Describe how the ideal gas constant R is defined.

Calculate the temperature of 0.100 mol of an ideal gas kept in a cylinder of volume 1.40×10^–3 m³ at a pressure of 2.32×10⁵ Pa.

The gas in (b) is kept in the cylinder by a freely moving piston. The gas is now heated at constant pressure until the volume occupied by the gas is 3.60×10^–3 m³. The increase in internal energy of the gas is 760 J. Determine the thermal energy given to the gas.

After heating, the gas is compressed rapidly to its original volume in (b). Outline why this compression approximates to an adiabatic change of state of the gas.

defined from the equation of state of an ideal gas PV=nRT;
all symbols (PVnT) correctly identified;

390/391 K;

work done\( = \left( {P\Delta V = 2.32 \times {{10}^5} \times 2.20 \times {{10}^{ - 3}} = } \right)510{\rm{J}}\);
thermal energy\( = \left( {760 + 510 = } \right)1.27 \times {10^3}{\rm{J}}\);
Award [1 max] if volume is taken as 3.6×10^–3, giving an answer of 1600 J.

an adiabatic change is one in which no (thermal/heat) energy is transferred between system and surroundings / no energy enters/leaves system;
a rapid compression means that there is insufficient time (for energy transfer) / OWTTE;

Using data from the graph above, identify which gas, A or B, undergoes the isothermal expansion.

Using the graph opposite, estimate the difference in work done by each gas.

Explain, with reference to the first law of thermodynamics, and without further calculation, the change in temperature of the gas undergoing the adiabatic change.

choice of two data points;
to show P×V constant for gas A / not constant for gas B;

identifies area between lines as work done;
counts squares (14 squares);
each square 12.5;
175 J; (allow answers in the range of 150 to 200 J)

Q=ΔU+W and Q=0; (both needed)
so ΔU+W=0;
work done by gas is positive;
so ΔU decreases therefore temperature decreases;

At point A in the cycle, the fuel-air mixture is at 18 °C. During process AB, the gas is compressed to 0.046 of its original volume and the pressure increases by a factor of 40. Calculate the temperature of the gas at point B.

State the nature of the change in the gas that takes place during process BC in the cycle.

Process CD is an adiabatic change. Discuss, with reference to the first law of thermodynamics, the change in temperature of the gas in the cylinder during process CD.

Explain how the diagram can be used to calculate the net work done during one cycle.

(i)     Calculate the volume of fuel injected into one cylinder during one cycle.

(ii)     Each of the four cylinders completes a cycle 18 times every second. Calculate the distance the car can travel on one litre of fuel at a speed of \({\text{56 m}}\,{{\text{s}}^{ - 1}}\).

A car accelerates uniformly along a straight horizontal road from an initial speed of \({\text{12 m}}\,{{\text{s}}^{ - 1}}\) to a final speed of \({\text{28 m}}\,{{\text{s}}^{ - 1}}\) in a distance of 250 m. The mass of the car is 1200 kg. Determine the rate at which the engine is supplying kinetic energy to the car as it accelerates.

(i)     Calculate the total resistive force acting on the car when it is travelling at a constant speed of \({\text{56 m}}\,{{\text{s}}^{ - 1}}\).

(ii)     The mass of the car is 1200 kg. The resistive force F is related to the speed v by \(F \propto {v^2}\). Using your answer to (g)(i), determine the maximum theoretical acceleration of the car at a speed of \({\text{28 m}}\,{{\text{s}}^{ - 1}}\).

(i)     Calculate the maximum speed of the car at which it can continue to move in the circular path. Assume that the radius of the path is the same for each tyre.

(ii)     While the car is travelling around the circle, the people in the car have the sensation that they are being thrown outwards. Outline how Newton’s first law of motion accounts for this sensation.

