Thermodynamics HL Papers 1& 2

B.4 — Thermodynamics
HL Practice: Paper 2

15 extended-response questions • 1 lab investigation • Expandable full solutions

This original IB-style practice bank covers the first and second laws, gas work, internal energy, thermodynamic processes, pressure–volume cycles, entropy, and heat engines. The first seven questions retain and develop the supplied examples. The later questions add quantitative adiabatic changes, Carnot-cycle analysis, microscopic entropy, uncertainty, and experimental evaluation.

Instructions, constants, and signs

Show full working, units, and explanations. Attempt each question before opening its solution.
Use $R=8.31\,\mathrm{J\,mol^{-1}\,K^{-1}}$ and $k_B=1.38\times10^{-23}\,\mathrm{J\,K^{-1}}$ .
Use $Q=\Delta U+W$ . Heat supplied to the gas is positive; work done by the gas is positive. Heat rejected and work done on the gas carry negative signs in this equation.
In heat-engine equations, $Q_h$ and $Q_c$ are positive magnitudes of absorbed and rejected heat. Therefore $Q_{\mathrm{net}}=Q_h-Q_c$ .
Assume closed ideal-gas systems unless stated otherwise. All quantitative gas internal-energy calculations here use monatomic gases.
Use kelvin for temperature ratios, Carnot efficiency, and entropy calculations. Give final calculated results to three significant figures unless uncertainty dictates otherwise.
Bracketed marks are suggested practice marks. This bank is not a complete official examination paper.

Key relationships and conditions

$\displaystyle Q=\Delta U+W,\qquad\Delta U=\frac32nR\Delta T$ For constant pressure during a quasistatic change: $\displaystyle W=p\Delta V$ For a reversible adiabatic change of a monatomic ideal gas: $\displaystyle pV^{5/3}=\mathrm{constant}$ For a cyclic heat engine: $\displaystyle \eta=\frac{W_{\mathrm{net}}}{Q_h}=1-\frac{Q_c}{Q_h},\qquad\eta_{\mathrm{Carnot}}=1-\frac{T_c}{T_h}$ Entropy relations: $\displaystyle \Delta S=\frac{Q_{\mathrm{rev}}}{T}\quad\text{for a reversible isothermal transfer},\qquad S=k_B\ln\Omega$ The entropy of an isolated system cannot decrease. Reversible processes leave its total entropy unchanged; irreversible processes increase it.

Extended-response questions

Question 1 — First law and signs [6 marks]

(a) State the first law of thermodynamics and define the sign convention used for heat and work. [2 marks]

(b) A gas absorbs $500\,\mathrm{J}$ of heat and does $200\,\mathrm{J}$ of work. Calculate its change in internal energy. [2 marks]

(c) In a separate process, the gas loses $70\,\mathrm{J}$ of heat while $120\,\mathrm{J}$ of work is done on it. Calculate its change in internal energy. [2 marks]

Show full solution and marking guidance

(a) [2 marks] The first law expresses conservation of energy for a closed system: $\displaystyle Q=\Delta U+W\quad\text{or}\quad\Delta U=Q-W$ Here $Q>0$ means heat supplied to the system, and $W>0$ means work done by the system.

(b) [2 marks] $\displaystyle \Delta U=500-200=+300\,\mathrm{J}$ Internal energy increases.

(c) [2 marks] Here $Q=-70\,\mathrm{J}$ and $W=-120\,\mathrm{J}$ . $\displaystyle \Delta U=-70-(-120)=+50.0\,\mathrm{J}$

Question 2 — Work from a pressure–volume graph [7 marks]

(a) A fixed gas sample expands quasistatically at constant pressure $1.2\times10^5\,\mathrm{Pa}$ from $1.0\times10^{-3}\,\mathrm{m^3}$ to $4.0\times10^{-3}\,\mathrm{m^3}$ . Calculate work done by the gas. [2 marks]

(b) Explain its representation on a pressure–volume graph and state the work for a constant-volume change. [2 marks]

(c) If the gas is monatomic and ideal, calculate the change in internal energy and the heat supplied during the expansion. [3 marks]

Show full solution and marking guidance

(a) [2 marks] $\displaystyle W=p\Delta V=(1.2\times10^5)(3.0\times10^{-3})=360\,\mathrm{J}$

(b) [2 marks] With $p$ vertical and $V$ horizontal, work is the signed area under the path. Expansion gives positive work. A constant-volume path has $\Delta V=0$ , so boundary work is zero.

