Calculate the thermal energy transferred from the sample during the first $30$ minutes.

Estimate the specific heat capacity of the oil in its liquid phase. State an appropriate unit for your answer.

The sample begins to freeze during the thermal energy transfer. Explain, in terms of the molecular model of matter, why the temperature of the sample remains constant during freezing.

Calculate the mass of the oil that remains unfrozen after $60$ minutes.

$«15\times 30\times 60»=27000 «\mathrm{J}»$ ✓

$27\times {10}^3=0.32\times c\times \left(290-250\right)$ OR $2100$ ✓

$\mathrm{J} {\mathrm{kg}}^{-1} {\mathrm{K}}^{-1}$ OR $\mathrm{J} {\mathrm{kg}}^{-1} { }^0{\mathrm{C}}^{-1}$ ✓

Allow any appropriate unit that is $\frac{energy}{mass\times termperature}$

«intermolecular» bonds are formed during freezing ✓

bond-forming process releases energy
OR
«intermolecular» PE decreases «and the difference is transferred as heat» ✓

«average random» KE of the molecules does not decrease/change ✓

temperature is related to «average» KE of the molecules «hence unchanged» ✓

To award MP3 or MP4 molecules/particles/atoms must be mentioned.

mass of frozen oil $«=\frac{27\times {10}^3}{130\times {10}^3}»=0.21 «\mathrm{kg}»$ ✓

unfrozen mass $«=0.32-0.21»=0.11 «\mathrm{kg}»$ ✓

The molar mass of water is 18 g mol⁻¹. Estimate the average speed of the water molecules in the vapor produced. Assume the vapor behaves as an ideal gas.

State one assumption of the kinetic model of an ideal gas.

Estimate the specific latent heat of vaporization of water. State an appropriate unit for your answer.

Explain why the temperature of water remains at 100 °C during this time.

The heater is removed and a mass of 0.30 kg of pasta at −10 °C is added to the boiling water.

Determine the equilibrium temperature of the pasta and water after the pasta is added. Other heat transfers are negligible.

Specific heat capacity of pasta = 1.8 kJ kg⁻¹ K⁻¹
Specific heat capacity of water = 4.2 kJ kg⁻¹ K⁻¹

Show that each resistor has a resistance of about 30 Ω.

Calculate the power transferred by the heater when both switches are closed.

E_k = « $\frac{3}{2}(1.38\times {10}^{-23})\left(373\right)$» = $7.7\times {10}^{-21}$ «J» ✓

v = «$\sqrt{\frac{3\times 1.38\times {10}^{-23}\times 6.02\times {10}^{23}\times 373}{0.018}}$» = 720 «m s⁻¹» ✓

particles can be considered points «without dimensions» ✓

no intermolecular forces/no forces between particles «except during collisions»✓

the volume of a particle is negligible compared to volume of gas ✓

collisions between particles are elastic ✓

time between particle collisions are greater than time of collision ✓

no intermolecular PE/no PE between particles  ✓

Accept reference to atoms/molecules for “particle”

«mL = P t» so «$L=\frac{1600\times 200}{0.14}$» = 2.3 x 10⁶ «J kg^-1» ✓

J kg⁻¹ ✓

«all» of the energy added is used to increase the «intermolecular» potential energy of the particles/break «intermolecular» bonds/OWTTE ✓

Accept reference to atoms/molecules for “particle” 

use of mcΔT ✓

0.86 × 4200 × (100 – T) = 0.3 × 1800 × (T +10) ✓

T_eq = 85.69«°C» ≅ 86«°C» ✓

Accept T_eq in Kelvin (359 K).

$P=\frac{v^2}{R} \mathrm{so} \frac{{220}^2}{1600} \mathrm{so} R=30.25$ «Ω» ✓

Must see either the substituted values OR a value for R to at least three s.f.

use of parallel resistors addition so R_eq = 15 «Ω» ✓

P = 3200 «W» ✓

Estimate the power input to the heating element. State an appropriate unit for your answer.

