ChemistryClass 9ICSE

A Gas Collected Over Water Is Never As Much As It Looks

Derive the combined gas equation from Boyle's Law and Charles's Law, solve for any unknown, find the volume of a given mass of gas at S.T.P., and correct a measurement made over water.

How do you handle a gas when pressure and temperature both change?

The two laws of the previous part each hold one thing still. Boyle's Law needs the temperature constant; Charles's Law needs the pressure constant.

Real measurements are not so obliging. A gas collected in a laboratory in the middle of the afternoon and then compared with a value quoted at standard conditions has changed its pressure and its temperature at the same time, and neither law on its own can cope with that.

The fix is to put the two laws together into a single equation that allows all three quantities to move at once. And it turns out that the combination follows from the two laws by pure algebra — nothing new has to be measured or assumed.

There is also a trap in the measuring itself. A gas collected over water in the laboratory is not pure gas. Some of the pressure in that jar belongs to water vapour, and unless you subtract it every answer comes out too large.

Both halves of this page are about the same discipline: get the conditions right before you calculate. Convert to kelvin, subtract the water vapour, and then the arithmetic is a single substitution.

This page covers the second part of the ICSE Class 9 Chemistry chapter on gas laws: deriving the combined gas equation, using it to find any unknown, standard temperature and pressure with the volume of a given mass of gas, and correcting a measurement made over water.

Formula

How is the combined gas equation derived from the two laws?

The combined gas equation is obtained by applying Boyle's Law and Charles's Law one after the other to the same sample of gas.

\frac{P _{1} V _{1}}{T _{1}} = \frac{P _{2} V _{2}}{T _{2}}

The derivation in two steps. Let a fixed mass of gas change from

P_{1}

V_{1}

T_{1}

P_{2}

V_{2}

T_{2}

. Imagine the change happening in two stages, so that only one quantity varies at a time.

Stage 1 — change the pressure at constant temperature. Take the gas from

P_{1}

P_{2}

while holding the temperature at

T_{1}

, and call the resulting volume

V

. Boyle's Law applies:

P_{1} V_{1} = P_{2} V \Rightarrow V = \frac{P _{1} V _{1}}{P _{2}}

Stage 2 — change the temperature at constant pressure. Now take the gas from

T_{1}

T_{2}

while holding the pressure at

P_{2}

, and the volume becomes

V_{2}

. Charles's Law applies:

\frac{V}{T _{1}} = \frac{V _{2}}{T _{2}} \Rightarrow V = \frac{V _{2} T _{1}}{T _{2}}

**Equate the two expressions for

V

**, since both describe the same intermediate state:

\frac{P _{1} V _{1}}{P _{2}} = \frac{V _{2} T _{1}}{T _{2}}

Cross-multiplying:

P_{1} V_{1} T_{2} = P_{2} V_{2} T_{1}

and dividing both sides by

T_{1} T_{2}

\frac{P _{1} V _{1}}{T _{1}} = \frac{P _{2} V _{2}}{T _{2}}

The same result read as a proportionality. Boyle gives

V \propto \frac{1}{P}

and Charles gives

V \propto T

, so combining them:

V \propto \frac{T}{P} \Rightarrow \frac{P V}{T} = constant

Notice that the intermediate state is imaginary and it does not matter. The gas need not actually pass through the volume

V

— it may go from one state to the other in any way at all. The two-stage route is a device for using two laws that each require something to be held constant, and the final equation has no trace of it. That freedom to invent a convenient path is used constantly in thermodynamics, and this is the first place in the syllabus it appears.

Each gas law is a special case of this one equation, which is the quickest way to remember all three.

- Put

T_{1} = T_{2}

and the temperatures cancel, leaving

P_{1} V_{1} = P_{2} V_{2}

— Boyle's Law
- Put

P_{1} = P_{2}

and the pressures cancel, leaving

\frac{V _{1}}{T _{1}} = \frac{V _{2}}{T _{2}}

— Charles's Law
- Put

V_{1} = V_{2}

and you get

\frac{P _{1}}{T _{1}} = \frac{P _{2}}{T _{2}}

, the pressure-temperature relation for a gas in a rigid container

So there is only one equation to learn, and the two laws are what it becomes when one quantity is held fixed.

How do you solve for an unknown pressure, volume or temperature?

Rearrange the combined gas equation for whichever quantity is missing, convert every temperature to kelvin, and substitute.

Worked example 1 — an unknown volume.

120 cm^{3}

of a gas is measured at

100^{\circ} C

and

725 mm

of mercury. Find its volume at

273 K

and

760 mm

.

