Gas laws

B.3 — Gas Laws

IB Physics HL • Topic B: The particulate nature of matter • SL and HL core content

Gas laws describe how the macroscopic properties of gases—pressure, volume, and temperature—arise from the microscopic behaviour of individual particles. By modelling gases as ideal systems, we can predict the behaviour of many real gases under suitable conditions.

Guiding ideas

Gas pressure comes from particle collisions with container walls.
The ideal gas model neglects particle volume and intermolecular forces between collisions.
Empirical gas laws connect $p$ , $V$ , and $T$ ; kinetic theory explains these relationships.
For a fixed amount of ideal monatomic gas, internal energy depends only on temperature.

B.3.1 What are ideal gases?

Learning objectives

Define pressure and explain it through particle momentum changes.
State the assumptions of the ideal gas model.
Explain when real gases behave approximately ideally.
Use the mole, the Avogadro constant, and the number of particles.

Pressure: macroscopic definition

Pressure is the perpendicular force applied per unit area:

$\displaystyle p=\frac{F}{A}$

Here, $p$ is pressure in pascals, $F$ is perpendicular force in newtons, and $A$ is area in square metres.

$\displaystyle 1\,\text{Pa}=1\,\text{N m}^{-2}$

For example, the same applied force concentrated on the point of one nail produces greater local pressure than when the force is distributed over many nails.

Pressure in gases: microscopic view

Gas particles move randomly in all directions.
Collisions with container walls change particle momentum.
The wall exerts a force on the particles, and the particles exert an equal and opposite force on the wall.
The average total force per unit area from many collisions is the gas pressure.

Other conditions being comparable, higher pressure can result from:

More frequent collisions with a given wall area.
A larger momentum transfer per collision.

The ideal gas model

An ideal gas is a simplified model based on the following assumptions:

The gas contains a very large number of particles.
The volume of the particles is negligible compared with the container volume.
There are no intermolecular forces between collisions.
Particles move randomly, with no preferred direction in equilibrium.
Collisions with one another and the container walls are perfectly elastic.
Collision duration is negligible compared with the time between collisions.
Intermolecular potential energy is taken to be zero.

No real gas follows this model perfectly, but many real gases behave approximately ideally under suitable conditions.

When is a real gas close to ideal?

A real gas behaves most like an ideal gas when it has:

A sufficiently high temperature, well away from condensation.
A low pressure.
A low density.

At low density, particles are widely separated, so their own volume occupies only a small fraction of the container volume. At sufficiently high temperature, their kinetic energy is large compared with the effects of attractive intermolecular interactions.

The mole and the Avogadro constant

Because gases contain enormous numbers of particles, we measure the amount of substance using the mole, with SI unit mol.

The Avogadro constant gives the number of specified particles per mole:

$\displaystyle N_A\approx6.02\times10^{23}\,\text{mol}^{-1}$

The relationship between amount of substance $n$ and number of particles $N$ is:

$\displaystyle n=\frac{N}{N_A}$

$\displaystyle N=nN_A$

The particles being counted may be atoms or molecules. These relationships also apply to substances in the liquid and solid states.

Worked example 1 — Pressure

A force of $250\,\text{N}$ acts uniformly on a surface area of $0.020\,\text{m}^2$ . Calculate the pressure.

Show answer and full working

$\displaystyle p=\frac{F}{A}=\frac{250}{0.020}=1.25\times10^4\,\text{Pa}$

Calculated pressure: $1.25\times10^4\,\text{Pa}$ .

To two significant figures, this is $1.3\times10^4\,\text{Pa}$ .

Worked example 2 — Moles and molecules

A glass contains $11\,\text{mol}$ of water. Calculate the number of water molecules in the glass.

Show answer and full working

$\displaystyle N=nN_A=11(6.02\times10^{23})$

$\displaystyle N\approx6.62\times10^{24}$

Answer: approximately $6.6\times10^{24}$ molecules to two significant figures.

Worked example 3 — Amount of substance

A sample contains $3.01\times10^{24}$ gas molecules. Calculate the amount of substance in moles.

Show answer and full working

$\displaystyle n=\frac{N}{N_A}=\frac{3.01\times10^{24}}{6.02\times10^{23}}=5.00\,\text{mol}$

Answer: $5.00\,\text{mol}$ .

B.3.2 How do ideal gases behave?

