IB DP Physics · Theme B · The particulate nature of matter

Gas laws.

B.3 · Pressure, the mole, the ideal gas equation and the kinetic theory of gases
Standard level and higher level

SL + HL · 6 hours B.4 Thermodynamics (HL) ↗

The journey

Three parts.

Part 01

Pressure and amount

\(P = \frac{F}{A}\), the mole and the Avogadro constant

Part 02

The gas laws

Three experiments, and one equation: \(PV = nRT\)

Part 03

The kinetic theory

Explaining pressure and temperature by the motion of molecules

Parts 1 and 2 describe what a gas does. Part 3 explains why.

B.3 GAS LAWS · SL + HL

Part 01

Pressure and amount.

Two quantities needed before any gas law can be written

B.3 GAS LAWS · SL + HL

Force per unit area

Pressure.

\(P = \dfrac{F}{A}\)

\(F\): the force at right angles to the surface
Unit: the pascal, \(1 \,\mathrm{Pa} = 1 \,\mathrm{N} \,\mathrm{m}^{-2}\)
Atmospheric pressure is about \(1.0 \times 10^{5} \,\mathrm{Pa}\)

The air pushes on a window 1.2 m by 0.80 m \(P = \dfrac{F}{A}\) so \(F = PA\) \(F = 1.0 \times 10^{5} \times (1.2 \times 0.80)\) \(F = \class{res}{9.6 \times 10^{4} \,\mathrm{N}}\) The weight of about ten tonnes

B.3 GAS LAWS · SL + HL

Why the window survives

The air inside the room pushes outwards on the glass with almost exactly the same force. Pressure in a gas acts equally in all directions, so what matters for any object is the difference in pressure across it. A sealed can crushes when the air is pumped out of it because the outward force has gone and the inward one has not.

Other units you will meet

One atmosphere is \(1.01 \times 10^{5}\) Pa. One bar is \(1.00 \times 10^{5}\) Pa. Tire and blood pressures are still often quoted in older units. Convert to pascals before using any gas equation.

Leaving the area in cm². \(1 \,\mathrm{cm}^{2} = 10^{-4} \,\mathrm{m}^{2}\).

Pressure is a scalar. The force it produces is always at right angles to whatever surface it acts on.

Where the force comes from

The pressure of a gas.

The molecules move at random and hit the walls
Each collision changes a molecule’s momentum
So the wall exerts a force on the molecule, and the molecule an equal and opposite force on the wall
Force on each unit area: the pressure

A BOX OF GAS

B.3 GAS LAWS · SL + HL

Two of Newton’s laws

From A.2, force is the rate of change of momentum. A molecule that bounces off a wall has had its momentum reversed, so the wall must have pushed it: that is the second law. By the third law the molecule pushed the wall just as hard, in the opposite direction. Pressure is this push, added up for all the molecules striking one square meter each second.

Why it feels steady

Each collision is a tiny, brief kick. But about \(10^{27}\) molecules strike each square meter of wall every second at room conditions, so the kicks merge into a constant force, just as heavy rain on a roof makes a steady roar and not separate taps.

A full answer to “explain how a gas exerts a pressure” has four steps: collisions with the wall; change of momentum; force from rate of change of momentum; pressure as force per unit area.

Counting molecules

Amount of substance.

\(n = \dfrac{N}{N_\mathrm{A}}\)

\(n\): the amount, in moles
\(N\): the number of molecules
\(N_\mathrm{A} = 6.02 \times 10^{23} \,\mathrm{mol}^{-1}\), the Avogadro constant

8.0 g of helium · molar mass 4.0 g mol⁻¹ \(n = \dfrac{\text{mass}}{\text{molar mass}} = \dfrac{8.0}{4.0} = 2.0 \,\mathrm{mol}\) \(N = n N_\mathrm{A} = 2.0 \times 6.02 \times 10^{23}\) \(N = \class{res}{1.2 \times 10^{24}}\) atoms

B.3 GAS LAWS · SL + HL

A mole is just a number

A dozen is 12 of anything. A mole is \(6.02 \times 10^{23}\) of anything. The number was chosen so that one mole of a substance has a mass in grams equal to its relative molecular mass: 4 g for helium, 28 g for nitrogen (N₂), 44 g for carbon dioxide.

