Gas Volumes at Different Conditions

Introduction

A reaction may produce a gas measured at one temperature and pressure, while a question asks for its volume under another set of conditions. There is no universal volume occupied by one mole of gas: volume depends on pressure and temperature. For an unchanged amount of ideal gas, compare the two states through PV/T. If the gas amount changes, return to PV = nRT for each state.

Core explanation

For one fixed gas sample, the ideal-gas equation gives P₁V₁ = nRT₁ and P₂V₂ = nRT₂. Divide each equation by its temperature: P₁V₁/T₁ = P₂V₂/T₂ = nR. Thus P₁V₁/T₁ = P₂V₂/T₂, and the desired final volume is V₂ = V₁(P₁/P₂)(T₂/T₁). The temperature ratio must use kelvin. Pressures must both be absolute and in the same units, but no explicit numerical R is needed because it cancels.

The formula gives a useful direction check. Heating at unchanged pressure increases volume; increasing pressure at unchanged temperature decreases volume. When both temperature and pressure change, compare their ratios rather than guessing which effect wins. For instance, an increase in Kelvin temperature by 20% and absolute pressure by 50% yields V₂/V₁ = 1.20/1.50 = 0.80, so volume decreases despite heating.

“Molar volume” is V m = V/n = RT/P for an ideal gas. It is a property of the stated conditions, not a permanent attribute of a substance. At 300 K and 1.00 atm, V m ≈ 24.6 L mol⁻¹. At 300 K and 2.00 atm, it is approximately 12.3 L mol⁻¹. An exam problem might provide a molar volume at a named reference state, but applying it to a different state without adjustment is incorrect. Standards named STP or standard conditions can have different conventions, so use the explicitly provided P and T.

The combined law requires constant n. A leak, chemical reaction, dissolution, condensation, or gas addition breaks that condition. If n differs between states, the more general relation is P₁V₁/(n₁T₁) = P₂V₂/(n₂T₂) = R for ideal gases. Often the cleanest route is to calculate actual moles from the reaction, then determine volume at the requested final state. For a wet collected gas, use the dry-gas pressure if following the dry target amount across states; total wet pressure includes water vapour and may change with temperature independently.

Measured volume should also be tied to the same gas identity or defined mixture. If water condenses during cooling, the remaining gas mixture has fewer moles even in a closed vessel, so a single fixed-n combined-law calculation for the gas phase fails. A phase-aware final inventory is needed.

Step-by-step reasoning

1. Check that the same gas amount remains in the gas phase between states. 2. Record P₁, V₁, T₁ and P₂, V₂, T₂ with pressure references and temperature units. 3. Convert each Celsius temperature to kelvin and use matching pressure units. 4. Rearrange P₁V₁/T₁ = P₂V₂/T₂ for the unknown and substitute. 5. Check whether the result follows the pressure and temperature trends; if n changes, solve each state with PV = nRT instead.

Visual explanation

Draw a movable-piston cylinder in state 1 and another in state 2, each containing the same number of gas dots. Write P₁, V₁, T₁ under the first and P₂, V₂, T₂ under the second. An arrow labelled “same n” connects them. Beneath the pictures, show V₂/V₁ = (P₁/P₂)(T₂/T₁) to make the competing effects visible.

Real-world analogy

Imagine the same group of people occupying a flexible room. A higher external squeeze reduces the available space, while more vigorous movement pushes the walls outward. The analogy gives only the direction of the ideal-gas effects; the exact volume ratio comes from absolute pressure and Kelvin temperature, not from a verbal comparison.

Real-world example

A measured cylinder of dry gas is brought from a cool laboratory into a warmer enclosure under a different ambient pressure. The technician predicts its new volume from the combined law before choosing a storage bag size. If the gas can liquefy or its container leaks, the prediction needs more than the fixed-amount model.

Why?

Why does R disappear from the two-state formula? Both states obey PV/T = nR. When n is unchanged, the right-hand side has the same value at both states. Equating the left-hand sides is simply comparing two applications of the same ideal-gas equation.

Common misconception

“One mole is always 22.4 L.” A quoted molar volume belongs to particular conditions and often a particular standard-state convention. At room temperature the ideal value at 1 atm is nearer 24.5 L. Read the pressure and temperature supplied instead of assuming a memorised constant.

Worked example

A fixed dry-gas sample occupies 2.50 L at 1.00 atm and 25.0 °C. What volume would it occupy at 1.20 atm and 350 K, assuming ideal behaviour? T₁ = 298.15 K, so V₂ = (2.50 L)(1.00/1.20)(350/298.15) = 2.45 L to three significant figures. Heating tends to expand it, but the 20% rise in pressure more than offsets the roughly 17% rise in Kelvin temperature. Check: V₂/V₁ ≈ 0.978, a slight decrease, which agrees with the two ratios.

Quick check

1. A fixed ideal-gas sample at constant temperature changes from 1.0 atm to 2.0 atm. What happens to its volume? Answer: Volume halves because V₂/V₁ = P₁/P₂ = 1.0/2.0 = 0.50.

Exam focus

State the fixed-n assumption. Convert temperatures to kelvin, check pressures are absolute, and resist an unstated molar-volume shortcut. If gas is produced or consumed between states, use a reaction inventory before the gas equation.

Advanced insight

The ratio form can compare different gas amounts as P₂V₂/(P₁V₁) = (n₂T₂)/(n₁T₁). This is useful when reaction stoichiometry gives n₂/n₁. It is not the ordinary combined gas law, because the amount ratio must be supplied or calculated independently.

Summary

For a fixed ideal-gas amount, P₁V₁/T₁ = P₂V₂/T₂. Use absolute pressure, Kelvin temperature and compatible units. Molar volume is state-dependent. When chemistry, leakage or phase change alters gas moles, include the changed amount through PV = nRT.

Practice questions

1. A gas has V₁ = 4.00 L at 300 K and 1.00 atm. At 300 K and 2.00 atm, what is V₂? Answer: V₂ = 4.00(1.00/2.00) = 2.00 L. 2. At fixed pressure, a sample grows from 2.00 L at 300 K to what volume at 450 K? Answer: V₂ = 2.00(450/300) = 3.00 L. 3. Why is the combined law invalid if water vapour condenses between two measured gas states? Answer: Condensation removes molecules from the gas phase, so gas-phase n is not constant.