Pressure-Temperature Gas Law

Introduction

Heating a gas trapped in a rigid sealed vessel cannot make its volume grow. Instead, its pressure rises. For a fixed amount of gas close to ideal behavior, pressure is proportional to absolute temperature when volume is fixed. This pressure-temperature law complements Charles's law: the same warming produces a different observable change because the boundary is rigid rather than movable.

Core explanation

From PV = nRT, hold volume V and amount n fixed. Rearranging gives P/T = nR/V, a constant. Two equilibrium states obey P₁/T₁ = P₂/T₂ with temperatures in kelvin and pressures absolute. If a gas warms from 300 K to 330 K, its ideal absolute pressure rises by the factor 330/300 = 1.10, or 10%. A fall from 400 K to 200 K would halve its pressure, provided it remains a gas and the ideal model remains valid.

The particle explanation begins with molecular kinetic energy. Higher absolute temperature means greater average translational kinetic energy. In a rigid container, molecules cannot gain more space; their impacts transfer more momentum on average, so the wall force per area rises. A real gas also has intermolecular interactions and finite particle size, so the ideal proportionality is approximate. For ordinary school problems, a sealed rigid vessel signals fixed n and V.

Pressure must be absolute. A tyre gauge or other ordinary gauge often reports pressure relative to the atmosphere, which itself exerts pressure outside. Suppose a gauge reads 200 kPa when atmospheric pressure is 100 kPa. The gas is at about 300 kPa absolute. If its Kelvin temperature increases by 10% at fixed V and n, ideal absolute pressure becomes 330 kPa, corresponding to a gauge reading near 230 kPa if the atmosphere stays at 100 kPa. Multiplying the original 200-kPa gauge number by 1.10 would wrongly predict 220 kPa gauge.

Temperature must likewise be absolute. A change from 20 °C to 40 °C is from 293.15 K to 313.15 K, a factor of about 1.068. It is not a factor of two. Pressure therefore rises about 6.8% in the ideal fixed-V,n model, not 100%. A Celsius ratio is invalid because zero Celsius is not zero molecular thermal state.

On a P-versus-T(K) graph for fixed V and n, the ideal relation is a straight line through the origin when extrapolated. Its slope is nR/V. Increasing gas amount makes the slope steeper; increasing container volume makes it shallower. On a P-versus-Celsius graph, the ideal extrapolation reaches zero near −273.15 °C. A real gas can condense or otherwise deviate before that extrapolated endpoint, and an actual container may have temperature-dependent expansion.

The law can be tested by measuring a sealed sample's pressure after it equilibrates at several temperatures. The container should be sufficiently rigid, and readings should be corrected to absolute pressure. If the gas leaks, n changes. If a reaction occurs, n may change even while the container remains closed. If measurements are made during rapid heating, gas temperature can be nonuniform; wait for a well-defined state or use a more detailed model.

Do not confuse pressure due to heating with pressure due to increased gas amount. Adding more gas to a fixed-volume tank at constant temperature can also raise pressure. The observation “pressure rose” alone does not identify the cause. State which variables were controlled before choosing P/T = constant.

Step-by-step reasoning

1. Check that the container is rigid and the gas amount remains fixed. 2. Convert state temperatures from °C to K. 3. Convert gauge readings to absolute pressure if needed. 4. Apply P₁/T₁ = P₂/T₂ and solve for the unknown. 5. Check that warming raises absolute pressure and that the model remains in the gas phase.

Visual explanation

Draw two identical rigid boxes with the same number of gas dots. In the warmer box, longer motion arrows lead to stronger wall-collision arrows, while volume labels match. Beside them plot absolute P against T(K) as a rising straight line through an ideal origin. Label its slope nR/V.

Real-world analogy

A group moving faster inside a room with fixed walls can hit those walls more forcefully. The room cannot expand, so the effect shows up at the boundary. The analogy suggests the direction of pressure change but cannot calculate it without the gas equation.

Real-world example

A sealed rigid laboratory vessel with gas can show a higher absolute pressure after being warmed. A flexible balloon may instead change volume substantially, so the pressure-temperature law is not automatically the right model for it. The apparatus, not merely the presence of heat, sets the fixed-variable conditions.

Why?

Why does warming a rigid vessel differ from warming a freely moving piston? The rigid vessel fixes V, so increased particle energy raises P. A moving piston can expand and maintain approximately fixed P, leading instead to Charles's volume-temperature relation.

Common misconception

“Double the Celsius reading and the gas pressure doubles.” The proportionality is to kelvin temperature and absolute pressure. Celsius and gauge-pressure zeros are arbitrary reference points for this calculation.

Worked example

A sealed rigid container holds gas at 150 kPa absolute and 300 K. It warms to 360 K without reaction or leakage. With V and n fixed, P₂ = P₁T₂/T₁ = 150×360/300 = 180 kPa absolute. The Kelvin temperature rose 20%, so the ideal absolute pressure rose 20%. If 150 kPa had been a gauge reading rather than absolute, atmospheric pressure would need to be added before the ratio.

Quick check

1. A rigid sealed gas sample cools from 400 K to 300 K. What fraction of its original ideal absolute pressure remains? Answer: P₂/P₁ = 300/400 = 3/4 under fixed V and n.

Exam focus

State fixed V and n, use kelvin and absolute pressure, and show P₁/T₁ = P₂/T₂. Distinguish a rigid vessel from a constant-pressure piston, and do not use Celsius or gauge readings directly as proportional variables.

Advanced insight

The ideal P/T line's slope nR/V can be used to estimate amount of gas in a known rigid volume. Real experimental data may show a nonzero intercept or curvature because of sensor offset, container expansion or nonideal gas interactions. Such deviations should be investigated rather than forced to fit a perfect line.

Summary

For a fixed amount of near-ideal gas in a rigid container, absolute pressure is proportional to Kelvin temperature. The relation P₁/T₁ = P₂/T₂ follows from PV = nRT. Temperature, pressure reference, leakage and real-gas conditions determine whether it applies.

Practice questions

1. A gas at 100 kPa absolute and 250 K is warmed to 300 K in a rigid sealed vessel. Find ideal final pressure. Answer: 100×300/250 = 120 kPa absolute. 2. Why is a tyre-gauge reading unsuitable for direct use in a P/T ratio? Answer: It is usually gauge pressure; add atmospheric pressure to obtain absolute pressure first. 3. How does the P-versus-Kelvin-temperature slope change if n doubles at fixed V? Answer: The ideal slope nR/V doubles.