IB Physics Quiz: Apply Thermodynamics
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Apply ThermodynamicsQuestion 1 of 20

An ideal gas undergoes an isothermal expansion. Which of the following statements about the gas is correct?

The heat supplied to the gas is zero.
The work done by the gas is zero.
The pressure of the gas remains constant.
The change in internal energy of the gas is zero.
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IB Physics Quiz

IB Physics Quiz: Apply Thermodynamics

Practice Apply Thermodynamics in IB Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Apply Thermodynamics, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Physics.

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Question 1

An ideal gas undergoes an isothermal expansion. Which of the following statements about the gas is correct?

  1. The heat supplied to the gas is zero.
  2. The work done by the gas is zero.
  3. The pressure of the gas remains constant.
  4. The change in internal energy of the gas is zero. (correct answer)
Explanation: Isothermal means the temperature of the gas remains constant. For an ideal gas, the internal energy is directly proportional to its absolute temperature (U ∝ T). Therefore, if the temperature is constant, the change in internal energy (ΔU) is zero. Since the gas expands, it does positive work (W > 0). From the first law (Q = ΔU + W), with ΔU=0, we have Q = W. This means heat must be supplied to the gas to allow it to expand isothermally.

Question 2

A thermally isolated box is divided by a partition. One side contains an ideal gas, and the other is a vacuum. The partition is removed, allowing the gas to expand freely and fill the entire box. What are the changes in the internal energy and the entropy of the gas?

  1. Internal energy decreases; entropy increases.
  2. Internal energy is constant; entropy is constant.
  3. Internal energy is constant; entropy increases. (correct answer)
  4. Internal energy increases; entropy increases.
Explanation: This is a free expansion. The gas expands into a vacuum, so it does no work (W=0). The box is thermally isolated, so no heat is exchanged (Q=0). According to the first law of thermodynamics (ΔU = Q - W), the change in internal energy is ΔU = 0. The expansion is irreversible, and the gas molecules now occupy a larger volume, which corresponds to a greater number of possible microscopic arrangements (microstates). Since entropy is a measure of this disorder, the entropy of the gas increases.

Question 3

An ideal gas is contained within a piston-cylinder assembly. 500 J of heat is supplied to the gas, and an external force compresses the gas, doing 200 J of work on it. What is the change in the internal energy of the gas?

  1. 300 J
  2. -300 J
  3. 700 J (correct answer)
  4. -700 J
Explanation: The first law of thermodynamics is Q = ΔU + W. Here, Q is the heat added to the system, so Q = +500 J. W is the work done by the system. Since 200 J of work is done on the gas, the work done by the gas is W = -200 J. Substituting these values: +500 J = ΔU + (-200 J). Solving for ΔU gives ΔU = 500 J + 200 J = 700 J.

Question 4

A fixed mass of an ideal monatomic gas is held in a container of constant volume. The absolute temperature of the gas is doubled. What is the ratio of the heat supplied to the gas to the initial internal energy of the gas?

  1. 1/2
  2. 1 (correct answer)
  3. 3/2
  4. 2
Explanation: For an isovolumetric (constant volume) process, the work done W = 0. From the first law, the heat supplied is Q = ΔU. The internal energy of an ideal monatomic gas is U = (3/2)nRT. The initial internal energy is U_i = (3/2)nRT. When the temperature doubles to 2T, the final internal energy is U_f = (3/2)nR(2T) = 2 * U_i. The change in internal energy is ΔU = U_f - U_i = 2U_i - U_i = U_i. Since Q = ΔU, we have Q = U_i. The ratio Q / U_i is therefore 1.

Question 5

A 0.50 kg block of ice at its melting point of 0°C absorbs heat and melts completely into water at 0°C. The specific latent heat of fusion of ice is 3.3 × 10⁵ J kg⁻¹. What is the change in entropy of the substance during this process?