Outline whether your answer to (a) is likely to overestimate or underestimate the power input.

Discuss, with reference to the molecules in the liquid, the difference between milk at 11 °C and milk at 84 °C.

State how energy is transferred from the inside of the metal pipe to the outside of the metal pipe.

The missing section of insulation is 0.56 m long and the external radius of the pipe is 0.067 m. The emissivity of the pipe surface is 0.40. Determine the energy lost every second from the pipe surface. Ignore any absorption of radiation by the pipe surface.

Describe one other method by which significant amounts of energy can be transferred from the pipe to the surroundings.

energy required for milk entering in 1 s = mass x specific heat x 73 ✓

16 kW OR 16000 W ✓

MP1 is for substitution into mcΔT regardless of power of ten.

Allow any correct unit of power (such as J s^-1 OR kJ s^-1) if paired with an answer to the correct power of 10 for MP2.

Underestimate / more energy or power required ✓

because energy transferred as heat / thermal energy is lost «to surroundings or electrical components» ✓

Do not allow general term “energy” or “power” for MP2.

the temperature has increased so the internal energy / « average » KE «of the molecules» has increased OR temperature is proportional to average KE «of the molecules». ✓

«therefore» the «average» speed of the molecules or particles is higher OR more frequent collisions « between molecules » OR spacing between molecules has increased OR average force of collisions is higher OR intermolecular forces are less OR intermolecular bonds break and reform at a higher rate OR molecules are vibrating faster. ✓

conduction/conducting/conductor «through metal» ✓

use of $P=e\sigma A{T}^4$ where T = 357 K ✓

use of $A=2\pi r l$ « = 0.236 m²» ✓

P = 87 «W» ✓

Allow 85 – 89 W for MP3.

Allow ECF for MP3.

convection «is likely to be a significant loss» ✓

«due to reduction in density of air near pipe surface» hot air rises «and is replaced by cooler air from elsewhere»

OR

«due to» conduction «of heat or thermal energy» from pipe to air ✓

Most candidates recognized that this was a specific heat question and set up a proper calculation, but many struggled to match their answer to an appropriate unit. A common mistake was to leave the answer in some form of an energy unit and others did not match the power of ten of the unit to their answer (e.g. 16 W).

Many candidates recognized that this was an underestimate of the total energy but failed to provide an adequate reason. Many gave generic responses (such as "some power will be lost"/not 100% efficient) without discussing the specific form of energy lost (e.g. heat energy).

This was generally well answered. Most HL candidates linked the increase in temperature to the increase in the kinetic energy of the molecules and were able to come up with a consequence of this change (such as the molecules moving faster). SL candidates tended to focus more on consequences, often neglecting to mention the change in KE.

Many candidates recognized that heat transfer by conduction was the correct response. This was a "state" question, so candidates were not required to go beyond this.

Candidates at both levels were able to recognize that this was a blackbody radiation question. One common mistake candidates made was not calculating the area of a cylinder properly. It is important to remind candidates that they are expected to know how to calculate areas and volumes for basic geometric shapes. Other common errors included the use of T in Celsius and neglecting to raise T ^4. Examiners awarded a large number of ECF marks for candidates who clearly showed work but made these fundamental errors.

A few candidates recognized that convection was the third source of heat loss, although few managed to describe the mechanism of convection properly for MP2. Some candidates did not read the question carefully and instead wrote about methods to increase the rate of heat loss (such as removing more insulation or decreasing the temperature of the environment).

Define internal energy.

0.46 mole of an ideal monatomic gas is trapped in a cylinder. The gas has a volume of 21 m³ and a pressure of 1.4 Pa.

(i) State how the internal energy of an ideal gas differs from that of a real gas.

(ii) Determine, in kelvin, the temperature of the gas in the cylinder.

