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Thermal Properties of Matter

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Chapter Notes

Overview
This chapter explores the fundamental concepts of heat and temperature, how they are measured, and how heat transfers between bodies. It covers thermal expansion of solids, liquids, and gases, specific heat capacity, calorimetry, changes of state with latent heat, and the three modes of heat transfer: conduction, convection, and radiation. The one big idea a student must walk away with is that heat is energy in transit due to temperature difference, and its effects—temperature change, expansion, and change of state—are governed by well-defined laws and measurable properties.
Key Concepts (25)
Temperature
A relative measure or indication of hotness or coldness of a body.
e.g. A hot utensil has a high temperature, an ice cube has a low temperature.
Heat
The form of energy transferred between two (or more) systems or a system and its surroundings by virtue of temperature difference.
e.g. Heat flows from a cup of hot tea to the environment, cooling the tea.
Celsius temperature scale
A temperature scale where the ice point is 0 °C and the steam point is 100 °C, with 100 equal intervals between them.
e.g. Water freezes at 0 °C and boils at 100 °C on this scale.
Fahrenheit temperature scale
A temperature scale where the ice point is 32 °F and the steam point is 212 °F, with 180 equal intervals between them.
e.g. Water freezes at 32 °F and boils at 212 °F on this scale.
Absolute zero
The absolute minimum temperature for an ideal gas, inferred by extrapolating the pressure versus temperature graph to zero pressure, found to be –273.15 °C.
e.g. 0 K is absolute zero, the foundation of the Kelvin scale.
Kelvin temperature scale
An absolute temperature scale where 0 K is absolute zero (–273.15 °C) and the size of one Kelvin is the same as one degree Celsius.
e.g. T (K) = tC + 273.15.
Thermal expansion
The increase in the dimensions of a body due to an increase in its temperature.
e.g. Mercury in a thermometer rises when placed in warm water.
Coefficient of linear expansion (αl)
The fractional change in length per unit change in temperature, characteristic of the material.
e.g. For copper, αl = 1.7 × 10⁻⁵ K⁻¹.
Coefficient of volume expansion (αv)
The fractional change in volume per unit change in temperature, characteristic of the substance.
e.g. For mercury, αv = 18.2 × 10⁻⁵ K⁻¹.
Specific heat capacity
The amount of heat per unit mass absorbed or given off by a substance to change its temperature by one unit (SI unit: J kg⁻¹ K⁻¹).
e.g. Water has a specific heat capacity of 4186 J kg⁻¹ K⁻¹.
Molar specific heat capacity
The heat capacity per mole of a substance (SI unit: J mol⁻¹ K⁻¹).
e.g. For helium, Cp = 20.8 J mol⁻¹ K⁻¹ and Cv = 12.5 J mol⁻¹ K⁻¹.
Calorimetry
Measurement of heat, based on the principle that heat lost by a hot body equals heat gained by a colder body in an isolated system.
e.g. Using a calorimeter to find the specific heat capacity of a solid.
Latent heat
The amount of heat per unit mass transferred during a change of state of a substance without any change in temperature (SI unit: J kg⁻¹).
e.g. Latent heat of fusion of ice is 3.33 × 10⁵ J kg⁻¹.
Latent heat of fusion (Lf)
The heat required per unit mass to change a substance from solid to liquid at its melting point.
e.g. Lf for water is 3.33 × 10⁵ J kg⁻¹.
Latent heat of vaporisation (Lv)
The heat required per unit mass to change a substance from liquid to vapour at its boiling point.
e.g. Lv for water is 22.6 × 10⁵ J kg⁻¹.
Melting point
The temperature at which the solid and liquid states of a substance coexist in thermal equilibrium.
e.g. Ice melts at 0 °C at standard atmospheric pressure.
Boiling point
The temperature at which the liquid and vapour states of a substance coexist in thermal equilibrium.
e.g. Water boils at 100 °C at standard atmospheric pressure.
Triple point
The temperature and pressure at which the fusion curve, vaporisation curve, and sublimation curve meet, and all three phases of a substance coexist.
e.g. The triple point of water is 273.16 K and 6.11 × 10⁻³ Pa.
Thermal conductivity (K)
The constant of proportionality in the heat conduction equation, representing the ability of a material to conduct heat (SI unit: W m⁻¹ K⁻¹).
e.g. Silver has a thermal conductivity of 406 W m⁻¹ K⁻¹.
Conduction
The mechanism of transfer of heat between two adjacent parts of a body because of their temperature difference.
