IB Physics Quiz: Understand Thermodynamics
20 questions · exam conditions
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Understand ThermodynamicsQuestion 1 of 20

A block of ice at 273 K melts into water at 273 K by absorbing heat from a room at 293 K. Let ΔS_sys be the entropy change of the ice/water, ΔS_sur be the entropy change of the room, and ΔS_uni be the entropy change of the universe. What are the signs of these entropy changes?

ΔS_sys > 0, ΔS_sur < 0, ΔS_uni > 0
ΔS_sys > 0, ΔS_sur < 0, ΔS_uni = 0
ΔS_sys > 0, ΔS_sur > 0, ΔS_uni > 0
ΔS_sys < 0, ΔS_sur > 0, ΔS_uni > 0
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IB Physics Quiz

IB Physics Quiz: Understand Thermodynamics

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

What this quiz covers

This quiz focuses on Understand 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

A block of ice at 273 K melts into water at 273 K by absorbing heat from a room at 293 K. Let ΔS_sys be the entropy change of the ice/water, ΔS_sur be the entropy change of the room, and ΔS_uni be the entropy change of the universe. What are the signs of these entropy changes?

  1. ΔS_sys > 0, ΔS_sur < 0, ΔS_uni > 0 (correct answer)
  2. ΔS_sys > 0, ΔS_sur < 0, ΔS_uni = 0
  3. ΔS_sys > 0, ΔS_sur > 0, ΔS_uni > 0
  4. ΔS_sys < 0, ΔS_sur > 0, ΔS_uni > 0
Explanation: Melting is a phase change from a more ordered solid to a less ordered liquid, so the entropy of the system increases (ΔS_sys > 0). The system absorbs heat Q from the surroundings (the room), so the surroundings lose entropy (ΔS_sur < 0). The entropy change is given by ΔS = Q/T. The system's entropy increases by Q/273, while the surroundings' entropy decreases by Q/293. Since 273 < 293, the magnitude of the entropy gain of the system is greater than the magnitude of the entropy loss of the surroundings. Thus, the total entropy of the universe increases (ΔS_uni = ΔS_sys + ΔS_sur > 0), as expected for an irreversible process.

Question 2

A proposed refrigerator requires no external work input to transfer heat from a cold object to a hot object in a continuous cycle. Which statement of the second law of thermodynamics does this device most directly violate?

  1. The Clausius statement, which forbids spontaneous heat flow from cold to hot. (correct answer)
  2. The Kelvin-Planck statement, which forbids 100% conversion of heat to work.
  3. The principle of increasing entropy, as this device would decrease total entropy.
  4. The first law of thermodynamics, as energy would need to be created.
Explanation: The Clausius statement of the second law of thermodynamics says it is impossible to construct a device operating in a cycle that produces no other effect than the transfer of heat from a colder body to a hotter body. Refrigerators require an input of work to achieve this. The proposed device violates this statement directly.

Question 3

Two different ideal gases, initially at the same temperature and pressure, are in separate, equal-volume containers. A valve connecting them is opened, and the gases mix. The system is thermally isolated. What happens to the total entropy of the system?

  1. It decreases, because the mixing process is spontaneous.
  2. It remains constant, because the temperature does not change.
  3. It increases, because the number of possible molecular arrangements increases. (correct answer)
  4. It remains constant, because no heat is exchanged with the surroundings.
Explanation: When the gases mix, the molecules of each gas can occupy the total volume of both containers. This increases the number of possible positions for each molecule, which leads to a large increase in the number of possible microscopic arrangements (microstates) of the system. According to the statistical definition of entropy (S = k_B ln Ω), an increase in the number of microstates (Ω) results in an increase in entropy. This is an example of an irreversible process where entropy is generated internally.

Question 4

A thermally isolated container 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 undergo free expansion. Which statement is correct?