(iii) The kinetic theory of ideal gases is one example of a scientific model. Identify one reason why scientists find such models useful.

mention of atoms/molecules/particles

sum/total of kinetic energy and «mutual/intermolecular» potential energy

Do not allow “kinetic energy and potential energy” bald.
Do not allow “sum of average ke and pe” unless clearly referring to total ensemble.

i
«intermolecular» potential energy/PE of an ideal gas is zero/negligible

ii
THIS IS FOR USE WITH AN ENGLISH SCRIPT ONLY
use of $T=\frac{PV}{nR}$ or $T=\frac{1.4\times 21}{0.46\times 8.31}$

Award mark for correct re-arrangement as shown here not for quotation of Data Booklet version.
Award [2] for a bald correct answer in K.
Award [2 max] if correct 7.7 K seen followed by –265°C and mark BOD. However, if only –265°C seen, award [1 max].

7.7 K
Do not penalise use of “°K”

ii
THIS IS FOR USE WITH A SPANISH SCRIPT ONLY
$T=\frac{PV}{nR}$

$T=\frac{1.4\times 2.1\times {10}^{-6}}{0.46\times 8.31}$

T = 7.7 ×10^-6 K
Award mark for correct re-arrangement as shown here not for quotation of Data Booklet version.
Uses correct unit conversion for volume
Award [2] for a bald correct answer in K. Finds solution. Allow an ECF from MP2 if unit not converted, ie candidate uses 21 m3 and obtains 7.7 K
Do not penalise use of “°K”

iii
models used to predict/hypothesize

explain

simulate

simplify/approximate
Allow similar responses which have equivalent meanings. Response needs to identify one reason.

Calculate the speed of light inside the ice cube.

Show that no light emerges from side AB.

Sketch, on the diagram, the subsequent path of the light ray.

Determine the energy required to melt all of the ice from –20 °C to water at a temperature of 0 °C.

Specific latent heat of fusion of ice  = 330 kJ kg^–1
Specific heat capacity of ice            = 2.1 kJ kg^–1 k^–1
Density of ice                                = 920 kg m^–3

Outline the difference between the molecular structure of a solid and a liquid.

«v = c$\frac{\text{sin }i}{\text{sin }r}$ =» $\frac{3\times {10}^8\times \text{sin}\left(33\right)}{\text{sin}\left(46\right)}$

2.3 x 10⁸ «m s^–1»

light strikes AB at an angle of 57°

critical angle is «sin^–1$\left(\frac{2.3}{3}\right)$ =» 50.1°

49.2° from unrounded value

angle of incidence is greater than critical angle so total internal reflection

OR

light strikes AB at an angle of 57°

calculation showing sin of “refracted angle” = 1.1

statement that since 1.1>1 the angle does not exist and the light does not emerge

[Max 3 marks]

total internal reflection shown

ray emerges at opposite face to incidence

Judge angle of incidence=angle of reflection by eye or accept correctly labelled angles

With sensible refraction in correct direction

mass = «volume x density» (0.75)³ x 920 «= 388 kg»

energy required to raise temperature = 388 x 2100 x 20 «= 1.63 x 10⁷ J»

energy required to melt = 388 x 330 x 10³ «= 1.28 x 10⁸ J»

1.4 x 10⁸ «J» OR 1.4 x 10⁵ «kJ»

Accept any consistent units

Award [3 max] for answer which uses density as 1000 kg^–3 (1.5× 10⁸ «J»)

in solid state, nearest neighbour molecules cannot exchange places/have fixed positions/are closer to each other/have regular pattern/have stronger forces of attraction

in liquid, bonds between molecules can be broken and re-form

OWTTE

Accept converse argument for liquids

[Max 1 Mark]

Write down the missing values in the nuclear equation for this decay.

Rutherford and Royds put some pure radium-226 in a small closed cylinder A. Cylinder A is fixed in the centre of a larger closed cylinder B.

At the start of the experiment all the air was removed from cylinder B. The alpha particles combined with electrons as they moved through the wall of cylinder A to form helium gas in cylinder B.

The wall of cylinder A is made from glass. Outline why this glass wall had to be very thin.