535 (K) / 262 (°C);

constant volume change / isochoric / isovolumetric / OWTTE;

Q/thermal energy transfer is zero;

\(\Delta U =  - W\);

as work is done by gas internal energy falls;

temperature falls as temperature is measure of average kinetic energy;

work done is estimated by evaluating area;

inside the loop / OWTTE;

(i)     \(1.6 \times {10^{ - 4}}{\text{ (litre)}}\);

(ii)     one litre \( = \left( {\frac{1}{{4 \times 18 \times 1.64 \times {{10}^{ - 4}}}} = } \right){\text{ }}87{\text{ s of travel}}\);

\((87 \times 56) = 4.7{\text{ (km)}}\);

Allow rounded 1.6 value to be used, giving 4.9 (km).

use of a kinematic equation to determine motion time \({\text{(}} = 12.5{\text{ s)}}\);

change in kinetic energy \( = \frac{1}{2} \times {\text{1200}} \times {\text{[2}}{{\text{8}}^2} - {\text{1}}{{\text{2}}^2}]{\text{ (}} = 384{\text{ kJ)}}\);

rate of change in kinetic energy \( = \frac{{{\text{384000}}}}{{{\text{12.5}}}}\); } (allow ECF of 16² from (28 \( - \) 12)² for this mark)

31 (kW);

or

use of a kinematic equation to determine motion time \(( = 12.5{\text{ s)}}\);

use of a kinematic equation to determine acceleration \(( = 1.28{\text{ m}}\,{{\text{s}}^{ - 2}})\);

work done \(\frac{{F \times s}}{{{\text{time}}}} = \frac{{1536 \times 250}}{{12.5}}\);

31 (kW);

(i)     \({\text{force}} = \frac{{{\text{power}}}}{{{\text{speed}}}}\);

2300 or 2.3k (N);

Award [2] for a bald correct answer.

(ii)     resistive force \( = \frac{{2300}}{4}\)\(\,\,\,\)or\(\,\,\,\)\(\frac{{2321}}{4}{\text{ }}( = {\text{575)}}\); (allow ECF)

so accelerating force \((2300 - 580 = ){\text{ }}1725{\text{ (N)}}\)\(\,\,\,\)or\(\,\,\,\)1741 (N);

\(a = \frac{{1725}}{{1200}} = 1.44{\text{ (m}}\,{{\text{s}}^{ - 2}})\)\(\,\,\,\)or\(\,\,\,\)\(a = \frac{{1741}}{{1200}} = 1.45{\text{ (m}}\,{{\text{s}}^{ - 2}})\);

Award [2 max] for an answer of 0.49 (m\(\,\)s^–2 (omits 2300 N).

(i)     centripetal force must be \( < 6000{\text{ (N)}}\); (allow force 6000 N)

\({v^2} = F \times \frac{r}{m}\);

\(31.6{\text{ (m}}\,{{\text{s}}^{ - 1}})\);

Allow [3] for a bald correct answer.

Allow [2 max] if 4\( \times \) is omitted, giving 15.8 (m\(\,\)s^–1).

(ii)     statement of Newton’s first law;

(hence) without car wall/restraint/friction at seat, the people in the car would move in a straight line/at a tangent to circle;

(hence) seat/seat belt/door exerts centripetal force;

(in frame of reference of the people) straight ahead movement is interpreted as “outwards”;

This simple gas law calculation was surprisingly badly done. Certainly similar questions have attracted better scores in previous examinations. Common errors included the inevitable failure to work in kelvin, and simple arithmetic errors.

Most candidates were able to describe the constant volume nature of the change in question.

Many candidates scored full credit in a question that has been well rehearsed in previous examinations. The zero change in thermal energy transfer was common and many were able to deduce that \(\Delta U\) is therefore equal to \( - W\). This led immediately to a deduction of temperature decrease.

Almost all recognised that the work done was related to some area under the graph. In a small minority of cases the exact specification of the area was too imprecise to gain the second mark.