  1. Zero, because the temperature is constant.
  2. An increase of 6.0 × 10² J K⁻¹. (correct answer)
  3. A decrease of 6.0 × 10² J K⁻¹.
  4. An increase of 1.7 × 10⁵ J K⁻¹.
Explanation: The entropy change for a reversible process at constant temperature is ΔS = Q / T. First, calculate the heat absorbed: Q = m * L = 0.50 kg * 3.3 × 10⁵ J kg⁻¹ = 1.65 × 10⁵ J. The temperature must be in Kelvin: T = 0°C + 273 = 273 K. Now, calculate the entropy change: ΔS = (1.65 × 10⁵ J) / 273 K ≈ 604.4 J K⁻¹. This is an increase, as melting represents an increase in disorder. The closest answer is an increase of 6.0 × 10² J K⁻¹.

Question 6

An inventor claims to have built a heat engine that, in each cycle, takes in 100 kJ of heat from a reservoir at 400 K, performs 30 kJ of mechanical work, and expels 70 kJ of heat to a reservoir at 250 K. Which statement correctly analyzes this claim?

  1. The claim is impossible as it violates the first law of thermodynamics.
  2. The claim is impossible as its efficiency exceeds the maximum Carnot efficiency.
  3. The claim is impossible as no real engine can operate at the Carnot efficiency.
  4. The claim is plausible as it satisfies both the first and second laws of thermodynamics. (correct answer)
Explanation: First, check the first law (conservation of energy): Q_H = W + Q_C. The claim states 100 kJ = 30 kJ + 70 kJ, which is true. The first law is satisfied. Second, check the second law by comparing the engine's efficiency to the maximum possible (Carnot) efficiency. The engine's efficiency is η = W / Q_H = 30 kJ / 100 kJ = 0.30. The Carnot efficiency is η_Carnot = 1 - T_C / T_H = 1 - 250 K / 400 K = 1 - 0.625 = 0.375. Since the engine's claimed efficiency (0.30) is less than the maximum possible efficiency (0.375), the claim is consistent with the second law. Thus, the claim is plausible.

Question 7

An ideal gas undergoes a cyclic process. It expands isobarically from a volume of 1.0 m³ to 3.0 m³ at a pressure of 200 kPa. It is then cooled isovolumetrically to a pressure of 100 kPa. Following this, it is compressed isobarically back to 1.0 m³, and finally heated isovolumetrically back to the initial state. What is the net work done by the gas in one complete cycle?

  1. 100 kJ
  2. 200 kJ (correct answer)
  3. 400 kJ
  4. 600 kJ
Explanation: Net work done in a cycle is the area enclosed on a P-V diagram. Work is only done during the volume changes (isobaric processes). Work done during expansion: W_exp = P₁ΔV₁ = (200 × 10³ Pa) × (3.0 m³ - 1.0 m³) = +400 kJ. Work done during compression: W_comp = P₂ΔV₂ = (100 × 10³ Pa) × (1.0 m³ - 3.0 m³) = -200 kJ. The net work is the sum: W_net = W_exp + W_comp = 400 kJ - 200 kJ = 200 kJ.

Question 8

An ideal gas in an initial state (P₀, V₀) expands to a final volume of 2V₀. Process 1 is an isothermal expansion. Process 2 is an adiabatic expansion from the same initial state. Let P₁ and P₂ be the final pressures for Process 1 and Process 2, respectively. Which statement correctly compares the final pressures?

  1. P₁ = P₂, because the final volume is the same for both processes.
  2. P₂ < P₁, because the gas cools as it expands adiabatically. (correct answer)
  3. P₂ > P₁, because no heat is supplied during the adiabatic process.
  4. The relationship between P₁ and P₂ cannot be determined without knowing the gas.
Explanation: In an isothermal expansion, temperature is constant. In an adiabatic expansion, the gas does work on the surroundings, so its internal energy decreases (since Q=0). A decrease in internal energy for an ideal gas means its temperature drops. According to the ideal gas law (PV=nRT), for the same final volume, the gas at the lower temperature (adiabatic process) will exert a lower pressure. Therefore, P₂ < P₁.