e.g. A metallic rod getting hot from one end to the other when placed in a flame.
Convection
A mode of heat transfer involving the movement of fluid (liquid or gas) due to temperature differences.
e.g. Wind at the beach reversing direction after sunset.
Radiation
A mode of heat transfer that does not require a medium and occurs via electromagnetic waves.
e.g. Heat from the Sun reaching Earth.
Newton’s law of cooling
The rate of cooling of a body is proportional to the difference in temperature between the body and its surroundings.
e.g. A hot cup of tea cools faster when the room is colder.
Regelation
The phenomenon of refreezing of water after melting under increased pressure, allowing a wire to pass through an ice slab without splitting it.
e.g. Skating is possible on snow due to formation of water under skates.
Sublimation
The change from solid state to vapour state without passing through the liquid state.
e.g. Dry ice (solid CO₂) sublimes.
Formulas & Equations (13)
tF – 32 / 180 = tC / 100
Reads as: Fahrenheit temperature minus 32 divided by 180 equals Celsius temperature divided by 100
Variables: tF = Fahrenheit temperature (°F), tC = Celsius temperature (°C)
Use when: Converting between Fahrenheit and Celsius scales
⚠ Students often forget to subtract 32 before dividing, or invert the fractions.
PV = μRT
Reads as: P V equals mu R T
Variables: P = pressure (Pa), V = volume (m³), μ = number of moles (mol), R = universal gas constant (8.31 J mol⁻¹ K⁻¹), T = absolute temperature (K)
Use when: Relating pressure, volume, and temperature of an ideal gas
⚠ Using Celsius temperature instead of Kelvin, or forgetting to convert mass to moles.
T = tC + 273.15
Reads as: T equals t C plus 273.15
Variables: T = absolute temperature (K), tC = temperature in Celsius (°C)
Use when: Converting Celsius to Kelvin
⚠ Using 273 instead of 273.15, or adding when subtracting is needed.
Δl / l = αl ΔT
Reads as: Delta l over l equals alpha l times delta T
Variables: Δl = change in length (m), l = original length (m), αl = coefficient of linear expansion (K⁻¹), ΔT = change in temperature (K or °C)
Use when: Calculating linear expansion of a solid rod
⚠ Forgetting that ΔT must be in the same unit as αl (usually Kelvin), or using diameter instead of length.
ΔA / A = αA ΔT = 2αl ΔT
Reads as: Delta A over A equals alpha A times delta T equals 2 alpha l times delta T
Variables: ΔA = change in area (m²), A = original area (m²), αA = coefficient of area expansion (K⁻¹), αl = coefficient of linear expansion (K⁻¹), ΔT = change in temperature (K)
Use when: Calculating area expansion of a solid sheet
⚠ Using αl directly without multiplying by 2.
ΔV / V = αv ΔT
Reads as: Delta V over V equals alpha v times delta T
Variables: ΔV = change in volume (m³), V = original volume (m³), αv = coefficient of volume expansion (K⁻¹), ΔT = change in temperature (K)
Use when: Calculating volume expansion of solids, liquids, or gases
⚠ Forgetting that αv for gases depends on temperature (αv = 1/T for ideal gas at constant pressure).
αv = 3αl
Reads as: Alpha v equals 3 alpha l
Variables: αv = coefficient of volume expansion (K⁻¹), αl = coefficient of linear expansion (K⁻¹)
Use when: Relating volume and linear expansion coefficients for isotropic solids
⚠ Using this for anisotropic materials or liquids where it does not hold.
αv = 1/T (for ideal gas at constant pressure)
Reads as: Alpha v equals 1 over T
Variables: αv = coefficient of volume expansion (K⁻¹), T = absolute temperature (K)
Use when: Calculating volume expansion of an ideal gas at constant pressure
⚠ Using Celsius temperature instead of Kelvin.
S = ΔQ / ΔT
Reads as: S equals delta Q over delta T
Variables: S = heat capacity (J K⁻¹), ΔQ = heat supplied (J), ΔT = change in temperature (K)
Use when: Defining heat capacity of a substance
⚠ Confusing heat capacity with specific heat capacity (which is per unit mass).
s = (1/m) (ΔQ / ΔT)
Reads as: s equals 1 over m times delta Q over delta T
Variables: s = specific heat capacity (J kg⁻¹ K⁻¹), m = mass (kg), ΔQ = heat supplied (J), ΔT = change in temperature (K)
Use when: Calculating specific heat capacity from experimental data
⚠ Forgetting to divide by mass, or using moles instead of kg.
C = (1/μ) (ΔQ / ΔT)
Reads as: C equals 1 over mu times delta Q over delta T
Variables: C = molar specific heat capacity (J mol⁻¹ K⁻¹), μ = number of moles (mol), ΔQ = heat supplied (J), ΔT = change in temperature (K)
Use when: Calculating molar specific heat capacity