(c) [3 marks] Use $U=\frac32nRT=\frac32pV$ : $\displaystyle \Delta U=\frac32p\Delta V=540\,\mathrm{J}$ $\displaystyle Q=\Delta U+W=540+360=900\,\mathrm{J}$

Question 3 — Isothermal expansion [8 marks]

(a) A fixed ideal gas undergoes isothermal expansion. Explain its internal-energy change. [2 marks]

(b) If it absorbs $600\,\mathrm{J}$ of heat, calculate the work done by the gas. [2 marks]

(c) Describe how an expansion can be made approximately isothermal. [2 marks]

(d) If the expansion is reversible at $300\,\mathrm{K}$ , calculate the entropy changes of the gas and its heat reservoir. [2 marks]

Show full solution and marking guidance

(a) [2 marks] Isothermal means temperature stays constant. For a fixed amount and composition of ideal gas, internal energy depends only on temperature, so $\Delta U=0$ .

(b) [2 marks] $\displaystyle W=Q-\Delta U=600-0=600\,\mathrm{J}$

(c) [2 marks] Expand slowly with good thermal contact to a large constant-temperature reservoir. Heat supplied compensates for energy transferred out as work.

(d) [2 marks] $\displaystyle \Delta S_{\mathrm{gas}}=\frac{Q_{\mathrm{rev}}}{T}=\frac{600}{300}=+2.00\,\mathrm{J\,K^{-1}}$ The reservoir loses the same heat at the same temperature, giving $\Delta S_{\mathrm{reservoir}}=-2.00\,\mathrm{J\,K^{-1}}$ . The total entropy change is zero.

Question 4 — Adiabatic compression [6 marks]

(a) A gas is compressed adiabatically, and $250\,\mathrm{J}$ of work is done on it. Calculate its change in internal energy. [2 marks]

(b) Explain the temperature change for a fixed ideal gas. [2 marks]

(c) If the gas is monatomic, contains $0.100\,\mathrm{mol}$ , and starts at $300\,\mathrm{K}$ , calculate its final temperature. [2 marks]

Show full solution and marking guidance

(a) [2 marks] Adiabatic means $Q=0$ ; work done by the gas is $W=-250\,\mathrm{J}$ . $\displaystyle \Delta U=0-(-250)=+250\,\mathrm{J}$

(b) [2 marks] Work transfers energy to the gas. Its internal energy and mean translational kinetic energy increase, so its temperature rises.

(c) [2 marks] $\displaystyle \Delta T=\frac{250}{(3/2)(0.100)(8.31)}\approx201\,\mathrm{K}$ $\displaystyle T_f=300+200.562\approx501\,\mathrm{K}$

Question 5 — Heat-engine efficiency and power [7 marks]

(a) A cyclic heat engine absorbs $1200\,\mathrm{J}$ from a hot reservoir and rejects $800\,\mathrm{J}$ to a cold reservoir each cycle. Calculate its net work output. [2 marks]

(b) Calculate its efficiency. [2 marks]

(c) It completes $25.0$ cycles per second. Calculate its mechanical output power and its heat-input rate. [2 marks]

(d) Explain why the working substance has zero net internal-energy change over a full cycle. [1 mark]

Show full solution and marking guidance

(a) [2 marks] For a cycle, $\Delta U=0$ : $\displaystyle W_{\mathrm{net}}=Q_h-Q_c=1200-800=400\,\mathrm{J}$ Here $Q_h$ and $Q_c$ are positive magnitudes.

(b) [2 marks] $\displaystyle \eta=\frac{W_{\mathrm{net}}}{Q_h}=\frac{400}{1200}=0.333=33.3\%$

(c) [2 marks] $\displaystyle P_{\mathrm{out}}=(400)(25.0)=1.00\times10^4\,\mathrm{W}$ $\displaystyle P_{\mathrm{heat\ input}}=(1200)(25.0)=3.00\times10^4\,\mathrm{W}$

(d) [1 mark] The working substance returns to its original thermodynamic state. Internal energy is a state function, so the net change is zero.

