B.4 Thermodynamics (HL) – IB Physics exam questions with mark scheme

1HardHL only8 marks

0.10 mol of a monatomic ideal gas undergoes the cycle ABCA shown on the $p$–$V$ diagram. AB is at constant pressure, BC is at constant volume and CA is a straight line. $R=8.31\ \mathrm{J\,mol^{-1}\,K^{-1}}$.

(a) Calculate the work done by the gas during AB. [1]

(b) Show that the change in internal energy of the gas during AB is 600 J. [2]

(d) Determine the net work done by the gas in one cycle. [2]

(e) Thermal energy is transferred into the gas only during AB. Calculate the efficiency of the cycle. [2]

$W=p\Delta V = 2.0\times10^5\times(3.0-1.0)\times10^{-3} = 400$ J (400 J)(1)
$\Delta U=\tfrac32 nR\Delta T$ with $T_A=p_AV_A/nR=241$ K and $T_B=3T_A=722$ K (or $\Delta U=\tfrac32\Delta(pV)=\tfrac32\times(6.0\times10^2-2.0\times10^2)$ J)(1)
$\Delta U=\tfrac32\times0.10\times8.31\times481 = 600$ J (600 J)(1)
$Q=\Delta U+W = 600+400 = 1000$ J (allow ECF from (a))(1)
Net work = area enclosed by the cycle (triangle $\tfrac12\times\Delta V\times\Delta p$) (allow summing the work for each stage: $W_{AB}+W_{BC}+W_{CA}=400+0-300$ J)(1)
$W_{net}=\tfrac12\times(2.0\times10^{-3})\times(1.0\times10^5) = 100$ J (100 J, cycle is clockwise so positive)(1)
Efficiency = net work / thermal energy input(1)
$=100/1000 = 0.10$ (10%) (allow ECF from (c) and (d))(1)
Answer: (a) 400 J (b) 600 J (c) 1000 J (d) 100 J (e) 10%
Examiner tip: Net work in a cycle is the enclosed area: check the direction of the cycle for the sign.

2CoreHL only5 marks

A fixed mass of ideal gas in a cylinder is compressed by a piston. During the compression, $250\ \mathrm{J}$ of work is done on the gas and $90\ \mathrm{J}$ of thermal energy is transferred from the gas to the surroundings.

(a) Using the first law of thermodynamics, $Q=\Delta U+W$, determine the change in internal energy of the gas. [2]

(b) State and explain what happens to the temperature of the gas. [2]

(c) The gas is now compressed again, this time slowly, so that its temperature stays constant. State the relationship between the thermal energy $Q$ transferred and the work $W$ done by the gas during this process. [1]

(a) correct signs: $Q=-90\ \mathrm{J}$ and $W=-250\ \mathrm{J}$ (work done by the gas is negative)(1)
$\Delta U=Q-W=-90-(-250)=+160\ \mathrm{J}$ (an increase of 160 J); award [1 max] for $-160$ J or $340$ J(1)
(b) temperature increases(1)
internal energy of an ideal gas is (only) the random kinetic energy of its particles / $\Delta U=\tfrac32Nk_B\Delta T$, so $\Delta U>0$ means $\Delta T>0$; allow ECF from a negative $\Delta U$ in (a)(1)
(c) $Q=W$ (both negative) / $\Delta U=0$ so the thermal energy given out equals the work done on the gas; accept $|Q|=|W|$(1)
Answer: (a) $+160\ \mathrm{J}$ (b) rises, since $\Delta U\propto\Delta T$ (c) $\Delta U=0$ so $Q=W$
Examiner tip: In the IB form $Q=\Delta U+W$, $W$ is work done BY the gas, so work done on the gas enters as a negative number.

3CoreHL only4 marks

A block of ice of mass $0.250\ \mathrm{kg}$ at $0\ ^\circ\mathrm{C}$ is left in a room whose air is at $20\ ^\circ\mathrm{C}$. The ice melts completely, and the water produced is still at $0\ ^\circ\mathrm{C}$. Specific latent heat of fusion of ice $=3.34\times10^{5}\ \mathrm{J\,kg^{-1}}$. Assume the room temperature does not change.