Rutherford and Royds expected 2.7 x 10¹⁵ alpha particles to be emitted during the experiment. The experiment was carried out at a temperature of 18 °C. The volume of cylinder B was 1.3 x 10^–5 m³ and the volume of cylinder A was negligible. Calculate the pressure of the helium gas that was collected in cylinder B.

Rutherford and Royds identified the helium gas in cylinder B by observing its emission spectrum. Outline, with reference to atomic energy levels, how an emission spectrum is formed.

The work was first reported in a peer-reviewed scientific journal. Outline why Rutherford and Royds chose to publish their work in this way.

222 AND 4

Both needed.

alpha particles highly ionizing
OR
alpha particles have a low penetration power
OR
thin glass increases probability of alpha crossing glass
OR
decreases probability of alpha striking atom/nucleus/molecule

conversion of temperature to 291 K

p = 4.5 x 10^–9 x 8.31 x «$\frac{2.91}{1.3\times {10}^{-5}}$»

OR

p = 2.7 x 10¹⁵ x 1.38 x 10^–23 x «$\frac{2.91}{1.3\times {10}^{-5}}$»

0.83 or 0.84 «Pa»

electron/atom drops from high energy state/level to low state

energy levels are discrete

wavelength/frequency of photon is related to energy change or quotes E = hf or E = $\frac{hc}{\lambda }$

and is therefore also discrete

peer review guarantees the validity of the work
OR
means that readers have confidence in the validity of work

OWTTE

Calculate the pressure of the gas.

Calculate, in kg, the mass of the gas.

Calculate the average kinetic energy of the particles of the gas.

Explain, with reference to the kinetic model of an ideal gas, how an increase in temperature of the gas leads to an increase in pressure.

«$\frac{3.0\times 8.31\times 290}{0.15}$»

48 «kPa»

[1 mark]

mass = «$\frac{860}{3100\times 23}$» 0.012 «kg»

Award [1] for a bald correct answer.

[1 mark]

$\frac{3}{2}$ 1.38 × 10^–23 × 313 = 6.5 × 10^–21 «J»

[1 mark]

larger temperature implies larger (average) speed/larger (average) KE of molecules/particles/atoms

increased force/momentum transferred to walls (per collision) / more frequent collisions with walls

increased force leads to increased pressure because P = F/A (as area remains constant)

Ignore any mention of PV = nRT.

[3 marks]

From A to B, 24 % of the gravitational potential energy transferred to kinetic energy. Show that the velocity at B is 12 m s^–1.

Some of the gravitational potential energy transferred into internal energy of the skis, slightly increasing their temperature. Distinguish between internal energy and temperature.

The dot on the following diagram represents the skier as she passes point B.
Draw and label the vertical forces acting on the skier.

The hill at point B has a circular shape with a radius of 20 m. Determine whether the skier will lose contact with the ground at point B.

The skier reaches point C with a speed of 8.2 m s^–1. She stops after a distance of 24 m at point D.

Determine the coefficient of dynamic friction between the base of the skis and the snow. Assume that the frictional force is constant and that air resistance can be neglected.

Calculate the impulse required from the net to stop the skier and state an appropriate unit for your answer.

Explain, with reference to change in momentum, why a flexible safety net is less likely to harm the skier than a rigid barrier.

$\frac{1}{2}{v}^2=0.24\text{gh}$

$v=11.9$ «m s^–1»

Award GPE lost = 65 × 9.81 × 30 = «19130 J»

Must see the 11.9 value for MP2, not simply 12.

Allow g = 9.8 ms^–2.

internal energy is the total KE «and PE» of the molecules/particles/atoms in an object

temperature is a measure of the average KE of the molecules/particles/atoms

Award [1 max] if there is no mention of molecules/particles/atoms.

arrow vertically downwards from dot labelled weight/W/mg/gravitational force/F_g/F_{gravitational} AND arrow vertically upwards from dot labelled reaction force/R/normal contact force/N/F_N

W > R

Do not allow gravity.
Do not award MP1 if additional ‘centripetal’ force arrow is added.
Arrows must connect to dot.
Ignore any horizontal arrow labelled friction.
Judge by eye for MP2. Arrows do not have to be correctly labelled or connect to dot for MP2.