  1. The temperature of the gas decreases because it does work on itself.
  2. The temperature of the gas increases because the molecules move faster in the larger volume.
  3. The temperature of the gas is constant, and the entropy of the gas is constant.
  4. The temperature of the gas is constant, and the entropy of the gas increases. (correct answer)
Explanation: In a free expansion, the gas expands into a vacuum, so it does no work (W=0). The container is thermally isolated, so no heat is exchanged (Q=0). By the first law, ΔU = Q - W = 0. For an ideal gas, internal energy is a function of temperature only, so ΔT=0. The gas now occupies a larger volume, increasing the number of available microstates. This corresponds to an increase in entropy.

Question 5

An ideal gas is in a thermally insulated container, separated from a vacuum by a partition. The partition is suddenly removed, and the gas expands to fill the entire container. Which row correctly identifies the change in internal energy (ΔU) of the gas and the work done (W) by the gas?

  1. ΔU = 0, W = 0 (correct answer)
  2. ΔU < 0, W > 0
  3. ΔU = 0, W > 0
  4. ΔU > 0, W = 0
Explanation: This process is called free expansion. The container is thermally insulated, so no heat is exchanged with the surroundings (Q=0). The gas expands into a vacuum, so there is no external pressure to push against. Therefore, the work done by the gas is zero (W=0). According to the first law of thermodynamics, ΔU = Q - W. Since both Q and W are zero, the change in internal energy (ΔU) is also zero.

Question 6

A fixed mass of an ideal monatomic gas is held at a constant pressure of 2.0×1052.0 \times 10^5 Pa. The gas expands from a volume of 5.0×10−35.0 \times 10^{-3} m³ to 8.0×10−38.0 \times 10^{-3} m³. During this process, 1500 J of heat is supplied to the gas. What is the change in the internal energy of the gas?

  1. 600 J
  2. 900 J (correct answer)
  3. 1500 J
  4. 2100 J
Explanation: First, calculate the work done by the gas during the isobaric (constant pressure) expansion: W = PΔV = (2.0 × 10⁵ Pa) × (8.0 × 10⁻³ m³ - 5.0 × 10⁻³ m³) = (2.0 × 10⁵) × (3.0 × 10⁻³) = 600 J. Next, apply the first law of thermodynamics, Q = ΔU + W. The change in internal energy is ΔU = Q - W = 1500 J - 600 J = 900 J.

Question 7

A fixed mass of an ideal gas is sealed in a rigid container. The container is placed in a warm bath, causing the gas temperature to increase. Which statement correctly describes the work done by the gas (W) and the change in its internal energy (ΔU)?

  1. W > 0 and ΔU > 0
  2. W = 0 and ΔU > 0 (correct answer)
  3. W = 0 and ΔU = 0
  4. W < 0 and ΔU > 0
Explanation: The container is rigid, so its volume is constant (an isovolumetric process). The work done by a gas is given by W = PΔV. Since ΔV = 0, the work done (W) is zero. Heat is transferred to the gas from the warm bath, increasing its temperature. For an ideal gas, internal energy is directly proportional to temperature, so if the temperature increases, the internal energy must also increase (ΔU > 0).

Question 8

A real heat engine has an efficiency of 35%. It operates between thermal reservoirs at 600 K and 300 K. The maximum possible efficiency for an engine operating between these reservoirs is 50%. What accounts for the difference between the actual and maximum efficiencies?

  1. The first law of thermodynamics is only an approximation for real engines.
  2. The second law of thermodynamics does not apply to the non-ideal gases used in real engines.
  3. Irreversible processes such as friction and uncontrolled heat loss occur in the real engine. (correct answer)
  4. The real engine operates at a faster rate than the ideal engine, reducing its efficiency.
Explanation: The maximum (Carnot) efficiency assumes all processes in the engine's cycle are perfectly reversible. Real engines always involve irreversible processes like friction, turbulence in the working fluid, and heat transfer across finite temperature differences. These processes generate entropy and dissipate energy, reducing the actual efficiency below the theoretical maximum.