List the conditions first:

-

P_{1} = 725 mm

V_{1} = 120 cm^{3}

T_{1} = 100 + 273 = 373 K

P_{2} = 760 mm

V_{2} = ?

T_{2} = 273 K

V_{2} = \frac{P _{1} V _{1} T _{2}}{T _{1} P _{2}} = \frac{725 \times 120 \times 273}{373 \times 760}

= \frac{23751000}{283480} = 83.78 cm^{3}

Check the sense of it. The gas was both cooled and compressed, and each of those shrinks a gas — so the answer must be smaller than

120 cm^{3}

. It is.

Worked example 2 — an unknown pressure.

500 cm^{3}

of a gas at

27^{\circ} C

and

1 atmosphere

is heated to

127^{\circ} C

and compressed to

250 cm^{3}

. Find the new pressure.

-

P_{1} = 1 atm

V_{1} = 500 cm^{3}

T_{1} = 300 K

P_{2} = ?

V_{2} = 250 cm^{3}

T_{2} = 400 K

P_{2} = \frac{P _{1} V _{1} T _{2}}{T _{1} V _{2}} = \frac{1 \times 500 \times 400}{300 \times 250} = \frac{200000}{75000} = 2.67 atm

Both changes push the pressure up — heating raises it and squeezing raises it — so an answer above

1 atm

is right.

Worked example 3 — an unknown temperature.

100 cm^{3}

of a gas at

300 K

and

750 mm

expands to

150 cm^{3}

600 mm

. Find the new temperature.

T_{2} = \frac{P _{2} V _{2} T _{1}}{P _{1} V _{1}} = \frac{600 \times 150 \times 300}{750 \times 100} = \frac{27000000}{75000} = 360 K

So the gas ended at

360 - 273 = 87^{\circ} C

.

Notice how to read the sense-check in example 3, because it is less obvious. The pressure fell, which on its own would expand the gas; the volume did rise, but by less than the pressure fell in proportion.

600

is four-fifths of

750

, which alone would have given

125 cm^{3}

— and the gas actually reached

150 cm^{3}

. The extra expansion had to come from heating, so

T_{2}

must exceed

300 K

. Working out which direction the answer should go before you calculate catches an inverted fraction every time, and it costs a few seconds.

Two rules for the working itself.

- Temperatures in kelvin, always. Write

T_{1}

and

T_{2}

in kelvin on their own line before you substitute
- Matching units for the other two. Pressures in the same unit as each other, volumes in the same unit as each other. They cancel, so any unit will do — but a problem that gives litres and asks for cubic centimetres needs one conversion

What is S.T.P., and how do you find the volume of a given mass of gas?

**Standard temperature and pressure, written S.T.P., means

0^{\circ} C

— that is

273 K

— and

760 mm

of mercury, which is one atmosphere.

Gas volumes vary with the conditions they were measured under, so a volume quoted without its conditions is useless. Agreeing on one standard set of conditions makes two measurements comparable, and S.T.P. is that agreement.

The molar volume. One mole of any** gas occupies

22.4 litres

— that is

22400 cm^{3}

— at S.T.P.

The identity of the gas does not matter. A mole of hydrogen, a mole of oxygen and a mole of carbon dioxide all occupy the same

22.4 litres

, even though their masses are

2 g

32 g

and

44 g

.

So the method has two steps. Convert the mass to moles by dividing by the relative molecular mass, then multiply by

22.4 litres

volume at S.T.P. = \frac{mass}{relative molecular mass} \times 22.4 litres

Worked example 1 — oxygen. Find the volume of

8 g

of oxygen at S.T.P. The relative molecular mass of

O_{2}

32

moles = \frac{8}{32} = 0.25 volume = 0.25 \times 22.4 = 5.6 litres

Worked example 2 — carbon dioxide. Find the volume of

11 g

of carbon dioxide at S.T.P., taking

CO_{2} = 44

moles = \frac{11}{44} = 0.25 volume = 0.25 \times 22.4 = 5.6 litres

Look at those two answers together.

8 g

of oxygen and

11 g

of carbon dioxide occupy the same volume, because both are a quarter of a mole. Equal volumes of gases at the same conditions contain equal numbers of molecules, not equal masses — and that is the single idea this whole section rests on.

Worked example 3 — hydrogen. Find the volume of

1 g

of hydrogen at S.T.P., taking

H_{2} = 2

moles = \frac{1}{2} = 0.5 volume = 0.5 \times 22.4 = 11.2 litres

One gram of hydrogen occupies twice the volume of eight grams of oxygen. A very light gas takes up a great deal of room for its mass, which is the same fact that made hydrogen so hard to keep in the earlier chapter.