Learning objectives

Relate experimental observations to Boyle’s law, Charles’ law, and the pressure law.
Apply the empirical gas laws and the combined gas law.
Use the ideal gas equation in macroscopic and molecular forms.
Explain gas pressure using force, momentum, and collisions.
Calculate internal energy for an ideal monatomic gas.

Use absolute pressure and kelvin

Gas laws use absolute pressure, measured relative to a vacuum. A gauge often measures pressure relative to the surrounding atmosphere.

$\displaystyle p_{\mathrm{absolute}}=p_{\mathrm{gauge}}+p_{\mathrm{atmospheric}}$

Temperatures must be in kelvin:

$\displaystyle T(\text{K})=\theta(^\circ\text{C})+273.15$

Doubling a Celsius temperature does not double the absolute temperature.

Useful conversions:

$\displaystyle 1\,\text{kPa}=10^3\,\text{Pa}$

$\displaystyle 1\,\text{L}=10^{-3}\,\text{m}^3$

$\displaystyle 1\,\text{cm}^3=10^{-6}\,\text{m}^3$

Empirical gas laws

For a fixed amount of gas, experimental observations show that pressure $p$ , volume $V$ , and absolute temperature $T$ are related by simple laws.

Boyle’s law: constant temperature

For a fixed amount of gas at constant temperature:

$\displaystyle pV=\mathrm{constant}$

$\displaystyle p\propto\frac{1}{V}$

Comparing two states:

$\displaystyle p_1V_1=p_2V_2$

Microscopic explanation: at constant temperature, average particle kinetic energy is unchanged. Decreasing the volume increases the collision rate per unit wall area, so pressure increases.

Charles’ law: constant pressure

For a fixed amount of gas at constant pressure:

$\displaystyle \frac{V}{T}=\mathrm{constant}$

$\displaystyle V\propto T$

Comparing two states:

$\displaystyle \frac{V_1}{T_1}=\frac{V_2}{T_2}$

Microscopic explanation: higher temperature means faster particles. The gas must expand to maintain the same pressure, reducing the collision rate per unit wall area sufficiently to compensate.

Pressure law: constant volume

For a fixed amount of gas at constant volume:

$\displaystyle \frac{p}{T}=\mathrm{constant}$

$\displaystyle p\propto T$

Comparing two states:

$\displaystyle \frac{p_1}{T_1}=\frac{p_2}{T_2}$

Microscopic explanation: higher temperature means faster particles. They collide with the walls more frequently and transfer more momentum per collision, increasing the average force and pressure.

Recognising gas-law graphs

At fixed temperature and amount of gas, pressure against volume is a decreasing hyperbola.
At fixed temperature and amount of gas, pressure against reciprocal volume is a straight line through the origin.
At fixed pressure and amount of gas, volume against kelvin temperature is a straight line through the origin in the ideal model.
At fixed volume and amount of gas, pressure against kelvin temperature is a straight line through the origin in the ideal model.
The corresponding Celsius-temperature lines extrapolate to approximately −273 °C, although real gases cease to behave ideally before reaching this limit.

Combined gas law: fixed amount of gas

When pressure, volume, and temperature all change for the same fixed amount of ideal gas:

$\displaystyle \frac{pV}{T}=\mathrm{constant}$

Comparing two states:

$\displaystyle \frac{p_1V_1}{T_1}=\frac{p_2V_2}{T_2}$

Worked example 4 — Pressure change at constant volume

A gas is contained in a sealed rigid cylinder of volume $0.80\,\text{m}^3$ at $20^\circ\text{C}$ and an absolute pressure of $7.2\,\text{kPa}$ . It is heated to $40^\circ\text{C}$ . Calculate the new pressure.

Show answer and full working

Step 1: Convert temperatures to kelvin.

$\displaystyle T_1=20+273.15=293.15\,\text{K}$

$\displaystyle T_2=40+273.15=313.15\,\text{K}$

Step 2: Use the pressure law.

$\displaystyle \frac{p_1}{T_1}=\frac{p_2}{T_2}$

$\displaystyle p_2=p_1\frac{T_2}{T_1}$

Step 3: Substitute.

$\displaystyle p_2=(7.2\,\text{kPa})\frac{313.15}{293.15}\approx7.69\,\text{kPa}$

Answer: $p_2\approx7.7\,\text{kPa}$ .

The cylinder volume does not enter the calculation because it remains constant.