Question 9

A Carnot engine performs 2.5×1032.5 \times 10^3 J of work per cycle while operating between a hot reservoir at 600 K and a cold reservoir at 350 K. What is the heat absorbed from the hot reservoir, Q_H, in one cycle?

  1. 2.5×1032.5 \times 10^3 J
  2. 3.5×1033.5 \times 10^3 J
  3. 4.3×1034.3 \times 10^3 J
  4. 6.0×1036.0 \times 10^3 J (correct answer)
Explanation: First, calculate the efficiency of the Carnot engine: η = 1 - T_C/T_H = 1 - 350 K / 600 K = 1 - 7/12 = 5/12. The efficiency of a heat engine is also defined as the ratio of work output to heat input, η = W/Q_H. Rearranging for Q_H gives Q_H = W/η = (2.5 × 10³ J) / (5/12) = (2.5 × 10³ J) × (12/5) = 6.0 × 10³ J.

Question 10

A hot block of metal and a cold block of metal are placed in thermal contact within a perfectly insulated box. They eventually reach a common equilibrium temperature. Why is this process considered thermodynamically irreversible?

  1. The total energy of the two-block system decreases as they reach equilibrium.
  2. The reverse process, where the block at equilibrium spontaneously separates into hot and cold parts, violates the first law of thermodynamics.
  3. The reverse process, though conserving energy, would result in a large decrease in total entropy, which is statistically impossible. (correct answer)
  4. Work must be done to separate the blocks after they have reached equilibrium.
Explanation: The process is irreversible because it leads to a net increase in the total entropy of the system. While the hot block's entropy decreases, the cold block's entropy increases by a larger amount, leading to a net increase in the entropy of the universe. The second law of thermodynamics states that the total entropy of an isolated system never decreases. The reverse process (spontaneous un-mixing of thermal energy) is not forbidden by energy conservation (the first law) but is statistically so improbable as to be considered impossible.

Question 11

An ideal gas engine operates between two thermal reservoirs at temperatures Th=500 KT_h = 500\text{ K} and Tc=300 KT_c = 300\text{ K}. In each cycle, the engine absorbs Qh=800 JQ_h = 800\text{ J} from the hot reservoir. If the actual efficiency of this engine is 75% of the theoretical maximum efficiency, what is the work output per cycle?

  1. 150 J, calculated using 75% directly applied to the temperature difference ratio
  2. 240 J, calculated using the actual efficiency of 30% applied to the absorbed heat (correct answer)
  3. 300 J, calculated by applying 75% efficiency directly to the maximum theoretical work
  4. 320 J, calculated using the Carnot efficiency of 40% applied to the absorbed heat
Explanation: The Carnot (theoretical maximum) efficiency is ηCarnot=1−Tc/Th=1−300/500=0.4=40%\eta_{Carnot} = 1 - T_c/T_h = 1 - 300/500 = 0.4 = 40\%. The actual efficiency is ηactual=0.75×0.4=0.3=30%\eta_{actual} = 0.75 \times 0.4 = 0.3 = 30\%. Therefore, the work output is W=ηactual×Qh=0.3×800 J=240 JW = \eta_{actual} \times Q_h = 0.3 \times 800\text{ J} = 240\text{ J}. Choice A incorrectly applies 75% to temperature ratios, C applies 75% to theoretical work instead of efficiency, and D uses the Carnot efficiency directly without the 75% reduction factor.

Question 12

A Carnot refrigerator operates between a cold reservoir at 5°C5°\text{C} and a hot reservoir at 35°C35°\text{C}. If the refrigerator removes 1200 J1200\text{ J} of heat from the cold reservoir in one cycle, what is the heat delivered to the hot reservoir?

