Ideal gases: use absolute scales and explicit constraints

MCAT content library / Chemistry & physics / Ideal gases: use absolute scales and explicit constraints

A heated gas may expand or press harder against a fixed wall. The container’s constraints determine which prediction the gas law makes.

About 19 minutes, including the worked examples and checks.

Start with a state and a set of constraints

A gas-law question is easier when its container is part of the description. A sealed rigid flask keeps volume and gas amount fixed. A movable piston can change volume, and a valve can change amount. The ideal-gas equation connects these possibilities, but it cannot decide which condition the apparatus enforces. AAMC category 4B includes gas relationships and pressure measurement.¹

For an ideal gas at equilibrium, PV = nRT. P is absolute pressure, V is gas volume, n is amount in moles, and T is absolute temperature in kelvin. R connects the chosen units. A useful combination is R = 8.314 kPa·L·mol⁻¹·K⁻¹; the numerical value is also 8.314 J·mol⁻¹·K⁻¹ because 1 kPa·L = 1 J.²

The equality relates state variables. It does not say that every change is isothermal, that the amount is constant, or that pressure is always proportional to volume. Identify the sealed boundaries, rigid walls, thermal conditions, and final mechanical equilibrium before cancelling any variable.

Use absolute scales, then choose compatible units

Temperature in kelvin is T = temperature in degrees Celsius +273.15. A difference of 1°C has the same size as a difference of 1 K, but the two zero points differ. Ratios such as T₂/T₁ therefore require kelvin. A gas warming from 300 to 360 K experiences a20% absolute-temperature increase; the corresponding Celsius values 26.85 and 86.85 do not provide that ratio.²

Absolute pressure is measured from vacuum. Gauge pressure is measured relative to a reference, usually the surrounding atmosphere. If a gauge reads 75 kPa above a local atmospheric pressure of 100 kPa, the pressure used in PV = nRT is 175 kPa. A zero gauge reading would mean equality with that atmosphere, not absence of gas pressure.

A liquid manometer supplies a pressure difference, ΔP = ρgh, between its connected sides. The higher-pressure gas depresses the liquid surface on its side. If that surface is 0.12 m lower than the atmosphere-connected surface, with liquid density 1000kg/m³ and g =10 m/s², the gas pressure is 1.2 kPa above atmosphere. With the same 100 kPa atmosphere, its absolute pressure is 101.2 kPa. The liquid height is not an absolute gas pressure until the reference is included.³

A simple mercury barometer uses an approximately evacuated space above the liquid column. Neglecting mercury vapor pressure, atmospheric absolute pressure is ρ_Hggh: the near-vacuum supplies the pressure reference. An open manometer instead compares the pressures connected to its two sides.³

Worked application: five independent piston trials

All trials and values below are original illustrative models. Begin each trial with 0.100 mol of an ideal gas at 300 K and 100 kPa. A frictionless piston is at mechanical equilibrium; the stated external pressure includes atmospheric pressure and any added load. The gas occupies V = nRT/P = (0.100)(8.314)(300)/100 =2.4942 L, displayed as 2.49 L.

Warming under the unchanged load increases volume by 360/300 =1.20. The piston moves until the gas pressure again matches the same external pressure. The resulting expansion is consistent with constant pressure; a20% pressure rise would instead describe a different constraint.

The increased-load trial holds temperature and amount fixed. Pressure rises by a factor 1.50, so volume falls to 1/1.50 of its reference value. Both variables change, but their product stays constant. When more gas is added under unchanged pressure and temperature, volume instead rises in proportion to n:0.150/0.100 =1.50.

Locking the piston makes a different experiment. Volume can no longer increase. With amount fixed, warming from 300 to 360 K increases pressure from 100 to 120 kPa. Saying “heating makes a gas expand” would miss the rigid constraint. Heating can change pressure, volume, or both depending on the apparatus.

Build the useful ratio from the complete equation

For two ideal-gas states, P₁V₁/(n₁T₁) = P₂V₂/(n₂T₂). When the sample is sealed and no reaction changes the number of gas particles, n cancels. This gives P₁V₁/T₁ = P₂V₂/T₂. A leak, injection, or gas-producing reaction can invalidate that cancellation even if the container appears unchanged.²

For example, suppose a rigid vessel loses 20% of its gas while temperature rises from 300 to 375 K. The final-to-initial pressure ratio is(n₂/n₁)(T₂/T₁) = (0.80)(1.25) =1. Pressure is unchanged despite changes in both amount and temperature. One unchanged measurement does not prove that the sample stayed sealed or unheated.

Molar volume belongs to specified conditions

Volume per mole follows V/n = RT/P. At273.15 K and 101.325 kPa, it is approximately 22.4 L/mol. At300 K and 100 kPa, it is 24.9 L/mol. Neither number is an intrinsic volume carried by one mole in every situation. Use a reference molar volume only when the stated pressure and temperature match its definition.²

Similarly, pressure units may be changed without changing the state:1 atm =101.325 kPa. Convert consistently with R instead of mixing liters with a pressure-volume constant expressed for cubic meters. A quick unit check should leave the requested unit after cancellation.

Make a gas-law test answer its intended question

To test the inverse pressure–volume relationship, use the same closed sample, change piston position, and let each reading reach the same measured temperature. A plot of absolute P against 1/V should be linear in this ideal model. A fast compression that also warms the gas changes two relevant variables and is not, by itself, a clean constant-temperature test.

The ideal model approximates real gases most effectively when particles are sufficiently dilute and the gas is away from condensation. A systematic failure of the predicted relationship can motivate checking leakage, thermal equilibration, pressure reference, and then nonideal behavior. The equation remains useful because it makes those assumptions and competing explanations visible.

Check your reasoning

Choose an answer before opening its feedback. Each option explains the reasoning behind that choice.

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• Pressure and buoyancy: distinguish depth, displacement, and interfaces
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Sources

1. AAMC: C/P Content Category4B

2. OpenStax Chemistry2e: The Ideal Gas Law

3. OpenStax Chemistry2e: Gas Pressure