Worked example 4 — working backwards. Find the mass of

2.24 litres

of nitrogen at S.T.P., taking

N_{2} = 28

moles = \frac{2.24}{22.4} = 0.1 mass = 0.1 \times 28 = 2.8 g

The molar volume applies only at S.T.P. A volume measured at any other temperature or pressure must first be converted to S.T.P. with the combined gas equation, and only then divided by

22.4 litres

. **Applying

22.4

to a room-temperature volume is the commonest error in this part of the chapter**, and the next section is about doing that conversion properly.

Why must you subtract the aqueous tension for a gas collected over water?

Because a gas collected over water is saturated with water vapour, so part of the measured pressure belongs to the vapour and not to the gas.

The pressure of that water vapour is called the aqueous tension, and it must be taken away before the gas laws are applied:

P_{dry gas} = P_{total} - aqueous tension

Why the vapour is there at all. Water evaporates into any space above it until the space is saturated. A gas jar standing over water in a trough has water underneath it and therefore water vapour inside it, mixed with the collected gas. The barometer records the total pressure of the mixture, and only part of that is the gas you are interested in.

Worked example 1.

200 cm^{3}

of a gas is collected over water at

27^{\circ} C

and a total pressure of

750 mm

. The aqueous tension at that temperature is given as

26.7 mm

. Find the volume of the dry gas at S.T.P.

Step 1 — correct the pressure.

P_{1} = 750 - 26.7 = 723.3 mm

Step 2 — convert to S.T.P. with

V_{1} = 200 cm^{3}

T_{1} = 300 K

P_{2} = 760 mm

T_{2} = 273 K

V_{2} = \frac{P _{1} V _{1} T _{2}}{T _{1} P _{2}} = \frac{723.3 \times 200 \times 273}{300 \times 760}

= \frac{39492180}{228000} = 173.21 cm^{3}

What skipping step 1 would have cost. Using the uncorrected

750 mm

gives

\frac{750 \times 200 \times 273}{300 \times 760} = 179.6 cm^{3}

— an answer about six cubic centimetres too large, and wrong for a reason no amount of careful arithmetic would reveal. An error in the data cannot be caught by checking the working, which is why the correction has to be the first line of the solution.

Worked example 2.

380 cm^{3}

of a gas is collected over water at

17^{\circ} C

and

745 mm

, the aqueous tension being given as

14 mm

. Find the volume of the dry gas at S.T.P.

P_{1} = 745 - 14 = 731 mm T_{1} = 17 + 273 = 290 K

V_{2} = \frac{731 \times 380 \times 273}{290 \times 760} = \frac{75833940}{220400} = 344.07 cm^{3}

Two facts about aqueous tension that are regularly examined.

- It depends only on the temperature — not on the gas collected, not on the total pressure, and not on how much water is in the trough. Warmer water evaporates more, so the aqueous tension rises with temperature
- It applies only when the gas is collected over water. A gas collected in a dry jar, or over mercury, needs no such correction

So the correction is a property of the method rather than of the gas. The same gas measured in a dry jar and over water at the same temperature gives two different total pressures, and the difference is the aqueous tension. A question that mentions water in the collection is asking for the subtraction, and a question that does not mention it is not — reading the wording is part of the skill.

One last point on the phrase "volume of the dry gas". Subtracting the aqueous tension does not change the volume you measured — the mixture filled the whole jar. It gives you the pressure the gas alone was exerting, and the combined gas equation then converts that correctly. You are correcting the pressure, not the volume, and writing

200 - 26.7

instead of

750 - 26.7

is a mistake worth guarding against.

Exam tip

Exam tip: correct the pressure, convert the temperature, then substitute

List the six quantities before you calculate —

P_{1}

V_{1}

T_{1}

and

P_{2}

V_{2}

T_{2}

— with the unknown marked. It turns every problem into one substitution.

Derive the combined gas equation in two stages when asked: Boyle from

P_{1}

P_{2}

T_{1}

, then Charles from

T_{1}

T_{2}

P_{2}

, then equate the two expressions for the intermediate volume.

Show that each law is a special case:

T_{1} = T_{2}

gives Boyle,

P_{1} = P_{2}

gives Charles,

V_{1} = V_{2}

gives

\frac{P _{1}}{T _{1}} = \frac{P _{2}}{T _{2}}

.

Temperatures in kelvin on their own line, every time.

Check the direction of your answer before moving on. Cooling and compressing both shrink a gas; heating and expanding both raise the pressure.

**S.T.P. is

0^{\circ} C

(that is

273 K

) and

760 mm

of mercury. Quote both.

