How to Use This Page
1. Solve first, check later — Attempt every exercise question on your own before reading the solutions below. 2. Follow the sequence — Work through sections I to VI in order; later sections build on earlier ones. 3. Understand the method — Focus on the reasoning and formula selection, not just the final answer. 4. Practise numericals — Re-work the worked examples with different values until you can set them up without hints. 5. Review mistakes — Read the "Common Mistakes to Avoid" section before every exam.
Key Concepts (Quick Recap)
Temperature Scales
| Scale | Freezing Point | Boiling Point | Relation |
|---|---|---|---|
| Celsius (°C) | 0 | 100 | C/5 = (F − 32)/9 |
| Fahrenheit (°F) | 32 | 212 | F = (9C/5) + 32 |
| Kelvin (K) | 273.15 | 373.15 | K = C + 273.15 |
| Absolute zero | −273.15°C | −459.67°F | 0 K |
Thermal Expansion
| Type | Formula | Coefficient |
|---|---|---|
| Linear expansion | ΔL = L₀αΔT | α (per °C or per K) |
| Area expansion | ΔA = A₀βΔT | β ≈ 2α |
| Volume expansion | ΔV = V₀γΔT | γ ≈ 3α |
| Relation | γ = 3α, β = 2α | — |
Calorimetry and Specific Heat
| Quantity | Formula | SI Unit |
|---|---|---|
| Heat exchanged | Q = mcΔT | J |
| Specific heat capacity | c = Q/(mΔT) | J/(kg·K) |
| Molar specific heat | C = Q/(nΔT) | J/(mol·K) |
| Latent heat (fusion) | Q = mL_f | J |
| Latent heat (vaporisation) | Q = mL_v | J |
| Heat lost = Heat gained | m₁c₁(T₁ − T) = m₂c₂(T − T₂) | — |
First Law of Thermodynamics
| Quantity | Formula | Sign Convention |
|---|---|---|
| First law | ΔQ = ΔU + ΔW | ΔQ > 0: heat absorbed |
| Work done by gas | ΔW = ∫P dV | ΔW > 0: gas expands |
| Internal energy | ΔU = nCᵥΔT | ΔU > 0: temperature rises |
Thermodynamic Processes
| Process | Condition | Work Done | Relation |
|---|---|---|---|
| Isothermal | ΔT = 0 | W = nRT ln(V₂/V₁) | PV = constant |
| Adiabatic | ΔQ = 0 | W = (P₁V₁ − P₂V₂)/(γ − 1) | PVᵞ = constant |
| Isobaric | P = constant | W = PΔV = nRΔT | V/T = constant |
| Isochoric | V = constant | W = 0 | P/T = constant |
Carnot Engine and Second Law
| Quantity | Formula |
|---|---|
| Carnot efficiency | η = 1 − T₂/T₁ |
| Carnot COP (refrigerator) | β = T₂/(T₁ − T₂) |
| Entropy change (reversible) | ΔS = ΔQ/T |
| Second law (Kelvin) | No engine can convert all heat into work |
| Second law (Clausius) | Heat cannot flow from cold to hot on its own |
I. Choose the Best Answer (MCQs)
1. The temperature at which Celsius and Fahrenheit scales read the same is:
- (a) −40°
- (b) 0°
- (c) −100°
- (d) 40°
2. In an isothermal process, the change in internal energy of an ideal gas is:
- (a) Positive
- (b) Negative
- (c) Zero
- (d) Depends on volume
3. The work done in an isochoric process is:
- (a) nRΔT
- (b) PΔV
- (c) Zero
- (d) nRT ln(V₂/V₁)