(i)     It was common to see a correct value for the volume of fuel used though not a correct unit.

(ii)     Many were able to arrive at a travel time for the fuel and therefore the distance travelled. However, routes were indirect and lengthy and few could see a direct way to the answer.

There were at least two routes to tackle this problem. Some solutions were so confused that it was difficult to decide which method had been used. Common errors included: forgetting that the initial speed was \({\text{12 m}}\,{{\text{s}}^{ - 1}}\) not zero, power of ten errors, and simple mistakes in the use of the kinematic equations, or failure to evaluate work done \( = {\text{force }} \times {\text{ distance}}\) correctly. However, many candidates scored partial credit. Scores of two or three out of the maximum four were common showing that many are persevering to get as far as they can.

(i)     Many correct solutions were seen. Candidates are clearly comfortable with the use of the equation force = power/speed.

(ii)     The method to be used here was obvious to many. What was missing was a clear appreciation of what was happening in terms of resistive force in the system. Many scored two out of three because they indicated a sensible method but did not use the correct value for the force. Scoring two marks does require that the explanation of the method is at least competent. Those candidates who give limited explanations of their method leading to a wrong answer will generally accumulate little credit. A suggestion (never seen in answers) is that candidates should have begun from a free-body force diagram which would have revealed the relationship of all the forces.

(i)     The major problem here was that most candidates did not recognise that 1500 N of force acting at each of four wheels will imply a total force of 6 kN. Again, partial credit was available only if it was clear what the candidate was doing and what the error was.

(ii)     Statements of Newton’s first law were surprisingly poor. As in previous examinations, few candidates appear to have learnt this essential rule by heart and they produce a garbled and incomplete version under examination pressure. The first law was then only loosely connected to the particular context of the question. Candidates have apparently not learnt to relate the physics they learn to everyday contexts.

Calculate the temperature of the gas in state B.

(i) Calculate the work done by the gas in expanding from state A to state B.

(ii) Determine the amount of thermal energy transferred during the expansion from state A to state B.

The gas is isothermally compressed from state B back to state A.

(i) Using the P–V diagram axes opposite, draw the variation of pressure with volume for this isothermal compression.

(ii) State and explain whether the magnitude of the thermal energy transferred in this case would be less than, equal to or greater than the thermal energy transferred in (b)(ii).

realization that since pV = constant, the temperature must be the same i.e. 400 K / full calculation using gas law to get 400 K;

(i) work done is area under curve;
and this is \(\left( {\frac{{4.0 + 8.0}}{2} \times 4 \times {{10}^2}} \right) = 2400{\rm{J}}\);
Award [2] for correct bald answer.

(ii) (Q=ΔU+W with) ΔU=0;
so Q=2400J;
Award [0] for correct answer with no or wrong argument.

(i) curve under given straight line starting at B and ending at A;

(ii) it would be less;
since the work done would be less / area under curve is less (and ΔU = 0 );

(i)     State the meanings of \(Q\) and \(W\).

\(Q\):

\(W\):

(ii)     Describe how the first law of thermodynamics applies in the operation of the heat exchanger.

(iii)     Discuss the entropy changes that take place in the gas and in the surroundings.

(i)     \(Q\): is the energy transferred between the system and surroundings;

\(W\): work done on/by system;

(ii)     \(Q\) transferred from reactor to gas;

no change in volume therefore \(W = 0\);

internal energy of gas increases;

\(Q\) transferred from gas to surroundings therefore internal energy of gas

decreases;

(iii)     entropy of the gas initially increases as energy transferred from the reactor;

entropy of the surroundings increases as energy transferred (from the gas);

entropy of gas decreases on cooling;

overall the entropy of the total system increases;

(i)     The meanings of \(Q\) and \(W\) were often expressed poorly and incompletely.

