Thermal Expansion and Gas Laws

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your WAEC exam.

Thermal Expansion — Key Formulas

• Linear expansion: $\dfrac{\Delta L}{L_0} = \alpha \Delta \theta$
• Area expansion: $\dfrac{\Delta A}{A_0} = \gamma \Delta \theta$ where $\gamma = 2\alpha$
• Volume expansion: $\dfrac{\Delta V}{V_0} = \beta \Delta \theta$ where $\beta = 3\alpha$

Gas Laws — Key Formulas

• Boyle’s Law: $PV = \text{constant}$ (at constant temperature)
• Charles’s Law: $\dfrac{V}{T} = \text{constant}$ (at constant pressure)
• Pressure Law: $\dfrac{P}{T} = \text{constant}$ (at constant volume)
• Ideal Gas Equation: $PV = nRT$ or $PV = NkT$

Quick Facts

• All materials expand when heated (with rare exceptions like water between 0–4°C)
• $\alpha_{\text{brass}} > \alpha_{\text{steel}}$ — brass expands more per degree rise in temperature
• For gases: $P$ in Pa or $\text{N m}^{-2}$, $V$ in $\text{m}^3$, $T$ in Kelvin (K)
• Convert °C to K: $T(\text{K}) = \theta(°C) + 273$
• Standard conditions: $P_0 = 1.01 \times 10^5 , \text{Pa}$, $T_0 = 273 , \text{K}$

⚡ WAEC Exam Tip: Always convert temperature to Kelvin before using gas law formulas. Failing to do this is the most common cause of errors in gas law questions. Also note: $\alpha$, $\beta$ and $\gamma$ are per °C (or per K) — check what unit the question uses.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for building a solid understanding of thermal expansion and gas laws.

Thermal Expansion of Solids

When a solid is heated, its atoms vibrate more vigorously and push apart, causing the material to expand. The expansion is small but measurable and can cause stress in structures like railway tracks and bridges, which is why expansion joints are built in.

Linear Expansion: If a rod of original length $L_0$ at temperature $\theta_0$ is heated to $\theta$, the new length is:

$$L = L_0(1 + \alpha \Delta \theta)$$

where $\alpha$ is the coefficient of linear expansion (units: $\text{K}^{-1}$ or °C$^{-1}$).

Area Expansion: $\gamma = 2\alpha$ (approximately), so $A = A_0(1 + \gamma \Delta \theta)$.

Volume Expansion: $\beta = 3\alpha$ (approximately), so $V = V_0(1 + \beta \Delta \theta)$.

Worked Example (Linear Expansion)

A steel rail is 20 m long at 20°C. If it expands to 20.04 m on a hot day, find the temperature. Given $\alpha_{\text{steel}} = 1.1 \times 10^{-5} , \text{°C}^{-1}$.

Solution: $\Delta L = 20.04 - 20 = 0.04 , \text{m}$; $L_0 = 20 , \text{m}$; $\Delta \theta = ?$

$\Delta L = L_0 \alpha \Delta \theta \implies 0.04 = 20 \times 1.1 \times 10^{-5} \times \Delta \theta$

$0.04 = 2.2 \times 10^{-4} \times \Delta \theta$

$\Delta \theta = \frac{0.04}{2.2 \times 10^{-4}} = 181.8 , \text{°C}$

$\theta = 20 + 181.8 \approx 202 , \text{°C}$

The Gas Laws

Gases respond to changes in pressure, volume, and temperature in predictable ways. At WAEC level, we treat gases as ideal — molecules are point particles with no intermolecular forces except during collisions.

Boyle’s Law: At constant temperature, the pressure of a fixed mass of gas is inversely proportional to its volume: $P \propto \frac{1}{V}$, so $PV = \text{constant}$.

Charles’s Law: At constant pressure, volume is directly proportional to absolute temperature: $V \propto T$, so $\frac{V}{T} = \text{constant}$.

Pressure Law: At constant volume, pressure is directly proportional to absolute temperature: $P \propto T$, so $\frac{P}{T} = \text{constant}$.

Absolute Temperature: The Kelvin scale sets $0 , \text{K} = -273.15 , \text{°C}$. This is the temperature at which all molecular motion ceases (theoretically). At $0 , \text{K}$, $P = 0$ and $V = 0$ for an ideal gas — hence the term absolute zero.

Combined Gas Law: Combining Boyle’s and Charles’s laws:

$$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} = \text{constant}$$

This is the most useful form for WAEC calculations involving changes in all three variables.

Worked Example (Gas Law)

A gas occupies $2.0 \times 10^{-3} , \text{m}^3$ at $27°C$ and $1.0 \times 10^5 , \text{Pa}$. It is heated to $127°C$ while expanding to $3.0 \times 10^{-3} , \text{m}^3$. Find the new pressure.