Question 9

A proposed device is claimed to extract 100 kJ of heat from a single heat reservoir and convert it entirely into 100 kJ of useful work, with no other energy transfers. Why is this device impossible to construct?

  1. It violates the first law of thermodynamics because energy cannot be converted.
  2. It violates the second law of thermodynamics because some heat must be rejected to a colder reservoir. (correct answer)
  3. It is practically impossible due to energy losses from friction, but it is theoretically possible.
  4. It violates the conservation of momentum as the device would need to accelerate.
Explanation: The first law (conservation of energy) is satisfied (100 kJ in = 100 kJ out). However, the Kelvin-Planck statement of the second law of thermodynamics states that it is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce a net amount of work. A heat engine must transfer some heat to a low-temperature reservoir.

Question 10

An ideal heat pump operates on a Carnot cycle to heat a building. It moves heat from a cold exterior at temperature T_C to a warm interior at temperature T_H. What is the expression for its coefficient of performance (COP), defined as the heat delivered to the interior divided by the work input?

  1. T_H / (T_H - T_C) (correct answer)
  2. T_C / (T_H - T_C)
  3. 1 - (T_C / T_H)
  4. (T_H - T_C) / T_H
Explanation: The coefficient of performance (COP) for a heat pump is defined as COP = Q_H / W. By conservation of energy, the heat delivered to the hot reservoir is Q_H = W + Q_C, where W is the work input and Q_C is the heat extracted from the cold reservoir. For a reversible (Carnot) cycle, the ratio of heats is equal to the ratio of absolute temperatures: Q_H / Q_C = T_H / T_C. So, Q_C = Q_H * (T_C / T_H). Substituting into the energy conservation equation: W = Q_H - Q_C = Q_H - Q_H * (T_C / T_H) = Q_H * (1 - T_C / T_H). Rearranging for COP = Q_H / W gives COP = 1 / (1 - T_C / T_H) = T_H / (T_H - T_C).

Question 11

An engineer is designing a heat engine to operate between a heat source at 227°C and a heat sink at 27°C. What is the maximum possible theoretical efficiency of this engine?

  1. 0.30
  2. 0.40 (correct answer)
  3. 0.88
  4. 1.00
Explanation: The maximum theoretical efficiency is the Carnot efficiency, given by η_Carnot = 1 - (T_C / T_H), where temperatures must be in Kelvin. The cold reservoir temperature is T_C = 27°C + 273 = 300 K. The hot reservoir temperature is T_H = 227°C + 273 = 500 K. The efficiency is η_Carnot = 1 - (300 K / 500 K) = 1 - 0.6 = 0.40.

Question 12

The air in a bicycle pump cylinder is compressed quickly. The temperature of the air rises because the compression is approximately adiabatic. Which statement provides the best explanation for this temperature rise?

  1. The rapid increase in pressure causes a proportional increase in temperature.
  2. Heat is generated by friction between the piston and the cylinder walls.
  3. Work is done on the air, and this energy increases the internal energy of the air. (correct answer)
  4. The density of the air increases, causing more frequent collisions which generate heat.
Explanation: In an adiabatic process, there is no heat exchange with the surroundings (Q=0). When the air is compressed, work is done on the air (W is negative). The first law of thermodynamics is ΔU = Q - W. For this process, ΔU = 0 - W = -W. Since W is negative, ΔU is positive. For an ideal gas, an increase in internal energy corresponds to an increase in temperature.

Question 13

Two identical blocks of metal, one at temperature T₁ and the other at T₂ (where T₁ > T₂), are brought into thermal contact and isolated from their surroundings. They reach a final equilibrium temperature T_f. What is the total change in entropy of the two-block system?

  1. Zero, because the system is isolated.
  2. Negative, because the final state is more ordered.
  3. Zero, because the entropy lost by the hot block equals the entropy gained by the cold block.
  4. Positive, because heat transfer from hot to cold is an irreversible process. (correct answer)
Explanation: The spontaneous flow of heat from a hotter body to a colder body is an irreversible process. According to the second law of thermodynamics, any irreversible process in an isolated system results in an increase in the total entropy of the system. While the hot block loses entropy and the cold block gains entropy, the magnitude of the entropy gain is larger than the magnitude of the entropy loss because entropy change is ΔQ/T, and the heat is transferred at lower temperatures for the gaining block.

