🧪 Contents · Interactive Lab Chapter 9 · Gases
Chemistry · Chapter 9

Gases

Gases are the easiest state of matter to predict. A gas has no shape and almost no structure. Its particles fly far apart and rarely touch. Because of that, one simple equation describes almost every gas you will meet. In this chapter you will learn what pressure really is and how we measure it. You will meet the simple gas laws, then fold them into the ideal gas law. You will find the volume of a gas a reaction makes. You will split the pressure of a mixture among its gases. At the end you will look at the moving particles themselves and see where the simple picture breaks down.

By the end of this chapter you can

  • Define pressure and convert among atm, torr, mmHg, kPa and bar.
  • Read a closed-end or open-end manometer and report the gas pressure.
  • Apply Boyle's, Charles's, Gay-Lussac's and Avogadro's laws, and the combined gas law.
  • Solve PV = nRT for any one missing variable, with the right value of R.
  • Use the molar volume 22.4 L/mol at STP, and find gas density and molar mass.
  • Find the volume of a gas made or used by a reaction.
  • Split a total pressure into partial pressures, and correct for water vapor.
  • Link temperature to particle speed, compare effusion rates, and say when a real gas strays from ideal.

Section 1Gas pressure and how we measure it

Pressure is a push spread over an area

A gas has no shape of its own. It spreads out to fill whatever holds it. Inside the container, the particles fly in every direction. Each one strikes the wall, bounces, and flies off again. Every hit is a tiny push. Add up all those pushes over a whole wall and you get pressure. Pressure is force divided by the area the force acts on.

P = F / A

The same force feels stronger on a small area. That is why a thumbtack sinks into wood but your thumb does not. The SI unit of pressure is the pascal (Pa), which is one newton per square meter. One pascal is a very small push, so we usually work in kilopascals.

You live at the bottom of an ocean of air. That air has weight, and it presses on you from all sides. This push is atmospheric pressure. At sea level it runs near 101 kPa. On a mountain top there is less air above you, so the push is smaller.

How a barometer and a manometer measure pressure BAROMETER 760 mm at sea level vacuum air presses on the open dish CLOSED-END gas vacuum Δh is the gas pressure in torr OPEN-END gas open air mercury higher on the open side, so Pgas = Patm + Δh
Three ways mercury reports a push. The barometer balances the whole atmosphere against a column of mercury. A closed-end manometer has a vacuum above the far arm, so the height gap is the gas pressure itself. An open-end manometer weighs the gas against the outside air.

Reading a barometer and a manometer

A barometer measures atmospheric pressure. Fill a long tube with mercury and tip it upside down into a dish. Some mercury drains out, but a tall column stays up. Air pressing on the dish is what holds it there. At sea level that column stands about 760 mm tall.

That height became a unit. One torr is the push of a mercury column 1 mm high, so 760 torr means the same thing as 760 mmHg. One atmosphere (atm) is defined as exactly 760 torr. Here are the units side by side.

UnitOne atm equalsWhere you meet it
atm1Gas law problems
torr or mmHg760Manometers, blood pressure
kPa101.325The SI choice, weather maps
Pa101,325Physics equations
bar1.01325Scuba, industry
psi14.696Tires in the United States

A manometer measures the pressure of gas trapped in a flask. A U-shaped tube of mercury joins the flask to the far side. In a closed-end manometer the far arm holds a vacuum. Nothing pushes back, so the height gap between the two mercury levels is the gas pressure in torr.

In an open-end manometer the far arm is open to the room. Now the gas and the air push against each other. If the mercury stands higher on the open side, the gas is stronger. Add the gap to air pressure. If the mercury stands higher on the gas side, the gas is weaker, so subtract the gap instead.

Worked example · pressure units and a manometer

A weather station reports 745 torr. Convert it to atm and to kPa. Then a flask of gas is joined to an open-end manometer in that same room. The mercury stands 68 mm higher on the open arm. What is the gas pressure in atm?

Step 1 · torr to atm

745 torr × (1 atm / 760 torr) = 0.980 atm

Step 2 · torr to kPa

745 torr × (101.325 kPa / 760 torr) = 99.3 kPa

Step 3 · read the manometer

Mercury is higher on the open arm, so the gas wins. Add the gap.

