Understand Thermodynamics - IB Physics

1. Which change to a thermodynamic cycle represented on a P-V diagram must result in an increase in the net work done by the gas per cycle?
   Answer: Modifying the cycle's path to enclose a larger area on the diagram. The net work done by a gas in a thermodynamic cycle is numerically equal to the area enclosed by the path of the cycle on a P-V diagram. To increase the net work done by the gas, the area enclosed by the loop must be increased. Other changes may or may not increase the work, but increasing the enclosed area guarantees it.
2. An ideal gas engine operates between two thermal reservoirs at temperatures $T_h = 500\text{ K}$ and $T_c = 300\text{ K}$. In each cycle, the engine absorbs $Q_h = 800\text{ J}$ 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?
   Answer: 240 J, calculated using the actual efficiency of 30% applied to the absorbed heat The Carnot (theoretical maximum) efficiency is $\eta_{Carnot} = 1 - T_c/T_h = 1 - 300/500 = 0.4 = 40\%$. The actual efficiency is $\eta_{actual} = 0.75 \times 0.4 = 0.3 = 30\%$. Therefore, the work output is $W = \eta_{actual} \times Q_h = 0.3 \times 800\text{ J} = 240\text{ J}$. 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.
3. A Carnot refrigerator operates between a cold reservoir at $5°\text{C}$ and a hot reservoir at $35°\text{C}$. If the refrigerator removes $1200\text{ J}$ of heat from the cold reservoir in one cycle, what is the heat delivered to the hot reservoir?
   Answer: 1329 J, calculated using the Carnot efficiency and energy conservation principles Convert to Kelvin: $T_c = 5 + 273 = 278\text{ K}$ and $T_h = 35 + 273 = 308\text{ K}$. For a Carnot refrigerator, $Q_h/Q_c = T_h/T_c$, so $Q_h = Q_c(T_h/T_c) = 1200 \times (308/278) = 1200 \times 1.108 = 1329\text{ J}$. 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.
4. 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?
   Answer: ΔU = 0, W = 0 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.
5. 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?
   Answer: The area enclosed by the rectangular path on the P-V diagram. 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.
6. A fixed mass of an ideal monatomic gas is held at a constant pressure of $2.0 \times 10^5$ Pa. The gas expands from a volume of $5.0 \times 10^{-3}$ m³ to $8.0 \times 10^{-3}$ m³. During this process, 1500 J of heat is supplied to the gas. What is the change in the internal energy of the gas?
   Answer: 900 J 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.
7. An ideal gas undergoes a thermodynamic cycle. Which statement is always true for the entire cycle?
   Answer: The net heat supplied to the gas equals the net work done by the gas. 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).
8. 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?
   Answer: The Clausius statement, which forbids spontaneous heat flow from cold to hot. 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.
9. 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?
   Answer: The temperature of the gas is constant, and the entropy of the gas increases. 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.
10. An inventor claims to have built an engine that takes in 1000 J of heat from a single thermal reservoir, performs 1000 J of net work, and completes a cycle. Why is this impossible?
    Answer: It violates the second law of thermodynamics, as a 100% efficient engine operating in a cycle is forbidden. This claim describes a heat engine with 100% efficiency (η = W/$Q_H$ = 1000/1000 = 1). 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. Some heat must be rejected to a colder reservoir. Therefore, the inventor's claim is impossible.
11. 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?
    Answer: Irreversible processes such as friction and uncontrolled heat loss occur in the real engine. 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.
12. The specific latent heat of fusion for water is $3.3 \times 10^5$ 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?
    Answer: 600 J K⁻¹ 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.
13. 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?
    Answer: The final temperature is lower in the adiabatic expansion because internal energy is used to perform work. 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.
14. 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?
    Answer: ΔS_sys > 0, ΔS_sur < 0, ΔS_uni > 0 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.
15. 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?
    Answer: The reverse process, though conserving energy, would result in a large decrease in total entropy, which is statistically impossible. 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.
16. 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?
    Answer: 62% 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%.
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)?
    Answer: Q < 0, ΔU = 0, W < 0 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.
18. 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?
    Answer: It increases, because the number of possible molecular arrangements increases. 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.
19. 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?
    Answer: Internal energy increases and temperature increases. 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.
20. A monatomic ideal gas undergoes a cyclic process consisting of three stages: isothermal expansion at temperature $T_1 = 400\text{ K}$ from volume $V_1 = 2.0\text{ L}$ to $V_2 = 6.0\text{ L}$, followed by isobaric cooling to temperature $T_3$, and finally isochoric heating back to the initial state. What is the temperature $T_3$ at the end of the isobaric process?
    Answer: 133 K, because the isobaric process reduces temperature proportionally to the volume ratio For the isothermal expansion, $P_2V_2 = P_1V_1$, so $P_2 = P_1(V_1/V_2) = P_1(2.0/6.0) = P_1/3$. For the isobaric cooling from state 2 to state 3, $V_3/T_3 = V_2/T_1$, and since the final isochoric process returns to $V_1$, we have $V_3 = V_1 = 2.0\text{ L}$. Therefore: $T_3 = T_1(V_3/V_2) = 400\text{ K}(2.0/6.0) = 133\text{ K}$. Choice B incorrectly assumes simple averaging, C uses circular reasoning, and D applies an incorrect expansion factor.

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