4. The efficiency of a Carnot engine operating between 500 K and 300 K is:
- (a) 20%
- (b) 40%
- (c) 60%
- (d) 80%
5. Which of the following is a path function (not a state function)?
- (a) Internal energy
- (b) Entropy
- (c) Enthalpy
- (d) Work
6. If 100 J of heat is supplied to a system and 40 J of work is done by the system, the change in internal energy is:
- (a) 140 J
- (b) 60 J
- (c) −60 J
- (d) −140 J
7. The coefficient of volume expansion of a solid is related to its coefficient of linear expansion by:
- (a) γ = α
- (b) γ = 2α
- (c) γ = 3α
- (d) γ = α/3
II. Fill in the Blanks
1. The SI unit of specific heat capacity is J/(kg·K). 2. In an adiabatic process, ΔQ = 0 (no heat is exchanged with the surroundings). 3. The relationship between Celsius and Kelvin scales is K = °C + 273.15. 4. For an ideal gas in an isothermal process, PV = constant. 5. The coefficient of linear expansion has the SI unit K⁻¹ (or °C⁻¹). 6. The Carnot cycle consists of two isothermal and two adiabatic processes. 7. In the first law ΔQ = ΔU + ΔW, if the gas expands, ΔW is positive. 8. Entropy is a state function. 9. The latent heat of fusion of ice is approximately 3.36 × 10⁵ J/kg (or 336 kJ/kg). 10. The ratio of specific heats γ = Cₚ/Cᵥ for a monatomic ideal gas is 5/3 (≈ 1.67). 11. According to the second law of thermodynamics, the total entropy of an isolated system always increases. 12. The work done by a gas in isothermal expansion from V₁ to V₂ is nRT ln(V₂/V₁).
III. True or False (with corrections)
IV. Match the Following
| Column A | Column B |
|---|---|
| 1. Isothermal process | (c) Temperature remains constant |
| 2. Adiabatic process | (e) No heat exchange with surroundings |
| 3. Isobaric process | (a) Pressure remains constant |
| 4. Isochoric process | (f) Volume remains constant |
| 5. Carnot cycle | (b) Most efficient cycle between two temperatures |
| 6. Entropy | (d) Measure of disorder; always increases in isolated system |
V. Answer Briefly (2-Mark Questions)
VI. Answer in Detail (5-Mark Questions)
1. Derive the relation between Cₚ and Cᵥ for an ideal gas (Mayer's relation) and solve a numerical.
For an ideal gas, the molar specific heat at constant pressure (Cₚ) exceeds the molar specific heat at constant volume (Cᵥ) because at constant pressure, the gas must do additional work (PΔV) as it expands.
At constant volume: ΔQ = nCᵥΔT = ΔU (since W = 0) At constant pressure: ΔQ = nCₚΔT = ΔU + ΔW
For an ideal gas: ΔU = nCᵥΔT (internal energy depends only on temperature) Work done at constant pressure: ΔW = PΔV = nRΔT (from PV = nRT)
Therefore:
Worked Example: A diatomic ideal gas at 300 K is heated at constant pressure until its volume doubles. Find the heat supplied, change in internal energy, and work done for 2 moles of the gas. (R = 8.31 J/(mol·K))
2. Describe the Carnot cycle with a PV diagram and derive its efficiency.
The Carnot cycle consists of four reversible processes:
1. Isothermal expansion (A → B): The gas absorbs heat Q₁ from the hot reservoir at temperature T₁ and expands isothermally. Work is done by the gas.
2. Adiabatic expansion (B → C): The gas expands adiabatically (no heat exchange). Temperature drops from T₁ to T₂.
3. Isothermal compression (C → D): The gas releases heat Q₂ to the cold reservoir at temperature T₂ and is compressed isothermally. Work is done on the gas.
4. Adiabatic compression (D → A): The gas is compressed adiabatically. Temperature rises from T₂ back to T₁.
Derivation of efficiency:
Worked Example: A Carnot engine operates between a source at 600 K and a sink at 300 K. If it absorbs 2000 J of heat per cycle, find the efficiency, work done per cycle, and heat rejected.