Pgas = 745 torr + 68 torr = 813 torr

Step 4 · convert the answer

813 torr × (1 atm / 760 torr) = 1.07 atm

Check it

The gas is a little above room pressure, so a value just over 1 atm makes sense. Going back the other way, 1.07 atm × 760 torr/atm = 813 torr.

Why mercury? Mercury is about 13.6 times denser than water. A water barometer would need a tube more than 10 m tall to balance the atmosphere. Mercury does the same job in under a meter.
Key points
  • Pressure is force per unit area, and gas pressure comes from particles striking the walls.
  • 1 atm = 760 torr = 760 mmHg = 101.325 kPa = 1.01325 bar.
  • A closed-end manometer reads the gas pressure straight off the height gap.
  • An open-end manometer adds the gap when mercury is higher on the open side, and subtracts it when mercury is higher on the gas side.
Interactive Read a manometer and convert atm, torr, mmHg and kPa. Open full ↑

Section 2The simple gas laws

Change one thing and watch another respond

A sample of gas has four numbers that matter. Pressure P, volume V, temperature T and amount in moles n. The simple gas laws each hold two of those steady and watch the other two trade. Every one of them was found by experiment before anyone knew why it worked.

First, a rule you cannot skip. Every gas law uses temperature in kelvin, never in degrees Celsius. Add 273.15 to a Celsius reading to get kelvin. The kelvin scale starts at absolute zero, the coldest point there is, so its numbers are never negative. Celsius numbers can be negative, and a negative volume is nonsense.

Boyle, Charles, Gay-Lussac and Avogadro

Boyle's law holds temperature and moles fixed. Squeeze a gas into half the space and its pressure doubles. The particles now hit each wall twice as often. Pressure and volume are inversely related, so their product stays the same.

P1V1 = P2V2

Charles's law holds pressure and moles fixed. Warm a gas and it expands. Hotter particles move faster and hit harder, so the walls move out until the push matches again. Volume divided by kelvin temperature stays the same.

V1 / T1 = V2 / T2

Gay-Lussac's law holds volume and moles fixed. Warm a gas in a sealed steel can and the pressure climbs. This is why a spray can carries a warning about heat.

P1 / T1 = P2 / T2

Avogadro's law holds pressure and temperature fixed. Pump in twice as many moles and the volume doubles. A surprising part follows. At the same pressure and temperature, equal volumes of any two gases hold equal numbers of particles. A liter of helium and a liter of carbon dioxide hold the same count, even though their masses differ.

V1 / n1 = V2 / n2
The shapes of the three simple gas laws BOYLE · P vs V at fixed T volume pressure PV is constant CHARLES · V vs T at fixed P 0 K 273 K kelvin temperature volume a straight line aimed at 0 K GAY-LUSSAC · P vs T at fixed V 0 K kelvin temperature pressure P/T is constant
Each law has its own shape. Boyle's law curves, because P and V are inverses. Charles's law and Gay-Lussac's law are straight lines that point back to 0 K. That shared aim is why gas work must use the kelvin scale.

One equation for all four

Stack Boyle, Charles and Gay-Lussac together and you get the combined gas law. Use it whenever the amount of gas does not change.

P1V1 / T1 = P2V2 / T2

Any one of the simple laws drops out of it. Hold temperature steady and the T terms cancel, leaving Boyle's law. That is worth noticing. You only have to remember one relationship.

Worked example · the combined gas law

A weather balloon holds 15.0 L of helium at 1.00 atm and 295 K. It rises until the pressure around it is 0.350 atm and the temperature is 240 K. What is its new volume?

Step 1 · solve for V2

V2 = V1 × (P1 / P2) × (T2 / T1)

Step 2 · put the numbers in

V2 = 15.0 L × (1.00 atm / 0.350 atm) × (240 K / 295 K)

Step 3 · multiply

V2 = 15.0 L × 2.857 × 0.8136 = 34.9 L

Check it

Two effects pull against each other. The lower pressure lets the balloon swell a lot. The colder air shrinks it a little. Swelling wins, so a bigger answer is right. Testing the ratio both ways: 1.00 × 15.0 / 295 = 0.0508, and 0.350 × 34.9 / 240 = 0.0509.