One mole of any gas occupies

22.4

litres at S.T.P.** — divide the mass by the relative molecular mass, then multiply by

22.4

.

**Never apply

22.4

to a volume measured away from S.T.P. Convert to S.T.P. first.

For a gas collected over water, subtract the aqueous tension from the PRESSURE** —

P_{dry} = P_{total} - aqueous tension

— and do it on the first line.

Aqueous tension depends only on temperature, and only applies when the gas is collected over water.

And remember you are correcting the pressure, never the volume — the mixture filled the whole jar.

Did you know

Why a mole of any gas takes up the same room

There is something quietly strange about the molar volume, and it is worth pausing on before it becomes routine.

A mole of hydrogen weighs two grams. A mole of carbon dioxide weighs forty-four — twenty-two times as much. The carbon dioxide molecule is far heavier, and it is also physically much larger, being three atoms rather than two.

At standard conditions they occupy exactly the same volume.

If gases were like grains of sand in a jar, this would be impossible. Bigger grains would take more room. But a gas is nothing like a jar of sand, and the reason is the one this chapter started with: the molecules are so far apart that their own size is irrelevant.

Picture ten large stones and ten small ones scattered across a cricket field, one every twenty paces. The field they need is the same in both cases, because what fixes the area is the spacing, not the stones. Shrink the stones to half their size and the field does not get smaller.

That is a gas. The volume is set almost entirely by the empty space between molecules, and the molecules themselves occupy so little of it that swapping small ones for large ones changes nothing measurable.

So the molar volume is not really a fact about hydrogen or carbon dioxide. It is a fact about how far apart molecules sit at a given temperature and pressure, and at standard conditions that spacing works out such that a mole of anything needs

22.4

litres.

And this is exactly where real gases eventually depart from the ideal picture. Squeeze a gas hard enough and the molecules are no longer far apart — their own size starts to matter, and gases with bigger molecules stop agreeing with each other. The molar volume holds because gases are mostly empty, and it fails under precisely the conditions where they stop being mostly empty — which is the subject Class 11 picks up as the deviation of real gases.

Exam relevance

How does the gas equation carry into JEE Main and NEET?

Because it becomes the ideal gas equation, and the ideal gas equation is used in more Class 11 and 12 calculations than any other single formula in chemistry.

This is the foundation for Class 11 Chemistry States of Matter and Some Basic Concepts of Chemistry, and Class 11 Physics Kinetic Theory, examined in both JEE Main and NEET. The combined gas equation is extended by Avogadro's Law into

P V = n R T

, where

n

is the number of moles and

R

the universal gas constant. **The constant you find here as

\frac{P V}{T}

n R

**, and the two-state form is what

P V = n R T

reduces to when you compare a gas with itself.

The molar volume becomes the centre of stoichiometry. Class 11 uses

22.4

litres per mole at S.T.P. to convert between masses and gas volumes in reaction calculations, and mole-concept numericals combining mass, volume and number of particles are among the most reliably examined items in JEE Main. The two-step method used here — mass to moles, moles to volume — is the method those questions want.

The aqueous-tension correction becomes Dalton's law of partial pressures. Class 11 states it formally: the total pressure of a mixture is the sum of the partial pressures of its components. Subtracting the aqueous tension on this page is Dalton's law applied to a two-component mixture of gas and water vapour, and States of Matter uses exactly this example. Partial-pressure and mole-fraction calculations are a recurring type in both JEE Main and NEET.

Gas density and molecular mass follow from the same equation. Rearranging

P V = n R T

gives the molecular mass from a measured density, and Class 11 examines that rearrangement directly. Finding the molar mass of an unknown gas from its density at given conditions is a standard numerical.

The limits of the equation become their own topic. Real gases deviate from

P V = n R T

at high pressure and low temperature, described by the compressibility factor and the van der Waals equation. The reason the molar volume works — molecules far apart and of negligible size — is precisely the assumption that breaks, so the qualitative account here is what makes the corrections intelligible.

For NEET Physics, the gas equation appears in kinetic theory numericals relating pressure, volume, temperature and molecular speeds. For NEET Biology, Breathing and Exchange of Gases uses partial pressures of oxygen and carbon dioxide throughout, and the transport of gases in blood is discussed entirely in terms of partial pressure — the same idea as aqueous tension, applied to a different mixture.

What the questions look like. For board work, expect derive the combined gas equation, solve for an unknown pressure, volume or temperature, define S.T.P., find the volume of a given mass of gas at S.T.P., and convert a volume measured over water to S.T.P. Working must show the kelvin conversion and, where relevant, the aqueous-tension subtraction. For JEE Main and NEET, expect

P V = n R T

calculations, partial pressures and mole fractions, gas density and molar mass, and real-gas deviations.