The mass of one molecule

Divide the molar mass by the Avogadro constant. For helium, \(\frac{4.0 \times 10^{-3}}{6.02 \times 10^{23}} = 6.6 \times 10^{-27}\) kg. You need this in Part 3, where the speed of a molecule depends on its mass.

Why gases are counted in moles

The pressure of a gas depends on how many molecules there are, not on what kind they are. A mole of helium and a mole of carbon dioxide in identical containers at the same temperature exert the same pressure, although one has eleven times the mass of the other.

Mixing grams and kilograms. Molar masses are usually given in g mol⁻¹; an answer in kilograms needs \(10^{-3}\).

Quick check

Pressure and amount.

01A 600 N force acts at right angles on 0.030 m². Pressure?

\(\frac{600}{0.030} = 2.0 \times 10^{4} \,\mathrm{Pa}\)

02What is 1 Pa in base SI units of force and area?

1 N m⁻²

03How many molecules are in 0.50 mol?

\(3.0 \times 10^{23}\)

04How many moles are in 56 g of nitrogen, N₂ (28 g mol⁻¹)?

2.0 mol

05What happens to a molecule’s momentum when it hits a wall?

It changes direction: a change of \(2mv\)

06Which of Newton’s laws gives the force on the wall?

The third (with the second)

PRESS → TO REVEAL EACH ANSWER

Part 02

The gas laws.

Three laws found by experiment, joined into one equation

B.3 GAS LAWS · SL + HL

The story · England, 1662

Background story · not assessed

Boyle, Hooke and the spring of the air.

Painted portrait of Robert Boyle, a man with long curled hair — **Robert Boyle**, painted by Johann Kerseboom about 1689

Engraving of an air pump: a tall box with a rack and handle on a wooden stand, joined by a pipe to a glass vessel with a small animal inside — His **air pump**, from a book of 1669

Boyle’s assistant Robert Hooke built the apparatus
Air trapped in the closed end of a J-shaped tube was squeezed by pouring in mercury
Double the pressure: half the volume. Published in 1662
Richard Towneley and Henry Power had noticed the pattern first; Edme Mariotte found it again in 1679

One of the first laws of physics to be established by measurement.

Portrait: Johann Kerseboom, about 1689, public domain. Engraving: Robert Boyle, 1669, Library of Congress, public domain (cropped).

B.3 GAS LAWS · SL + HL

First law · constant temperature

Pressure and volume.

\(PV = \text{constant}\)

For a fixed mass of gas at constant temperature
\(P \propto \dfrac{1}{V}\): halve the volume, double the pressure

SQUEEZING A GAS SLOWLY

B.3 GAS LAWS · SL + HL

Why in terms of molecules

At a constant temperature the molecules keep the same average speed, so each collision with a wall is as hard as before. In half the volume, though, there are twice as many molecules in every cubic centimeter, so twice as many strike each square centimeter of wall every second. Twice the collisions at the same strength give twice the pressure.

Why “slowly”

Pushing a piston in does work on the gas and warms it, as a bicycle pump shows. The law needs the temperature to stay the same, so the compression must be slow enough for the extra energy to leak away through the walls.

To test the law with data, plot \(P\) against \(\frac{1}{V}\). A straight line through the origin confirms \(P \propto \frac{1}{V}\); a curve of \(P\) against \(V\) could be many things.

Boyle’s law as a graph

Pressure against volume.

Each curve is for one temperature: an isotherm
Along it, \(P \times V\) has the same value at every point
A hotter gas has its curve further from the origin

TWO TEMPERATURES

B.3 GAS LAWS · SL + HL

A quick test for an isotherm

Pick two points on a curve and multiply the pressure by the volume at each. On the slide, \(2P \times V\) and \(P \times 2V\) are equal. If the products differ, the temperature was not constant, however smooth the curve looks. The same test settles exam questions that ask whether a change shown on a graph was isothermal.