⚠ Confusing Cp and Cv for gases.
Q = mL
Reads as: Q equals m L
Variables: Q = heat transferred during phase change (J), m = mass (kg), L = latent heat (J kg⁻¹)
Use when: Calculating heat required for melting, freezing, vaporisation, or condensation
⚠ Using the wrong latent heat (fusion vs vaporisation) for the process.
H = K A (TC – TD) / L
Reads as: H equals K A times T C minus T D divided by L
Variables: H = rate of heat flow or heat current (W), K = thermal conductivity (W m⁻¹ K⁻¹), A = cross-sectional area (m²), TC – TD = temperature difference (K), L = length (m)
Use when: Calculating steady-state heat conduction through a bar
⚠ Forgetting that the temperature difference must be in Kelvin (though °C difference is numerically same).
How It Works
Measurement of temperature using a constant-volume gas thermometer
  1. A fixed volume of gas is kept in a bulb.
  2. The bulb is placed in thermal contact with the object whose temperature is to be measured.
  3. The pressure of the gas is measured; since volume is constant, P ∝ T.
  4. The temperature is read from a calibrated pressure scale.
💡 Real gases deviate from ideal gas law at low temperatures, but the linear relationship holds over a large range.
Thermal expansion of a solid rod (linear expansion)
  1. A rod of length l is heated, causing its temperature to increase by ΔT.
  2. The fractional change in length Δl/l is directly proportional to ΔT.
  3. The constant of proportionality is the coefficient of linear expansion αl.
  4. The new length is l' = l (1 + αl ΔT).
💡 Δl is small compared to l, so higher-order terms are neglected.
Thermal expansion of a rectangular sheet (area expansion)
  1. A rectangular sheet of length a and breadth b is heated by ΔT.
  2. Length increases by Δa = αl a ΔT, breadth increases by Δb = αl b ΔT.
  3. The increase in area ΔA = a Δb + b Δa + (Δa)(Δb).
  4. Neglecting the product term (Δa)(Δb) since it is very small, ΔA = 2αl A ΔT, so αA = 2αl.
💡 The product term (Δa)(Δb) is negligible because αl is of order 10⁻⁵ K⁻¹.
Calorimetry: determining specific heat capacity of a solid
  1. A solid of known mass m1 is heated to a known high temperature (e.g., 100 °C in boiling water).
  2. It is quickly transferred to a calorimeter containing water of known mass m2 and known initial temperature.
  3. The mixture is stirred and the final steady temperature is recorded.
  4. Using the principle of calorimetry: heat lost by solid = heat gained by water + heat gained by calorimeter.
  5. Solve for the specific heat capacity of the solid.
💡 The calorimeter itself absorbs heat, so its heat capacity must be accounted for.
Change of state: melting of ice
  1. Ice at a temperature below 0 °C is heated.
  2. Temperature rises until it reaches 0 °C (melting point).
  3. Further heating does not raise temperature; the heat is used to melt ice (latent heat of fusion).
  4. Once all ice is melted, temperature of water begins to rise again.
💡 During phase change, temperature remains constant even though heat is being added or removed.
Change of state: boiling of water
  1. Water is heated; temperature rises until it reaches 100 °C (boiling point).
  2. Further heating does not raise temperature; the heat is used to convert water to steam (latent heat of vaporisation).
  3. Boiling point increases with increase in pressure and decreases with decrease in pressure.
💡 Steam at 100 °C carries more heat than water at 100 °C, causing more serious burns.
Heat conduction through a composite bar (series)
  1. Two rods of different materials are joined end to end.
  2. The free ends are maintained at different temperatures.
  3. In steady state, the heat current through each rod is the same.
  4. The temperature at the junction is found by equating the heat currents: K1A1(T1 – T0)/L1 = K2A2(T0 – T2)/L2.
  5. Solve for T0.
💡 The heat current is constant through every cross-section in steady state.
Regelation
  1. A wire with heavy weights at its ends is placed over an ice slab.
  2. The pressure under the wire lowers the melting point of ice, causing it to melt locally.
  3. The wire passes through the melted water.
  4. Above the wire, the pressure is released, and the water refreezes (regelation).
💡 This phenomenon explains why skating is possible on snow.
Exam Traps — Don't Make These Mistakes
⚠ Confusing heat and temperature
✗ Heat and temperature are the same thing.
→ Heat is energy in transit due to temperature difference, while temperature is a measure of hotness. They are distinct concepts.