Question 6 — Carnot limit and reservoir entropy [7 marks]

(a) A heat engine operates between reservoirs at $600\,\mathrm{K}$ and $300\,\mathrm{K}$ . Calculate the maximum theoretical efficiency. [2 marks]

(b) An actual cyclic engine between these reservoirs absorbs $1200\,\mathrm{J}$ and rejects $800\,\mathrm{J}$ per cycle. Compare its efficiency with the limit. [2 marks]

(c) Calculate the total entropy change of the two reservoirs per cycle and explain its significance. [3 marks]

Show full solution and marking guidance

(a) [2 marks] $\displaystyle \eta_{\mathrm{Carnot}}=1-\frac{T_c}{T_h}=1-\frac{300}{600}=0.500=50.0\%$ Use absolute temperatures.

(b) [2 marks] The actual efficiency is $(1200-800)/1200=33.3\%$ , below the $50.0\%$ reversible limit.

(c) [3 marks] $\displaystyle \Delta S_h=-\frac{1200}{600}=-2.00\,\mathrm{J\,K^{-1}}$ $\displaystyle \Delta S_c=\frac{800}{300}=2.67\,\mathrm{J\,K^{-1}}$ $\displaystyle \Delta S_{\mathrm{total}}=+0.667\,\mathrm{J\,K^{-1}}$ The cyclic working substance has zero net entropy change. Positive total entropy production is consistent with an irreversible engine operating below the Carnot limit.

Question 7 — Rectangular cycle and direction [7 marks]

(a) A gas follows a clockwise rectangular cycle on a pressure–volume diagram. Its pressure difference is $2.0\times10^5\,\mathrm{Pa}$ and its volume difference is $2.0\times10^{-3}\,\mathrm{m^3}$ . Calculate the net work done by the gas. [2 marks]

(b) Determine the work if the same cycle is traversed anticlockwise. [1 mark]

(c) State the net heat transfer to the gas in the clockwise cycle and explain why. [2 marks]

(d) If the clockwise engine absorbs a total of $1400\,\mathrm{J}$ per cycle, calculate its efficiency and rejected heat. [2 marks]

Show full solution and marking guidance

(a) [2 marks] Net work is the signed enclosed area: $\displaystyle W_{\mathrm{net}}=\Delta p\,\Delta V=(2.0\times10^5)(2.0\times10^{-3})=+400\,\mathrm{J}$ The rectangle’s height is a pressure difference, not a single branch pressure.

(b) [1 mark] Anticlockwise traversal reverses the sign: $W_{\mathrm{net}}=-400\,\mathrm{J}$ , so $400\,\mathrm{J}$ of net work is supplied to the gas.

(c) [2 marks] For the clockwise cycle, $\Delta U_{\mathrm{cycle}}=0$ , so $Q_{\mathrm{net}}=W_{\mathrm{net}}=+400\,\mathrm{J}$ .

(d) [2 marks] $\displaystyle \eta=\frac{400}{1400}\approx28.6\%$ $\displaystyle Q_c=1400-400=1000\,\mathrm{J}$

Question 8 — Constant-volume heating [7 marks]

(a) A rigid sealed container holds $0.500\,\mathrm{mol}$ of a monatomic ideal gas. Its temperature rises from $300\,\mathrm{K}$ to $450\,\mathrm{K}$ . Determine the work done by the gas. [1 mark]

(b) Calculate the internal-energy increase and heat supplied to the gas. [3 marks]

(c) Determine the final-to-initial pressure ratio. [1 mark]

(d) Explain why the gas entropy change for the whole heating process cannot generally be calculated by substituting the total heat into $\Delta S=Q/T$ with one gas temperature. [2 marks]

Show full solution and marking guidance

(a) [1 mark] Volume is constant, so $W=0$ .

(b) [3 marks] $\displaystyle \Delta U=\frac32nR\Delta T=\frac32(0.500)(8.31)(150)\approx935\,\mathrm{J}$ $\displaystyle Q=\Delta U+W=935\,\mathrm{J}$

(c) [1 mark] At fixed $n,V$ , $p\propto T$ : $\displaystyle \frac{p_f}{p_i}=\frac{450}{300}=1.50$

(d) [2 marks] The gas temperature changes during heating. The expression $\Delta S=Q_{\mathrm{rev}}/T$ with one temperature applies to a reversible isothermal transfer. A changing-temperature process requires the entropy contributions to be evaluated along a reversible path with the appropriate temperatures.