Key points
  • Always convert temperature to kelvin first. Add 273.15 to the Celsius value.
  • Boyle: P and V are inverse. Charles: V and T are direct. Gay-Lussac: P and T are direct.
  • Avogadro's law says equal volumes of any gases hold equal moles at the same P and T.
  • The combined gas law, P1V1/T1 = P2V2/T2, contains all three of them.
Interactive Hold two variables and see Boyle, Charles and Avogadro appear. Open full ↑

Section 3The ideal gas law

Four variables, one equation

The combined gas law needs a before and an after. The ideal gas law needs neither. It ties all four gas variables together at one moment in time.

PV = nRT

Here P is pressure, V is volume in liters, n is moles and T is kelvin temperature. The letter R is the ideal gas constant. It is the number that makes the units agree, and its value depends on the pressure unit you choose.

Value of RUnitsUse it when
0.08206L·atm/mol·KP in atm, V in liters
8.314J/mol·KYou want energy, or P in Pa and V in m3
62.36L·torr/mol·KP already in torr

Most gas problems use 0.08206 L·atm/mol·K. Pick that one and the rest of your units are decided for you. Pressure must be in atm, volume in liters, temperature in kelvin. Convert anything that does not fit before you start.

An ideal gas is a gas that obeys this equation exactly. Real gases follow it closely at ordinary room conditions. The match gets worse at very high pressure and very low temperature, which is the last section of this chapter.

Molar volume and STP

Because equal volumes hold equal moles, one mole of any gas takes up the same space under the same conditions. Chemists picked a reference point so they could compare. Standard temperature and pressure, or STP, means 0 °C (273.15 K) and 1 atm. Put one mole into the ideal gas law at STP and out comes the molar volume.

V = (1 mol)(0.08206 L·atm/mol·K)(273.15 K) / (1 atm) = 22.4 L

So 22.4 L/mol at STP is a shortcut worth memorizing. One mole of oxygen, one mole of neon and one mole of methane each fill about 22.4 L at STP. That is roughly the volume of three basketballs.

One mole of any gas fills 22.4 liters at STP SAME VOLUME · SAME COUNT · DIFFERENT MASS He 22.4 L · 1 mol 4.00 g density 0.179 g/L N2 22.4 L · 1 mol 28.02 g density 1.25 g/L CO2 22.4 L · 1 mol 44.01 g density 1.96 g/L
Equal volumes, equal counts. At STP each box holds one mole, which is 6.022 × 1023 particles, in 22.4 L. Only the mass on the balance tells the three gases apart, so density rises with molar mass.

Worked example · solving for the missing variable

A steel cylinder with a volume of 2.00 L holds 0.0820 mol of argon. The gauge reads 1.10 atm. What is the temperature inside, in kelvin and in degrees Celsius?

Step 1 · rearrange

T = PV / nR

Step 2 · check the units

Pressure is in atm and volume is in liters. So use R = 0.08206 L·atm/mol·K.

Step 3 · compute

T = (1.10 atm)(2.00 L) / [(0.0820 mol)(0.08206 L·atm/mol·K)]

T = 2.20 / 0.006729 = 327 K

Step 4 · convert to Celsius

327 K − 273.15 = 54 °C

Check it

Feed 327 K back in. P = nRT/V = (0.0820)(0.08206)(327)/2.00 = 1.10 atm. It matches.

Two standards, two numbers. Older tables set standard pressure at 1 atm, which gives 22.4 L/mol. Modern IUPAC tables set it at exactly 1 bar, which gives 22.7 L/mol. This book uses 1 atm and 22.4 L/mol, so check which standard a table assumes before you trust it.
Key points
  • PV = nRT describes one gas sample at one moment, with no before and after needed.
  • R = 0.08206 L·atm/mol·K demands atm, liters, moles and kelvin. R = 8.314 J/mol·K is for energy work.
  • STP is 0 °C and 1 atm. At STP one mole of any gas fills 22.4 L.
  • Rearrange the one equation rather than memorizing four versions of it.
Interactive Solve PV = nRT for any missing variable. Open full ↑

Section 4Gas density, molar mass and stoichiometry

Density follows from PV = nRT

Density is mass divided by volume. For a gas, both of those change with conditions, so gas density is not a fixed property the way it is for a solid. You can still predict it. Start from the ideal gas law and swap moles for mass.

