Collecting Gas over Water

Introduction

A common laboratory collection method displaces water from an inverted container. The gas produced by a reaction occupies the container together with water vapour. Its measured pressure therefore is not all due to the target gas. Before using PV = nRT to find the target amount, subtract the water-vapour partial pressure at the measured temperature, provided the gas is saturated with water vapour and other vapours are negligible.

Core explanation

For a wet gas consisting of a target gas and water vapour, Dalton's law gives P total = P target + P H2O. Rearranging, P target = P total − P H2O. The vapour pressure of water is a temperature-dependent equilibrium property; it is not a fixed fraction of total pressure. At approximately 25 °C, water's saturated vapour pressure is about 23.8 mmHg, also about 3.17 kPa. A problem normally supplies the needed value or table. The target's dry amount is then n target = P target V/(RT), using the collection volume V and temperature T in compatible units.

Why is water vapour present? Liquid water and vapour exchange molecules until a saturated equilibrium is established, assuming enough time and liquid water are available. The gas container need not visibly contain much liquid water: contact with a water surface can provide vapour. A “dry gas” calculation without the correction generally overestimates target-gas moles because it assigns water's share of pressure to the target.

When water levels inside and outside the inverted container are equal, the gas pressure inside equals the external atmospheric pressure, neglecting small surface-tension effects. If levels differ, a hydrostatic pressure difference must be added or subtracted before removing P H2O. Do not automatically equate a barometer reading with inside total pressure when a water-column difference is reported. The sign follows which side's water level is higher: a higher water level inside indicates lower gas pressure inside than external pressure.

The simple two-component relation requires the collected target gas to remain chemically intact and not dissolve appreciably or react with water. Water-soluble gases can be undercounted because some product stays dissolved, even when pressure correction is perfect. Also, if another volatile substance is present, its vapour contributes another partial pressure. These are experimental limitations, separate from the arithmetic of subtracting water vapour.

Pressure units must match. If atmospheric and water-vapour pressures are in mmHg, subtract them there and then convert to atm or kPa for a chosen R. For example, 760 mmHg total minus 23.8 mmHg water vapour equals 736.2 mmHg dry target pressure; 736.2/760 = 0.9687 atm. Do not subtract 23.8 directly from 1.00 atm. Temperature must also be absolute in the ideal-gas equation, so 25.0 °C becomes 298.15 K.

Step-by-step reasoning

1. Identify the measured total pressure inside the collector; correct any water-level difference if specified. 2. Obtain water's saturated vapour pressure at the gas temperature from the supplied data. 3. Subtract P H2O from inside total pressure to find the target's dry partial pressure. 4. Convert pressure and volume to units consistent with R, and convert temperature to kelvin. 5. Use n = P target V/(RT), then compare with the reaction's predicted gas amount if requested.

Visual explanation

Sketch an inverted gas cylinder above a water bath. Inside the upper space write “target gas + H₂O vapour”; beside it draw a pressure bar split into a large target segment and a small water-vapour segment. Label the whole bar P total and the target segment P total − P H2O. Mark matching water levels inside and outside for the simplest pressure case.

Real-world analogy

A box on a scale contains a requested sample plus its packaging. To find the sample alone, subtract the known packaging contribution from the total. A wet-gas pressure has a similar two-contribution accounting step, though partial pressures add because each gas species contributes molecular collisions with the walls.

Real-world example

In an electrolysis experiment, hydrogen may be collected by water displacement. A student measuring 0.500 L of wet gas at room temperature corrects for water vapour before comparing calculated hydrogen moles with the electrical charge passed. A large discrepancy after the correction may point to leaks, dissolved gas, incomplete collection or a mistaken pressure reference.

Why?

Why does subtracting water vapour matter even if the volume seems unchanged? Both gases occupy the same container volume. The total pressure includes collisions from both species. PV = nRT with total pressure would give the total moles of target plus water vapour, not the target amount alone.

Common misconception

“Water vapour pressure is whatever pressure remains after collecting gas.” Saturated P H2O is set mainly by temperature at equilibrium, not by how much target gas was generated. Another mistake is using the dry-gas pressure with a temperature still written in Celsius.

Worked example

A gas is collected over water at 25.0 °C. Its volume is 0.500 L, the water levels are equal, and atmospheric pressure is 760.0 mmHg. Use P H2O = 23.8 mmHg. Dry target pressure is 760.0 − 23.8 = 736.2 mmHg = 0.9687 atm. With T = 298.15 K and R = 0.08206 L atm mol⁻¹ K⁻¹, n target = (0.9687 atm)(0.500 L)/[(0.08206 L atm mol⁻¹ K⁻¹)(298.15 K)] = 0.0198 mol. If water vapour had been ignored, the calculated amount would be about 0.0204 mol, an overestimate.

Quick check

1. Total wet-gas pressure is 100.0 kPa and water-vapour pressure is 3.0 kPa. What pressure belongs to the dry target gas? Answer: P target = 100.0 − 3.0 = 97.0 kPa, provided these are the only two gas components.

Exam focus

Write the pressure equation before inserting numbers, check water levels, and keep all pressures in one unit during subtraction. Use the dry pressure with the total measured gas volume in PV = nRT; do not subtract a “water volume” without evidence.

Advanced insight

If the target gas dissolves substantially in water, a gas-phase calculation measures only what remains in the headspace. A full material balance would add dissolved target moles, often estimated through a solubility relation and water volume. The water-vapour correction alone cannot recover dissolved product.

Summary

Collected gas over water is usually a wet mixture. Find pressure inside the vessel, subtract water's temperature-dependent vapour pressure, and use the resulting dry partial pressure in the ideal-gas equation. Check hydrostatic level differences and possible dissolution when interpreting the experimental result.

Practice questions

1. At a given temperature, P total = 745 mmHg and P H2O = 20 mmHg. What is dry-gas pressure? Answer: 745 − 20 = 725 mmHg. 2. Does doubling target-gas moles at constant temperature double saturated P H2O while liquid water remains? Answer: No. Saturated water-vapour pressure is determined mainly by the temperature, not the target amount. 3. If the inside water level is higher than outside, is inside total gas pressure above or below atmospheric pressure? Answer: Below atmospheric pressure by the hydrostatic head represented by that level difference.