Why the hotter curve is higher

At any given volume a hotter gas has the greater pressure, so its curve lies above. The product \(PV\) is larger all the way along it: Part 2 ends by showing that \(PV\) is proportional to the kelvin temperature.

Drawing an isotherm that touches the axes. The pressure never reaches zero, however large the volume, and the volume never reaches zero.

Second law · constant volume

Pressure and temperature.

\(\dfrac{P}{T} = \text{constant}\)

For a fixed mass of gas at constant volume
\(P \propto T\), with \(T\) in kelvin
The line points back to −273 °C

A SEALED FLASK, HEATED

B.3 GAS LAWS · SL + HL

Finding absolute zero in a school laboratory

Heat a sealed flask of air in a water bath and read the pressure at several temperatures between 0 °C and 100 °C. The points lie on a straight line. Nothing in the experiment goes anywhere near −273 °C, but extending the line back shows where the pressure would fall to zero. Whatever gas is used, the line reaches the axis at the same temperature. That is how absolute zero was first located.

Why in terms of molecules

A hotter gas has faster molecules. In a fixed volume they reach the walls more often, and each collision changes the momentum by more. Both effects raise the pressure.

Saying that pressure is proportional to the Celsius temperature. Heating a gas from 10 °C to 20 °C does not double its pressure: it goes from 283 K to 293 K, a rise of 3.5%.

Third law · constant pressure

Volume and temperature.

\(\dfrac{V}{T} = \text{constant}\)

For a fixed mass of gas at constant pressure
\(V \propto T\), with \(T\) in kelvin
The same intercept: −273 °C

A GAS FREE TO EXPAND, HEATED

B.3 GAS LAWS · SL + HL

Why in terms of molecules

Heat a gas behind a freely moving piston. The molecules speed up and, for a moment, the pressure inside exceeds the pressure outside, so the piston is pushed out. As the volume grows, the molecules have further to travel between collisions and strike the walls less often. The piston stops when the pressure inside has fallen back to the outside pressure. The gas ends up hotter and larger, at the same pressure as before.

A hot-air balloon

The balloon is open at the bottom, so the air inside stays at atmospheric pressure. Heating it from 290 K to 390 K makes each kilogram of air take up \(\frac{390}{290} = 1.34\) times the volume, so about a quarter of the air spills out. What remains weighs less than the cooler air it displaces, and the balloon rises.

A real gas would turn to liquid long before −273 °C. The straight line all the way to zero volume belongs to the ideal gas of Part 3.

The story · Paris, 1783

Background story · not assessed

Jacques Charles goes up in a balloon.

Painted portrait of Jacques Charles, a man in a powdered wig and dark coat — **Jacques Charles**, about 1798

Painting of a striped balloon carrying two men in a boat-shaped basket, rising above a crowd in a garden with statues — The ascent from the **Tuileries**, 1 December 1783

27 August 1783: Charles and the Robert brothers launch the first hydrogen balloon
1 December: Charles and Nicolas-Louis Robert fly in one for 2 hours, covering 36 km
About 1787 he heated balloons of different gases and found that they expanded equally
He never published. Gay-Lussac did, in 1802, and named the law after Charles

Portrait: Adélaïde Labille-Guiard, about 1798, public domain. Painting: anonymous, 1783, public domain.

B.3 GAS LAWS · SL + HL

One relation for a fixed mass of gas

Combining the three laws.