Moles equal mass divided by molar mass, so n = m/M. Put that in place of n and rearrange until mass over volume stands alone.

PV = (m/M)RT  →  d = m/V = PM / RT

Read that equation like a sentence. Heavier gas particles make a denser gas. More pressure packs them closer, so density rises. Higher temperature spreads them out, so density falls. Warm air rising above a radiator is this equation at work.

Turn it around and you have a way to weigh a molecule. Measure the density of an unknown gas at a known pressure and temperature, and its molar mass drops out.

M = dRT / P

Worked example · density, then molar mass

Part one: find the density of carbon dioxide at 1.00 atm and 25 °C. Part two: an unknown gas has a density of 1.34 g/L at 1.00 atm and 0 °C. Find its molar mass.

Step 1 · molar mass of CO2

12.01 + 2(16.00) = 44.01 g/mol

Step 2 · temperature in kelvin

25 + 273.15 = 298.15 K

Step 3 · use d = PM/RT

d = (1.00 atm)(44.01 g/mol) / [(0.08206 L·atm/mol·K)(298.15 K)]

d = 44.01 / 24.47 = 1.80 g/L

Step 4 · now the unknown, using M = dRT/P

M = (1.34 g/L)(0.08206 L·atm/mol·K)(273.15 K) / (1.00 atm) = 30.0 g/mol

Check it

The second gas is at STP, so the shortcut works: M = d × 22.4 L/mol = 1.34 × 22.4 = 30.0 g/mol. A molar mass of 30.0 g/mol fits nitrogen monoxide, NO.

Volumes as mole ratios

Avogadro's law gives gas stoichiometry a shortcut. At the same temperature and pressure, the volume ratio of two gases equals their mole ratio. Look at ammonia synthesis.

N2(g) + 3 H2(g) → 2 NH3(g)

One liter of nitrogen needs three liters of hydrogen and makes two liters of ammonia. You can read the coefficients as liters directly, with no mole math at all. This only works when every substance is a gas at the same conditions.

When a solid or liquid is involved, take the long road. Convert the known amount to moles. Cross the balanced equation with the mole ratio. Then use PV = nRT to turn moles of gas into a volume.

The road map from grams of reactant to liters of gas GRAMS → MOLES → MOLES → LITERS grams Zn what you weigh out ÷ 65.38 moles Zn use molar mass 1 : 1 moles H2 use the mole ratio PV = nRT liters H2 the answer Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g) Only the middle arrow needs the balanced equation. The last arrow needs P and T. At STP you may replace the last arrow with a simple × 22.4 L/mol.
The gas stoichiometry road map. Mass and volume never connect directly. Both must pass through moles, and the mole ratio from the balanced equation is the bridge in the middle.

Worked example · volume of a gas from a reaction

Zinc metal reacts with hydrochloric acid to make hydrogen gas. You use 5.00 g of zinc and plenty of acid. The hydrogen is collected at 745 torr and 298 K. What volume do you get?

Step 1 · the balanced equation

Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g)

Step 2 · grams of zinc to moles

5.00 g ÷ 65.38 g/mol = 0.0765 mol Zn

Step 3 · mole ratio

The ratio of Zn to H2 is 1 to 1, so you make 0.0765 mol H2.

Step 4 · convert the pressure

745 torr ÷ 760 torr/atm = 0.980 atm

Step 5 · use PV = nRT

V = nRT/P = (0.0765)(0.08206)(298) / 0.980 = 1.91 L

Check it

At STP that many moles would fill 0.0765 × 22.4 = 1.71 L. The real sample is warmer and at slightly lower pressure, so a bit more than 1.71 L is just what you expect.

Key points
  • d = PM/RT, so gas density rises with molar mass and pressure and falls with temperature.
  • M = dRT/P lets a density measurement identify an unknown gas.
  • At STP, density equals molar mass divided by 22.4 L/mol.
  • For gas to gas, coefficients act as volume ratios. Otherwise go grams to moles to moles to liters.
Interactive Run a reaction that makes a gas and find its volume. Open full ↑

Section 5Dalton's law of partial pressures

Each gas acts as if it were alone

Gas particles are far apart and ignore each other. So a gas in a mixture does not care what else is in the flask. It pushes on the walls exactly as hard as it would if the other gases were removed. That push is its partial pressure.