Question 9 — Reversible adiabatic expansion [9 marks]

(a) A fixed monatomic ideal gas expands reversibly and adiabatically from $p_1=4.00\times10^5\,\mathrm{Pa}$ , $V_1=2.00\times10^{-3}\,\mathrm{m^3}$ , and $T_1=400\,\mathrm{K}$ to $V_2=4.00\times10^{-3}\,\mathrm{m^3}$ . Calculate $p_2$ using $pV^{5/3}=\mathrm{constant}$ . [2 marks]

(b) Calculate the final temperature. [2 marks]

(c) Calculate the work done by the gas using its internal-energy change. [3 marks]

(d) Compare the final pressure with an isothermal expansion from the same initial state to the same final volume. Explain which curve falls more steeply. [2 marks]

Show full solution and marking guidance

(a) [2 marks] $\displaystyle p_2=p_1\left(\frac{V_1}{V_2}\right)^{5/3}=(4.00\times10^5)\left(\frac12\right)^{5/3}\approx1.26\times10^5\,\mathrm{Pa}$

(b) [2 marks] Combine $pV=nRT$ with the adiabatic relation: $\displaystyle T_2=T_1\left(\frac{V_1}{V_2}\right)^{2/3}=400\left(\frac12\right)^{2/3}\approx252\,\mathrm{K}$

(c) [3 marks] $\displaystyle \Delta U=\frac32(p_2V_2-p_1V_1)=\frac32(503.968-800)\approx-444\,\mathrm{J}$ Since $Q=0$ : $\displaystyle W=-\Delta U\approx+444\,\mathrm{J}$

(d) [2 marks] For the isothermal path, $p_2=p_1V_1/V_2=2.00\times10^5\,\mathrm{Pa}$ . The adiabatic pressure is lower because the gas also cools as it expands. From the same initial point, the adiabat falls more steeply than the isotherm.

Question 10 — Triangular cycle with a sloping path [10 marks]

(a) A monatomic ideal gas follows A → B → C → A, with states given below. The A–B and C–A paths are respectively constant-volume and constant-pressure; B–C is a straight line on the pressure–volume graph. Calculate the work on each branch. [3 marks]

States for the triangular cycle
State	Pressure / Pa	Volume / m³
A	1.00 × 10⁵	1.00 × 10⁻³
B	3.00 × 10⁵	1.00 × 10⁻³
C	1.00 × 10⁵	3.00 × 10⁻³

(b) Calculate the net work per cycle and identify the direction of traversal. [2 marks]

(c) Determine the net internal-energy change and net heat supplied over the cycle. [2 marks]

(d) B and C have the same temperature. Show that this does not make the straight-line B–C path isothermal by evaluating the pressure–volume product at its midpoint. [3 marks]

Show full solution and marking guidance

(a) [3 marks] $\displaystyle W_{AB}=0$ Along the straight sloping branch, use the trapezium area: $\displaystyle W_{BC}=\frac{p_B+p_C}{2}(V_C-V_B)=\frac{3.00\times10^5+1.00\times10^5}{2}(2.00\times10^{-3})=400\,\mathrm{J}$ $\displaystyle W_{CA}=(1.00\times10^5)(-2.00\times10^{-3})=-200\,\mathrm{J}$

(b) [2 marks] $\displaystyle W_{\mathrm{net}}=0+400-200=+200\,\mathrm{J}$ The cycle is clockwise; its triangular enclosed area is $\tfrac12\Delta p\Delta V=200\,\mathrm{J}$ .

(c) [2 marks] The gas returns to its starting state, so $\Delta U_{\mathrm{cycle}}=0$ . The first law gives $Q_{\mathrm{net}}=W_{\mathrm{net}}=+200\,\mathrm{J}$ .

(d) [3 marks] At both B and C, $pV=300\,\mathrm{J}$ , so their temperatures agree. At the midpoint, $p=2.00\times10^5\,\mathrm{Pa}$ and $V=2.00\times10^{-3}\,\mathrm{m^3}$ , giving $pV=400\,\mathrm{J}$ . The midpoint temperature is $4/3$ of the endpoint temperature. Temperature therefore varies along the straight path; an isotherm would have constant $pV$ .