\(\dfrac{PV}{T} = \text{constant}\)

Between two states: \(\dfrac{P_{1} V_{1}}{T_{1}} = \dfrac{P_{2} V_{2}}{T_{2}}\)
Hold one quantity constant and it reduces to one of the three laws

A weather balloon: 2.0 m³ at \(1.0 \times 10^{5}\) Pa and 290 K It rises to where \(P = 2.5 \times 10^{4}\) Pa and \(T = 232\) K \(V_{2} = \dfrac{P_{1} V_{1} T_{2}}{T_{1} P_{2}}\) \(V_{2} = \dfrac{1.0 \times 10^{5} \times 2.0 \times 232}{290 \times 2.5 \times 10^{4}}\) \(V_{2} = \class{res}{6.4 \,\mathrm{m}^{3}}\)

B.3 GAS LAWS · SL + HL

Reading the answer

Two things happened to the balloon. The pressure fell to a quarter, which alone would have made the volume four times as large: 8.0 m³. The gas also cooled, from 290 K to 232 K, which shrinks it by the factor \(\frac{232}{290} = 0.80\). Together: \(8.0 \times 0.80 = 6.4\) m³. Splitting a problem into one change at a time is a good check on the algebra.

Why balloons burst

A weather balloon is released only partly filled. As it climbs, the falling pressure lets the gas expand until, at about 30 km, the rubber can stretch no further.

Only \(T\) must be in a particular unit, the kelvin. \(P\) and \(V\) can be in any units, as long as they are the same on both sides.

Putting a value on the constant

The ideal gas equation.

\(PV = nRT = N k_\mathrm{B} T\)

\(R = 8.31 \,\mathrm{J} \,\mathrm{K}^{-1} \,\mathrm{mol}^{-1}\), the gas constant
\(k_\mathrm{B} = 1.38 \times 10^{-23} \,\mathrm{J} \,\mathrm{K}^{-1}\), the Boltzmann constant

Counting in moles

\(PV = nRT\)

Use it when the amount is given in moles or as a mass

Counting molecules

\(PV = N k_\mathrm{B} T\)

Use it when the number of molecules is wanted

The link

\(k_\mathrm{B} = \frac{R}{N_\mathrm{A}}\)

The gas constant for one molecule

B.3 GAS LAWS · SL + HL

From “constant” to \(nR\)

For a fixed mass of gas, \(\frac{PV}{T}\) is constant. Experiment shows what the constant depends on: double the amount of gas, at the same pressure and temperature, and the volume doubles. So the constant is proportional to \(n\), and the same for every gas. Writing it as \(nR\) gives \(PV = nRT\).

The two forms are the same equation

Since \(n = \frac{N}{N_\mathrm{A}}\), \(nR = N \frac{R}{N_\mathrm{A}} = N k_\mathrm{B}\). Check the numbers: \(\frac{8.31}{6.02 \times 10^{23}} = 1.38 \times 10^{-23}\).

A number worth knowing

One mole of any gas at 273 K and \(1.01 \times 10^{5}\) Pa occupies \(V = \frac{nRT}{P} = \frac{1 \times 8.31 \times 273}{1.01 \times 10^{5}} = 0.0225\) m³: 22.5 liters, the size of a large bucket.

Volumes in liters or cm³. The equation needs m³: \(1 \,\mathrm{liter} = 10^{-3} \,\mathrm{m}^{3}\) and \(1 \,\mathrm{cm}^{3} = 10^{-6} \,\mathrm{m}^{3}\).

Worked example

A cylinder of oxygen.

\(V = 0.020 \,\mathrm{m}^{3}\) · \(P = 1.5 \times 10^{7} \,\mathrm{Pa}\) · \(T = 293 \,\mathrm{K}\) \(PV = nRT\) so \(n = \dfrac{PV}{RT}\) \(n = \dfrac{1.5 \times 10^{7} \times 0.020}{8.31 \times 293}\) \(n = \class{res}{123 \,\mathrm{mol}}\)

Oxygen, O₂: 32 g mol⁻¹ \(m = n \times \text{molar mass} = 123 \times 0.032\) \(m = \class{res}{3.9 \,\mathrm{kg}}\) Number of molecules \(N = n N_\mathrm{A} = 123 \times 6.02 \times 10^{23} = \class{res}{7.4 \times 10^{25}}\)

B.3 GAS LAWS · SL + HL

What the cylinder is worth

Let the same gas out into a room at atmospheric pressure and the same temperature. By Boyle’s law the volume grows by the ratio of the pressures: \(0.020 \times \frac{1.5 \times 10^{7}}{1.0 \times 10^{5}} = 3.0\) m³. A cylinder the size of a rucksack holds a small room’s worth of oxygen.