John Dalton turned this idea into a law. The law of partial pressures says the total pressure of a mixture is the sum of the partial pressures of its gases.

Ptotal = PA + PB + PC + ...

Air is the everyday example. At 1.00 atm, nitrogen supplies about 0.78 atm and oxygen about 0.21 atm. Argon, carbon dioxide and the rest make up the last sliver. Your lungs respond to the partial pressure of oxygen, not to the total.

Mole fractions split the total

There is a faster route when you know the amounts. The mole fraction of a gas, written X, is its moles divided by the total moles. It has no units and all the mole fractions in a mixture add to 1.

XA = nA / ntotal   and   PA = XA × Ptotal

So a gas that is one quarter of the particles owns one quarter of the pressure. Mass does not enter into it. Four grams of helium and four grams of oxygen are very different counts, so they claim very different shares.

Worked example · splitting a total pressure

You seal 2.00 g of helium and 8.00 g of oxygen in a 10.0 L flask at 298 K. Find each partial pressure, the total pressure and the mole fraction of helium.

Step 1 · moles of each gas

He: 2.00 g ÷ 4.003 g/mol = 0.500 mol

O2: 8.00 g ÷ 32.00 g/mol = 0.250 mol

Step 2 · each partial pressure from PV = nRT

PHe = (0.500)(0.08206)(298) / 10.0 = 1.22 atm

PO2 = (0.250)(0.08206)(298) / 10.0 = 0.611 atm

Step 3 · add them

Ptotal = 1.22 + 0.611 = 1.83 atm

Step 4 · mole fraction of helium

XHe = 0.500 / (0.500 + 0.250) = 0.667

Check it

Use the mole fraction the other way. 0.667 × 1.83 atm = 1.22 atm, which is the helium value from step 2. Notice that helium is only a fifth of the mass but two thirds of the pressure.

Collecting a gas over water and correcting for water vapor COLLECTING A GAS OVER WATER water trough H2 + H2O vapor gas from the flask bottle filled with water, then pushed out by gas SPLITTING THE TOTAL Level the water inside and out, and the total inside equals room pressure. P of the dry gas water vapor Pgas = Ptotal − Pwater At 25 °C water vapor adds 23.8 torr, about 3% of 1 atm. Warmer water adds more, so the correction grows.
Water always sends some of itself along. A gas bubbled up through water carries water vapor with it. The vapor owns part of the measured pressure, so subtract its tabulated value to get the dry gas pressure.

Gas collected over water

Many lab gases are collected by bubbling them into an upturned bottle full of water. The gas pushes the water out and takes its place. One detail is easy to miss. Liquid water keeps evaporating into the space above it, so the trapped sample is a mixture.

The water contributes its vapor pressure, a value that depends only on temperature. Tables list it, and it climbs fast as water warms. At 20 °C it is 17.5 torr. At 25 °C it is 23.8 torr. Subtract it to get the pressure of the gas you actually want.

Pgas = Ptotal − Pwater

Worked example · moles of a gas collected over water

You collect 0.250 L of hydrogen over water at 25 °C. The barometer reads 755 torr, and the water levels are even. How many moles of hydrogen do you have?

Step 1 · subtract the water vapor

PH2 = 755 torr − 23.8 torr = 731 torr

Step 2 · convert to atm

731 torr ÷ 760 torr/atm = 0.962 atm

Step 3 · solve for moles

n = PV/RT = (0.962)(0.250) / [(0.08206)(298.15)]

n = 0.2405 / 24.47 = 9.83 × 10−3 mol

Check it

Skipping the water step would have given 1.02 × 10−2 mol. That is about 3% too high, which is exactly the share water vapor was holding.

Key points
  • Each gas in a mixture pushes as if it were alone, and the pushes add up.
  • PA = XA × Ptotal, where XA is moles of A over total moles.
  • Mole fractions split pressure by particle count, never by mass.
  • A gas collected over water shares the space with water vapor, so subtract the tabulated vapor pressure.
Interactive Mix gases in a flask and split the total pressure. Open full ↑

Section 6Kinetic molecular theory, effusion and real gases

What the particles are really doing

The gas laws say what happens. Kinetic molecular theory says why. It is a short list of claims about the particles themselves.