ALTERNATIVE 1
recognition that centripetal force is required / $\frac{m{v}^2}{r}$ seen

= 468 «N»

W/640 N (weight) is larger than the centripetal force required, so the skier does not lose contact with the ground

ALTERNATIVE 2

recognition that centripetal acceleration is required / $\frac{v^2}{r}$ seen

a = 7.2 «ms^–2»

g is larger than the centripetal acceleration required, so the skier does not lose contact with the ground

ALTERNATIVE 3

recognition that to lose contact with the ground centripetal force ≥ weight

calculation that v ≥ 14 «ms^–1»

comment that 12 «ms^–1» is less than 14 «ms^–1» so the skier does not lose contact with the ground

ALTERNATIVE 4

recognition that centripetal force is required / $\frac{m{v}^2}{r}$ seen

calculation that reaction force = 172 «N»

reaction force > 0 so the skier does not lose contact with the ground

Do not award a mark for the bald statement that the skier does not lose contact with the ground.

ALTERNATIVE 1
0 = 8.2² + 2 × a × 24 therefore a = «−»1.40 «m s⁻²»

friction force = ma = 65 × 1.4 = 91 «N»

coefficient of friction = $\frac{91}{65\times 9.81}$ = 0.14

ALTERNATIVE 2
KE = $\frac{1}{2}$mv² = 0.5 x 65 x 8.2² = 2185 «J»

friction force = KE/distance = 2185/24 = 91 «N»

coefficient of friction = $\frac{91}{65\times 9.81}$ = 0.14

Allow ECF from MP1.

«76 × 9.6»= 730
Ns OR kg ms^–1

safety net extends stopping time

F = $\frac{\mathrm{\Delta }p}{\mathrm{\Delta }t}$ therefore F is smaller «with safety net»

OR

force is proportional to rate of change of momentum therefore F is smaller «with safety net»

Accept reverse argument.

State what is meant by an ideal gas.

Calculate the number of atoms in the gas.

Calculate, in J, the internal energy of the gas.

Calculate, in Pa, the new pressure of the gas.

Explain, in terms of molecular motion, this change in pressure.

a gas in which there are no intermolecular forces

OR

a gas that obeys the ideal gas law/all gas laws at all pressures, volumes and temperatures

OR

molecules have zero PE/only KE

Accept atoms/particles.

[1 mark]

N = «$\frac{pV}{kT}=\frac{5.3\times {10}^5\times 2.1\times {10}^{-4}}{1.38\times {10}^{-23}\times 310}$» 2.6 × 10²²

[1 mark]

«For one atom U = $\frac{3}{2}$kT» $\frac{3}{2}$ × 1.38 × 10^–23 × 310 / 6.4 × 10^–21 «J»

U = «2.6 × 10²² × $\frac{3}{2}$ × 1.38 × 10^–23 × 310» 170 «J»

 Allow ECF from (a)(ii)

Award [2] for a bald correct answer

Allow use of U = $\frac{3}{2}$pV

[2 marks]

p₂ = «5.3 × 10⁵ × $\frac{2.1\times {10}^{-4}}{6.8\times {10}^{-4}}$» 1.6 × 10⁵ «Pa»

[1 mark]

«volume has increased and» average velocity/KE remains unchanged

«so» molecules collide with the walls less frequently/longer time between collisions with the walls

«hence» rate of change of momentum at wall has decreased

«and so pressure has decreased»

The idea of average must be included

Decrease in number of collisions is not sufficient for MP2. Time must be included.

Accept atoms/particles.

[2 marks]

Distinguish between the internal energy of the oxygen at the boiling point when it is in its liquid phase and when it is in its gas phase.

Calculate, in kW, the heater power required.

Calculate the volume of the oxygen produced in one second when it is allowed to expand to a pressure of 0.11 MPa and to reach a temperature of 260 K.

State one assumption of the kinetic model of an ideal gas that does not apply to oxygen.