The ideal gas equation

The empirical gas laws are represented by the ideal gas equation:

$\displaystyle pV=nRT$

$p$ — absolute pressure in $\text{Pa}$ .
$V$ — volume in $\text{m}^3$ .
$n$ — amount of substance in $\text{mol}$ .
$T$ — absolute temperature in $\text{K}$ .
$R$ — molar gas constant.

$\displaystyle R\approx8.31\,\text{J mol}^{-1}\text{K}^{-1}$

Molecular form of the ideal gas equation

Using the number of gas particles $N$ , which may be atoms or molecules:

$\displaystyle pV=Nk_BT$

Here, $k_B$ is the Boltzmann constant:

$\displaystyle k_B\approx1.38\times10^{-23}\,\text{J K}^{-1}$

The constants are related by:

$\displaystyle R=N_Ak_B$

Pressure and molecular speed

Kinetic theory links pressure to particle motion:

$\displaystyle p=\frac{1}{3}\rho\overline{v^2}$

Here, $\rho$ is gas density and $\overline{v^2}$ is the mean square speed of the particles. At fixed density, a larger mean square speed produces a higher pressure.

Mean square speed and root mean square speed

Mean square speed is the average of the squared particle speeds:

$\displaystyle \overline{v^2}=\frac{v_1^2+v_2^2+\cdots+v_N^2}{N}$

Root mean square speed is the square root of this quantity:

$\displaystyle v_{\mathrm{rms}}=\sqrt{\overline{v^2}}$

Therefore:

$\displaystyle p=\frac{1}{3}\rho v_{\mathrm{rms}}^2$

$\displaystyle v_{\mathrm{rms}}=\sqrt{\frac{3p}{\rho}}=\sqrt{\frac{3k_BT}{m_0}}$

Here, $m_0$ is the mass of one particle.

Important: root mean square speed is not the same as mean speed. Also, $\overline{v^2}$ is not generally equal to $(\overline{v})^2$ .

Internal energy of an ideal monatomic gas

In the ideal monatomic gas model, intermolecular potential energy is neglected. Internal energy is the total random translational kinetic energy of the atoms.

The average translational kinetic energy per particle is:

$\displaystyle \overline{E_k}=\frac{3}{2}k_BT$

Therefore:

$\displaystyle U=\frac{3}{2}Nk_BT=\frac{3}{2}nRT$

For a fixed amount of ideal monatomic gas, internal energy depends only on temperature.

Ideal molecular gases may also store rotational and vibrational energy, so the factor $3/2$ does not generally give their total internal energy.

Worked example 5 — Ideal gas equation

A gas occupies $0.040\,\text{m}^3$ at an absolute pressure of $1.20\times10^5\,\text{Pa}$ and a temperature of $300\,\text{K}$ . Calculate the amount of gas.

Show answer and full working

$\displaystyle pV=nRT$

$\displaystyle n=\frac{pV}{RT}$

$\displaystyle n=\frac{(1.20\times10^5)(0.040)}{(8.31)(300)}=\frac{4800}{2493}$

$\displaystyle n\approx1.93\,\text{mol}$

Answer: approximately $1.9\,\text{mol}$ to two significant figures.

Worked example 6 — Heating at constant volume

A fixed amount of gas in a rigid container has an absolute pressure of $2.0\times10^5\,\text{Pa}$ at $300\,\text{K}$ . It is heated to $450\,\text{K}$ . Calculate the new pressure.

Show answer and full working

$\displaystyle \frac{p_1}{T_1}=\frac{p_2}{T_2}$

$\displaystyle p_2=p_1\frac{T_2}{T_1}$

$\displaystyle p_2=(2.0\times10^5)\frac{450}{300}=3.0\times10^5\,\text{Pa}$

Answer: $3.0\times10^5\,\text{Pa}$ .

Worked example 7 — Internal energy

Calculate the internal energy of $2.0\,\text{mol}$ of an ideal monatomic gas at $400\,\text{K}$ .

Show answer and full working

$\displaystyle U=\frac{3}{2}nRT$

$\displaystyle U=\frac{3}{2}(2.0)(8.31)(400)=9972\,\text{J}$

Answer: $U\approx1.0\times10^4\,\text{J}$ .