✓ Heat is the form of energy transferred between systems due to temperature difference; temperature is a measure of the hotness of a body.
⚠ Using Celsius instead of Kelvin in gas law
✗ PV = μRT with T in °C.
→ The ideal gas equation requires absolute temperature in Kelvin. Using Celsius gives an incorrect result.
✓ Always convert Celsius to Kelvin by adding 273.15 before using PV = μRT.
⚠ Forgetting to account for calorimeter heat capacity
✗ Heat lost by hot body = heat gained by water only.
→ The calorimeter vessel also absorbs heat, so its heat capacity must be included in the heat gained.
✓ Heat lost by hot body = heat gained by water + heat gained by calorimeter material.
⚠ Misapplying αv = 3αl for liquids
✗ αv for mercury is 3 times its αl.
→ The relation αv = 3αl is derived for isotropic solids that expand equally in all directions. Liquids do not have a linear expansion coefficient in the same sense.
✓ For isotropic solids, αv = 3αl. For liquids, αv is measured directly and is not related to αl.
⚠ Ignoring the sign of ΔT in expansion
✗ Δl = αl l ΔT, and then using a negative ΔT gives a negative Δl, but they forget to interpret it as contraction.
→ The formula works for both expansion and contraction, but students often forget that a negative ΔT means contraction.
✓ For a decrease in temperature (ΔT negative), Δl is negative, indicating contraction.
⚠ Confusing latent heat of fusion and vaporisation
✗ Using Lv = 3.33 × 10⁵ J kg⁻¹ for melting ice.
→ Lf for ice is 3.33 × 10⁵ J kg⁻¹, while Lv for water is 22.6 × 10⁵ J kg⁻¹. Using the wrong value gives an incorrect answer.
✓ For melting/freezing, use Lf = 3.33 × 10⁵ J kg⁻¹. For boiling/condensation, use Lv = 22.6 × 10⁵ J kg⁻¹.
Compare & Contrast
Definition and nature
Heat
Temperature
Heat is energy in transit due to temperature difference; temperature is a measure of the hotness or coldness of a body.
Zero point and unit size
Celsius scale
Kelvin scale
Celsius scale has 0 °C at ice point and 100 °C at steam point; Kelvin scale starts at absolute zero (0 K = –273.15 °C) and has the same unit size as Celsius.
Effect on temperature
Specific heat capacity
Latent heat
Specific heat capacity relates heat to temperature change (ΔQ = m s ΔT); latent heat relates heat to phase change with no temperature change (Q = m L).
Medium and mechanism
Conduction
Convection
Conduction occurs in solids via molecular vibration without bulk movement; convection occurs in fluids via actual movement of heated particles.
Exam Focus
Marks pattern: 2 MCQs (1 mark each), 1 SA-I (2 marks), 1 SA-II (3 marks), 1 LA (5 marks) — total approximately 11 marks
Most asked: Thermal expansion (linear, area, volume) and relation αv = 3αl · Specific heat capacity and calorimetry numericals · Latent heat and change of state problems · Heat conduction through composite bars · Ideal gas equation and absolute temperature
Don't skip: Numericals on calorimetry involving mixture of ice and water; Heat conduction through composite rods (series/parallel); Conversion between temperature scales and absolute zero concept
One-Line Revision (read aloud before the exam)
  1. Temperature is a measure of hotness; heat is energy transferred due to temperature difference.
  2. Celsius to Kelvin conversion: T(K) = t(°C) + 273.15.
  3. Absolute zero is –273.15 °C or 0 K, the minimum possible temperature.
  4. Ideal gas equation: PV = μRT, where R = 8.31 J mol⁻¹ K⁻¹.
  5. Linear expansion: Δl = αl l ΔT.
  6. Area expansion: ΔA = 2αl A ΔT.
  7. Volume expansion: ΔV = αv V ΔT, and for isotropic solids αv = 3αl.
  8. Water has anomalous expansion: maximum density at 4 °C.
  9. Specific heat capacity s = ΔQ / (m ΔT); SI unit J kg⁻¹ K⁻¹.
  10. Calorimetry principle: heat lost = heat gained in an isolated system.
  11. Latent heat of fusion of ice: Lf = 3.33 × 10⁵ J kg⁻¹.
  12. Latent heat of vaporisation of water: Lv = 22.6 × 10⁵ J kg⁻¹.
  13. During phase change, temperature remains constant.
  14. Heat conduction rate: H = K A (ΔT) / L.
  15. Good conductors (metals) have high K; insulators (wood, air) have low K.
  16. Newton’s law of cooling: rate of cooling ∝ temperature difference with surroundings.
  17. Triple point of water: 273.16 K and 6.11 × 10⁻³ Pa.
  18. Boiling point increases with pressure; cooking is faster in a pressure cooker.
  19. Steam burns are more serious than boiling water burns due to higher latent heat.
  20. Regelation: ice melts under pressure and refreezes when pressure is released.

