were a pile of resting objects.

Temperature held fixed is essential. A rapid physical compression can change the temperature; maintaining it may require energy exchange with the surroundings. The widget compares imposed equilibrium states and does not calculate this transient process. A slider movement is not a safe experimental procedure for a real pressurized vessel.

The ideal model neglects particle volume compared with the container and neglects intermolecular forces except idealized collisions. Its usefulness depends on conditions and the substance. The model here deliberately remains ideal across its control range; a displayed state is not evidence that every real gas stays gaseous or ideal there.

Derive the inverse relationship with units

The ideal-gas relation is PV = nRT. P is absolute pressure in kilopascals (kPa), V is volume in litres (L), n is amount in moles (mol), and T is absolute temperature in kelvin (K). For these units the rounded teaching value of R is 8.314 L kPa/(mol K). R links the selected unit system; changing pressure or volume units without converting R would break the calculation.

For the starting state, P = nRT/V = (1 × 8.314 × 300)/20 = 124.71 kPa. The mol and K units cancel, leaving kPa. The lab keeps n = 1 mol. Multiplying the resulting pressure by its volume gives 2,494.2 L kPa for every 300 K comparison.

With n and T fixed, nRT is constant. Therefore P1V1 = P2V2, and P2/P1 = V1/V2. Halving volume doubles pressure; doubling volume halves it. This is an inverse relationship, not a straight line in a pressure-versus-volume graph. Plotting pressure against 1/V would give a straight line for these ideal fixed-temperature states.

Absolute pressure is not a gauge reading

Absolute pressure uses a vacuum reference. Gauge pressure commonly reports a difference from the surrounding atmosphere. Suppose the surroundings are assigned 100 kPa for a teaching case. A gauge reading of 50 kPa then corresponds to 150 kPa absolute. If the gas doubles its volume at constant temperature and amount, the new absolute pressure is 75 kPa and the gauge pressure is −25 kPa. Halving the gauge reading to 25 kPa would give the wrong answer.

A negative gauge reading can mean pressure below the surroundings; it does not mean a negative absolute pressure. Atmospheric pressure varies with conditions, so a real conversion requires the applicable reference. The widget reports absolute pressure only.

Challenge the model with evidence

A disagreement between a measurement and PV = nRT can have several explanations: instrument offset, an incorrect temperature, a leak that changed n, or actual nonideal behavior. Do not call every mismatch “experimental error” or immediately assume the model is wrong. Specify a discriminating check, such as an independent pressure reference or a repeated equilibrium temperature measurement.

At high density, finite molecular size becomes more relevant. Attractions between particles also matter, especially as condensation becomes important. Real-gas departures depend on the substance and state; there is no universal pressure cutoff that makes the ideal law suddenly valid or invalid. This lab has no fitted real-gas correction, so it cannot quantify those departures.

Transfer: an ideal sample initially occupies 15 L at 180 kPa absolute. Its volume becomes 30 L at the same temperature and amount. Calculate the final pressure. Then explain why the answer would be insufficient if the final temperature were unknown.

Check the controlled expansion

P2 = 180 × 15/30 = 90 kPa absolute. If temperature changes, the ratio also includes T2/T1. The volume change alone would no longer determine the pressure. Amount must also remain fixed.

OpenStax ideal-gas relationships and nonideal behavior provide reference explanations. Continue with absolute temperature to compare different process constraints.

Apply

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

Explain the collision-based pressure argument, distinguish absolute from gauge pressure, and identify one physical condition the ideal-gas calculation omits.

Your explanation and limitations

Continue

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

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Previous: Design a separation and account for every fraction Next: Use absolute temperature and specify what is held fixed

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