Ideal gases and gas pressure: force and momentum

Microscopic explanation using momentum

A particle of mass $m_0$ has velocity component $v_x$ perpendicular to a wall.
In an elastic collision with a stationary smooth wall, the perpendicular velocity component reverses.
Velocity components parallel to the wall remain unchanged.
The magnitude of momentum transferred to the wall is $2m_0|v_x|$ .
The average force is the total momentum transferred to the wall per unit time.

$\displaystyle F=\frac{\Delta p_{\mathrm{mom}}}{\Delta t}$

The subscript “mom” distinguishes momentum from the symbol $p$ used for pressure.

Making connections

Mechanics: momentum change explains the force on container walls.
Thermal physics: absolute temperature is proportional to average translational kinetic energy.
Energy: internal energy can change through heating or work.
Applications: expanding gases can push a piston and do work.

Kinetic theory — deriving the pressure equation

Consider $N$ particles, each of mass $m_0$ , moving in a cube of side length $l$ . For one particle, let $v_x$ be its velocity component perpendicular to a chosen wall.

Each elastic collision transfers momentum of magnitude $2m_0|v_x|$ to that wall. In the elementary model, the particle travels a perpendicular distance $2l$ between successive collisions with the same wall.

$\displaystyle \Delta t=\frac{2l}{|v_x|}$

The time-averaged force contributed by this particle is:

$\displaystyle F_i=\frac{2m_0|v_x|}{2l/|v_x|}=\frac{m_0v_x^2}{l}$

Adding the contributions of all particles and dividing by the wall area, $l^2$ , gives:

$\displaystyle p=\frac{Nm_0\overline{v_x^2}}{l^3}=\frac{Nm_0\overline{v_x^2}}{V}$

Random motion is isotropic, meaning that no direction is preferred:

$\displaystyle \overline{v_x^2}=\overline{v_y^2}=\overline{v_z^2}=\frac{1}{3}\overline{v^2}$

Using the density relation:

$\displaystyle \rho=\frac{Nm_0}{V}$

We obtain:

$\displaystyle p=\frac{1}{3}\rho\overline{v^2}$

This elementary derivation models journeys between opposite walls. Collisions between particles redistribute velocities but do not change the equilibrium result.

Connecting kinetic theory to temperature

Multiply the pressure equation by volume and compare it with the molecular ideal gas equation:

$\displaystyle pV=\frac{1}{3}Nm_0\overline{v^2}=Nk_BT$

Therefore:

$\displaystyle \frac{1}{2}m_0\overline{v^2}=\frac{3}{2}k_BT$

Thus, average translational kinetic energy is proportional to absolute temperature.

At the same temperature, particles of different ideal gases have the same average translational kinetic energy. Lighter particles have greater root mean square speeds.

Internal energy — useful equivalent forms

For an ideal monatomic gas:

$\displaystyle U=\frac{3}{2}nRT=\frac{3}{2}Nk_BT=\frac{3}{2}pV$

For a fixed amount:

$\displaystyle \Delta U=\frac{3}{2}nR\Delta T$

At constant temperature, its internal energy remains unchanged.

During constant-pressure heating, the energy supplied is shared between increasing internal energy and work done by the expanding gas. It is therefore not generally equal to $\Delta U$ alone.

B.3 Summary and key terms

Key ideas

Gas pressure arises from particle collisions with container walls.
The ideal gas model neglects particle volume and intermolecular forces between collisions.
Real gases behave approximately ideally at sufficiently high temperature and low density.
Boyle’s law, Charles’ law, and the pressure law describe relationships between pressure, volume, and temperature.
The ideal gas equation can be written as $pV=nRT$ or $pV=Nk_BT$ .
For a fixed amount of ideal monatomic gas, internal energy depends only on temperature.

Key terms

Pressure, absolute pressure, gauge pressure, mole, Avogadro constant, Boyle’s law, Charles’ law, pressure law, combined gas law, ideal gas equation, Boltzmann constant, internal energy, elastic collision, momentum, mean square speed, and root mean square speed.

IB exam tips

Always use kelvin in gas-law calculations.
Use absolute pressure rather than gauge pressure.
For explanation questions, link macroscopic changes to collision frequency and momentum transfer.
Use SI units when applying $pV=nRT$ : pressure in pascals and volume in cubic metres.
In two-state ratio calculations, consistent pressure and volume units can cancel.
State what remains constant when selecting a gas law.
Use $U=\frac{3}{2}nRT$ for an ideal monatomic gas.
For a fixed amount of ideal monatomic gas, use $\Delta U=\frac{3}{2}nR\Delta T$ .
Do not confuse mean square speed with root mean square speed.