MCQs — 10 Questions with Answers

Q1. What is the SI unit of heat?

  • A. Joule (J) ✓
  • B. Calorie (cal)
  • C. Watt (W)
  • D. Kelvin (K)

Answer: A — Joule is the SI unit of energy, including heat energy; other units are either non-SI or measure different quantities (W measures power, K measures temperature).

Q2. The ice point and steam point of water on the Celsius scale are:

  • A. −40°C and 60°C
  • B. 0°C and 100°C ✓
  • C. 32°C and 212°C
  • D. 273 K and 373 K

Answer: B — By definition, pure water freezes at 0°C and boils at 100°C under standard atmospheric pressure on the Celsius scale.

Q3. Using the relation (tF − 32)/180 = tC/100, what is 32°F in Celsius?

  • A. 0°C ✓
  • B. 180°C
  • C. 32°C
  • D. 100°C

Answer: A — Substituting tF = 32: (32 − 32)/180 = tC/100 gives 0/180 = tC/100, so tC = 0°C (ice point).

Q4. If a gas at constant volume has pressure P₁ at temperature T₁, what is its pressure at temperature 2T₁?

  • A. P₁/2
  • B. P₁
  • C. 2P₁ ✓
  • D. 4P₁

Answer: C — From ideal gas law at constant volume: P/T = constant, so P₁/T₁ = P₂/2T₁, giving P₂ = 2P₁.

Q5. The value of the universal gas constant R in SI units is:

  • A. 0.082 L atm mol⁻¹ K⁻¹
  • B. 8.31 J mol⁻¹ K⁻¹ ✓
  • C. 1.987 cal mol⁻¹ K⁻¹
  • D. 0.0821 bar L mol⁻¹ K⁻¹

Answer: B — R = 8.31 J mol⁻¹ K⁻¹ is the SI value used in the ideal gas equation PV = µRT; other options use non-SI units.

Q6. A gas thermometer reads the same for all low-density gases because:

  • A. All gases have the same boiling point
  • B. All gases obey Boyle's and Charles' laws identically ✓
  • C. All gases have the same density
  • D. All gases conduct heat at the same rate

Answer: B — All low-density gases exhibit identical expansion behaviour following PV/T = constant, making gas thermometers universal; other factors are not related to thermometric consistency.

Q7. Which statement about absolute zero is NOT correct?