Internal energy is the sum of all the PEs and KEs of the molecules (of the oxygen) ✔

PE of molecules in gaseous state is zero ✔

(At boiling point) average KE of molecules in gas and liquid is the same ✔

gases have a higher internal energy ✔

Molecules/particles/atoms must be included once, if not, award [1 max]

ALTERNATIVE 1:

flow rate of oxygen = 8 «g s⁻¹» ✔

«2.1 ×10⁵ × 8 × 10⁻³» = 1.7 «kW» ✔

ALTERNATIVE 2:

Q = «0.25 × 32 ×10⁻³ × 2.1 × 10⁵ =» 1680 «J» ✔

power = «1680 W =» 1.7 «kW» ✔

V = «$\frac{nRT}{p}=$» 4.9 × 10⁻³ «m³»

ideal gas has point objects ✔

no intermolecular forces ✔

non liquefaction ✔

ideal gas assumes monatomic particles ✔

the collisions between particles are elastic ✔

Allow the opposite statements if they are clearly made about oxygen eg oxygen/this can be liquified

Deduce whether helium behaves as an ideal gas over the temperature range 250 K to 500 K.

Helium has a molar mass of 4.0 g. Calculate the mass of gas in the container.

A second container, of the same volume as the original container, contains twice as many helium atoms. The graph of the variation of P with T is determined for the gas in the second container.

Predict how the graph for the second container will differ from the graph for the first container.

«He behaves as ideal gas if» $p\propto T$ «at constant V» ✓

uses two points to show that $p\propto T$ ✓

MP1 can also be described as $\frac{p}{T}=k \mathit{\text{OR }}\frac{p}{T}\text{=}\frac{nR}{V}$

$\frac{100\times {10}^3\times {10}^{-3}}{250\times 8.31}=«0.048 \text{mol}»$ ✓

$«0.048\times 4=» 0.19 «\text{g}»$ ✓

Allow any correct data point to be used.

Allow ECF from MP1

recognizes that pressure will double ✓

graph will be steeper OR gradient will be larger ✓

graph will still go through the origin ✓

MP1 can be expressed as e.g.“$p\propto n$” OR “$\frac{nR}{V}$will double”.

Accept $pv=2nRT$ for MP1.

State what is meant by the internal energy of an ideal gas.

Calculate the pressure of the gas.

The temperature of the gas is increased to 500 K. Sketch, on the axes, a graph to show the variation with temperature T of the pressure P of the gas during this change.

A container is filled with 1 mole of helium (molar mass 4 g mol⁻¹) and 1 mole of neon (molar mass 20 g mol⁻¹). Compare the average kinetic energy of helium atoms to that of neon atoms.

the total «random» kinetic energy of the molecules/atoms/particles ✓

$p=«\frac{nRT}{V}=\frac{0.24\times 8.31\times 300}{0.20}=»3.0\times {10}^3$«Pa» ✓

straight line joining (300, 3) and (500, 5) ✓

drawn only in the range from 300 to 500 K ✓

Allow ECF from (b)(i) for incorrect initial pressure.
Allow tolerance of ± one grid square for the endpoints.

temperature is the same for both gases ✓

«average» kinetic energy is the same «because $E_k=\frac{3}{2}kT$  OR  $E_k$ depends on $T$ only» ✓

Award [1 max] for a bald statement that kinetic energy is the same.

The molar mass of helium is 4.0 g mol^-1. Show that the mass of a helium atom is 6.6 × 10^-27 kg.

Estimate the average speed of the helium atoms in the container.

Show that the number of helium atoms in the container is about 4 × 10²⁰.

Calculate the ratio  $\frac{\text{total volume of helium atoms}}{\text{volume of helium gas}}$.

Explain, using your answer to (d)(i) and with reference to the kinetic model, why this sample of helium can be assumed to be an ideal gas.