If the cylinder is left in the sun

The volume is fixed, so the pressure law applies. At 333 K the pressure becomes \(1.5 \times 10^{7} \times \frac{333}{293} = 1.7 \times 10^{7}\) Pa. Gas cylinders are built with a safety margin for this.

A density can be found the same way: \(\rho = \frac{m}{V} = \frac{3.9}{0.020} = 200 \,\mathrm{kg} \,\mathrm{m}^{-3}\), over a hundred times that of the air around you.

Showing a change of state

Changes on a pressure–volume diagram.

A point is one state of the gas: \(P\), \(V\) and so \(T\)
A line is a change from one state to another
Larger \(PV\): hotter

Each line can be followed in either direction.

THREE CHANGES FROM ONE STATE

B.3 GAS LAWS · SL + HL

Reading temperature from the diagram

For a fixed amount of gas, \(T = \frac{PV}{nR}\), so the temperature at any point is proportional to the product of its two coordinates. Moving right along the constant-pressure line, \(V\) grows and so does \(T\): the gas is being heated. Moving up the constant-volume line, \(P\) grows and so does \(T\). Along the curve, \(PV\) and the temperature stay the same.

Finding an unknown state

Given \(P\), \(V\) and \(T\) at one point of a diagram, you can find the temperature anywhere else on it from \(\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}\), without ever knowing how much gas there is.

B.4 (HL) gives these changes their names, isobaric, isovolumetric and isothermal, adds a fourth, and shows that the area under a line is the work done by the gas.

Quick check

The gas laws.

01A gas at constant temperature is squeezed to a third of its volume. Pressure?

Three times as great

02A sealed tire at \(2.0 \times 10^{5}\) Pa warms from 280 K to 308 K. New pressure?

\(2.0 \times 10^{5} \times \frac{308}{280} = 2.2 \times 10^{5} \,\mathrm{Pa}\)

03Where does a graph of \(V\) against \(\theta\) in °C meet the temperature axis?

−273 °C

04What is the name of a constant-temperature curve on a \(P\)–\(V\) diagram?

An isotherm

05How are \(R\) and \(k_\mathrm{B}\) related?

\(k_\mathrm{B} = \frac{R}{N_\mathrm{A}}\)

06Volume of 2.0 mol at 300 K and \(1.0 \times 10^{5}\) Pa?

\(\frac{2.0 \times 8.31 \times 300}{1.0 \times 10^{5}} = 0.050 \,\mathrm{m}^{3}\)

PRESS → TO REVEAL EACH ANSWER

Part 03

The kinetic theory.

A model of molecules in motion that explains the gas laws

B.3 GAS LAWS · SL + HL

The assumptions of the kinetic model

The ideal gas.

Many, and random

A very large number of identical molecules in random motion

So averages are reliable

Tiny

The volume of the molecules is negligible

Compared with the volume of the container

No forces

No forces between molecules except during collisions

So no intermolecular potential energy

Elastic

All collisions are elastic

No kinetic energy is lost

Brief

A collision takes a negligible time

Compared with the time between collisions

Newtonian

The molecules obey Newton’s laws

Straight lines at constant speed between collisions

B.3 GAS LAWS · SL + HL

An ideal gas is a model, not a substance

No real gas satisfies these assumptions exactly: real molecules take up some room and attract each other slightly. An ideal gas is the imaginary gas that satisfies them perfectly, and it obeys \(PV = nRT\) at every pressure and temperature. Real gases follow the equation closely whenever the assumptions are nearly true.