Question 14

An ideal gas is taken from an initial equilibrium state A to a final equilibrium state B by two different processes. Process 1 is reversible, and Process 2 is irreversible. Which statement is correct regarding the heat added to the gas (Q) and the work done by the gas (W) for these two processes?

  1. The difference Q - W must be identical for both processes. (correct answer)
  2. The sum Q + W must be identical for both processes.
  3. Both Q and W must be identical for both processes.
  4. The ratio Q / W must be identical for both processes.
Explanation: According to the first law of thermodynamics, the change in internal energy is given by ΔU = Q - W. Internal energy (U) is a state function, meaning its value depends only on the state of the system (e.g., its pressure, volume, and temperature) and not on the path taken to reach that state. Since both processes start at state A and end at state B, the change in internal energy (ΔU) must be the same for both. Therefore, the quantity Q - W must be identical for both processes.

Question 15

A thermally insulated container holds an ideal gas. The gas is agitated by a paddle wheel which does 100 J of work on the gas. What are the changes in heat added to the gas (Q), work done by the gas (W), and internal energy of the gas (ΔU)?

  1. Q = 0 J, W = +100 J, ΔU = -100 J
  2. Q = +100 J, W = -100 J, ΔU = +200 J
  3. Q = 0 J, W = -100 J, ΔU = +100 J (correct answer)
  4. Q = -100 J, W = 0 J, ΔU = -100 J
Explanation: The container is thermally insulated, so no heat is exchanged with the surroundings, meaning Q = 0 J. The paddle wheel does 100 J of work on the gas. The variable W in the first law of thermodynamics (Q = ΔU + W) represents the work done by the gas. Therefore, W = -100 J. Applying the first law: 0 J = ΔU + (-100 J), which gives ΔU = +100 J. The internal energy of the gas increases.

Question 16

A Stirling engine operates with 0.1 mol0.1\text{ mol} of helium gas between volumes V1=100 cm3V_1 = 100\text{ cm}^3 and V2=400 cm3V_2 = 400\text{ cm}^3, and temperatures T1=300 KT_1 = 300\text{ K} and T2=600 KT_2 = 600\text{ K}. The cycle consists of two isothermal and two isochoric processes. What is the ratio of heat absorbed to heat rejected during one complete cycle?

  1. 1.51.5
  2. 2.02.0 (correct answer)
  3. 2.52.5
  4. 4.04.0
Explanation: A Stirling cycle has four processes: isothermal expansion at T2T_2, isochoric cooling, isothermal compression at T1T_1, and isochoric heating. Heat absorbed occurs during: (1) isothermal expansion at T2T_2: Qh=nRT2ln⁡(V2/V1)=(0.1)(8.314)(600)ln⁡(4)=692 JQ_h = nRT_2 \ln(V_2/V_1) = (0.1)(8.314)(600)\ln(4) = 692\text{ J}, and (2) isochoric heating: Qheat=nCV(T2−T1)=(0.1)(32)(8.314)(300)=374 JQ_{\text{heat}} = nC_V(T_2 - T_1) = (0.1)(\frac{3}{2})(8.314)(300) = 374\text{ J}. Heat rejected occurs during: (1) isothermal compression at T1T_1: Qc=nRT1ln⁡(V2/V1)=(0.1)(8.314)(300)ln⁡(4)=346 JQ_c = nRT_1 \ln(V_2/V_1) = (0.1)(8.314)(300)\ln(4) = 346\text{ J}, and (2) isochoric cooling: same 374 J. In a Stirling engine, the isochoric heat transfers are internal (via regenerator), so external heat absorbed = 692 J, external heat rejected = 346 J. Ratio = 692/346=2.0692/346 = 2.0. Other ratios result from including/excluding regenerator effects incorrectly.