$m=\frac{4.0\times {10}^{-3}}{6.02\times {10}^{23}}$«kg»

OR

6.64 × 10⁻²⁷ «kg» ✔

$\frac{1}{2}m{v}^2=\frac{3}{2}kT/v=\sqrt{\frac{3kT}{m}}/\sqrt{\frac{3\times 1.38\times {10}^{-23}\times 320}{6.6\times {10}^{-27}}}$  ✔

v = 1.4 × 10³ «ms⁻¹» ✔ 

$N=\frac{pV}{kT}/\frac{5.1\times {10}^5\times 3.2\times {10}^{-6}}{1.38\times {10}^{-23}\times 320}$

OR

$N=\frac{pV{N}_A}{RT}/\frac{5.1\times {10}^5\times 3.2\times {10}^{-6}\times 6.02\times {10}^{23}}{8.31\times 320}$   ✔

N = 3.7 × 10²⁰ ✔

«$\frac{4\times {10}^{20}\times 4.9\times {10}^{-31}}{3.2\times {10}^{-6}}=»6\times {10}^{-5}$  ✔

«For an ideal gas» the size of the particles is small compared to the distance between them/size of the container/gas

OR

«For an ideal gas» the volume of the particles is negligible/the volume of the particles is small compared to the volume of the container/gas

OR

«For an ideal gas» particles are assumed to be point objects ✔

calculation/ratio/result in (d)(i) shows that volume of helium atoms is negligible compared to/much smaller than volume of helium gas/container «hence assumption is justified» ✔

The mark was awarded for a clear substitution or an answer to at least 3sf. Many gained the mark for a clear substitution with a conversion from g to kg somewhere in their response. Fewer gave the answer to the correct number of sf.

At HL this was very well answered but at SL many just worked out E=3/2kT and left it as a value for KE.

Again at HL this was very well answered with the most common approach being to calculate the number of moles and then multiply by N_A to calculate the number of atoms. At SL many candidates calculated n but stopped there. Also at SL there was some evidence of candidates working backwards and magically producing a value for ‘n’ that gave a result very close to that required after multiplying by N_A.

This was well answered with the most common mistake being to use the volume of a single atom rather than the total volume of the atoms.

In general this was poorly answered at SL. Many other non-related gas properties given such as no / negligible intermolecular forces, low pressure, high temperature. Some candidates interpreted the ratio as meaning it is a low density gas. At HL candidates seemed more able to focus on the key part feature of the question, which was the nature of the volumes involved. Examiners were looking for an assumption of the kinetic theory related to the volume of the atoms/gas and then a link to the ratio calculated in ci). The command terms were slightly different at SL and HL, giving slightly more guidance at SL.

A solid cylinder of height h and density ρ rests on a flat surface.

Show that the pressure p_c exerted by the cylinder on the surface is given by p_c = ρgh.

Show that (p_o + p_m) × 0.190 = $\frac{nRT}{A}$  where

p_o = atmospheric pressure

p_m = pressure due to the mercury column

T = temperature of the trapped gas

n = number of moles of the trapped gas

A = cross-sectional area of the tube. 

Determine the atmospheric pressure. Give a suitable unit for your answer.

Outline why the gas particles in the tube hit the mercury surface less often after the tube has been rotated.

weight of cylinder = Ahg ρ  ✔

pressure = $\frac{F}{A}$ = $\frac{Ahg\rho }{A}$ ✔

Allow use of A = $\pi r^2$ in MP1. 

use of PV = nRT and V = Area × (0.190) seen ✔

substitution of P = p_o + p_m «re-arrangement to give answer»✔

recognition that $\frac{\text{nRT}}{\text{A}}$ is constant OR 190p_o + 190p_m = 208p_o − 208p_m

OR  p_o = $\frac{398}{18}$ p_m ✔

pressure due to mercury p_m = 0.035 × 1.36 × 10⁴ × 9.81(= 4.67 × 10³ Pa) ✔

1.03 × 10⁵ ✔

Pa OR Nm^-2 OR kgm^-1s^-2 ✔

Do not award for a bald correct answer. Working must be shown to award MP3.