Essential equations — B.3

Use absolute pressure and kelvin. The two-state gas laws require a fixed amount of gas.

$\displaystyle p=\frac{F}{A}$

$\displaystyle n=\frac{N}{N_A}$

$\displaystyle \frac{pV}{T}=\mathrm{constant}$

$\displaystyle pV=nRT=Nk_BT$

$\displaystyle p=\frac{1}{3}\rho\overline{v^2}$

$\displaystyle U=\frac{3}{2}nRT=\frac{3}{2}Nk_BT$

Constants: $R\approx8.31\,\text{J mol}^{-1}\text{K}^{-1}$ , $k_B\approx1.38\times10^{-23}\,\text{J K}^{-1}$ , and $N_A\approx6.02\times10^{23}\,\text{mol}^{-1}$ .

Practice — check your understanding

1. Boyle’s law

A fixed amount of gas occupies $2.0\,\text{L}$ at an absolute pressure of $150\,\text{kPa}$ . It is slowly compressed at constant temperature to $0.75\,\text{L}$ . Calculate its new pressure.

Show answer and working

$\displaystyle p_1V_1=p_2V_2$

$\displaystyle p_2=\frac{p_1V_1}{V_2}=\frac{(150)(2.0)}{0.75}=400\,\text{kPa}$

The volume units cancel because the same units are used on both sides.

Answer: $4.0\times10^5\,\text{Pa}$ .

2. Charles’ law

A fixed amount of gas occupies $2.40\,\text{L}$ at $27.0^\circ\text{C}$ . It is heated at constant pressure to $127.0^\circ\text{C}$ . Calculate the final volume.

Show answer and working

$\displaystyle T_1=300.15\,\text{K},\qquad T_2=400.15\,\text{K}$

$\displaystyle V_2=V_1\frac{T_2}{T_1}=(2.40)\frac{400.15}{300.15}\approx3.20\,\text{L}$

Answer: $3.20\,\text{L}$ .

Using Celsius directly would give an incorrect temperature ratio.

3. Combined gas law

A sealed sample initially has absolute pressure $100\,\text{kPa}$ , volume $2.0\,\text{L}$ , and temperature $300\,\text{K}$ . Its final volume is $1.5\,\text{L}$ and final temperature is $360\,\text{K}$ . Calculate the final pressure.

Show answer and working

$\displaystyle \frac{p_1V_1}{T_1}=\frac{p_2V_2}{T_2}$

$\displaystyle p_2=p_1\frac{V_1T_2}{V_2T_1}$

$\displaystyle p_2=(100)\frac{(2.0)(360)}{(1.5)(300)}=160\,\text{kPa}$

Answer: $160\,\text{kPa}$ . Both the smaller volume and higher temperature increase the pressure.

4. Root mean square speed

A gas has density $1.20\,\text{kg m}^{-3}$ and absolute pressure $1.00\times10^5\,\text{Pa}$ . Calculate its root mean square speed.

Show answer and working

$\displaystyle v_{\mathrm{rms}}=\sqrt{\frac{3p}{\rho}}$

$\displaystyle v_{\mathrm{rms}}=\sqrt{\frac{3(1.00\times10^5)}{1.20}}=500\,\text{m s}^{-1}$

Answer: $5.00\times10^2\,\text{m s}^{-1}$ to three significant figures.

5. Change in internal energy

A sealed sample contains $1.00\,\text{mol}$ of an ideal monatomic gas. Its temperature rises from $300\,\text{K}$ to $450\,\text{K}$ . Calculate the change in internal energy.

Show answer and working

$\displaystyle \Delta T=450-300=150\,\text{K}$

$\displaystyle \Delta U=\frac{3}{2}nR\Delta T=\frac{3}{2}(1.00)(8.31)(150)$

$\displaystyle \Delta U\approx1.87\times10^3\,\text{J}$

The change is positive because the gas warms. This does not by itself determine the heat supplied unless the work done is also known.

6. Lab-style question — testing Boyle’s law

A student traps air in a gas syringe connected to a pressure sensor. Describe how to test Boyle’s law, identify the quantities to control, state a suitable graph, and explain two sources of error.