(ii)     Some candidates are confused about the statement of the first law of thermodynamics and quoted the second. Others gave vague and uncreditworthy accounts that showed that they were not answering in the context of the reactor heat exchanger. There was no real attempt by most candidates to arrive at four points in the answer, despite the fact that there are two exchanges going on in the reactor.

(iii)     As in part (ii) some candidates did not focus on the context intended. The failure to identify all exchanges and entropy changes was common as in part (ii).

Calculate the number of moles of air in the cylinder.

The cork leaves the toy after the air is compressed to a pressure of 1.9×10⁵Pa and a volume of 4.0×10^–4 m³.

(i) Deduce that the compression of the gas is not isothermal.

(ii) Outline why the compression might be adiabatic.

(iii) The work needed to compress the air in (a) is 15J. Determine, with reference to the first law of thermodynamics, the change in the internal energy of the air in the cylinder.

(iv) Calculate the change in average kinetic energy of an air molecule as a result of the compression.

The piston is now pushed in slowly so that the compression is isothermal. Discuss the entropy changes that take place in the air of the toy and in its cylinder as the air is compressed.

\(n = \frac{{pV}}{{RT}} = \frac{{1.1 \times {{10}^5} \times 6.0 \times {{10}^{ - 4}}}}{{8.31 \times 290}}\);
0.027;

(i) calculate pV correctly for both states: 66 and 76; } (do not penalize 66 k/K and 76 k/K as k may be a constant)
isothermal change would mean that p₁V₁=p₂V₂;
so not isothermal

(ii) no heat/thermal energy transferred;
(because change/compression) occurs (too) quickly/fast; [2]

(iii) Q=0;
W=−15; (minus sign is required)
so ΔU=(+)15J; } (symbols must be defined)
(allow ECF from second marking point)

(iv) number of air molecules=0.0274×6.0×10²³(=1.64×10²²);
9.13×10^–22J;

entropy is a property that indicates degree of disorder in the system / OWTTE;
gas occupies a smaller volume; (do not allow “compressed”)
entropy of gas/air in toy decreases because more ordered;
energy reaches surroundings/cylinder / entropy cannot decrease;
so entropy of surroundings/cylinder increases; } (do not allow ECF from “entropy of gas increases”)

Outline, with reference to charge carriers, what is meant by the internal resistance of a cell.

On the graph, sketch the variation of \(V\) with \(I\) for the cell.

Using the graph, determine the current in the circuit.

The graph shows how the pressure \(P\) of a fixed mass of gas varies with volume \(V\). The lines show the state of the gas sample during adiabatic expansion and during isothermal expansion.

State and explain whether line A or line B represents an adiabatic expansion.

Determine the work done during the change represented by line A.

Outline, with reference to the first law of thermodynamics, the direction of change in temperature during the adiabatic expansion.

charge carriers/electrons move through cell;

transfer energy to the components of the cell (which is not therefore available to external circuit);

energy dissipated in cell equivalent to dissipation in a resistance;

causes potential difference of cell to be less than its emf;

straight line of negative gradient; (allow any straight line of negative gradient)

intercept at \({\text{1.25 V (}} \pm 0.05{\text{)}}\) and position/gradient as shown;

Watch for ECF from (c)(iii).

uses a point at which lines intersect;

reads correct current value from their own graph;

\(0.28{\text{A}} \pm 0.1{\text{A}}\); (award for correct answer only)

\(PV\) is a constant for A / B is a steeper curve / final temperature/pressure lower for B than A;

(hence) B;

number of small squares below \({\text{A}} = 180 \pm 20\);

area of 1 square \( = 6.25{\text{ J}}\);

adds additional 64 squares below false origin;

answer within the range of 1400 to 1650 J; } (allow ECF for an area that excludes the area below the false origin – giving an answer within the range of 1000 to 1250 J)

Award [2] for a mean P of \(4.65 \times {10^5} \times \Delta V\) of \(4 \times {10^{ - 3}}\) giving an answer of 1860 (J).

