Exam study guide
Cambridge A Level Physics Thermodynamics Practice Questions
Welcome to our comprehensive set of practice questions for Cambridge A Level Physics Thermodynamics. This essential topic explores the relationship between heat, work, and energy, forming a cornerstone of physical science. Mastering these concepts is crucial for success in your examinations. This resource provides original problems, detailed solutions, and strategic insights to enhance your understanding and problem-solving skills.
Quick revision summary
- First Law of Thermodynamics: ΔU = Q - W, where ΔU is change in internal energy, Q is heat supplied *to* the system, and W is work done *by* the system.
- Ideal Gas Equation: pV = nRT or pV = NkT. Remember T must be in Kelvin.
- Work Done by a Gas: W = pΔV for constant pressure processes, or the area under the p-V curve for variable pressure.
- Internal energy of an ideal gas depends only on its temperature.
Core Concepts: Internal Energy and Heat Transfer
Thermodynamics begins with understanding internal energy, which is the sum of the random kinetic and potential energies of the molecules within a system. For an ideal gas, internal energy is solely dependent on its temperature, as intermolecular forces are considered negligible. Heat is the transfer of thermal energy due to a temperature difference, while temperature itself is a measure of the average kinetic energy of the particles. Key quantities include specific heat capacity (energy required to raise the temperature of 1 kg of a substance by 1 K) and specific latent heat (energy required to change the state of 1 kg of a substance without changing its temperature).
The First Law of Thermodynamics and Gas Processes
The First Law of Thermodynamics is a statement of the conservation of energy: ΔU = Q - W. This law relates the change in a system's internal energy (ΔU) to the heat added to the system (Q) and the work done by the system (W). Understanding the sign conventions for Q and W is critical. Different thermodynamic processes—isobaric (constant pressure), isochoric (constant volume), isothermal (constant temperature), and adiabatic (no heat exchange)—each have distinct characteristics and implications for how Q, W, and ΔU relate. The ideal gas model simplifies these interactions, allowing for calculations using the ideal gas equation and the kinetic theory of gases.
Practice questions
Question 1 (3 marks)
A cylinder contains 0.50 mol of an ideal monatomic gas at a pressure of 1.0 × 10⁵ Pa and a volume of 1.2 × 10⁻³ m³. The gas is then heated, causing its temperature to rise from 293 K to 353 K while its volume is kept constant. Calculate the change in internal energy of the gas.
Answer: ΔU = 374 J
Explanation: For a monatomic ideal gas, the internal energy U = (3/2)nRT. The change in internal energy is ΔU = (3/2)nRΔT. n = 0.50 mol R = 8.31 J mol⁻¹ K⁻¹ ΔT = 353 K - 293 K = 60 K ΔU = (3/2) × 0.50 mol × 8.31 J mol⁻¹ K⁻¹ × 60 K ΔU = 1.5 × 0.50 × 8.31 × 60 ΔU = 373.95 J ≈ 374 J (to 3 significant figures). Since the volume is kept constant, no work is done by the gas (W=0). By the First Law of Thermodynamics, ΔU = Q - W, so ΔU = Q. The heat supplied goes entirely into increasing the internal energy.
Question 2 (1 marks)
Which of the following statements about the internal energy of an ideal gas is correct?
- A. It depends on the pressure and volume of the gas.
- B. It is proportional to the absolute temperature of the gas.
- C. It includes the potential energy of interaction between gas molecules.
- D. It is zero at 0 °C.
Answer: B
Explanation: For an ideal gas, it is assumed that there are no intermolecular forces, so there is no potential energy associated with molecular interactions. Therefore, the internal energy is solely due to the kinetic energy of the molecules. The average kinetic energy of gas molecules is directly proportional to the absolute temperature (in Kelvin). Hence, the total internal energy of an ideal gas is proportional to its absolute temperature.
Question 3 (3 marks)
A gas expands from state A to state B as shown on the p-V diagram below. The path taken is a straight line from (1.0 × 10⁵ Pa, 2.0 × 10⁻³ m³) to (2.0 × 10⁵ Pa, 5.0 × 10⁻³ m³). Calculate the work done by the gas during this expansion.
Answer: W = 450 J
Explanation: The work done by the gas is the area under the p-V graph. Since the path is a straight line, the area is that of a trapezium. Height 1 (p₁) = 1.0 × 10⁵ Pa Height 2 (p₂) = 2.0 × 10⁵ Pa Base (ΔV) = V₂ - V₁ = (5.0 × 10⁻³ m³) - (2.0 × 10⁻³ m³) = 3.0 × 10⁻³ m³ Area = (1/2) × (p₁ + p₂) × ΔV Area = (1/2) × (1.0 × 10⁵ + 2.0 × 10⁵) Pa × (3.0 × 10⁻³ m³) Area = (1/2) × (3.0 × 10⁵ Pa) × (3.0 × 10⁻³ m³) Area = (1/2) × 900 J Work done W = 450 J. Since the gas is expanding, work is done *by* the gas, so it is positive.
Question 4 (2 marks)
An ideal gas is confined in a cylinder. It absorbs 300 J of heat from its surroundings. During this process, the gas expands and does 120 J of work on the surroundings. Calculate the change in the internal energy of the gas.
Answer: ΔU = 180 J
Explanation: We use the First Law of Thermodynamics: ΔU = Q - W. Heat absorbed by the gas (Q) = +300 J (positive because heat is supplied to the system). Work done *by* the gas (W) = +120 J (positive because the system does work). ΔU = 300 J - 120 J ΔU = 180 J. The internal energy of the gas increases by 180 J.