Question 17

Two identical containers each hold nn moles of an ideal gas at the same temperature TT. Container A holds a monatomic gas and container B holds a diatomic gas. Both gases undergo adiabatic expansion to twice their original volume. What is the ratio of the final temperature of gas A to the final temperature of gas B?

  1. (12)2/3\left(\frac{1}{2}\right)^{2/3}
  2. (12)2/5\left(\frac{1}{2}\right)^{2/5}
  3. (12)4/15\left(\frac{1}{2}\right)^{4/15} (correct answer)
  4. (12)1/3\left(\frac{1}{2}\right)^{1/3}
Explanation: For adiabatic processes, TVγ−1=constantTV^{\gamma-1} = \text{constant}. For monatomic gas (A): γA=5/3\gamma_A = 5/3, so TA,f/T=(1/2)(5/3−1)=(1/2)2/3T_{A,f}/T = (1/2)^{(5/3-1)} = (1/2)^{2/3}. For diatomic gas (B): γB=7/5\gamma_B = 7/5, so TB,f/T=(1/2)(7/5−1)=(1/2)2/5T_{B,f}/T = (1/2)^{(7/5-1)} = (1/2)^{2/5}. The ratio is TA,fTB,f=(1/2)2/3(1/2)2/5=(1/2)2/3−2/5=(1/2)10/15−6/15=(1/2)4/15\frac{T_{A,f}}{T_{B,f}} = \frac{(1/2)^{2/3}}{(1/2)^{2/5}} = (1/2)^{2/3-2/5} = (1/2)^{10/15-6/15} = (1/2)^{4/15}. Choice A uses only monatomic exponent. Choice B uses only diatomic exponent. Choice D uses incorrect calculation.

Question 18

A heat engine absorbs 2500 J of energy from a hot reservoir and expels 1500 J of energy to a cold reservoir in each cycle. What is the efficiency of this engine?

  1. 0.40 (correct answer)
  2. 0.60
  3. 0.67
  4. 1.67
Explanation: The heat absorbed from the hot reservoir is Q_H = 2500 J. The heat expelled to the cold reservoir is Q_C = 1500 J. By the first law, the work done by the engine is W = Q_H - Q_C = 2500 J - 1500 J = 1000 J. The efficiency (η) is the ratio of the work done to the heat absorbed from the hot reservoir: η = W / Q_H = 1000 J / 2500 J = 0.40.

Question 19

An ideal monatomic gas at pressure P expands isobarically from an initial volume V to a final volume 2V. In terms of P and V, how much heat Q was supplied to the gas?

  1. PV
  2. (3/2)PV
  3. (5/2)PV (correct answer)
  4. 3PV
Explanation: The heat supplied is given by the first law: Q = ΔU + W. For an isobaric process, the work done by the gas is W = PΔV = P(2V - V) = PV. The change in internal energy for a monatomic ideal gas is ΔU = (3/2)nRΔT. From the ideal gas law, PV = nRT, so P(2V) = nR(T_f) and PV = nR(T_i). This means T_f = 2T_i, so ΔT = T_i. Therefore, ΔU = (3/2)nR(T_i) = (3/2)PV. Combining these, Q = (3/2)PV + PV = (5/2)PV.

Question 20

An ideal gas undergoes a very slow compression at constant temperature. Which statement correctly describes the energy transfers for the gas during this process?

  1. Work is done on the gas, and an equal amount of energy is removed from the gas as heat. (correct answer)
  2. Work is done on the gas, causing a corresponding increase in its internal energy.
  3. No work is done because the process is slow, but heat is removed from the gas.
  4. The process is adiabatic, so no energy is exchanged with the surroundings.
Explanation: The process is at constant temperature, so it is isothermal. For an ideal gas, internal energy (U) depends only on temperature. Since temperature is constant, the change in internal energy is zero (ΔU = 0). From the first law, Q = ΔU + W. With ΔU = 0, we have Q = W. Since the gas is compressed, work is done on the gas, so W is negative. Therefore, Q must also be negative, meaning heat is removed from the gas. The amount of heat removed is equal to the amount of work done on the gas.