teaching-notes — Physics (THERMAL PHYSICS)

Topic: THERMAL PHYSICS

Subtopic: Expansion of Solids, Liquids and Gases

QUALITATIVELY THE THERMAL EXPANSION OF SOLIDS, LIQUIDS AND GASES

Thermal expansion is the increase in the size of a substance when its temperature is raised. When a substance is heated, its particles gain kinetic energy and move faster. In solids, the particles vibrate more violently about their fixed positions, pushing each other slightly further apart. In liquids and gases, the particles move faster and slide or fly apart, increasing the average distance between them. This causes the entire substance to occupy a larger space (expand). When a substance is cooled, its particles lose kinetic energy, move closer together, and the substance contracts (decreases in size).

Thermal expansion can be described in three ways depending on the dimensions of the substance:

1. Linear Expansion: This is the expansion in one dimension (length). It is most noticeable in long, thin solids such as railway lines, metal rods, and overhead telephone wires.

2. Area (Superficial) Expansion: This is the expansion in two dimensions (length and width), which leads to an increase in surface area. An example is the heating of a flat metal sheet.

3. Volume (Cubical) Expansion: This is the expansion in three dimensions (length, width, and height), resulting in an increase in the overall volume. This type of expansion is significant in solids, liquids, and gases.

An important application of linear expansion is the bimetallic strip. A bimetallic strip consists of two different metal strips (usually brass and iron) bonded firmly together. Brass expands more than iron when heated through the same temperature rise. Therefore, when heated, the bimetallic strip bends with the brass on the outside of the curve. When cooled, the brass contracts more than the iron, causing the strip to bend in the opposite direction.

Bimetallic strip Image: ToobEZ · CC0 · Wikimedia Commons

Example: Bimetallic strips are widely used as switches in automatic heating appliances such as electric laundry irons, refrigerators, and fire alarms. In an electric iron, the strip bends away from a contact point when it gets too hot, breaking the circuit and turning off the heating element. As it cools, it bends back to restore contact, turning the heater on again.

EFFECTS OF EXPANSION OF WATER ON AQUATIC LIFE

Most liquids contract continuously when they are cooled, becoming denser. However, water behaves in a strange and unique way when cooled within a specific temperature range. This unusual behaviour is called the anomalous expansion of water.

When water is cooled from 100°C, it contracts normally until its temperature reaches 4°C. At 4°C, water has its minimum volume and maximum density. However, when cooled further from 4°C down to 0°C, water expands instead of contracting. This means its volume increases and its density decreases. When it freezes at 0°C to form ice, it expands even more, making ice less dense than liquid water. This is why ice floats on water.

This anomalous expansion of water has a critical biological importance for the survival of aquatic life (such as fish and water plants) in cold regions or during extremely cold nights in places like the highlands of Zambia:

1. As the air temperature drops, the surface water of a lake cools. This cold surface water becomes denser and sinks to the bottom, while warmer, less dense water rises to the surface to be cooled in turn.

2. This circulation continues until all the water in the lake reaches 4°C.

3. If the air temperature drops below 4°C, the surface water cools to 3°C, 2°C, and then 0°C. Because water expands below 4°C, this colder water is less dense than the 4°C water at the bottom, so it remains at the surface.

4. Eventually, the surface water freezes to form a layer of ice at 0°C. Since ice is a poor conductor of heat and floats at the top, it insulates the water underneath from the cold air outside. This keeps the water at the bottom of the lake liquid at 4°C, allowing fish and other aquatic organisms to survive comfortably.

Example: In a deep lake during a cold season, the water temperature at the bottom remains at 4°C even when the surface is frozen. This is why aquatic organisms do not freeze to death in winter.

DIFFERENT RATES OF EXPANSIONS OF MATTER

Solids, liquids, and gases do not expand by the same amount when subjected to the same temperature increase. They expand at different rates because of the differences in their internal structures and intermolecular forces.

1. Solids: Solids have very strong intermolecular forces holding their particles in fixed positions. Therefore, they expand the least. The rate of expansion also varies among different solids (e.g., brass expands more than iron).

2. Liquids: Liquids have weaker intermolecular forces than solids. The particles are free to move around. As a result, liquids expand much more than solids for the same temperature rise (about 10 times more than solids).

3. Gases: Gases have negligible intermolecular forces between their particles. The particles are very far apart and free to move in all directions. Consequently, gases expand the most when heated (about 100 times more than liquids and 1000 times more than solids).

The comparative rates of expansion are summarized below:

Example: When a glass bottle completely filled with cold water is heated, the water spills out because the liquid water expands at a faster rate than the solid glass bottle.