Those five claims explain every law in this chapter. Shrink the volume and each particle reaches a wall sooner, so hits come more often and pressure rises. That is Boyle. Raise the temperature and the particles move faster, so they hit harder and more often. That is Charles and Gay-Lussac. Add more particles and you add more hits. That is Avogadro.

Speed, mass and temperature

The last claim hides something important. At a given temperature, every gas has the same average kinetic energy per mole.

KEavg = (3/2)RT

Kinetic energy is one half mass times speed squared. So if the energy is equal but the masses differ, the speeds cannot be equal. Light particles must move faster to carry the same energy. Solve for speed and you get the root-mean-square speed.

urms = √(3RT/M)

Use R = 8.314 J/mol·K here, and put molar mass in kilograms per mole. Then the answer comes out in meters per second. This is the one place in the chapter where 0.08206 is the wrong constant.

Not every particle moves at that speed. The speeds spread out into a curve called the Maxwell-Boltzmann distribution. Heat the gas and the peak slides right and flattens out, because the range of speeds widens. Swap in a heavier gas at the same temperature and the peak slides left and gets taller and narrower.

How temperature and molar mass reshape the speed distribution THE SPREAD OF MOLECULAR SPEEDS 500 1000 1500 2000 2500 3000 particle speed (m/s) share of particles N2 at 300 K · urms = 517 m/s N2 at 1000 K · urms = 943 m/s He at 300 K · urms = 1367 m/s Dotted lines mark urms for each curve. Every curve covers the same total area.
Same gas, hotter: the peak slides right and flattens. Lighter gas, same temperature: the peak slides much further right. Each curve encloses the same area, because it still describes one whole sample. The tail on the right is the small group of very fast particles.

Effusion and Graham's law

Effusion is the escape of a gas through a tiny hole into a vacuum. A faster gas finds the hole more often, so it leaks out faster. Since speed goes as one over the square root of molar mass, so does the effusion rate. That is Graham's law.

rateA / rateB = √(MB / MA)

Watch the subscripts. The lighter gas is on top of the rate ratio but on the bottom inside the square root. A helium balloon goes limp in a day for this reason, while an air-filled one lasts much longer.

Worked example · speed and effusion

Find the root-mean-square speed of nitrogen at 298 K. Then compare how fast helium and oxygen effuse.

Step 1 · molar mass in kg/mol

N2 is 28.02 g/mol, which is 0.02802 kg/mol.

Step 2 · use the 8.314 value of R

urms = √[3(8.314 J/mol·K)(298 K) / 0.02802 kg/mol]

urms = √(265,300) = 515 m/s

Step 3 · now Graham's law

rateHe / rateO2 = √(32.00 / 4.003) = √7.994 = 2.83

Check it

Helium effuses 2.83 times faster than oxygen. Oxygen is about eight times heavier, and the square root of 8 is close to 2.8. The speed answer of 515 m/s is faster than a jet, which fits how quickly a smell crosses a room.

Where real gases stray

No gas is truly ideal. Two of the five claims are not quite true. Particles do take up space, and they do attract each other a little. At ordinary conditions those errors are too small to notice. Squeeze hard or cool far down and they show up.

Chemists track the error with the compressibility factor, Z = PV/nRT. An ideal gas has Z = 1 at every condition. Attractions pull particles away from the walls, so pressure falls below ideal and Z drops under 1. At very high pressure the particles' own volume takes over, so Z climbs above 1.

The van der Waals equation patches both problems. The constant a adds back the pressure lost to attraction. The constant b removes the space the particles themselves fill.

(P + an2/V2)(V − nb) = nRT

Both constants are measured for each gas. Large, sticky molecules get large values. Carbon dioxide has a = 3.592 L2·atm/mol2 and b = 0.04267 L/mol. Helium, which is tiny and barely sticky, has a = 0.0346 and b = 0.0238.

Worked example · how far off is ideal?

Squeeze 1.00 mol of carbon dioxide into 0.500 L at 273.15 K. Compare the ideal pressure with the van der Waals pressure.