Solution: $T_1 = 27 + 273 = 300 , \text{K}$; $T_2 = 127 + 273 = 400 , \text{K}$

Using $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$:

$P_2 = P_1 \times \frac{V_1}{V_2} \times \frac{T_2}{T_1} = 1.0 \times 10^5 \times \frac{2.0 \times 10^{-3}}{3.0 \times 10^{-3}} \times \frac{400}{300}$

$P_2 = 1.0 \times 10^5 \times \frac{2}{3} \times \frac{4}{3} = 1.0 \times 10^5 \times \frac{8}{9} = 8.9 \times 10^4 , \text{Pa}$

⚡ WAEC Exam Tip: Questions often state “a fixed mass of gas” — this confirms the amount $n$ doesn’t change, so you can apply the gas laws. If the question says “closed container,” volume is constant. If it says “flexible container” (like a balloon), pressure is constant. Read carefully to identify which law applies.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage including derivations, molecular interpretation, and WAEC past question patterns.

Derivation of the Ideal Gas Equation

From the kinetic theory of gases, we assume:

1. Molecules are point particles (volume negligible compared to container)
2. No intermolecular forces except collisions
3. Collisions with walls are perfectly elastic
4. Large number of molecules — statistical average is meaningful

Pressure $P$ arises from molecules colliding with container walls. From Newton’s second law applied to molecular collisions with a wall, and combining with the mean squared velocity $\overline{c^2}$:

$$PV = NkT$$

where $N$ is the number of molecules, $n$ is the number of moles, and $R = N_A k = 8.31 , \text{J mol}^{-1}\text{K}^{-1}$ is the molar gas constant, giving the familiar form:

$$PV = nRT$$

This is the ideal gas equation. Real gases obey this approximately at low pressure and high temperature (far above liquefaction conditions).

Molecular Interpretation of Absolute Temperature

From $PV = NkT$, we can write $P = \frac{2}{3}N \times \frac{1}{2}m\overline{c^2}$ per unit volume (from kinetic theory). This means:

$$\frac{1}{2}m\overline{c^2} = \frac{3}{2}kT$$

Absolute temperature $T$ is directly proportional to the average kinetic energy of gas molecules. This explains why absolute zero is physically meaningful — at $0 , \text{K}$, molecular kinetic energy would be zero.

Coefficient of Volume Expansion for Gases

From the ideal gas equation $PV = nRT$, for a fixed mass of gas at constant pressure:

$$\frac{V}{T} = \frac{nR}{P} = \text{constant}$$

This confirms Charles’s Law, and the coefficient of volume expansion at constant pressure is $\beta = \frac{1}{T}$ (in $\text{K}^{-1}$). Note: for solids and liquids, $\beta$ is approximately constant over modest temperature ranges, but for gases it is better expressed as $1/T$ at the reference temperature.

Anomalous Expansion of Water

Between 0°C and 4°C, water contracts as temperature increases — this is anomalous expansion. Water has its maximum density at 4°C ($1000 , \text{kg m}^{-3}$). This is biologically crucial: lakes freeze from the surface downward, allowing aquatic life to survive beneath the ice layer. This anomaly does not appear directly in WAEC questions but is important background knowledge.

Bimetallic Strips

Two different metals with different $\alpha$ values are bonded together. On heating, the metal with higher $\alpha$ expands more, causing the strip to bend toward the metal with lower $\alpha$. This principle is used in thermostats — the bending strip makes or breaks an electrical circuit, controlling heating systems. This application frequently appears in WAEC multiple-choice questions.

WAEC Past Question Patterns

From WAEC Physics papers (2020–2024):

1. Thermal expansion — Usually asks: “A metal rod expands by X m when heated from Y°C to Z°C. Find $\alpha$” — rearrange $\Delta L = L_0 \alpha \Delta \theta$. Or: “Which material expands most?” — compare $\alpha$ values.

2. Gas laws — Common format: “A fixed mass of gas at $P_1, V_1, T_1$ undergoes a change to $P_2, V_2, T_2$. Find the unknown.” — Use $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.

3. Conversions — “27°C = X K” — add 273. Never use Celsius in gas law formulas.

4. Units — $P$ must be in Pa (or $\text{N m}^{-2}$), $V$ in $\text{m}^3$. Convert $\text{cm}^3$ to $\text{m}^3$ by dividing by $10^6$.

Key Formulas Summary Table

Law	Condition	Formula
Linear expansion	—	$L = L_0(1 + \alpha \Delta \theta)$
Area expansion	$\gamma = 2\alpha$	$A = A_0(1 + \gamma \Delta \theta)$
Volume expansion	$\beta = 3\alpha$	$V = V_0(1 + \beta \Delta \theta)$
Boyle’s Law	$T$ constant	$P_1 V_1 = P_2 V_2$
Charles’s Law	$P$ constant	$\frac{V_1}{T_1} = \frac{V_2}{T_2}$
Pressure Law	$V$ constant	$\frac{P_1}{T_1} = \frac{P_2}{T_2}$
Combined Gas Law	fixed $n$	$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$
Ideal Gas	—	$PV = nRT$

⚡ Common Pitfalls to Avoid

• Mixing up $\alpha$, $\beta$, and $\gamma$ — remember $\beta = 3\alpha$ and $\gamma = 2\alpha$
• Using °C instead of K in gas law calculations — always add 273
• Forgetting to convert $\text{cm}^3$ to $\text{m}^3$ (divide by $10^6$) or $\text{mmHg}$ to Pa (multiply by 133.3)
• Not identifying whether the container is rigid (constant volume) or flexible (constant pressure)
• Using $g = 10 , \text{m s}^{-2}$ in pressure calculations where $P = \rho g h$ — check the question

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