HOW TO DETERMINE THE BOILING AND MELTING POINT OF DIFFERENT SUBSTANCES

Every pure substance has a unique temperature at which it changes state. These temperatures are physical constants that can be used to identify pure substances.

1. Melting Point: The constant temperature at which a solid substance changes into a liquid state.

2. Boiling Point: The constant temperature at which a liquid substance changes into a gas state rapidly throughout the liquid.

To determine these points experimentally, we heat or cool a substance and record its temperature at regular time intervals. We then plot a graph of Temperature against Time. This graph is called a heating curve (if heat is added) or a cooling curve (if heat is lost).

During the state changes (melting and boiling), the temperature remains constant (represented by the flat horizontal sections on the heating curve). This is because the heat energy being supplied is not used to raise the temperature; instead, it is absorbed to break the intermolecular bonds holding the particles together. This hidden heat is called latent heat.

Example: In a laboratory experiment to find the melting point of naphthalene, naphthalene powder is heated in a boiling tube placed in a water bath. The temperature is recorded every minute. The temperature rises until it reaches 80°C, where it remains constant as the powder melts into liquid. Once all the naphthalene has melted, the temperature begins to rise again. The melting point is therefore determined to be 80°C.

EFFECTS OF PRESSURE ON THE MELTING AND BOILING POINTS

The melting and boiling points of substances are highly affected by changes in the external pressure surrounding the substance.

1. Effect of Pressure on Melting Point:

• For substances that expand when they melt (most substances), an increase in pressure raises the melting point.
• For substances that contract when they melt (such as water/ice), an increase in pressure lowers the melting point. This explains why ice melts under the weight of a heavy object. When the pressure is released, the ice refreezes. This process of melting under pressure and refreezing when pressure is released is called regelation.

2. Effect of Pressure on Boiling Point:

• An increase in pressure raises the boiling point of a liquid. This is because higher external pressure makes it more difficult for liquid molecules to escape from the liquid surface to become gas. More thermal energy is required to build up enough vapour pressure to match the external pressure.
• A decrease in pressure lowers the boiling point of a liquid.

Example: Inside a pressure cooker, the lid seals the cooker tightly, trapping steam. This increases the internal pressure above the water surface. The high pressure raises the boiling point of water from 100°C to about 120°C. Because the water is much hotter before it boils, food cooks much faster. Conversely, on top of a high mountain where atmospheric pressure is low, water boils at temperatures lower than 100°C (e.g., 90°C), making it take longer to cook food like hard-boiled eggs.

EFFECTS OF IMPURITIES ON THE MELTING AND BOILING POINTS OF SUBSTANCES

A pure substance has sharp, well-defined melting and boiling points. The presence of impurities (foreign substances dissolved in the pure substance) alters these values significantly.

1. Effect on Melting Point:

Impurities lower the melting point of a substance. They also cause the substance to melt over a range of temperatures rather than at a single sharp temperature.

2. Effect on Boiling Point:

Impurities raise the boiling point of a liquid. The liquid will also boil over a range of temperatures above its normal boiling point.

Example: In cold climates, salt is thrown onto icy roads to melt the ice. The salt acts as an impurity that lowers the melting point of the ice below the surrounding winter air temperature, turning the solid ice back into liquid water to prevent cars from sliding.

EFFECT OF VARYING PRESSURE ON VOLUME OF A GAS

Gases are highly compressible. The relationship between the pressure and volume of a gas was first investigated by Robert Boyle. His findings are summarized in Boyle's Law.

Boyle's Law states that the volume (V) of a fixed mass of gas is inversely proportional to its pressure (P), provided the temperature remains constant.

Mathematically:

V ∝ 1P

P × V = Constant

If a gas has an initial pressure P₁ and volume V₁, and is changed to a new pressure P₂ and volume V₂ without any temperature change, then:

P₁V₁ = P₂V₂

Worked Example: A cylinder contains 300 cm³ of air at a pressure of 100 kPa. If the piston is pushed in so that the volume of the gas is compressed to 120 cm³ at constant temperature, what is the new pressure of the gas?

Step 1: Identify given quantities and the target variable.

P₁ = 100 kPa

V₁ = 300 cm³

V₂ = 120 cm³

P₂ = ?

Step 2: State the formula and rearrange it for the target variable.

P₁V₁ = P₂V₂

P₂ = P₁V₁V₂

Step 3: Substitute the values and calculate the final answer.

P₂ = 100 kPa × 300 cm³120 cm³

P₂ = 30000120

P₂ = 250 kPa

Answer: The new pressure of the gas is 250 kPa.

RELATIONSHIP BETWEEN TEMPERATURE AND VOLUME OF A GAS

The relationship between the volume and the temperature of a gas was established by Jacques Charles and is known as Charles's Law.