Why each assumption is needed

Without elastic collisions the molecules would gradually lose kinetic energy and the gas would cool by itself in a sealed, insulated box, which does not happen. Without “no forces” the molecules would slow down as they approached a wall, pulled back by the others, and the pressure would be lower. Without “negligible volume” the space available to each molecule would be less than the volume of the container.

“State the assumptions of the kinetic model” is a standard question. Learn them as short phrases, not as the cards’ full sentences.

One molecule and one wall

Pressure from collisions.

One collision with the right-hand wall · right is positive \(\Delta p_\mathrm{molecule} = m(-v) - mv = -2mv\) Time between its collisions with that wall \(\Delta t = \dfrac{2L}{v}\) Average force on the wall \(F_\mathrm{wall} = -\dfrac{\Delta p_\mathrm{molecule}}{\Delta t} = \dfrac{m v^{2}}{L}\)

BACK AND FORTH

B.3 GAS LAWS · SL + HL

Why \(v^{2}\)

Use final minus initial for the molecule: its momentum changes from \(+mv\) to \(-mv\), giving \(\Delta p_\mathrm{molecule} = -2mv\). The impulse on the wall is opposite, \(+2mv\). Speed counts twice. A faster molecule hits the wall harder: the magnitude of its momentum change, \(2mv\), is proportional to \(v\). It also returns sooner: the time for the round trip, \(\frac{2L}{v}\), is inversely proportional to \(v\). Harder and more often together make the force proportional to \(v^{2}\).

From one molecule to a gas

Take a cubical box of side \(L\) holding \(N\) molecules. On average a third of the motion is along each of the three directions, so the total force on one wall is \(\frac{1}{3} N \frac{m v^{2}}{L}\). The wall has area \(L^{2}\), so the pressure is \(\frac{1}{3} \frac{N m v^{2}}{L^{3}}\). \(Nm\) is the mass of the gas and \(L^{3}\) its volume, so \(\frac{Nm}{L^{3}}\) is the density, and the result is \(P = \frac{1}{3} \rho v^{2}\).

You will not be asked to reproduce the derivation in full, but questions do ask for its steps: momentum change, time between collisions, force, then pressure.

The result for the whole gas

Pressure and molecular speed.

\(P = \dfrac{1}{3} \rho v^{2}\)

\(\rho\): the density of the gas
\(v\): the average translational speed of the molecules
Links what a gauge reads to what the molecules are doing

Air: \(P = 1.01 \times 10^{5} \,\mathrm{Pa}\) · \(\rho = 1.2 \,\mathrm{kg} \,\mathrm{m}^{-3}\) \(P = \dfrac{1}{3} \rho v^{2}\) so \(v = \sqrt{\dfrac{3P}{\rho}}\) \(v = \sqrt{\dfrac{3 \times 1.01 \times 10^{5}}{1.2}}\) \(v = \class{res}{5.0 \times 10^{2} \,\mathrm{m} \,\mathrm{s}^{-1}}\)

B.3 GAS LAWS · SL + HL

Molecules from a pressure gauge

Nothing in the worked example involves a single molecule: the pressure comes from a barometer and the density from weighing a flask of air. Yet the answer is the speed of the molecules, about 500 m s⁻¹, half as fast again as sound. This was one of the first successes of the kinetic theory.

Which average

The molecules have a wide range of speeds. The \(v^{2}\) in the equation is the average of the squares of their speeds, and \(v\) is its square root, the “root-mean-square” speed. It is a little larger than the ordinary mean speed, because squaring gives extra weight to the fastest molecules.

If they move so fast, why do smells spread slowly?

A molecule travels less than a ten-thousandth of a millimeter before it hits another and changes direction. It makes billions of collisions a second and wanders, never going far in a straight line.

Sound in air travels at about 340 m s⁻¹. It cannot outrun the molecules that carry it.

Joining the two equations

Temperature and molecular energy.