(a) Calculate the change in entropy of the ice as it melts. [2]

(b) Calculate the change in entropy of the room air. [1]

(a) $Q=mL=0.250\times3.34\times10^{5}=8.35\times10^{4}\ \mathrm{J}$(1)
$\Delta S=\dfrac{Q}{T}=\dfrac{8.35\times10^{4}}{273}=+306\ \mathrm{J\,K^{-1}}$ (accept 305–306); award [0] for this mark if $T=0$ or 20 used(1)
(b) $\Delta S=-\dfrac{8.35\times10^{4}}{293}=-285\ \mathrm{J\,K^{-1}}$; sign required; allow ECF for $Q$ from (a)(1)
(c) total entropy change $=306-285=+21\ \mathrm{J\,K^{-1}}$ (accept 20–21) which is positive, so the entropy of the isolated system (ice + room) increases, as the second law requires; must quote a total(1)
Answer: (a) $+306\ \mathrm{J\,K^{-1}}$ (b) $-285\ \mathrm{J\,K^{-1}}$ (c) total $+21\ \mathrm{J\,K^{-1}}>0$
Examiner tip: Use kelvin temperatures in $\Delta S=\Delta Q/T$, and give the room's entropy change a minus sign because it loses energy.

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4StandardHL only6 marks

A model gas has six identical but distinguishable particles in a box divided into two equal halves, L and R. Each particle is equally likely to be in either half. A macrostate is described by how many particles are in L; a microstate says which particular particles are in L.

(a) Show that the macrostate “3 in L, 3 in R” has 20 microstates. [1]

(b) Calculate the entropy of the system when it is in the macrostate “3 in L, 3 in R”. $k_B=1.38\times10^{-23}\ \mathrm{J\,K^{-1}}$ [1]

(c) State the entropy of the macrostate “6 in L, 0 in R”, and explain why the system is more likely to be found in the 3–3 macrostate. [2]

(d) One mole of an ideal gas is initially confined to one half of an insulated container; the other half is a vacuum. The partition is removed and the gas expands to fill the whole container. Using $S=k_B\ln\Omega$, determine the change in entropy of the gas. $N_A=6.02\times10^{23}\ \mathrm{mol^{-1}}$ [2]

(a) $\Omega=\dfrac{6!}{3!\,3!}=20$ / $^6C_3=20$ / systematic listing of the 20 arrangements(1)
(b) $S=k_B\ln\Omega=1.38\times10^{-23}\times\ln20=4.1\times10^{-23}\ \mathrm{J\,K^{-1}}$ (accept $4.13\times10^{-23}$)(1)
(c) $\Omega=1$ so $S=0$(1)
all 64 microstates are equally likely, and the 3–3 macrostate has the most microstates (20 out of 64 compared with 1 out of 64) / the highest-entropy macrostate is the most probable(1)
(d) each of the $N$ particles now has twice as many positions available, so $\Omega$ increases by a factor $2^{N}$(1)
$\Delta S=k_B\ln2^{N_A}=N_Ak_B\ln2=6.02\times10^{23}\times1.38\times10^{-23}\times\ln2=5.8\ \mathrm{J\,K^{-1}}$ (accept $R\ln2=5.76$); award [1 max] for a bare $5.8\ \mathrm{J\,K^{-1}}$ with no reasoning(1)
Answer: (b) $4.1\times10^{-23}\ \mathrm{J\,K^{-1}}$ (c) $S=0$; 3–3 has most microstates (d) $+5.8\ \mathrm{J\,K^{-1}}$
Examiner tip: In (d) the number of microstates is multiplied by $2^N$, not 2; the logarithm turns this into $N k_B \ln 2$.

5StandardHL only6 marks

In a model of a diesel engine, $4.0\times10^{-4}\ \mathrm{m^3}$ of a monatomic ideal gas at a pressure of $1.0\times10^{5}\ \mathrm{Pa}$ and a temperature of $300\ \mathrm{K}$ is compressed adiabatically to a volume of $5.0\times10^{-5}\ \mathrm{m^3}$.

(a) Show that the final pressure of the gas is about $3\times10^{6}\ \mathrm{Pa}$. [2]

(b) Calculate the final temperature of the gas. [2]