Award MP4 for any correct unit of pressure (eg “mm of mercury / Hg”).

same number of particles to collide with a larger surface area OR greater volume with constant rms speed decreases collision frequency ✔

Look for a correct statement that connects pressure to molecular movement/collisions.

This question was fairly well answered with a not insignificant number of candidates making clearly wrong substitutions (such as F=mgh) to make the equation work out. As a “show that” question the derivation should be neatly laid out with the fundamental equations written out and the substitutions/cancelations clearly shown.

This question was generally well answered. Most candidates took the time to show the set up and substitutions they used to derive the given expression. A small number of candidates attempted to “show that” by making unit substitutions - this is not acceptable for a question like this.

This question was left blank by many candidates. Of those who attempted a solution, few appreciated the difference made to the derived equation from 4bi when the tube was rotated. It should be noted that this was also the “unit question” on this exam, and a candidate could have been awarded a mark for clearly writing any correct unit of pressure without doing any calculations. Candidates should be reminded to keep an eye out for this opportunity and to at least write a unit even if the question seems unapproachable.

This was generally poorly answered with the candidates split between answers that generally demonstrated some good understanding of physics (such as connecting the increase in volume AND constant rms speed of particles with the rate of collisions with the mercury) and answers that did not (such as gases want to rise so the gas it will hit the mercury less).

Calculate the number of gas particles in the cylinder.

Discuss, for this process, the changes that occur in the density of the gas.

Discuss, for this process, the changes that occur in the internal energy of the gas.

Correct conversion of T «T = 310 K» seen ✓

« use of = $N=\frac{pV}{kT}$ to get » 2.3 × 10²³ ✓

Allow ECF from MP1 i.e., T in Celsius (Result is 2.7 x 10²⁴)

Allow use of n, R and N_A

density decreases ✓

volume is increased AND mass/number of particles remains constant ✓

internal energy is constant ✓

internal energy depends on kinetic energy/temperature «only»

OR

since temperature/kinetic energy is constant ✓

Do not award MP2 for stating that “temperature is constant” unless linked to the correct conclusion, as that is mentioned in the stem.

Award MP2 for stating that kinetic energy remains constant.

a) This was well answered with the majority converting to K. Quite a few found the number of moles but did not then convert to molecules.

bi) Well answered. It was pleasing to see how many recognised the need to state that the mass/number of molecules stayed the same as well as stating that the volume increased. At SL this recognition was less common so only 1 mark was often awarded.

bii) This was less successfully answered. A surprising number of candidates said that the internal energy of an ideal gas increases during an isothermal expansion. Many recognised that constant temp meant constant KE but then went on to state that the PE must increase and so the internal energy would increase.

With the door open the air in the refrigerator is initially at the same temperature and pressure as the air in the kitchen. Calculate the number of molecules of air in the refrigerator.

Determine the pressure of the air inside the refrigerator.

The door of the refrigerator has an area of 0.72 m². Show that the minimum force needed to open the refrigerator door is about 4 kN.

Comment on the magnitude of the force in (b)(ii).

$N=\frac{pV}{kT}$ OR $N=\frac{1.0\times {10}^5\times 0.36}{1.38\times {10}^{-23}\times 295}$ ✔

$N=8.8\times {10}^{24}$ ✔

NOTE: Allow [1 max] for substitution with T in Celsius.
Allow [1 max] for a final answer of n = 14.7 or 15
Award [2] for bald correct answer.

use of $\frac{p}{T}$ = constant  OR  $p=\frac{nRT}{V}$  OR  $\frac{NkT}{V}$ ✔
$p=9.4\times {10}^4$« Pa »✔

NOTE: Allow ECF from (a)
Award [2] for bald correct answer

$F=A\times ∆p$ ✔

$F=0.72\times \left(1.0-0.94\right)\times {10}^5$OR 4.3 × 10³ « N »✔

NOTE: Allow ECF from (b)(i)
Allow ECF from MP1

force is «very» large ✔

there must be a mechanism that makes this force smaller
OR
assumption used to calculate the force/pressure is unrealistic ✔