Show a suggested answer

Method: change the gas volume gradually, wait for temperature and pressure to stabilise, and record several pairs of absolute pressure and volume. Repeat measurements where practical.

Controls: keep the amount of gas fixed by preventing leaks. Keep temperature constant by compressing slowly and allowing thermal equilibrium with the surroundings.

Graph: plot $p$ against $1/V$ . A straight line through the origin supports $pV=\mathrm{constant}$ , within experimental uncertainty. Its gradient is $nRT$ .

Error 1: rapid compression temporarily heats the gas, producing a pressure above the isothermal prediction.

Error 2: uncounted gas volume in the tubing and sensor causes a systematic error. Use the total enclosed volume, not just the syringe reading.

Check whether the sensor measures absolute or gauge pressure. Convert gauge values before testing the proportionality.

Faith connection — trusting God under pressure

“God is our refuge and strength, a very present help in trouble.”
Psalm 46:1 — KJV

Creation — order in the smallest details

Gas laws connect the motion of countless tiny particles with measurable pressure, volume, and temperature. As Christians, we appreciate the order of creation and the opportunity to understand it through careful study. The air around us supports life, and learning its properties can deepen our gratitude and guide us to use scientific knowledge responsibly.

The Fall — responding to life’s pressures

In a world affected by sin and brokenness, we experience pressures such as fear, conflict, illness, and disappointment. Psalm 46 invites us to find refuge in God. We can bring our concerns to Him in prayer and respond to others’ struggles with patience, honesty, and compassion.

Redemption — peace through Jesus Christ

Jesus Christ came to reconcile us to God through His death and resurrection. Turn to Him in repentance and faith, receive Him as your personal God, Saviour, and Lord, and trust His grace. His invitation in Matthew 11:28 encourages those who are weary and burdened to come to Him for rest.

Restoration — living with hope

God promises a renewed creation where suffering and death will end, as described in Revelation 21:4. That hope helps us live faithfully today: persevere in learning, use knowledge to serve others, and seek God’s guidance in difficult situations. Connect with other believers, read and study the Bible, and spend time in prayer so that your faith and knowledge of God continue to grow.

Reflect and pray

Reflect: What pressure are you facing today? What would it look like to bring that concern to God and seek support from someone you trust?

Prayer: Lord Jesus, thank You for the order of Your creation and for being my refuge and strength. Forgive my sins and help me trust You as my God, Saviour, and Lord. Give me patience in learning, peace in difficulty, and a willing heart to serve others. Help me grow through Your Word, prayer, and fellowship. Amen.

IB Physics HL Book 2025; Papers 1 & 2 Practice 5 Sections65 Lessons54 Weeks

Curriculum

Gas laws

B.3 — Gas Laws

Guiding ideas

B.3.1 What are ideal gases?

Learning objectives

Pressure: macroscopic definition

Pressure in gases: microscopic view

The ideal gas model

When is a real gas close to ideal?

The mole and the Avogadro constant

Worked example 1 — Pressure

Worked example 2 — Moles and molecules

Worked example 3 — Amount of substance

B.3.2 How do ideal gases behave?

Learning objectives

Use absolute pressure and kelvin

Empirical gas laws

Boyle’s law: constant temperature

Charles’ law: constant pressure

Pressure law: constant volume

Recognising gas-law graphs

Combined gas law: fixed amount of gas

Worked example 4 — Pressure change at constant volume

The ideal gas equation

Molecular form of the ideal gas equation

Pressure and molecular speed

Mean square speed and root mean square speed

Internal energy of an ideal monatomic gas

Worked example 5 — Ideal gas equation

Worked example 6 — Heating at constant volume

Worked example 7 — Internal energy

Ideal gases and gas pressure: force and momentum

Microscopic explanation using momentum

Making connections

Kinetic theory — deriving the pressure equation

Connecting kinetic theory to temperature

Internal energy — useful equivalent forms

B.3 Summary and key terms

Key ideas

Key terms

IB exam tips

Essential equations — B.3

Practice — check your understanding

1. Boyle’s law

2. Charles’ law

3. Combined gas law

4. Root mean square speed

5. Change in internal energy

6. Lab-style question — testing Boyle’s law

Faith connection — trusting God under pressure

Creation — order in the smallest details

The Fall — responding to life’s pressures

Redemption — peace through Jesus Christ

Restoration — living with hope

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