\(N\) molecules, each of mass \(m\), in a volume \(V\) \(P = \dfrac{1}{3} \rho v^{2}\) with \(\rho = \dfrac{Nm}{V}\) \(PV = \dfrac{1}{3} N m v^{2}\) But also \(PV = N k_\mathrm{B} T\) \(\dfrac{1}{3} m v^{2} = k_\mathrm{B} T\) \(\dfrac{1}{2} m v^{2} = \class{res}{\dfrac{3}{2} k_\mathrm{B} T}\)

The average kinetic energy of a molecule depends only on the kelvin temperature.

Helium at 300 K · \(m = 6.64 \times 10^{-27}\) kg \(v = \sqrt{\dfrac{3 k_\mathrm{B} T}{m}} = \class{res}{1.4 \times 10^{3} \,\mathrm{m} \,\mathrm{s}^{-1}}\)

B.3 GAS LAWS · SL + HL

What has been achieved

\(P = \frac{1}{3} \rho v^{2}\) came from Newton’s laws applied to molecules. \(PV = N k_\mathrm{B} T\) came from experiments with gauges and thermometers. Setting them equal explains temperature: it is a measure of the average kinetic energy of the molecules. This is the equation that B.1 quoted without proof.

The helium numbers

\(v = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 300}{6.64 \times 10^{-27}}} = \sqrt{1.87 \times 10^{6}} = 1.4 \times 10^{3}\) m s⁻¹. Air molecules, about seven times heavier, move at about 500 m s⁻¹ at the same temperature: the same kinetic energy with more mass means less speed.

Why the Earth has no helium

At the top of the atmosphere a small fraction of helium atoms move faster than the escape speed, 11 km s⁻¹, and leave for good. Over millions of years nearly all of it has gone. Nitrogen and oxygen molecules are too slow to escape.

Thinking that doubling the temperature doubles the speed. It doubles \(v^{2}\), so the speed rises by \(\sqrt{2}\).

Macroscopic behavior from molecules

Explaining the gas laws.

Squeeze at constant temperature

Pressure rises

Same speed, but more molecules in each unit volume: more frequent collisions with the walls

Heat at constant volume

Pressure rises

Faster molecules: collisions are more frequent and each gives a larger change of momentum

Heat at constant pressure

Volume rises

Faster molecules push the walls outwards until the collisions are rare enough to restore the pressure

Always two questions: how often do molecules hit the wall, and how hard?

B.3 GAS LAWS · SL + HL

A model answer

“Explain, in terms of molecules, why the pressure of a gas in a sealed container rises when it is heated.” The average kinetic energy, and so the average speed, of the molecules increases. They therefore collide with the walls more frequently, and the change of momentum in each collision is greater. The rate of change of momentum at the walls is greater, so the force on the walls is greater. The area is unchanged, so the pressure is greater.

The link back to \(v^{2}\)

Frequency and momentum change are each proportional to \(v\), which is why pressure goes as \(v^{2}\), and \(v^{2}\) is proportional to \(T\). The qualitative answer and \(P = \frac{1}{3} \rho v^{2}\) say the same thing.

Writing that the molecules “collide more” without saying with what. It is collisions with the walls that give the pressure, not collisions with each other.

All of it is kinetic

Internal energy of an ideal monatomic gas.

\(U = \dfrac{3}{2} N k_\mathrm{B} T = \dfrac{3}{2} nRT\)

No forces between molecules: no potential energy
So \(U\) = number of molecules × average kinetic energy of each
Monatomic: single atoms, such as helium, neon, argon

2.0 mol of helium at 300 K \(U = \dfrac{3}{2} nRT\) \(U = \dfrac{3}{2} \times 2.0 \times 8.31 \times 300\) \(U = \class{res}{7.5 \times 10^{3} \,\mathrm{J}}\)

B.3 GAS LAWS · SL + HL

Temperature alone decides it

For a fixed amount of ideal gas, the internal energy depends on nothing but the temperature. Compress the gas or let it expand: if the temperature ends where it started, the internal energy is unchanged. In B.1 the internal energy of a real substance also had a potential energy part that changed with the spacing of the molecules. An ideal gas has none.