Step 1 · the ideal answer

P = nRT/V = (1.00)(0.08206)(273.15) / 0.500 = 44.8 atm

Step 2 · the van der Waals form, solved for P

P = nRT/(V − nb) − an2/V2

Step 3 · the volume term

0.500 L − (1.00)(0.04267) = 0.4573 L, so 22.41 / 0.4573 = 49.0 atm

Step 4 · the attraction term

3.592 / (0.500)2 = 3.592 / 0.250 = 14.4 atm

Step 5 · combine

P = 49.0 − 14.4 = 34.6 atm

Check it

The real pressure is about 23% below the ideal guess. Attraction wins at this squeeze, so Z = 34.6/44.8 = 0.77, well under 1.

A useful rule of thumb. Gases behave most ideally when they are hot and spread out. They behave least ideally near the temperature and pressure at which they would turn into a liquid.
Key points
  • Average kinetic energy depends only on kelvin temperature, so lighter gases move faster.
  • urms = √(3RT/M), with R = 8.314 J/mol·K and M in kg/mol.
  • Heating widens and flattens the speed curve. Extra mass narrows it and shifts it left.
  • Graham's law: rateA/rateB = √(MB/MA).
  • Real gases stray at high pressure and low temperature. The van der Waals constants a and b correct for attraction and particle volume.
Interactive Change temperature and mass and watch the speed distribution shift. Open full ↑

The chapter in one breath

  • Pressure is particles drumming on a wall, and 1 atm = 760 torr = 101.325 kPa.
  • Manometers compare two pushes. A closed end reads the gas pressure outright, while an open end adds or subtracts the room.
  • Kelvin or nothing. Every gas law needs absolute temperature, so add 273.15 to a Celsius value first.
  • The combined gas law holds the simple ones. Boyle, Charles and Gay-Lussac all fall out of P1V1/T1 = P2V2/T2.
  • One equation does most of the work. PV = nRT with R = 0.08206 L·atm/mol·K handles any single state.
  • At STP one mole fills 22.4 L, whatever the gas, so density at STP is molar mass over 22.4.
  • Gas stoichiometry is normal stoichiometry with PV = nRT bolted on at the end.
  • Pressure splits by count. PA = XAPtotal, and a gas over water loses the water's vapor pressure.
  • Temperature sets energy, mass sets speed. Light gases move faster, effuse faster, and all gases stray from ideal when cold and squeezed.

ReferenceWords worth knowing

These are the terms this chapter leans on. Come back to them when a problem uses a word you are not sure of.

Avogadro's law
At fixed pressure and temperature, gas volume is proportional to moles. Equal volumes hold equal counts.
barometer
A tube of mercury inverted in a dish, used to measure atmospheric pressure from the height of the column.
Boyle's law
At fixed temperature and amount, pressure and volume are inversely related, so PV stays constant.
Charles's law
At fixed pressure and amount, volume is proportional to kelvin temperature.
combined gas law
The single relation P1V1/T1 = P2V2/T2, used when the amount of gas does not change.
Dalton's law of partial pressures
The total pressure of a gas mixture equals the sum of the partial pressures of its gases.
effusion
The escape of gas through a very small opening into a vacuum.
Gay-Lussac's law
At fixed volume and amount, pressure is proportional to kelvin temperature.
Graham's law
Effusion rate varies as one over the square root of molar mass, so rateA/rateB = √(MB/MA).
ideal gas constant (R)
The proportionality constant in PV = nRT. It is 0.08206 L·atm/mol·K or 8.314 J/mol·K.
kinetic molecular theory
The particle-level model of a gas: tiny, far apart, in constant random motion, with elastic collisions and energy set by temperature.
manometer
A U-shaped mercury tube that reports the pressure of a gas sample, either against a vacuum or against room air.
Maxwell-Boltzmann distribution
The curve showing how many particles in a sample hold each speed at a given temperature.
molar volume
The volume one mole of gas fills. At STP it is 22.4 L/mol.
mole fraction (X)
Moles of one gas divided by total moles in the mixture. It sets that gas's share of the pressure.
partial pressure
The pressure one gas in a mixture would exert if it alone filled the container.
root-mean-square speed
The speed matching the average kinetic energy of a sample, equal to √(3RT/M).
standard temperature and pressure (STP)
The reference state 0 °C and 1 atm, where one mole of gas fills 22.4 L.
van der Waals equation
A corrected gas equation that adds a for particle attraction and subtracts b for particle volume.
vapor pressure
The pressure of vapor above a liquid at a given temperature. Water's value must be subtracted from a gas collected over water.