Question 11 — The four stages of a Carnot cycle [10 marks]

(a) Describe the four successive stages of a clockwise Carnot heat-engine cycle, starting with heat absorption from the hot reservoir. [4 marks]

(b) A Carnot engine operates between $500\,\mathrm{K}$ and $300\,\mathrm{K}$ and absorbs $1000\,\mathrm{J}$ per cycle. Calculate its work output and rejected heat. [2 marks]

(c) Calculate the working gas’s entropy changes on the two isothermal stages and state the entropy changes on the reversible adiabatic stages. [2 marks]

(d) Explain why practical engines do not achieve this reversible limit. [2 marks]

Show full solution and marking guidance

(a) [4 marks] First: reversible isothermal expansion at $T_h$ , absorbing heat and doing work. Second: reversible adiabatic expansion, cooling from $T_h$ to $T_c$ . Third: reversible isothermal compression at $T_c$ , rejecting heat. Fourth: reversible adiabatic compression, warming from $T_c$ to $T_h$ and returning to the initial state.

(b) [2 marks] $\displaystyle \eta=1-\frac{300}{500}=0.400$ $\displaystyle W_{\mathrm{net}}=(0.400)(1000)=400\,\mathrm{J},\qquad Q_c=1000-400=600\,\mathrm{J}$

(c) [2 marks] $\displaystyle \Delta S_{\mathrm{hot\ stage}}=\frac{1000}{500}=+2.00\,\mathrm{J\,K^{-1}}$ $\displaystyle \Delta S_{\mathrm{cold\ stage}}=-\frac{600}{300}=-2.00\,\mathrm{J\,K^{-1}}$ Each reversible adiabatic stage has $\Delta S=0$ . Net gas entropy change over the cycle is zero.

(d) [2 marks] Real engines have entropy-producing effects such as friction, heat transfer across finite temperature differences, and non-quasistatic changes. The Carnot limit requires an ideal reversible cycle; it is an efficiency limit, not a guarantee of maximum output power.

Question 12 — Entropy and the direction of heat transfer [7 marks]

(a) $600\,\mathrm{J}$ transfers directly from a large reservoir at $400\,\mathrm{K}$ to one at $300\,\mathrm{K}$ . Calculate the entropy change of each reservoir and the total. [3 marks]

(b) Use the second law to explain why the exact reverse transfer cannot occur as the sole change in an isolated composite system. [2 marks]

(c) Explain why adiabatic conditions alone do not guarantee constant entropy. [2 marks]

Show full solution and marking guidance

(a) [3 marks] $\displaystyle \Delta S_h=-\frac{600}{400}=-1.50\,\mathrm{J\,K^{-1}}$ $\displaystyle \Delta S_c=\frac{600}{300}=+2.00\,\mathrm{J\,K^{-1}}$ $\displaystyle \Delta S_{\mathrm{total}}=+0.500\,\mathrm{J\,K^{-1}}$ The reservoirs remain at their stated temperatures.

(b) [2 marks] The second law requires the entropy of an isolated system not to decrease. Reversing this transfer with no other change would give $\Delta S_{\mathrm{total}}=-0.500\,\mathrm{J\,K^{-1}}$ , so it cannot occur spontaneously. A refrigerator can transfer energy from cold to hot when work and other associated changes are included.

(c) [2 marks] Adiabatic means no entropy transfer with heat, but irreversible processes can still generate entropy internally. Entropy is constant for a reversible adiabatic process; it can increase in an irreversible adiabatic process.

Question 13 — Entropy and microscopic states [8 marks]

(a) Explain the distinction between a macrostate and a microstate, and state Boltzmann’s entropy equation using multiplicity $\Omega$ . [3 marks]

(b) For a small model system, multiplicity changes from $2.00\times10^4$ to $8.00\times10^4$ . Calculate the entropy change using $k_B=1.38\times10^{-23}\,\mathrm{J\,K^{-1}}$ . [2 marks]

(c) If these are the only two possible macrostates and all compatible microstates are equally probable, calculate the probability of the higher-multiplicity macrostate. [1 mark]

(d) Explain why increasing entropy does not mean that energy has been destroyed. [2 marks]