Why “monatomic”

A single atom can only move from place to place. A molecule of two or more atoms can also rotate and vibrate, and those motions hold energy too, so its internal energy is more than \(\frac{3}{2} N k_\mathrm{B} T\). The course calculates only with monatomic gases.

Another form

Since \(nRT = PV\), \(U = \frac{3}{2} PV\). The internal energy can be read straight from a pressure–volume diagram.

B.4 (HL) uses \(\Delta U = \frac{3}{2} nR \Delta T\) in the first law of thermodynamics.

When the model works

Real gases and the ideal gas.

Low pressure

Molecules far apart

Their own volume is negligible compared with the container

Low density

Forces rarely act

Molecules are seldom close enough to attract each other

High temperature

Fast molecules

Their kinetic energy is far greater than any potential energy of attraction

A real gas is close to ideal at low pressure, low density and high temperature: well away from the conditions at which it would turn to liquid.

B.3 GAS LAWS · SL + HL

How an ideal gas differs from a real one

An ideal gas can never be liquefied: with no forces between its molecules, nothing could hold a liquid together. It obeys \(PV = nRT\) under all conditions, and its internal energy is purely kinetic. A real gas has molecules of finite size that attract each other at short range, turns to liquid when cold or compressed enough, and departs from \(PV = nRT\) as it nears those conditions.

Where real gases go wrong

Compress a gas until its molecules are nearly touching and they resist further squeezing: the pressure is higher than the equation predicts. Cool it until the molecules move slowly and the attractions between them start to matter, pulling them back from the walls: the pressure is lower than predicted.

Ordinary air is nearly ideal

At room temperature and atmospheric pressure the molecules occupy about a thousandth of the volume, and room temperature is far above the boiling points of nitrogen and oxygen, near 80 K. The equation is good to a fraction of a per cent.

Three conditions, three reasons. Give the reason with each condition when a question says “explain”.

Quick check

The kinetic theory.

01A molecule of mass \(m\) and speed \(v\) rebounds from a wall. Change of momentum?

\(2mv\)

02The speed of the molecules doubles at constant density. Pressure?

Four times as great

03What kind of energy is the internal energy of an ideal gas?

Kinetic only

04The kelvin temperature of an ideal gas doubles. Its internal energy?

It doubles

05Hydrogen and oxygen at the same temperature. Which molecules are faster?

Hydrogen: the lighter ones

06Under which conditions is a real gas most nearly ideal?

Low pressure, low density, high temperature

PRESS → TO REVEAL EACH ANSWER

Summary

Gas laws on one slide.

Pressure and amount

\(P = \frac{F}{A}\) · \(n = \frac{N}{N_\mathrm{A}}\)

Pascals and moles

Fixed mass of gas

\(\frac{PV}{T} = \text{constant}\)

\(T\) in kelvin
Contains all three gas laws

Ideal gas equation

\(PV = nRT = N k_\mathrm{B} T\)

\(k_\mathrm{B} = \frac{R}{N_\mathrm{A}}\)

Kinetic theory

\(P = \frac{1}{3} \rho v^{2}\)

Pressure from the change of momentum of molecules at the walls

Internal energy

\(U = \frac{3}{2} nRT\)

Monatomic ideal gas: kinetic only

Real gases

Nearly ideal when thin and hot

Low pressure, low density, high temperature

B.3 GAS LAWS · SL + HL

Summary

Key words.

PressureForce at right angles to a surface, per unit area

MoleThe amount containing \(6.02 \times 10^{23}\) particles

Avogadro constantThe number of particles in one mole

IsothermA constant-temperature curve on a \(P\)–\(V\) diagram

Absolute zero0 K, −273 °C: where an ideal gas would have no pressure

Ideal gasThe model gas that obeys \(PV = nRT\) exactly

Kinetic theoryGas behavior explained by molecules in random motion

Elastic collisionNo kinetic energy is lost

MonatomicMade of single atoms

B.3 GAS LAWS · SL + HL