Charles's Law states that the volume (V) of a fixed mass of gas is directly proportional to its absolute temperature (T), provided the pressure remains constant.

Mathematically:

V ∝ T

VT = Constant

If a gas has an initial volume V₁ and absolute temperature T₁, and changes to a new volume V₂ and absolute temperature T₂ at constant pressure, then:

V₁T₁ = V₂T₂

CRITICAL RULE: In all gas law calculations, temperatures must always be converted to the absolute scale (Kelvin). To convert from degrees Celsius (°C) to Kelvin (K), use the formula:

T = θ + 273 (where T is in Kelvin and θ is in Celsius).

Worked Example: A gas has a volume of 120 cm³ at 27°C. What will its volume be at 127°C, assuming the pressure remains constant?

Step 1: Identify given quantities and convert temperatures to Kelvin.

V₁ = 120 cm³

T₁ = 27 + 273 = 300 K

T₂ = 127 + 273 = 400 K

V₂ = ?

Step 2: State the formula and rearrange it for the target variable.

V₁T₁ = V₂T₂

V₂ = V₁ × T₂T₁

Step 3: Substitute values and calculate.

V₂ = 120 cm³ × 400 K300 K

V₂ = 48000300

V₂ = 160 cm³

Answer: The new volume of the gas is 160 cm³.

KELVIN SCALE; VOLUME- TEMPERATURE CHANGE (CONSTANT PRESSURE ) GRAPHICAL EXTRAPOLATION

If we perform an experiment to measure the volume of a gas at different Celsius temperatures (while keeping pressure constant), and plot a graph of Volume (V) against Temperature (θ in °C), we obtain a straight line showing that volume increases linearly with temperature.

If we extend (extrapolate) this straight line backwards to the point where the volume of the gas theoretically shrinks to zero, the line cuts the temperature axis at exactly -273.15°C (commonly rounded to -273°C).

This temperature (-273.15°C) is known as Absolute Zero. It is the lowest possible temperature in the universe. At absolute zero:

• The volume of a gas would theoretically become zero (though in reality, all gases liquefy or solidify before reaching this temperature).
• The kinetic energy of the particles becomes zero, meaning all molecular motion stops entirely.

Lord Kelvin proposed a new temperature scale starting at absolute zero, known as the Kelvin Scale or Absolute Temperature Scale. On this scale:

• Absolute zero is written as 0 K (not 0°K). No negative temperatures exist on the Kelvin scale.
• The size of 1 K is exactly equal to the size of 1°C.

The formulas to convert between the scales are:

T (Kelvin) = θ (Celsius) + 273

θ (Celsius) = T (Kelvin) - 273

Example: Convert the boiling point of pure water (100°C) into Kelvin:

T = 100 + 273 = 373 K

THE IDEAL GAS EQUATION (P1V1/T1=P2V2/T2) AND NUMERICAL PROBLEMS

In many real-life situations, the pressure, volume, and temperature of a gas change at the same time. By combining Boyle's Law and Charles's Law, we obtain a single formula called the Ideal Gas Equation (or General Gas Equation):

P₁V₁T₁ = P₂V₂T₂

Where:

• P₁, V₁, and T₁ are the initial pressure, volume, and Kelvin temperature.
• P₂, V₂, and T₂ are the final pressure, volume, and Kelvin temperature.

CRITICAL REMINDER: T₁ and T₂ MUST be in Kelvin (K). However, P and V can be in any units as long as the same units are used on both sides of the equation.

Worked Example 1: A weather balloon has a volume of 2.0 m³ at sea level where the pressure is 101 kPa and the temperature is 27°C. The balloon rises to an altitude where the pressure is 30 kPa and the temperature is -23°C. Find the volume of the balloon at this altitude.

Step 1: Write down the given values and convert temperatures to Kelvin.

P₁ = 101 kPa

V₁ = 2.0 m³

T₁ = 27 + 273 = 300 K

P₂ = 30 kPa

T₂ = -23 + 273 = 250 K

V₂ = ?

Step 2: State the formula and rearrange it to solve for V₂.

P₁V₁T₁ = P₂V₂T₂

V₂ = P₁V₁T₂T₁P₂

Step 3: Substitute the values and calculate.

V₂ = 101 × 2.0 × 250300 × 30

V₂ = 505009000

V₂ = 5.61 m³

Answer: The volume of the balloon at high altitude is 5.61 m³.