Question 5 (2 marks)
Describe two key assumptions made in the kinetic theory of gases for an ideal gas.
Answer: 1. The gas consists of a large number of identical particles (atoms or molecules) that are in constant, random motion. 2. The volume occupied by the gas particles themselves is negligible compared to the volume of the container; particles are considered point masses.
Explanation: Other valid assumptions include: no intermolecular forces acting between particles except during collisions, collisions between particles and with the container walls are perfectly elastic, and the time duration of collisions is negligible compared to the time between collisions.
Question 6 (2 marks)
A fixed mass of gas undergoes an isothermal compression. State and explain what happens to the internal energy of the gas.
Answer: During an isothermal compression, the temperature of the gas remains constant. For an ideal gas, internal energy is directly proportional to its absolute temperature. Therefore, if the temperature remains constant, the internal energy of the gas remains unchanged (ΔU = 0).
Explanation: In an isothermal process, the definition requires temperature to be constant. Since for an ideal gas, internal energy is solely a function of temperature, constant temperature implies constant internal energy. According to the First Law (ΔU = Q - W), if ΔU = 0, then Q = W. This means that during isothermal compression, the work done *on* the gas (W is negative, so -W is positive) must be equal to the heat released *by* the gas (Q is negative, so heat flows out).
Question 7 (4 marks)
An air bubble released from the bottom of a lake at a depth of 10.0 m has a volume of 1.0 cm³. What is its volume just before it reaches the surface, assuming the temperature remains constant and the atmospheric pressure is 1.0 × 10⁵ Pa? The density of water is 1000 kg m⁻³ and g = 9.81 m s⁻².
Answer: V₂ = 1.98 cm³
Explanation: Since the temperature remains constant, we can use Boyle's Law: p₁V₁ = p₂V₂. At the bottom of the lake (state 1): V₁ = 1.0 cm³ = 1.0 × 10⁻⁶ m³ P₁ = Atmospheric pressure + Pressure due to water column P₁ = 1.0 × 10⁵ Pa + (ρgh) = 1.0 × 10⁵ Pa + (1000 kg m⁻³ × 9.81 m s⁻² × 10.0 m) P₁ = 1.0 × 10⁵ Pa + 9.81 × 10⁴ Pa = 1.0 × 10⁵ Pa + 0.981 × 10⁵ Pa = 1.981 × 10⁵ Pa. At the surface (state 2): P₂ = Atmospheric pressure = 1.0 × 10⁵ Pa V₂ = ? Using Boyle's Law: p₁V₁ = p₂V₂ V₂ = (p₁V₁) / p₂ V₂ = (1.981 × 10⁵ Pa × 1.0 × 10⁻⁶ m³) / (1.0 × 10⁵ Pa) V₂ = 1.981 × 10⁻⁶ m³ V₂ = 1.981 cm³ ≈ 1.98 cm³ (to 3 significant figures) Units: Check that pressures cancel and volume unit is correct.
Common mistakes
- Incorrectly applying the sign convention for work done (W) and heat transfer (Q) in the First Law of Thermodynamics (ΔU = Q - W). Remember: Q is positive if heat is supplied to the system, negative if heat leaves. W is positive if work is done *by* the system, negative if work is done *on* the system.
- Confusing temperature in Celsius with Kelvin in gas law calculations. Always convert temperature to Kelvin (T_K = T_°C + 273.15) for calculations involving the ideal gas equation (pV=nRT) or kinetic theory formulae.
Exam tips
- Always draw a p-V diagram when dealing with gas processes. It helps visualise the process, determine if work is done (area under the curve), and identify the type of process (isobaric, isochoric, etc.).
- Pay close attention to units! Ensure consistency, especially for pressure (Pascals), volume (m³), and temperature (Kelvin). Remember that 1 atmosphere is approximately 1.01 × 10⁵ Pa, and 1 cm³ = 10⁻⁶ m³.
Why this matters for exams
Thermodynamics is a fundamental topic in Cambridge A Level Physics, frequently appearing in Paper 2 (AS Level) and Paper 4 (A2 Level). Questions often assess understanding of the First Law of Thermodynamics, ideal gas behaviour, kinetic theory, specific heat capacities, and the interpretation of p-V diagrams. Strong problem-solving skills in this area are essential for achieving top grades.
Frequently asked questions
What is the key difference between heat and temperature?
Temperature is a measure of the average kinetic energy of the particles within a substance, indicating its degree of hotness or coldness. Heat, on the other hand, is the transfer of thermal energy between objects or systems due to a temperature difference.
When is work done by a gas considered positive or negative?
In the standard A Level convention (ΔU = Q - W), work done *by* the gas (e.g., expansion against an external pressure) is positive. Work done *on* the gas (e.g., compression) is negative. This means if the gas expands, W is positive; if it's compressed, W is negative.
What are the characteristics of an ideal gas?
An ideal gas is a theoretical gas composed of randomly moving point particles that do not interact with each other except for perfectly elastic collisions. Key characteristics include: negligible volume of particles, no intermolecular forces, random motion, and perfectly elastic collisions. Real gases approximate ideal gas behaviour at high temperatures and low pressures.
How do I approach problems involving the First Law of Thermodynamics?
1. Identify the system and the process it undergoes. 2. Determine the given values for Q, W, or ΔU, paying close attention to their signs (heat added vs. removed, work done by vs. on the gas). 3. Use the formula ΔU = Q - W, or rearrange it to solve for the unknown quantity. 4. If work is done, calculate it (e.g., W = pΔV for constant pressure, or area under p-V graph). If temperature changes, relate it to internal energy change.