Gas Collected over Water

Introduction

Collecting a reaction gas over water is convenient, but the gas space above the water contains both the desired product and water vapour. A pressure measurement usually reports their combined pressure. Using that total directly in PV = nRT would overestimate the moles of dry product gas. Dalton's law supplies the correction: subtract water-vapour pressure at the collection temperature before calculating product amount.

Core explanation

In an ideal-mixture approximation, Ptotal = Pdry gas + Pwater vapour. Therefore Pdry gas = Ptotal − Pwater vapour. The water-vapour pressure depends primarily on temperature when liquid water is present and equilibrium has been reached. A problem normally supplies its value or a table. The subtraction must use pressure values in the same units and at the same collection temperature.

Suppose total gas pressure is 101.3 kPa and water-vapour pressure at the collection temperature is 3.2 kPa. The dry product gas pressure is 98.1 kPa. Use 98.1 kPa, not 101.3 kPa, in n = PV/(RT) for the product. The difference may be modest at some temperatures but can be significant for precise yield calculations or warm water, whose vapour pressure is higher.

The physical reason is that water molecules evaporate into the collected gas space. Their collisions contribute to measured pressure just as product-gas molecules do. The product's partial pressure accounts for its own amount. The two gas species occupy the same volume; they are not stacked in separate layers with separate measured volumes in the ideal model.

Be careful with hydrostatic pressure when water levels differ inside and outside an inverted collection tube. If levels are equal, total gas pressure inside is approximately atmospheric pressure. If levels differ, the gas pressure differs from atmospheric pressure by the liquid-column pressure. A problem may give the corrected total gas pressure directly; if it does not, the level difference and density may be needed before subtracting vapour pressure. Simply setting Ptotal = Patm regardless of levels can be wrong.

Water-vapour pressure is not an arbitrary fixed “water correction” for all experiments. It increases strongly with temperature and depends on equilibrium with liquid water. Use the stated value at the measured temperature. If the gas is later dried chemically or collected by another method, the water contribution may no longer be present, so subtracting a vapour pressure without reason would undercount the product.

The ideal equation requires the gas volume and temperature corresponding to the same state as corrected pressure. If the collected gas cools or is transferred to a different vessel, use the appropriate final P, V and T or a state conversion. A reaction's balanced equation can then turn dry product moles into reactant consumption or yield. This calculation chain combines mixture pressure, gas law and stoichiometry; each step answers a different question.

Some product gases dissolve appreciably in water or react with it. Their measured gas volume may then be less than the amount formed, so the simple pressure correction alone does not recover total production. A suitable collection medium must be chosen for the gas and the experiment. In classroom exercises, the problem often assumes the desired gas is negligibly soluble and unreactive, but a scientific answer should identify that assumption.

If more than one dry gas product is present, subtracting water vapour yields pressure of the dry mixture , not automatically one named product. Further composition information is needed to assign component partial pressures. The label “dry gas” should therefore be connected to the actual contents of the collected sample.

Step-by-step reasoning

1. Determine total gas pressure inside the collector, correcting for liquid-level difference if required. 2. Look up or use the supplied water-vapour pressure at the collection temperature. 3. Subtract it from total pressure to obtain dry gas partial pressure. 4. Use dry pressure with collection volume and Kelvin temperature in PV = nRT. 5. Apply reaction stoichiometry and consider gas solubility or additional components if relevant.

Visual explanation

Draw an inverted tube filled with product-gas dots and water-vapour dots above a water surface. A pressure bar beside it is split into Pdry and PH₂O portions. Show an arrow “subtract PH₂O” before a box labelled n = PdryV/(RT). Mark equal water levels for the simplest Patm = Ptotal case.

Real-world analogy

A scale holding a bag and its contents reads their combined mass; subtracting the bag's mass reveals the contents' mass. The pressure reading likewise includes two contributions. Gas partial pressures arise from shared molecular motion, not physically separate containers, so the analogy is only about subtraction.

Real-world example

Hydrogen formed in a metal-acid reaction can be collected over water in a teaching experiment. Its volume, temperature and total pressure are measured, and water-vapour pressure is subtracted to estimate hydrogen moles. That amount can be compared with a stoichiometric prediction from the metal mass.

Why?

Why does ignoring water vapour overestimate dry gas moles? PV = nRT makes n proportional to the pressure assigned to that gas at fixed V and T. Using the larger total pressure incorrectly attributes water-vapour collisions to dry product molecules.

Common misconception

“The pressure in a wet gas collection is all product-gas pressure because most of the collected volume looks like gas.” Water vapour is invisible and occupies the same space. Its partial pressure must be included in the total and subtracted for dry-product calculations.

Worked example

An idealised gas is collected over water at 298 K. The gas occupies 0.500 L, total corrected pressure is 100.0 kPa and water-vapour pressure is 3.2 kPa. Dry-gas pressure is 96.8 kPa. With R = 8.314 kPa·L mol⁻¹ K⁻¹, n = 96.8×0.500/(8.314×298) ≈ 0.0195 mol. Using 100.0 kPa instead would give about 0.0202 mol and overstate the dry gas. Both numbers use the same measured volume, but only the first assigns the correct partial pressure.

Quick check

1. Total wet-gas pressure is 102 kPa and water vapour contributes 4 kPa. What is dry-gas pressure? Answer: 102 − 4 = 98 kPa.

Exam focus

Write Pdry = Ptotal − PH₂O and use water-vapour pressure at the measured temperature. Check water-level pressure corrections, shared units and solubility assumptions. Use the corrected dry pressure in PV = nRT.

Advanced insight

The correction is a partial-pressure application, not a fixed volume subtraction. If water vapour condenses during cooling, both total pressure and gas volume can change. A rigorous transfer between collection and measurement states tracks the dry gas moles separately from the changing water-vapour contribution.

Summary

Gas collected over water is usually a wet mixture. Dalton's law separates total pressure into dry-gas and water-vapour contributions, and the dry partial pressure belongs in the ideal-gas mole calculation. Water levels, temperature, gas solubility and additional components can add further conditions.

Practice questions

1. A wet gas has total pressure 99.0 kPa and water-vapour pressure 2.5 kPa. Find dry-gas pressure. Answer: 99.0 − 2.5 = 96.5 kPa. 2. Why must the water-vapour pressure correspond to the collection temperature? Answer: Water's equilibrium vapour pressure changes with temperature, so using a value from another state gives the wrong partial-pressure correction. 3. If the desired gas dissolves substantially in water, will vapour-pressure subtraction alone recover total gas produced? Answer: No. Some product remains dissolved and is absent from the measured gas volume.