B.3 Gas laws – IB Physics exam questions with mark scheme

1Standard6 marks

A fixed mass of an ideal gas occupies a rigid container of volume $0.0250\ \mathrm{m^3}$ at a pressure of $3.0\times10^5\ \mathrm{Pa}$ and a temperature of $300\ \mathrm{K}$. $R=8.31\ \mathrm{J\,mol^{-1}\,K^{-1}}$, $N_A=6.02\times10^{23}\ \mathrm{mol^{-1}}$, $k_B=1.38\times10^{-23}\ \mathrm{J\,K^{-1}}$.

(a) Calculate the amount of gas in moles. [2]

(b) Determine the number of molecules in the container. [1]

(d) Calculate the average translational kinetic energy of a molecule at $300\ \mathrm{K}$. [1]

$n=pV/RT = (3.0\times10^5\times0.0250)/(8.31\times300)$(1)
$n=3.01$ mol (3.0)(1)
$N=nN_A = 1.81\times10^{24}$ ($1.8\times10^{24}$)(1)
Constant volume and amount so $p/T$ is constant: $p_2=p_1T_2/T_1$(1)
$p_2=4.0\times10^5$ Pa ($4.0\times10^5$ Pa)(1)
$\bar E_K=\tfrac32 k_BT = 6.21\times10^{-21}$ J ($6.2\times10^{-21}$ J)(1)
Answer: (a) 3.0 mol (b) $1.8\times10^{24}$ (c) $4.0\times10^5$ Pa (d) $6.2\times10^{-21}$ J
Examiner tip: Temperatures in gas-law equations must be in kelvin.

2Core5 marks

A cylinder fitted with a frictionless piston contains $12.0\ \mathrm{g}$ of helium gas. The molar mass of helium is $4.00\ \mathrm{g\,mol^{-1}}$. $R=8.31\ \mathrm{J\,mol^{-1}\,K^{-1}}$, $k_B=1.38\times10^{-23}\ \mathrm{J\,K^{-1}}$, $N_A=6.02\times10^{23}\ \mathrm{mol^{-1}}$.

(a) Calculate the number of helium atoms in the cylinder. [2]

(b) The piston has a cross-sectional area of $4.5\times10^{-3}\ \mathrm{m^2}$ and the pressure of the gas is $2.4\times10^{5}\ \mathrm{Pa}$. Calculate the force exerted on the piston by the gas. [1]

(a) $n=\dfrac{12.0}{4.00}=3.00\ \mathrm{mol}$(1)
$N=nN_A=3.00\times6.02\times10^{23}=1.8\times10^{24}$ (accept $1.81\times10^{24}$); award [1 max] for $7.2\times10^{24}$ (mass in g used as the amount in mol)(1)
(b) $F=PA=2.4\times10^5\times4.5\times10^{-3}=1.1\times10^{3}\ \mathrm{N}$ (accept 1080 N)(1)
(c) $T=290\ \mathrm{K}$ and $V=\dfrac{nRT}{P}$ (or $V=\dfrac{Nk_BT}{P}$) used(1)
$V=\dfrac{3.00\times8.31\times290}{2.4\times10^5}=0.030\ \mathrm{m^3}$ (accept $3.0\times10^{-2}\ \mathrm{m^3}$; ECF from (a)); award [0] for this mark if 17 used for $T$ ($1.8\times10^{-3}\ \mathrm{m^3}$)(1)
Answer: (a) $1.8\times10^{24}$ (b) $1.1\times10^3$ N (c) $0.030\ \mathrm{m^3}$
Examiner tip: Convert grams to moles with the molar mass before multiplying by Avogadro's constant, and convert 17 °C to 290 K.

3Core4 marks

The kinetic model of an ideal gas is based on a number of assumptions about the molecules.