Show full solution and marking guidance

(a) [3 marks] A macrostate is specified by macroscopic quantities such as temperature, pressure, and volume. A microstate is a particular microscopic arrangement compatible with a macrostate. If $\Omega$ counts compatible microstates: $\displaystyle S=k_B\ln\Omega$

(b) [2 marks] $\displaystyle \Delta S=k_B\ln\left(\frac{\Omega_2}{\Omega_1}\right)=(1.38\times10^{-23})\ln4\approx1.91\times10^{-23}\,\mathrm{J\,K^{-1}}$

(c) [1 mark] $\displaystyle P_{\mathrm{higher}}=\frac{8.00\times10^4}{2.00\times10^4+8.00\times10^4}=0.800$

(d) [2 marks] The first law still conserves energy. Entropy concerns the number of compatible microscopic arrangements and the constraints on energy transfer; entropy production can reduce the work obtainable from energy without destroying that energy.

Question 14 — Testing an impossible engine claim [8 marks]

(a) A proposed cyclic engine interacts only with reservoirs at $500\,\mathrm{K}$ and $300\,\mathrm{K}$ . It absorbs $1000\,\mathrm{J}$ and claims to produce $450\,\mathrm{J}$ of work per cycle. Calculate the rejected heat required by the first law. [2 marks]

(b) Compare its claimed efficiency with the Carnot limit. [2 marks]

(c) Calculate the total reservoir entropy change and determine whether the claim is physically possible. [3 marks]

(d) State the largest possible work output for that heat input and those reservoir temperatures. [1 mark]

Show full solution and marking guidance

(a) [2 marks] $\displaystyle Q_c=Q_h-W=1000-450=550\,\mathrm{J}$ This allocation satisfies energy conservation.

(b) [2 marks] $\displaystyle \eta_{\mathrm{claimed}}=\frac{450}{1000}=45.0\%$ $\displaystyle \eta_{\mathrm{Carnot}}=1-\frac{300}{500}=40.0\%$ The claim exceeds the reversible limit.

(c) [3 marks] $\displaystyle \Delta S_{\mathrm{total}}=-\frac{1000}{500}+\frac{550}{300}\approx-0.167\,\mathrm{J\,K^{-1}}$ The cyclic engine has no net entropy change, so the negative composite entropy change violates the second law. The claimed engine is impossible under the stated conditions, despite satisfying the first law.

(d) [1 mark] $\displaystyle W_{\max}=(0.400)(1000)=400\,\mathrm{J}$ It must reject at least $600\,\mathrm{J}$ per cycle.

Question 15 — Carnot efficiency, uncertainty, and design [9 marks]

(a) A heat engine uses reservoirs at $T_h=(600\pm5)\,\mathrm{K}$ and $T_c=(300\pm3)\,\mathrm{K}$ . Calculate the nominal Carnot efficiency. [2 marks]

(b) Estimate its maximum absolute uncertainty by adding fractional uncertainty contributions for the temperature ratio. State the efficiency with its uncertainty. [3 marks]

(c) The measured engine efficiency is $0.450\pm0.015$ . Use the quoted uncertainty bounds to decide whether it clearly exceeds the Carnot limit. [2 marks]

(d) With a cold reservoir at $300\,\mathrm{K}$ , find the minimum hot-reservoir temperature that could permit efficiency $0.600$ . Explain the equality case. [2 marks]

Show full solution and marking guidance

(a) [2 marks] $\displaystyle \eta_C=1-\frac{300}{600}=0.500$

(b) [3 marks] Let $r=T_c/T_h=0.500$ . $\displaystyle \frac{\delta r}{r}\approx\frac{3}{300}+\frac{5}{600}=0.01833$ $\displaystyle \delta\eta_C=\delta r\approx(0.500)(0.01833)=0.00917$ Thus $\eta_C=0.500\pm0.009$ , or approximately $(50.0\pm0.9)\%$ . The uncertainty in efficiency is not found by applying the ratio’s fractional uncertainty directly to $1-r$ in general.

(c) [2 marks] The measured efficiency lies from $0.435$ to $0.465$ . The estimated Carnot bound lies from about $0.491$ to $0.509$ . The measurement is clearly below the limit on these quoted bounds. This comparison alone does not validate every aspect of the experiment.

(d) [2 marks] $\displaystyle 0.600\leq1-\frac{300}{T_h}\quad\Rightarrow\quad T_h\geq750\,\mathrm{K}$ Equality would require a reversible Carnot engine. An irreversible engine reaching the same efficiency would require a hotter reservoir.