Accept working in large squares (= 16 small squares) using equivalent tolerances.

no thermal energy enters or leaves / \(Q = 0\);

work done by the gas / \(W\) is positive;

so internal energy decreases / \(\Delta U\) is negative;

temperature is a measure of internal energy and so temperature falls;

This was a 3 mark question about internal resistance, so reasonable detail was expected. Candidates needed to refer to electrons/charge dissipating energy in moving through a cell and the effect on terminal potential difference.

Very few correct answers were seen. The equation of the line required is \(V = E - Ir\). Hence a line of negative gradient \({\text{(}}-{\text{1.9 }}\Omega )\) was required with a y intercept of 1.24 V.

If a candidate drew any kind of line, ECF was used when the coordinate of the intersection of the two lines was used to determine a current.

Most candidates were able to identify B as the adiabatic – either by referring to the gradients or lower pressures and temperatures compared to A, or by showing PV was constant for A but not B.

This was a 4 mark question and so candidates needed to give a detailed response. Many stated the 1st law, identified the three symbols and explained the change in each during an adiabatic expansion. Others were far less systematic. Most eventually predicted that temperature would decrease, but lost marks in their explanation.

This was a 4 mark question and so candidates needed to give a detailed response. Many stated the 1st law, identified the three symbols and explained the change in each during an adiabatic expansion. Others were far less systematic. Most eventually predicted that temperature would decrease, but lost marks in their explanation.

State which of the processes is isothermal, isochoric (isovolumetric) or isobaric.

Process AB:

Process BC:

Process CA:

The temperature of the gas at A is 300K. Calculate the temperature of the gas at B.

The increase in internal energy of the gas during process AB is 4100J. Determine the heat transferred to the gas from the surroundings during the process AB.

The gas is compressed at constant temperature. Explain what changes, if any, occur to the entropy of

(i) the gas.

(ii) the surroundings.

(iii) the universe.

process AB: isobaric;
process BC: isochoric / isovolumetric;
process CA: isothermal;

use of \(\frac{{{V_2}}}{{{V_1}}} = \frac{{{T_2}}}{{{T_1}}}\);
to give 750 K;

W=(PΔV=9×10⁵×[5-2]×10^-3=)2700;
Q=ΔU+W=4100+2700=6800J;

(i) same number of molecules occupy smaller volume, so disorder and hence entropy decrease;

(ii) heat flows from gas to surroundings, so temperature of surroundings increases, so entropy increases;

(iii) from second law, entropy of universe increases during any process;

Award [0] for simple statement of entropy increases/decreases.

With respect to a gas, explain the meaning of the terms thermal energy and internal energy.

The graph shows how the pressure P of a sample of a fixed mass of an ideal gas varies with volume V. The gas is taken through a cycle ABCD.

V / 10^–6 m³

(i) Estimate the net work done during the cycle.

(ii) Explain whether the net work is done on the gas or by the gas.

(iii) Deduce, using the data from the graph, that the change C is isothermal.

(iv) Isothermal change A occurs at a temperature of 450 K. Calculate the temperature at which isothermal change C occurs.

(v) Describe the changes B and D.

(Q) energy transferred between two objects (at different temperatures);
(U) (total) potential energy and (random) kinetic energy of the molecules/particles (of the gas);

(i) use of area within cycle;
each large square has work value of 250 (J);
estimate (16 x 250= )4000 (J); (allow 3600 − 4100)
Award [3] for same outcome with small squares of area 10 (J).

(ii) (work is done by the gas because) area under expansion is greater than that under compression/pressure during expansion is greater than during compression;

(iii) clear attempt to compare two PV values;
evaluate two PV values correctly eg 75 x 80= 6000 and 200 x 30= 6000;  

(iv) use of PV =nRT or equivalent;
1350/1330 (K);

(v) both changes are isochoric/isovolumetric/constant volume changes;
B: temperature/internal energy increases, D: temperature/internal energy decreases;
B: thermal energy/heat input (to system), D: thermal energy/heat output (from system);
B: pressure increases, D: pressure decreases;

Few candidates were able to explain thermal energy was the energy transfer between two objects at different temperatures. Many knew the definition of internal energy but a high percentage omitted to mention the potential energy (probably assuming that the gas was ideal).