  • A. Absolute zero is −273.15°C on the Celsius scale
  • B. Absolute zero is 0 K on the Kelvin scale
  • C. Absolute zero is the temperature where molecular motion stops completely and can be achieved experimentally ✓
  • D. Absolute zero is the lower limit of the Kelvin temperature scale

Answer: C — Absolute zero cannot be practically achieved due to thermodynamic laws; it is a theoretical limit, not an experimentally reachable temperature.

Q8. If pressure P of a fixed mass of gas at constant volume is plotted against Celsius temperature, the line extrapolates to meet the temperature axis at −273.15°C because:

  • A. This is the freezing point of water
  • B. The P vs T relationship is linear, and P = 0 at this absolute temperature ✓
  • C. All gases liquefy at this temperature
  • D. This is the boiling point of all substances

Answer: B — From PV = µRT (constant V), P ∝ T absolutely; the line must pass through origin in P vs T(K) graph, which corresponds to −273.15°C on Celsius scale where P = 0.

Q9. Which pair correctly matches a temperature scale property with its value?

  • A. Fahrenheit: 180 equal intervals between ice and steam points
  • B. Celsius: 273.15 equal intervals between ice and steam points
  • C. Kelvin: Ice point at 273.15 K and steam point at 373.15 K, 100 equal intervals ✓
  • D. All scales: Same absolute zero temperature

Answer: C — Kelvin scale has ice point = 273.15 K and steam point = 373.15 K with 100 equal intervals, matching Celsius; Fahrenheit has 180 intervals; only Kelvin has true absolute zero at 0 K.

Q10. A constant-pressure gas thermometer measures temperature by reading the volume change of gas. If volume V₁ at temperature T₁ (in Kelvin) increases to V₂ at temperature T₂, derive the relation between V₁, V₂, T₁, and T₂ using Charles' law, and calculate T₂ if V₁ = 100 cm³ at T₁ = 273 K and V₂ = 150 cm³.

  • A. V₁/V₂ = T₁/T₂; T₂ = 409.5 K
  • B. V₁/T₁ = V₂/T₂; T₂ = 409.5 K ✓
  • C. V₁T₁ = V₂T₂; T₂ = 182 K
  • D. V₁/T₂ = V₂/T₁; T₂ = 150 K

Answer: B — Charles' law at constant pressure: V/T = constant, so V₁/T₁ = V₂/T₂; substituting values: T₂ = T₁(V₂/V₁) = 273 × (150/100) = 273 × 1.5 = 409.5 K.

Flashcards

What is the SI unit of heat energy?

Joule (J) is the SI unit of heat energy.

Define temperature in physics.

Temperature is a relative measure of the hotness or coldness of a body, indicating the average kinetic energy of its particles.

State Boyle's law for gases.

At constant temperature, the product of pressure and volume of a fixed mass of gas is constant: PV = constant.

What is the relationship between Celsius and Kelvin scales?

T(K) = tC + 273.15, where both scales have the same unit size but different zero points.

Define heat in thermodynamics.

Heat is the form of energy transferred between two systems or a system and surroundings due to temperature difference.

What are the two fixed reference points for temperature scales?

Ice point (freezing point of water at 0°C) and steam point (boiling point of water at 100°C) under standard atmospheric pressure.

State Charles' law for gases.

At constant pressure, the volume of a fixed mass of gas is directly proportional to absolute temperature: V/T = constant.

What is the ideal gas equation?

PV = µRT, where P is pressure, V is volume, µ is number of moles, R is universal gas constant (8.31 J mol⁻¹ K⁻¹), and T is absolute temperature.

Convert 98.6°F to Celsius.

Using (tF - 32)/180 = tC/100: tC = (98.6 - 32) × 100/180 = 37°C.

What is absolute zero and why is it important?

Absolute zero (-273.15°C or 0 K) is the lowest possible temperature where all molecular motion theoretically stops, defining the Kelvin scale's zero point.

Important Board Questions

Define heat and temperature. Explain why heat flows from a body at higher temperature to one at lower temperature. [2 marks]

State that temperature measures hotness (average kinetic energy of particles) and heat is energy transferred due to temperature difference; explain that systems spontaneously move toward thermal equilibrium.

Derive the ideal gas equation starting from Boyle's law and Charles' law. Show all steps and explain what each law represents physically. [5 marks]

Begin with PV = const (Boyle's law at constant T) and V/T = const (Charles' law at constant P); combine these relations to show PV/T = constant for a given mass; introduce moles and universal gas constant R to obtain PV = µRT; explain that Boyle's law represents inverse pressure-volume relationship and Charles' law represents direct volume-temperature proportionality.