Lab investigation — Testing a constant-pressure energy balance

Investigation 1 — Heat, work, and the apparatus [15 marks]

A cylinder contains a sealed $0.0100\,\mathrm{mol}$ sample of monatomic ideal gas. A movable piston maintains a constant absolute gas pressure of $1.20\times10^5\,\mathrm{Pa}$ during slow heating. The gas and relevant apparatus are at a common temperature for each reading. Electrical energy is measured cumulatively from the initial state. Use the idealised data below; neglect the effect of measurement uncertainty in the numerical calculations.

Heating data at constant absolute pressure
Temperature / K	Gas volume / mL	Electrical energy supplied / J
300	207.75	0
320	221.60	35.0
340	235.45	70.0
360	249.30	105

(a) Describe an apparatus and procedure to obtain these data. Identify variables and important controls. [3 marks]

(b) Calculate the work done by the gas from 300 K to 360 K. [2 marks]

(c) Calculate the internal-energy change of the gas over this interval. [2 marks]

(d) Use the first law to determine the heat transferred to the gas. Compare it with the recorded electrical energy and explain why a difference need not violate energy conservation. [3 marks]

(e) Assume external heat loss is negligible and the remaining difference heats apparatus that shares the gas temperature. Use the energy–temperature gradient to estimate the apparatus heat capacity. [3 marks]

(f) Suggest two specific experimental improvements to identify or reduce the discrepancy. [2 marks]

Show investigation solution and evaluation

(a) [3 marks] Use a sealed cylinder containing a known amount of monatomic gas, a low-friction piston with a fixed load, a calibrated absolute-pressure sensor, a temperature probe, and an electrical heater. Record temperature and volume as measured electrical energy is supplied, using E = VIt for steady voltage and current or a calibrated energy meter. Electrical energy input is the controlled independent quantity; temperature and volume are measured responses. Maintain constant external pressure, gas amount, and ambient conditions. Heat slowly, check temperature uniformity and pressure equilibration, and test for leaks and piston friction. Repeat the measurements and respect apparatus ratings.

(b) [2 marks] $\displaystyle \Delta V=(249.30-207.75)\times10^{-6}=4.155\times10^{-5}\,\mathrm{m^3}$ $\displaystyle W=p\Delta V=(1.20\times10^5)(4.155\times10^{-5})\approx4.99\,\mathrm{J}$

(c) [2 marks] $\displaystyle \Delta U=\frac32nR\Delta T=\frac32(0.0100)(8.31)(60.0)\approx7.48\,\mathrm{J}$

(d) [3 marks] $\displaystyle Q_{\mathrm{gas}}=\Delta U+W=7.479+4.986=12.465\,\mathrm{J}\approx12.5\,\mathrm{J}$ Recorded electrical energy is $105\,\mathrm{J}$ , leaving $105-12.465\approx92.5\,\mathrm{J}$ for other transfers or stores. Electrical energy supplied to the whole apparatus is not automatically heat received by the gas; the cylinder, heater, and probe can warm, and energy can reach the surroundings.

(e) [3 marks] The observed gradient is $105/60.0=1.75\,\mathrm{J\,K^{-1}}$ . The gas’s constant-pressure heat capacity is: $\displaystyle C_{p,\mathrm{gas}}=\frac{Q_{\mathrm{gas}}}{\Delta T}=\frac52nR=0.20775\,\mathrm{J\,K^{-1}}$ Under the supplied negligible-loss assumption: $\displaystyle C_{\mathrm{apparatus}}=1.75-0.20775\approx1.54\,\mathrm{J\,K^{-1}}$ Heat loss would confound this estimate if it were not negligible.

(f) [2 marks] Calibrate the heat capacity of the cylinder, heater, and probe so their energy increase can be included separately. Improve insulation and quantify residual heat loss, for example with a calibrated cooling measurement. Other valid improvements include checking leaks, calibrating pressure and volume sensors, and reducing or measuring piston friction; credit two specific improvements relevant to the energy balance.