Check yourself

1. A gauge reads 0.875 atm. Give that pressure in torr and in kPa.
665 torr and 88.7 kPa. Multiply by the conversion factor each time. 0.875 atm × 760 torr/atm = 665 torr. 0.875 atm × 101.325 kPa/atm = 88.7 kPa.
2. A closed-end manometer shows mercury 148 mm higher on the closed arm than on the gas arm. What is the gas pressure in torr and in atm?
148 torr, or 0.195 atm. The closed arm holds a vacuum, so nothing pushes back and the height gap is the pressure itself. Then 148 torr ÷ 760 torr/atm = 0.195 atm.
3. A syringe holds 350 mL of air at 1.00 atm. You push the plunger until the pressure reads 2.50 atm, keeping the temperature steady. What is the new volume?
140 mL. Use Boyle's law. V2 = P1V1/P2 = (1.00)(350)/2.50 = 140 mL. Pressure went up by a factor of 2.5, so volume dropped by the same factor.
4. A balloon holds 1.75 L at 295 K. You warm it to 350 K at constant pressure. What is the new volume?
2.08 L. Charles's law gives V2 = V1T2/T1 = 1.75 L × (350/295) = 1.75 × 1.186 = 2.08 L.
5. Why do gas laws fail if you put temperature in degrees Celsius?
Because the Celsius zero is not a real zero. Volume and pressure are proportional to temperature measured from absolute zero. Going from 10 °C to 20 °C does not double anything, since in kelvin that is 283 K to 293 K. Celsius values can also be negative, which would predict a negative volume.
6. What volume does 0.500 mol of neon fill at 2.00 atm and 350 K?
7.18 L. V = nRT/P = (0.500)(0.08206)(350)/2.00. The top is 14.36, and 14.36 ÷ 2.00 = 7.18 L.
7. What volume does 3.00 mol of oxygen fill at STP?
67.2 L. At STP the molar volume is 22.4 L/mol, so 3.00 mol × 22.4 L/mol = 67.2 L. The gas identity never enters the calculation.
8. Find the density of oxygen gas at STP.
1.43 g/L. One mole of O2 has a mass of 32.00 g and fills 22.4 L. So d = 32.00 g ÷ 22.4 L = 1.43 g/L. Room air is near 1.2 g/L, and oxygen is a little heavier than average air.
9. A flask holds 0.200 mol of nitrogen and 0.300 mol of oxygen. The total pressure is 2.50 atm. What is the partial pressure of nitrogen?
1.00 atm. First the mole fraction: X = 0.200 / 0.500 = 0.400. Then P = 0.400 × 2.50 atm = 1.00 atm. The oxygen holds the other 1.50 atm.
10. You collect oxygen over water at 25 °C. The total pressure is 762 torr, and the vapor pressure of water at that temperature is 23.8 torr. What is the partial pressure of the oxygen?
738 torr. Subtract the water vapor: 762 torr − 23.8 torr = 738.2 torr, which rounds to 738 torr. Skipping this step would overstate your oxygen by about 3%.
11. Methane (CH4, 16.04 g/mol) and sulfur dioxide (SO2, 64.06 g/mol) sit beside the same pinhole. How much faster does the methane effuse?
About 2.00 times faster. Graham's law gives rateCH4/rateSO2 = √(64.06/16.04) = √3.994 = 2.00. Sulfur dioxide is four times heavier, and the square root of 4 is 2.
12. Two samples of carbon dioxide sit at 1 atm and 300 K, and at 200 atm and 220 K. Which one strays further from PV = nRT, and why?
The cold, squeezed sample at 200 atm and 220 K. High pressure crowds the particles, so their own volume stops being negligible. Low temperature makes them slow, so weak attractions get a chance to pull. Both effects push Z = PV/nRT away from 1.