(a) $pV^{5/3}=\text{constant}$ used: $p_2=1.0\times10^{5}\times\left(\dfrac{4.0\times10^{-4}}{5.0\times10^{-5}}\right)^{5/3}$(1)
$=1.0\times10^{5}\times8^{5/3}=3.2\times10^{6}\ \mathrm{Pa}$ (must see value to at least 2 s.f.)(1)
(b) $\dfrac{p_1V_1}{T_1}=\dfrac{p_2V_2}{T_2}$ / $TV^{2/3}=\text{constant}$ used(1)
$T_2=300\times\dfrac{3.2\times10^{6}\times5.0\times10^{-5}}{1.0\times10^{5}\times4.0\times10^{-4}}=1200\ \mathrm{K}$ (accept $1.2\times10^{3}$ K); allow ECF from (a)(1)
(c) adiabatic so $Q=0$ and $W_{\text{on}}=\Delta U$(1)
$\Delta U=\tfrac32\Delta(pV)=\tfrac32(160-40)=180\ \mathrm{J}$ (accept $\tfrac32nR\Delta T$ with $n=0.016$ mol, giving 180 J); accept $-180$ J if clearly identified as work done by the gas; do not accept $p\Delta V$ with a single pressure(1)
Answer: (a) $3.2\times10^{6}\ \mathrm{Pa}$ (b) $1200\ \mathrm{K}$ (c) $180\ \mathrm{J}$
Examiner tip: $W=p\Delta V$ only works at constant pressure; for an adiabatic change use $Q=0$ so the work equals $\Delta U=\tfrac32\Delta(pV)$.

6StandardHL only5 marks

An inventor claims that her engine takes in $1500\ \mathrm{J}$ of thermal energy from a reservoir at $600\ \mathrm{K}$ in each cycle, rejects $600\ \mathrm{J}$ to a reservoir at $300\ \mathrm{K}$ and does the rest as useful work.

(a) Calculate the efficiency claimed for the engine. [2]

(b) Determine whether the claim is possible. [2]

(a) $W=1500-600=900\ \mathrm{J}$(1)
$\eta=\dfrac{900}{1500}=0.60$ / 60%(1)
(b) $\eta_{\text{Carnot}}=1-\dfrac{T_c}{T_h}=1-\dfrac{300}{600}=0.50$(1)
claimed efficiency (0.60) is greater than the Carnot (maximum possible) efficiency between these temperatures, so the claim is impossible; must compare with a calculated Carnot value; allow ECF from (a)(1)
(c) it is impossible for a cyclic process to extract thermal energy from a single reservoir and convert it completely into work / no heat engine operating in a cycle can have an efficiency of 1 (100%); do not accept “entropy always increases”(1)
Answer: (a) 0.60 (b) impossible: exceeds $\eta_{\text{Carnot}}=0.50$ (c) no cyclic engine can convert heat from one reservoir entirely into work
Examiner tip: To judge an engine claim, always calculate the Carnot efficiency from the two reservoir temperatures (in kelvin) and compare.

7HardHL only10 marks

A monatomic ideal gas is taken round the cycle ABCA shown on the $p$–$V$ diagram.

AB: isothermal expansion at $800\ \mathrm{K}$ from $1.0\times10^{-3}\ \mathrm{m^3}$ to $8.0\times10^{-3}\ \mathrm{m^3}$. At A the pressure is $1.6\times10^{6}\ \mathrm{Pa}$.
BC: change at constant volume.
CA: adiabatic compression back to A.

(a) Show that the pressure at C is $5.0\times10^{4}\ \mathrm{Pa}$. [2]

(b) Calculate the thermal energy transferred during BC, and state its direction. [2]

(c) The work done by the gas during AB is $3.33\ \mathrm{kJ}$. State the thermal energy transferred to the gas during AB. [1]

(d) Determine the efficiency of the cycle. [3]

(e) The gas is used in an engine operating between reservoirs at the highest and lowest temperatures reached in the cycle. Calculate the Carnot efficiency for these reservoirs and suggest why the efficiency in (d) is lower. [2]