SUMMARY

• Thermal expansion occurs when substances are heated, causing particles to gain kinetic energy and spread out.
• Expansion rates follow the order: Gases > Liquids > Solids.
• Anomalous expansion of water: Water contracts as it cools down to 4°C, but expands when cooled from 4°C to 0°C. This permits ice to float on top of water bodies, insulating the aquatic life below.
• A heating curve shows constant temperature plateaus during phase changes (melting and boiling).
• Increasing pressure lowers the melting point of ice but raises the boiling point of liquids.
• Impurities lower the melting point of solids and raise the boiling point of liquids.
• Boyle's Law: P₁V₁ = P₂V₂ (at constant temperature).
• Charles's Law: V₁T₁ = V₂T₂ (at constant pressure).
• Absolute zero is -273.15°C (0 K), where all molecular motion ceases.
• Ideal Gas Equation: P₁V₁T₁ = P₂V₂T₂.

KEY TERMS

REVISION QUESTIONS

1. Explain why concrete bridges are built with roller supports and small gaps left at one end.

2. Describe how a bimetallic strip made of brass and iron works when it is cooled below room temperature. (Hint: Brass contracts more than iron).

3. Why does ice float on water? Explain why this behaviour is beneficial to fish during cold nights in Zambia.

4. State the main difference in the rates of expansion of solids, liquids, and gases when they are heated equally.

5. In an experiment to determine the boiling point of pure water, the thermometer reads 102°C when the water is boiling. Give two possible reasons why the boiling point is higher than 100°C.

6. Explain how a pressure cooker is able to cook food much faster than an ordinary open pot.

7. State Boyle's law and write its mathematical equation.

8. What is meant by the term "Absolute Zero" and what is its value on the Celsius scale?

PRACTICE EXERCISE

1. Convert the following temperatures to Kelvin:
a) -23°C
b) 27°C
c) 100°C

2. Convert the following Kelvin temperatures to Celsius:
a) 0 K
b) 293 K
c) 373 K

3. A cylinder with a movable piston contains 400 cm³ of gas at a pressure of 100 kPa. If the volume is decreased to 250 cm³ at a constant temperature, what will be the new pressure of the gas?

4. A sample of oxygen gas has a volume of 150 cm³ at a temperature of 27°C. If the pressure is kept constant, calculate the volume of the gas when the temperature is raised to 177°C.

5. A mass of gas has a volume of 50.0 cm³ at a pressure of 120 kPa and a temperature of 27°C. Calculate its volume at standard temperature and pressure (S.T.P., which is 0°C and 101.3 kPa).

6. An air bubble at the bottom of a lake has a volume of 1.2 cm³ where the pressure is 300 kPa and the temperature is 7°C. The bubble rises to the surface where the pressure is 100 kPa and the temperature is 27°C. Calculate the volume of the bubble at the surface of the lake.

ANSWERS TO PRACTICE EXERCISE

1. Converting to Kelvin (T = θ + 273):
a) T = -23 + 273 = 250 K
b) T = 27 + 273 = 300 K
c) T = 100 + 273 = 373 K

2. Converting to Celsius (θ = T - 273):
a) θ = 0 - 273 = -273°C
b) θ = 293 - 273 = 20°C
c) θ = 373 - 273 = 100°C

3. Solving Boyle's Law:
Given: P₁ = 100 kPa, V₁ = 400 cm³, V₂ = 250 cm³
Formula: P₁V₁ = P₂V₂
Working:
P₂ = 100 kPa × 400 cm³250 cm³
P₂ = 40000250
P₂ = 160 kPa

4. Solving Charles's Law:
Given: V₁ = 150 cm³, T₁ = 27 + 273 = 300 K, T₂ = 177 + 273 = 450 K
Formula: V₁T₁ = V₂T₂
Working:
V₂ = V₁ × T₂T₁
V₂ = 150 cm³ × 450 K300 K
V₂ = 67500300
V₂ = 225 cm³

5. Solving with the Ideal Gas Equation:
Given:
V₁ = 50.0 cm³, P₁ = 120 kPa, T₁ = 27 + 273 = 300 K
P₂ = 101.3 kPa, T₂ = 0 + 273 = 273 K
Formula: P₁V₁T₁ = P₂V₂T₂
Working:
V₂ = P₁V₁T₂T₁P₂
V₂ = 120 × 50.0 × 273300 × 101.3
V₂ = 163800030390
V₂ = 53.9 cm³

6. Solving with the Ideal Gas Equation:
Given:
V₁ = 1.2 cm³, P₁ = 300 kPa, T₁ = 7 + 273 = 280 K
P₂ = 100 kPa, T₂ = 27 + 273 = 300 K
Formula: P₁V₁T₁ = P₂V₂T₂
Working:
V₂ = P₁V₁T₂T₁P₂
V₂ = 300 × 1.2 × 300280 × 100
V₂ = 10800028000
V₂ = 3.86 cm³ (or 3.9 cm³)