  1. 1329 J, calculated using the Carnot efficiency and energy conservation principles (correct answer)
  2. 1450 J, calculated by applying the temperature ratio to the extracted heat directly
  3. 1500 J, calculated using the coefficient of performance for the refrigeration cycle
  4. 1800 J, calculated by using the temperature difference to determine heat transfer
Explanation: Convert to Kelvin: Tc=5+273=278 KT_c = 5 + 273 = 278\text{ K} and Th=35+273=308 KT_h = 35 + 273 = 308\text{ K}. For a Carnot refrigerator, Qh/Qc=Th/TcQ_h/Q_c = T_h/T_c, so Qh=Qc(Th/Tc)=1200×(308/278)=1200×1.108=1329 JQ_h = Q_c(T_h/T_c) = 1200 \times (308/278) = 1200 \times 1.108 = 1329\text{ J}. Choice B uses an incorrect temperature ratio application, C misapplies the COP formula, and D uses temperature differences instead of ratios. The correct answer uses the fundamental Carnot relationship between heat transfers and absolute temperatures.

Question 13

A monatomic ideal gas undergoes a cyclic process consisting of three stages: isothermal expansion at temperature T1=400 KT_1 = 400\text{ K} from volume V1=2.0 LV_1 = 2.0\text{ L} to V2=6.0 LV_2 = 6.0\text{ L}, followed by isobaric cooling to temperature T3T_3, and finally isochoric heating back to the initial state. What is the temperature T3T_3 at the end of the isobaric process?

  1. 133 K, because the isobaric process reduces temperature proportionally to the volume ratio (correct answer)
  2. 200 K, because the average temperature during cooling equals half the initial temperature
  3. 267 K, because the isochoric process requires this temperature to close the cycle
  4. 300 K, because the isobaric cooling reduces temperature by the expansion factor
Explanation: For the isothermal expansion, P2V2=P1V1P_2V_2 = P_1V_1, so P2=P1(V1/V2)=P1(2.0/6.0)=P1/3P_2 = P_1(V_1/V_2) = P_1(2.0/6.0) = P_1/3. For the isobaric cooling from state 2 to state 3, V3/T3=V2/T1V_3/T_3 = V_2/T_1, and since the final isochoric process returns to V1V_1, we have V3=V1=2.0 LV_3 = V_1 = 2.0\text{ L}. Therefore: T3=T1(V3/V2)=400 K(2.0/6.0)=133 KT_3 = T_1(V_3/V_2) = 400\text{ K}(2.0/6.0) = 133\text{ K}. Choice B incorrectly assumes simple averaging, C uses circular reasoning, and D applies an incorrect expansion factor.

Question 14

An ideal gas undergoes a thermodynamic cycle. Which statement is always true for the entire cycle?

  1. The net heat supplied to the gas equals the net work done by the gas. (correct answer)
  2. The change in internal energy of the gas is positive if the cycle is clockwise on a P-V diagram.
  3. The net work done by the gas is zero because the gas returns to its initial state.
  4. The net heat supplied to the gas is zero because temperature is a state function.
Explanation: For any complete thermodynamic cycle, the system returns to its initial state. Since internal energy (U) is a state function, the net change in internal energy (ΔU) over one cycle is zero. According to the first law of thermodynamics, Q = ΔU + W. If ΔU = 0, it follows that the net heat supplied (Q_net) must equal the net work done (W_net).

Question 15

A heat engine operates between a hot reservoir at 500 °C and a cold reservoir at 20 °C. What is the maximum possible theoretical efficiency of this engine?

  1. 38%
  2. 62% (correct answer)
  3. 87%
  4. 96%
Explanation: The maximum possible efficiency is the Carnot efficiency, η_Carnot = 1 - (T_C / T_H). Temperatures must be converted to Kelvin. T_H = 500 + 273 = 773 K. T_C = 20 + 273 = 293 K. η_Carnot = 1 - (293 / 773) ≈ 1 - 0.379 = 0.621, or 62.1%.

Question 16

An ideal gas is taken through a clockwise rectangular cycle on a P-V diagram. What is the net heat transferred to the gas during one full cycle?