(a) $p_AV_A^{5/3}=p_CV_C^{5/3}$ used (C and A lie on the same adiabatic)(1)
$p_C=1.6\times10^{6}\times\left(\dfrac{1.0}{8.0}\right)^{5/3}=\dfrac{1.6\times10^{6}}{32}=5.0\times10^{4}\ \mathrm{Pa}$(1)
(b) $W=0$ so $Q=\Delta U=\tfrac32V\Delta p=\tfrac32\times8.0\times10^{-3}\times(5.0\times10^{4}-2.0\times10^{5})$ (seeing $p_B=2.0\times10^{5}$ Pa from $p_AV_A=p_BV_B$)(1)
$Q=-1.8\times10^{3}\ \mathrm{J}$, i.e. $1800\ \mathrm{J}$ transferred out of the gas(1)
(c) $\Delta U=0$ (isothermal) so $Q=W=3.33\ \mathrm{kJ}$ into the gas(1)
(d) CA: $Q=0$ so work done on the gas $=\Delta U=\tfrac32(p_AV_A-p_CV_C)=\tfrac32(1600-400)=1800\ \mathrm{J}$ (allow ECF from (b): the magnitudes are equal because $\Delta U$ round a cycle is zero)(1)
net work $=3330-1800=1530\ \mathrm{J}$ (accept 1.5 kJ)(1)
$\eta=\dfrac{1530}{3330}=0.46$ (accept 0.45–0.46); efficiency must use the thermal energy input during AB only(1)
(e) $T_C=T_A\dfrac{p_CV_C}{p_AV_A}=200\ \mathrm{K}$ so $\eta_{\text{Carnot}}=1-\dfrac{200}{800}=0.75$(1)
thermal energy is rejected during BC while the gas temperature falls from 800 K to 200 K, i.e. not all at the lowest temperature / the transfer in BC involves a finite temperature difference with the cold reservoir, so it is irreversible; only a cycle of two isothermals and two adiabatics (Carnot) reaches the maximum(1)
Answer: (a) $5.0\times10^{4}$ Pa (b) 1800 J out of the gas (c) 3.33 kJ (d) 0.46 (e) 0.75; heat rejected over a range of temperatures, not reversibly at $T_c$
Examiner tip: Over a complete cycle $\Delta U=0$; the efficiency uses only the thermal energy taken IN, not the net thermal energy.

8StretchHL only6 marks

A hot reservoir at $500\ \mathrm{K}$ and a cold reservoir at $250\ \mathrm{K}$ are connected by a metal rod. In a certain time, $1200\ \mathrm{J}$ of thermal energy flows along the rod from the hot reservoir to the cold reservoir. The temperatures of the reservoirs do not change.

(a) Calculate the total change in entropy of the two reservoirs. [2]

(b) The rod is replaced by a reversible (Carnot) engine that takes the same $1200\ \mathrm{J}$ from the hot reservoir. By considering entropy, determine the maximum work that the engine can do. [3]

(c) Explain, using entropy, why no engine working between these reservoirs could do more work than your answer to (b). [1]

(a) $\Delta S_{\text{hot}}=-\dfrac{1200}{500}=-2.4\ \mathrm{J\,K^{-1}}$ and $\Delta S_{\text{cold}}=+\dfrac{1200}{250}=+4.8\ \mathrm{J\,K^{-1}}$(1)
total $=+2.4\ \mathrm{J\,K^{-1}}$(1)
(b) for a reversible cycle the total entropy change is zero (the engine's own entropy is unchanged over a cycle), so $\dfrac{Q_c}{250}=\dfrac{1200}{500}$(1)
$Q_c=600\ \mathrm{J}$(1)
$W=1200-600=600\ \mathrm{J}$; accept the alternative route $W=\eta_{\text{Carnot}}Q_h=0.50\times1200$ for [2 max] unless the entropy argument is also given(1)
(c) more work means less than 600 J rejected, so the entropy gained by the cold reservoir is less than 2.4 J K⁻¹ and the total entropy of the (isolated) system would decrease, which the second law forbids(1)
Answer: (a) $+2.4\ \mathrm{J\,K^{-1}}$ (b) $600\ \mathrm{J}$ (c) total entropy would decrease
Examiner tip: A reversible engine creates no entropy: set the entropy lost by the hot reservoir equal to that gained by the cold one to find $Q_c$.

9HardHL only6 marks

A domestic refrigerator keeps its interior at $4\ ^\circ\mathrm{C}$ in a kitchen at $22\ ^\circ\mathrm{C}$. It does this by continuously transferring thermal energy out of the interior and into the kitchen, using electrical energy to drive a compressor.

Discuss how the operation of the refrigerator is consistent with the second law of thermodynamics. In your answer refer to the Clausius statement of the second law and to the entropy changes of the interior, the kitchen and the universe. [6]

Clausius statement: thermal energy cannot flow spontaneously (by itself) from a colder body to a hotter body(1)
the refrigerator does not break this, because the transfer is not spontaneous: work (electrical energy) is supplied by the compressor(1)
energy rejected to the kitchen $Q_h$ = energy removed from the interior $Q_c$ + work input $W$ (first law), so $Q_h>Q_c$(1)
entropy of the interior decreases by $Q_c/T_c$ (with $T_c=277$ K)(1)
entropy of the kitchen increases by $Q_h/T_h$ (with $T_h=295$ K)(1)
$Q_h/T_h\ge Q_c/T_c$ because the extra energy $W$ is also rejected, so the total entropy of the universe does not decrease (increases for a real, irreversible refrigerator)(1)
entropy can decrease locally (inside the refrigerator) only because a larger increase occurs in the surroundings(1)
the minimum work needed rises as the temperature difference between interior and kitchen increases / if the power is switched off, heat flows back in spontaneously and the interior warms(1)

