Apply Thermodynamics - IB Physics

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)?
Answer: Q = 0 J, W = -100 J, ΔU = +100 J 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.
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?
Answer: 0.40 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.
Two identical containers each hold $n$ moles of an ideal gas at the same temperature $T$ . 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?
Answer: $\left(\frac{1}{2}\right)^{4/15}$ For adiabatic processes, $TV^{\gamma-1} = \text{constant}$ . For monatomic gas (A): $\gamma_A = 5/3$ , so $T_{A,f}/T = (1/2)^{(5/3-1)} = (1/2)^{2/3}$ . For diatomic gas (B): $\gamma_B = 7/5$ , so $T_{B,f}/T = (1/2)^{(7/5-1)} = (1/2)^{2/5}$ . The ratio is $\frac{T_{A,f}}{T_{B,f}} = \frac{(1/2)^{2/3}}{(1/2)^{2/5}} = (1/2)^{2/3-2/5} = (1/2)^{10/15-6/15} = (1/2)^{4/15}$ . Choice A uses only monatomic exponent. Choice B uses only diatomic exponent. Choice D uses incorrect calculation.
An ideal gas undergoes a very slow compression at constant temperature. Which statement correctly describes the energy transfers for the gas during this process?
Answer: Work is done on the gas, and an equal amount of energy is removed from the gas as heat. 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.
An ideal gas is used as the working substance in a heat engine that operates in a thermodynamic cycle. Which of the following quantities is zero for the gas after one complete cycle?
Answer: The net change in internal energy of the gas. A thermodynamic cycle is a process in which the system returns to its initial state. Internal energy is a state function, meaning it depends only on the state of the system (P, V, T). Since the final state is identical to the initial state after a complete cycle, the net change in internal energy (ΔU) must be zero. The net work done and net heat supplied are non-zero (their difference is ΔU=0), and the heat expelled is also non-zero for any engine with efficiency less than 100%.
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?
Answer: 700 J 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.
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?
Answer: 0.40 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.
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?
Answer: An increase of 6.0 × 10² J K⁻¹. 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.
A Stirling engine operates with $0.1\text{ mol}$ of helium gas between volumes $V_1 = 100\text{ cm}^3$ and $V_2 = 400\text{ cm}^3$ , and temperatures $T_1 = 300\text{ K}$ and $T_2 = 600\text{ K}$ . The cycle consists of two isothermal and two isochoric processes. What is the ratio of heat absorbed to heat rejected during one complete cycle?
Answer: $2.0$ A Stirling cycle has four processes: isothermal expansion at $T_2$ , isochoric cooling, isothermal compression at $T_1$ , and isochoric heating. Heat absorbed occurs during: (1) isothermal expansion at $T_2$ : $Q_h = nRT_2 \ln(V_2/V_1) = (0.1)(8.314)(600)\ln(4) = 692\text{ J}$ , and (2) isochoric heating: $Q_{\text{heat}} = nC_V(T_2 - T_1) = (0.1)(\frac{3}{2})(8.314)(300) = 374\text{ J}$ . Heat rejected occurs during: (1) isothermal compression at $T_1$ : $Q_c = nRT_1 \ln(V_2/V_1) = (0.1)(8.314)(300)\ln(4) = 346\text{ J}$ , 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 = $692/346 = 2.0$ . Other ratios result from including/excluding regenerator effects incorrectly.
A container of liquid water is placed in a freezer, where it turns into ice. The entropy of the water decreases during this process. How is this consistent with the second law of thermodynamics?
Answer: The freezer transfers heat to the surroundings, causing a larger entropy increase in the surroundings. The second law states that the total entropy of an isolated system (or the universe) must increase or remain constant. The water is not an isolated system. The freezer is a heat pump that removes heat from the water and expels it into the warmer surroundings (the room). This transfer of heat causes the entropy of the surroundings to increase. The second law requires that this entropy increase in the surroundings must be greater than the entropy decrease of the water, so the total entropy of the universe increases.
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?
Answer: The difference Q - W must be identical for both processes. 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.
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?
Answer: $T_H$ / ( $T_H$ - $T_C$ ) 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 - $T_C$ / $T_H$ ) = $T_H$ / ( $T_H$ - $T_C$ ).
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?
Answer: It violates the second law of thermodynamics because some heat must be rejected to a colder reservoir. 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.
An ideal gas undergoes an isothermal expansion. Which of the following statements about the gas is correct?
Answer: The change in internal energy of the gas is zero. 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.
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?
Answer: 200 kJ 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.
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?
Answer: (5/2)PV 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.
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?
Answer: Positive, because heat transfer from hot to cold is an irreversible process. 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.
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?
Answer: The claim is plausible as it satisfies both the first and second laws of thermodynamics. 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.
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?
Answer: Work is done on the air, and this energy increases the internal energy of the air. 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.
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?
Answer: 1 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.

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