3. Derive the expression for work done in an isothermal process and an adiabatic process, and compare them.
Isothermal process (T = constant):
For an ideal gas, PV = nRT. Since T is constant, P = nRT/V.
Adiabatic process (ΔQ = 0):
For an adiabatic process, PVᵞ = constant = P₁V₁ᵞ = P₂V₂ᵞ.
Comparison:
| Property | Isothermal | Adiabatic |
|---|---|---|
| Heat exchange | ΔQ ≠ 0 | ΔQ = 0 |
| Temperature | Constant | Changes |
| PV relation | PV = constant | PVᵞ = constant |
| Slope on PV diagram | Less steep | Steeper (γ > 1) |
| Internal energy change | ΔU = 0 | ΔU = nCᵥΔT |
| Work done | nRT ln(V₂/V₁) | (P₁V₁ − P₂V₂)/(γ − 1) |
Worked Example: 3 moles of an ideal gas at 400 K expand isothermally from 2 L to 4 L. Calculate the work done. If the same expansion occurs adiabatically (γ = 1.4), find the final temperature and work done.
Common Mistakes to Avoid
1. Using °C instead of Kelvin in gas laws — All gas law calculations (ideal gas equation, Carnot efficiency, etc.) require absolute temperature in Kelvin. Always convert: K = °C + 273.15.
2. Mixing up sign conventions in the first law — In the first law ΔQ = ΔU + ΔW, heat absorbed by the system is positive, work done by the system is positive. Students often get the sign of ΔW wrong when the gas is compressed (work is done on the gas, so ΔW is negative).
3. Confusing Cₚ and Cᵥ — Cₚ is always greater than Cᵥ by exactly R (Mayer's relation: Cₚ = Cᵥ + R). Cₚ is used for constant-pressure processes; Cᵥ for constant-volume processes. Using Cₚ where Cᵥ is needed (or vice versa) is a common exam error.
4. Applying wrong PV relation for adiabatic process — For isothermal: PV = constant. For adiabatic: PVᵞ = constant. The adiabatic curve is steeper because γ > 1. Do not substitute PV = nRT into the adiabatic relation incorrectly.
5. Forgetting latent heat in calorimetry problems — When a substance changes state (ice to water, water to steam), heat is absorbed or released at constant temperature using Q = mL. Do not use Q = mcΔT during a phase change.
6. Assuming efficiency can reach 100% — Even a Carnot engine has efficiency less than 100% unless T₂ = 0 K (impossible). No real engine can exceed Carnot efficiency. Always express efficiency as a percentage and check if it is between 0 and 100%.
7. Incorrectly calculating work from area under PV curve — In isobaric process, W = PΔV (rectangle). In isothermal process, W = nRT ln(V₂/V₁) (curve). Do not approximate the area as a simple rectangle for isothermal or adiabatic processes.
8. Confusing heat (Q) and internal energy (U) — Heat is energy transferred due to temperature difference; internal energy is the total energy within the system. They are related but distinct. ΔU = nCᵥΔT for an ideal gas regardless of the process.
Exam Checklist
- [ ] I can convert between Celsius, Fahrenheit, and Kelvin temperature scales
- [ ] I can calculate linear, area, and volume expansion using α, β, and γ
- [ ] I can apply Q = mcΔT and Q = mL in calorimetry problems
- [ ] I can state the first law of thermodynamics and apply the correct sign convention
- [ ] I can derive and use Mayer's relation Cₚ = Cᵥ + R and know Cᵥ, Cₚ values for mono/diatomic gases
- [ ] I can calculate work done in isothermal, adiabatic, isobaric, and isochoric processes
- [ ] I can derive Carnot efficiency and solve numerical problems involving T₁, T₂, Q₁, Q₂, and W
- [ ] I can state the second law in Kelvin-Planck and Clausius forms and explain entropy
- [ ] I can distinguish between isothermal and adiabatic processes on a PV diagram
- [ ] I can correctly identify state functions (U, S, H) vs path functions (W, Q)