How board and competitive emphasis differ. A board paper rewards the derivation and a clean, fully shown substitution. A competitive paper assumes both and asks for a partial pressure, a molar mass from a density, or which gas deviates most from ideal behaviour.

The single trap that costs the most marks. Forgetting the aqueous-tension subtraction, or subtracting it from the volume instead of the pressure. The mixture of gas and vapour filled the whole jar, so the volume is right as measured — it is the pressure that has to be split between the two components. The defence is to read the question for the word "water" before doing anything else, and to write the corrected pressure as the first line of the answer so it cannot be forgotten later.

Key takeaways

The gas equation, S.T.P. and aqueous tension: quick revision

- Combined gas equation:

\frac{P _{1} V _{1}}{T _{1}} = \frac{P _{2} V _{2}}{T _{2}}

, equivalently

\frac{P V}{T} = constant

and

V \propto \frac{T}{P}

.
- Derivation: apply Boyle from

P_{1}

P_{2}

at constant

T_{1}

to get

V = \frac{P _{1} V _{1}}{P _{2}}

; apply Charles from

T_{1}

T_{2}

at constant

P_{2}

to get

V = \frac{V _{2} T _{1}}{T _{2}}

; equate and rearrange.
- The intermediate state is imaginary and leaves no trace in the result.
- Each law is a special case:

T_{1} = T_{2}

gives Boyle;

P_{1} = P_{2}

gives Charles;

V_{1} = V_{2}

gives

\frac{P _{1}}{T _{1}} = \frac{P _{2}}{T _{2}}

.
-

120 cm^{3}

373 K

and

725 mm

becomes

\frac{725 \times 120 \times 273}{373 \times 760} = 83.78 cm^{3}

at S.T.P.
-

500 cm^{3}

300 K

and

1 atm

, heated to

400 K

and squeezed to

250 cm^{3}

, reaches

\frac{1 \times 500 \times 400}{300 \times 250} = 2.67 atm

.
-

100 cm^{3}

300 K

and

750 mm

, expanding to

150 cm^{3}

600 mm

, reaches

\frac{600 \times 150 \times 300}{750 \times 100} = 360 K

, that is

87^{\circ} C

.
- Always convert temperatures to kelvin, and keep pressures and volumes in matching units.
- S.T.P. is

0^{\circ} C

—

273 K

— and

760 mm

of mercury, one atmosphere.
- **One mole of any gas occupies

22.4

litres (

22400 cm^{3}

) at S.T.P., whatever the gas.
- Volume at S.T.P.**

= \frac{mass}{relative molecular mass} \times 22.4 litres

.
-

8 g

of oxygen gives

\frac{8}{32} \times 22.4 = 5.6 L

;

11 g

CO_{2}

gives the same

5.6 L

;

1 g

of hydrogen gives

11.2 L

;

2.24 L

of nitrogen weighs

2.8 g

.
- Equal volumes at the same conditions hold equal numbers of molecules, not equal masses.
- **Never apply

22.4

away from S.T.P. — convert with the gas equation first.
- A gas collected over water is saturated with water vapour**, so

P_{dry} = P_{total} - aqueous tension

.
-

200 cm^{3}

over water at

300 K

and

750 mm

with aqueous tension

26.7 mm

P_{1} = 723.3 mm

, giving

\frac{723.3 \times 200 \times 273}{300 \times 760} = 173.21 cm^{3}

at S.T.P. Skipping the correction would have given

179.6 cm^{3}

.
-

380 cm^{3}

over water at

290 K

and

745 mm

with aqueous tension

14 mm

P_{1} = 731 mm

, giving

344.07 cm^{3}

at S.T.P.
- Aqueous tension depends only on the temperature, rises as the temperature rises, and applies only for collection over water.
- The correction is to the pressure, never to the volume — the mixture filled the whole jar.

Take any volume measured over water at room temperature and convert it to S.T.P. in three lines — correct the pressure, convert the temperature, substitute — and see whether you can do it without looking at the order.

How do you handle a gas when pressure and temperature both change?

How is the combined gas equation derived from the two laws?

How do you solve for an unknown pressure, volume or temperature?

What is S.T.P., and how do you find the volume of a given mass of gas?

Why must you subtract the aqueous tension for a gas collected over water?

Exam tip: correct the pressure, convert the temperature, then substitute

Why a mole of any gas takes up the same room

How does the gas equation carry into JEE Main and NEET?

The gas equation, S.T.P. and aqueous tension: quick revision

Related study notes

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