Level	Marks	What it looks like
Level 3	5–6	States the Clausius statement correctly and explains that the work input makes the transfer non-spontaneous. Uses the first law to show more energy enters the kitchen than leaves the interior, and compares the entropy changes ($Q/T$) of interior and kitchen to conclude that total entropy does not decrease. Reasoning is logical and uses correct terminology throughout.
Level 2	3–4	States the Clausius statement or the entropy form of the second law and links it to the work done by the compressor. Describes the entropy of the interior falling and the kitchen rising, but the comparison is incomplete or not justified with $Q_h>Q_c$ or temperatures. Some lapses in clarity.
Level 1	1–2	Makes relevant but isolated points, e.g. that heat normally flows from hot to cold, or that the refrigerator uses energy. Little or no use of entropy, or entropy statements are vague or incorrect.
No marks	0	No relevant content.

Answer: Heat moves from cold to hot only because work is done; interior entropy falls by $Q_c/T_c$, kitchen entropy rises by $Q_h/T_h$ with $Q_h=Q_c+W$, so total entropy does not decrease.
Examiner tip: Top-band answers link the work input to $Q_h>Q_c$ and then compare $Q_h/T_h$ with $Q_c/T_c$, rather than just saying “entropy increases”.

10StandardHL only7 marks

A student tests the relationship $pV^{\gamma}=\text{constant}$ for argon, a monatomic gas. Argon in a sealed syringe connected to a fast-response pressure sensor is compressed rapidly. The results are shown in the table and on the graph of $\ln(p/\mathrm{kPa})$ against $\ln(V/\mathrm{cm^3})$, with a line of best fit.

$V$ / cm³	$p$ / kPa
60	100
50	133
40	190
30	299
24	425

(a) Explain why the compression must be done rapidly. [1]

(b) Show that, if $pV^{\gamma}=\text{constant}$, the graph should be a straight line of gradient $-\gamma$. [1]

(d) The theoretical value for a monatomic gas is $\tfrac53$. Suggest one reason why the experimental value is smaller. [1]

(e) The initial temperature of the argon is $295\ \mathrm{K}$. Calculate its temperature at the smallest volume. [2]

(a) so that there is no time for (significant) thermal energy transfer to the surroundings / so that the process is (approximately) adiabatic, $Q\approx0$(1)
(b) $\ln p+\gamma\ln V=\ln(\text{constant})$ so $\ln p=-\gamma\ln V+\text{constant}$, which has the form $y=mx+c$ with $m=-\gamma$(1)
(c) gradient from two well-separated points on the line, e.g. $\dfrac{6.05-4.61}{3.18-4.09}$ (triangle must span at least half the line)(1)
$\gamma=1.58$ (accept 1.54–1.62); answer must be positive; ignore units(1)
(d) some thermal energy still leaves the gas through the syringe walls during compression, so the temperature (and pressure) rise less than for a truly adiabatic change / gas leaks out past the plunger, so the pressure at small volumes is lower than expected; do not accept friction alone(1)
(e) $\dfrac{p_1V_1}{T_1}=\dfrac{p_2V_2}{T_2}$: $T_2=295\times\dfrac{425\times24}{100\times60}$ or $T_2=295\times\left(\dfrac{60}{24}\right)^{\gamma-1}$ using their $\gamma$ from (c) (award [1 max] for 543 K from $\gamma=\tfrac53$)(1)
$T_2=5.0\times10^{2}\ \mathrm{K}$ (accept 500–503 K; allow ECF for the $TV^{\gamma-1}$ route with their $\gamma$, e.g. 484–521 K for $\gamma$ = 1.54–1.62); award [1 max] if 22 °C or Celsius used(1)
Answer: (c) $\gamma\approx1.58$ (d) heat leaks out, so not fully adiabatic (e) $5.0\times10^{2}\ \mathrm{K}$
Examiner tip: The gradient of a log graph has no units; read two points from the LINE, not from the table, and drop the minus sign to state $\gamma$.

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