  1. Zero, because the internal energy change is zero.
  2. Negative, because heat must be exhausted to complete the cycle.
  3. The sum of the heat added during the expansion and heating stages.
  4. The area enclosed by the rectangular path on the P-V diagram. (correct answer)
Explanation: For any complete cycle, the change in internal energy (ΔU) is zero. The first law of thermodynamics states Q_net = ΔU + W_net. With ΔU = 0, Q_net = W_net. The net work done by the gas (W_net) in a cycle is the area enclosed by the path on the P-V diagram. Therefore, the net heat transferred to the gas is also equal to this enclosed area.

Question 17

An ideal gas is compressed at constant temperature. What are the signs of the heat transfer Q (to the gas), the change in internal energy ΔU, and the work done W (by the gas)?

  1. Q > 0, ΔU = 0, W > 0
  2. Q < 0, ΔU < 0, W = 0
  3. Q = 0, ΔU > 0, W < 0
  4. Q < 0, ΔU = 0, W < 0 (correct answer)
Explanation: The process is isothermal (constant temperature). For an ideal gas, internal energy depends only on temperature, so ΔU = 0. The gas is compressed, so its volume decreases, and work is done on the gas. This means the work done by the gas, W, is negative. From the first law, Q = ΔU + W = 0 + W. Since W is negative, Q must also be negative, meaning heat flows out of the gas.

Question 18

A gas in a cylinder is rapidly compressed by a piston. The process is approximately adiabatic. What are the resulting changes in the internal energy and temperature of the gas?

  1. Internal energy increases and temperature increases. (correct answer)
  2. Internal energy is constant and temperature is constant.
  3. Internal energy decreases and temperature decreases.
  4. Internal energy increases and temperature is constant.
Explanation: The process is adiabatic, so no heat is exchanged (Q = 0). The gas is compressed, so work is done on the gas, meaning the work done by the gas (W) is negative. From the first law, ΔU = Q - W = 0 - W. Since W is negative, ΔU is positive. The internal energy of an ideal gas is proportional to its absolute temperature, so an increase in internal energy corresponds to an increase in temperature.

Question 19

The specific latent heat of fusion for water is 3.3×1053.3 \times 10^5 J kg⁻¹. What is the approximate change in entropy when 0.50 kg of ice at 0 °C melts to become water at 0 °C?

  1. 0 J K⁻¹
  2. 600 J K⁻¹ (correct answer)
  3. 1200 J K⁻¹
  4. 1.7×1051.7 \times 10^5 J K⁻¹
Explanation: First, calculate the heat absorbed during melting: Q = mL = (0.50 kg) × (3.3 × 10⁵ J kg⁻¹) = 1.65 × 10⁵ J. The process occurs at a constant temperature T = 0 °C = 273 K. The change in entropy is ΔS = Q/T = (1.65 × 10⁵ J) / 273 K ≈ 604 J K⁻¹, which is approximately 600 J K⁻¹.

Question 20

An ideal gas expands from an initial state (P₁, V₁) to the same final volume V₂ via two different paths: one isothermal and one adiabatic. Which statement correctly compares the two processes?

  1. The work done by the gas is greater in the adiabatic expansion because no energy is lost as heat.
  2. The final temperature is lower in the adiabatic expansion because internal energy is used to perform work. (correct answer)
  3. The change in internal energy is zero for both expansions because the gas is ideal.
  4. More heat is supplied to the gas during the adiabatic expansion than the isothermal expansion.
Explanation: In an adiabatic expansion, Q = 0, so the work done by the gas comes from its internal energy (ΔU = -W). This causes the internal energy and thus the temperature to decrease. In an isothermal expansion, ΔT = 0, so ΔU = 0, and heat must be supplied to equal the work done (Q = W). Because the adiabatic path results in a lower temperature at every point during the expansion (after the start), the pressure is also lower. Therefore, the area under the P-V curve (work done) is smaller for the adiabatic case, and its final temperature is lower.