(i) Many candidates appeared to attempt to calculate area without actually saying what they were doing; although this was obvious when they referred to the area of a square, in many case it was not obvious and marks were lost when the candidates technique produced an answer out of tolerance. In examples like this there will be a reasonable tolerance for the area and it is not expected that candidates will waste considerable time in counting the small squares.

(ii) Although some candidates were aware that a clockwise cycle applies to net work done by the gas, this does not explain the choice. Simply saying that the area under the expansion was greater than the area under the compression was all that was needed.

(iii) This part was mostly well done by candidates. It is accepted, in line with SL A1, that showing constancy of two PV values does not prove that the change is isothermal; however in terms of deducing that the change is isothermal this technique is fine – that is, the candidates are told that it is isothermal and they are simply illustrating that this is the case. Often examiners will expect three values to be taken in questions such as this.

(iv) This part was well done by those many candidates who used any appropriate variant of the ideal gas equation to calculate the temperature.

(v) The large majority of candidates did well here although a minority were deducted marks when they used contradictory statements such as isochoric and compression or expansion.

For the cycle identify, with the letter I, an isochoric (isovolumetric) change.

The temperature at point X is 310 K. Calculate the temperature at point Y.

The shaded area WXYZ is 610 J. The total thermal energy transferred out of the gas in one cycle is 1.3 kJ.

(i) State what is represented by the shaded area WXYZ.

(ii) Determine the efficiency of cycle WXYZ.

(iii) Explain why the total thermal energy transferred out of the gas is degraded.

The work done on the gas during the adiabatic compression XY is 210 J. Determine the change in internal energy during the change from X to Y.

vertical line identified;
Do not penalize candidate who identifies both vertical lines.
Award [0] if vertical and another line identified.

(i) (net) work done by engine in one cycle; } (must imply “by engine” and “one cycle”)

(ii) efficiency=\(\frac{{610}}{{1300}}\);
0.47 or 47%; 
Award [2] for a bald correct answer.
Award [1] for \(\frac{{610}}{{\left( {610 + 1300} \right)}}\)=0.32.
Award [0] for any fraction that would exceed 1 even if later fudged.
Award [0] for efficiency \(=\frac{{{T_C} - {T_H}}}{{{T_H}}}\)=38%, question does not imply that this is a Carnot cycle.

ΔQ = 0;
210 (J); (do not accept negative sign)

A gas undergoes a thermodynamic cycle. The P–V diagram for the cycle is shown below.

In the changes of state B to C and D to A, the gas behaves as an ideal gas and the changes in state are adiabatic.

(i) State the circumstances in which the behaviour of a gas approximates to ideal gas behaviour.

(ii) State what is meant by an adiabatic change of state.

With reference to the first law of thermodynamics, explain for the change of state A to B, why energy is transferred from the surroundings to the gas.

Estimate the total work done in the cycle.

Three ice cubes at a temperature of 0ºC are dropped into a container of water at a temperature of 22ºC. The mass of each ice cube is 25 g and the mass of the water is 330 g. The ice melts, so that the temperature of the water decreases. The thermal capacity of the container is negligible.

(i) The following data are available.

Specific latent heat of fusion of ice = 3.3×10⁵ J kg^–1
Specific heat capacity of water = 4.2×10³ J kg^–1 K^–1

Calculate the final temperature of the water when all of the ice has melted. Assume that no thermal energy is exchanged between the water and the surroundings.

(ii) Explain how the first law of thermodynamics applies to the water when the ice cubes are dropped into it.