A blacksmith heats an iron ring before fitting it onto the wooden rim of a cart wheel. Explain this technique using the concept of thermal expansion and derive the linear expansion formula. Why cannot the ring be forced on at room temperature? [6 marks]

Explain that heating increases atomic spacing, causing the ring's diameter to expand; derive ΔL = αL₀ΔT by considering that fractional expansion is proportional to temperature change and initial length; show that at room temperature the ring's diameter is smaller than the wheel rim, making fitting impossible without expansion; at elevated temperature, the expanded ring diameter matches the rim, allowing easy fitting and subsequent tight grip upon cooling.

Assertion-Reason Questions for Thermal Properties of Matter

Q1. Assertion (A):

Heat is defined as the form of energy transferred between two systems due to their temperature difference.

Reason (R):

Temperature is a measure of the average kinetic energy of molecules in a body.

Answer: B — Both A and R are true, but R is NOT the correct explanation of A — Both statements are true. The assertion correctly defines heat (energy transfer due to ΔT), and the reason correctly explains what temperature represents. However, the reason does not explain why heat is transferred—it merely defines temperature. The explanation for heat transfer is the temperature difference, not the definition of temperature itself. Hence, reason is not the correct explanation of the assertion.

Q2. Assertion (A):

The Kelvin scale is called an absolute temperature scale.

Reason (R):

The Kelvin scale has its zero point at the temperature where all molecular motion ceases, and it is derived by extrapolating gas laws to zero pressure.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true. The Kelvin scale is absolute because zero (0 K = -273.15 °C) represents a fundamental thermodynamic limit where pressure becomes zero in ideal gas extrapolation. The reason correctly explains why Kelvin is absolute—it's based on a physical limit, not an arbitrary reference point like water's freezing point. The reason fully explains the assertion.

Q3. Assertion (A):

At constant pressure, the volume of an ideal gas is directly proportional to its absolute temperature.

Reason (R):

This relationship is known as Boyle's law.

Answer: C — A is true but R is false — The assertion is correct (Charles' law: V ∝ T at constant P). However, the reason is false. Boyle's law describes the inverse relationship between P and V at constant T, not the V-T relationship. The correct name for the assertion is Charles' law. Students often confuse the names of different gas laws.

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Case-Based Questions (CBQ) for Thermal Properties of Matter

Case Study:

A hospital in Mumbai uses a constant-volume helium gas thermometer to monitor temperature in a vaccine storage facility. The thermometer is calibrated at the ice point (273.15 K) with a pressure reading of 80 kPa, and at the steam point (373.15 K) with a pressure reading of 108.6 kPa. The storage facility maintains the vaccine vials at a temperature between these fixed points. One morning, the pressure gauge reads 94.3 kPa. Using the ideal gas law and assuming constant volume, calculate whether the storage temperature is within acceptable limits (between 0 °C and 100 °C).

Q1. What is the temperature reading of the gas thermometer when the pressure is 94.3 kPa?

  • A. 308.15 K (35 °C)
  • B. 323.15 K (50 °C) ✓
  • C. 338.15 K (65 °C)
  • D. 353.15 K (80 °C)

Answer: B

Q2. Which principle of ideal gas behavior is used to solve this problem?

  • A. Charles' law (V/T = constant at constant P)
  • B. Gay-Lussac's law (P/T = constant at constant V) ✓
  • C. Boyle's law (PV = constant at constant T)
  • D. Universal gas law (PV = nRT)

Answer: B

Q3. If the thermometer reading in the morning was 323.15 K and by noon it rises to 333.15 K, what is the new pressure reading expected?

  • A. 97.5 kPa
  • B. 99.8 kPa
  • C. 102.1 kPa ✓
  • D. 105.3 kPa

Answer: C

Q4. Why is helium preferred over other gases for this constant-volume thermometer in a hospital?

  • A. Helium is the cheapest gas available
  • B. Helium is non-toxic and safe for medical environments
  • C. Helium has the highest atomic mass among all gases
  • D. Helium exhibits ideal gas behavior over a wide range of temperatures and pressures ✓

Answer: D

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