Exam tips and common pitfalls

Write the first law and assign the signs of $Q$ and $W$ before calculating.
For a fixed ideal gas, isothermal means $\Delta U=0$ ; adiabatic means $Q=0$ ; constant volume means no boundary work.
Use $p\Delta V$ only when pressure is constant. For a straight sloping path, use average pressure times volume change.
Cycle work is the signed enclosed area: clockwise positive, anticlockwise negative on the usual axes.
A complete cycle gives zero net change in internal energy and entropy of the working substance; total reservoir entropy can still increase.
Carnot efficiency is a reversible upper limit. Use kelvin and distinguish a limit from actual efficiency.
Do not calculate changing-temperature entropy changes using total heat divided by an arbitrary single gas temperature.
Adiabatic does not automatically mean isentropic; reversibility matters.
State the system boundary in experimental energy balances. Heater electrical energy can warm the apparatus as well as the gas.

Faith connection — Sustaining grace and the hope of renewal

Hebrews 1:1–3 presents Jesus as the One who sustains creation and brings purification for sins. In 1 John 1:7, His blood is described as cleansing us from sin; Revelation 21:3–5 points to God dwelling with His people and making all things new. These passages invite us to trust Jesus’ love, forgiveness, and promised restoration.

Thermodynamics helps us examine physical processes with honesty and care. As you learn, consider how gratitude, faithful use of resources, and integrity in your work can reflect your trust in God.

Further reading: OpenStax — The Carnot cycle; OpenStax — Entropy.

B.4 — Thermodynamics
HL Practice: Paper 2

Instructions, constants, and signs

Key relationships and conditions

Extended-response questions

Question 1 — First law and signs [6 marks]

Question 2 — Work from a pressure–volume graph [7 marks]

Question 3 — Isothermal expansion [8 marks]

Question 4 — Adiabatic compression [6 marks]

Question 5 — Heat-engine efficiency and power [7 marks]

Question 6 — Carnot limit and reservoir entropy [7 marks]

Question 7 — Rectangular cycle and direction [7 marks]

Question 8 — Constant-volume heating [7 marks]

Question 9 — Reversible adiabatic expansion [9 marks]

Question 10 — Triangular cycle with a sloping path [10 marks]

Question 11 — The four stages of a Carnot cycle [10 marks]

Question 12 — Entropy and the direction of heat transfer [7 marks]

Question 13 — Entropy and microscopic states [8 marks]

Question 14 — Testing an impossible engine claim [8 marks]

Question 15 — Carnot efficiency, uncertainty, and design [9 marks]

Lab investigation — Testing a constant-pressure energy balance

Investigation 1 — Heat, work, and the apparatus [15 marks]

Exam tips and common pitfalls

Faith connection — Sustaining grace and the hope of renewal

Related

Company

Links

Support

Recommend

alphy school publishers

IB Physics HL Book 2025; Papers 1 & 2 Practice 5 Sections65 Lessons54 Weeks

Curriculum

Thermodynamics HL Papers 1& 2

B.4 — ThermodynamicsHL Practice: Paper 2

Instructions, constants, and signs

Key relationships and conditions

Extended-response questions

Question 1 — First law and signs [6 marks]

Question 2 — Work from a pressure–volume graph [7 marks]

Question 3 — Isothermal expansion [8 marks]

Question 4 — Adiabatic compression [6 marks]

Question 5 — Heat-engine efficiency and power [7 marks]

Question 6 — Carnot limit and reservoir entropy [7 marks]

Question 7 — Rectangular cycle and direction [7 marks]

Question 8 — Constant-volume heating [7 marks]

Question 9 — Reversible adiabatic expansion [9 marks]

Question 10 — Triangular cycle with a sloping path [10 marks]

Question 11 — The four stages of a Carnot cycle [10 marks]

Question 12 — Entropy and the direction of heat transfer [7 marks]

Question 13 — Entropy and microscopic states [8 marks]

Question 14 — Testing an impossible engine claim [8 marks]

Question 15 — Carnot efficiency, uncertainty, and design [9 marks]

Lab investigation — Testing a constant-pressure energy balance

Investigation 1 — Heat, work, and the apparatus [15 marks]

Exam tips and common pitfalls

Faith connection — Sustaining grace and the hope of renewal

Related

Leave A Reply Cancel reply

Company

Links

Support

Recommend

alphy school publishers

Customise Consent Preferences

Log in with your account

Modal title

B.4 — Thermodynamics
HL Practice: Paper 2