Explain how the pattern demonstrates that electrons have wave properties.

Electrons are accelerated to a speed of 3.6×10⁷ ms⁻¹ by the electric field.

(i) Calculate the de Broglie wavelength of the electrons.

(ii) The cathode and anode are 22 mm apart and the field is uniform.
The potential difference between the cathode and the anode is 3.7 kV.
Show that the acceleration of the electrons is approximately 3×10¹⁶ms⁻².

State what can be deduced about an electron from the amplitude of its associated wavefunction.

An electron reaching the central bright spot on the fluorescent screen has a small uncertainty in its position. Outline what the Heisenberg uncertainty principle is able to predict about another property of this electron.

(i) use of M×4.2×103×∆θ;
ml = 75×10–3×3.3×105 / 24750 J;
recognition that melted ice warms and water cools to common final temperature;
3.4°C;

(ii) work done on water by dropping cubes / negligible work done;
W negative or unchanged;
water gives thermal energy to ice;
Q negative;
water cools to a lower temperature;
∆ U negative / U decreases;

square of amplitude (of wavefunction);
(proportional to) probability of finding an electron (at a particular point);

relates position to momentum (or velocity);
large uncertainty in momentum / most information on momentum is lost;

Calculate the temperature of the gas at point C.

Compare, without any calculation, the work done and the thermal energy supplied along route ABC and route ADC.

use of \(\frac{{PV}}{T} = \) constant or use of \(T \propto PV\) or via intermediate calculation of \(n\) in \(PV = nRT\);

\(\frac{{1.95}}{{358}} = \frac{{8.55}}{{{T_{\text{c}}}}}\) or \(n = 6.55 \times {10^{ - 3}}{\text{ (mol)}}\); } (allow power of ten omission provided same omission on both sides)

1570 K or 1300 °C;

Award [3] for a bald correct answer ignoring small rounding errors.

Omitting conversion to Kelvin yields answer of 373 – award [2 max] as one error.

Award [1 max] if 273 subsequently added (to give 646).

same temperature change so same change in internal energy/\(\Delta U\);

work done along ABC is larger/ADB is smaller because area under ABC is greater than area under ADC/\(\Delta V\) same in both, \(P\) greater for ABC so \(P\Delta V\) also greater for ABC;

because \(\Delta Q = \Delta U + W\) thermal energy transferred is greater for route ABC/smaller for route ADB;

Must see reference to first law for MP3.

It is sad that many young physicists at this level cannot negotiate their way correctly through a straightforward gas-law calculation. As usual, many failed to convert from degree Celsius to Kelvin before carrying out the numerical manipulation. This was an excellent way to lose marks.

A good number were able to give answers to parts of this question but few could pull all the strands together convincingly. Frequent omissions were: to show that the internal energy changes are identical because the endpoint temperatures are the same, and to use the first law to confirm the final linking statement.

(i) State what is meant by an isothermal process.

(ii) Show that process AB is isothermal.

State the nature of process BC.

During the cycle ABCD, the net work done by the gas is 550J. Calculate the net thermal energy absorbed by the gas.

Explain why it is not possible for this engine, operating in this cycle, to be 100% efficient.

(i) a process in which temperature remains constant;

(ii) calculation to show that pV=16(×10²) at any point from A to B;
calculation to show that pV=16(×10²) at any other point from A to B;
pV is constant;

isochoric / isovolumetric / occurs at constant volume;

\(\Delta U = 0\);
Q=W=550(J);

the gas/system must return to original conditions (P, V, and T);
(to do that) some of the thermal energy absorbed by gas must be given off to the surroundings;
hence not all thermal energy absorbed by gas can be converted to work;
it is the second law of thermodynamics;

Estimate the total work done in the cycle.

The change AB is isothermal and occurs at a temperature of 420K. Calculate the number of moles of gas.

Identify and explain the change, if any, in the entropy of the gas when it has completed one cycle.