(a) State two of these assumptions. [2]

(b) A cylinder of nitrogen is stored at a very high pressure. A separate sample of nitrogen is cooled to $-180\ ^\circ\mathrm{C}$, close to its boiling point. Explain why the behaviour of nitrogen departs from the ideal gas model under each of these conditions. [2]

(a) Any two of: molecules have negligible volume compared with the volume of the gas/container; there are no intermolecular forces except during collisions; collisions (with each other and with the walls) are elastic; the duration of a collision is negligible compared with the time between collisions; molecules move randomly / in random directions; a very large number of identical molecules; molecules obey Newton's laws of motion. First assumption(1)
Second assumption (from the list above; do not accept 'molecules are far apart' alone or 'no collisions')(1)
(b) High pressure: molecules are close together so the volume of the molecules is no longer negligible compared with the volume of the gas (accept: intermolecular forces become significant because molecules are close)(1)
Low temperature: molecules move slowly / have low kinetic energy, so the intermolecular attractive forces are no longer negligible compared with their kinetic energy (accept: the gas is close to liquefying)(1)
Answer: (a) e.g. negligible molecular volume; no intermolecular forces except in collisions. (b) High P: molecular volume not negligible; low T: intermolecular forces significant relative to KE.
Examiner tip: For (b), link each condition to the specific assumption that fails; 'the gas is not ideal' scores nothing.

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4Hard6 marks

A student investigates how the volume of a fixed mass of air varies with pressure. Air is trapped in a sealed syringe connected by a short tube to a pressure gauge. The temperature of the air is $20\ ^\circ\mathrm{C}$. The student reads the volume $V$ from the scale on the syringe and records the pressure $P$.

$P$ / kPa	100	125	160	200	250
$V$ / $\mathrm{cm^3}$	47.9	37.9	29.2	23.0	18.0

The graph shows $V$ plotted against $\dfrac{1}{P}$ with the line of best fit. $R=8.31\ \mathrm{J\,mol^{-1}\,K^{-1}}$.

(a) After each change of pressure the student waits for a minute before taking the reading. Suggest why. [1]

(b) Determine the gradient of the line and hence the amount of air, in mol, in the apparatus. [3]

(c) The line does not pass through the origin. State the value of the intercept on the $V$ axis and suggest a reason for it. [2]

(a) Compressing the gas does work on it / warms it, so the student waits for the air to return to room temperature (so that $T$ is constant); accept 'to reach thermal equilibrium'(1)
(b) Gradient calculated from two well-separated points on the line (not data points unless they lie on the line): e.g. $\dfrac{(52.9-(-1.9))\times10^{-6}}{1.1\times10^{-5}}$; gradient $=5.0\ \mathrm{J}$ (accept 4.8 to 5.2; accept $\mathrm{Pa\,m^3}$ or $\mathrm{N\,m}$); award [0] for this mark if powers of ten are not handled(1)
Gradient $=nRT$ (since $V=nRT\cdot\dfrac{1}{P}-V_0$) and $T=293\ \mathrm{K}$ used(1)
$n=\dfrac{5.0}{8.31\times293}=2.0\times10^{-3}\ \mathrm{mol}$ (accept $2.0$ to $2.1\times10^{-3}$; ECF from gradient)(1)
(c) Intercept $\approx-2\ \mathrm{cm^3}$ (accept $-1.5$ to $-2.5\ \mathrm{cm^3}$)(1)
Some air is in the connecting tube / pressure gauge (or syringe nozzle) and is not included in the scale reading, so the measured volume is less than the true volume by a constant amount (systematic error); do not accept 'random error' or 'zero error on the gauge'(1)
Answer: (b) gradient $\approx5.0$ J, $n\approx2.0\times10^{-3}$ mol (c) $\approx-2\ \mathrm{cm^3}$: unrecorded air volume in tube/gauge
Examiner tip: The gradient of V against 1/P is nRT, not n: convert cm³ to m³ and kPa to Pa before dividing by RT.

5Standard6 marks

A fixed mass of gas is held in a sealed flask of constant volume. The pressure $P$ of the gas is measured at different Celsius temperatures $\theta$. The graph shows the results and the line of best fit through them.

(a) Use the graph to estimate the value of absolute zero in $^\circ\mathrm{C}$. Show your method on the graph. [2]

(b) Calculate the pressure of the gas at $250\ ^\circ\mathrm{C}$. [2]

(c) The experiment is repeated with twice the mass of the same gas in the same flask. Describe how the line obtained differs from the original line. [2]

(a) Line extrapolated back (straight, using the same gradient) to $P=0$(1)
Absolute zero $\approx-270\ ^\circ\mathrm{C}$ (accept $-250$ to $-290\ ^\circ\mathrm{C}$)(1)
(b) $\dfrac{P}{T}$ constant with temperatures in kelvin: $P=100\times\dfrac{523}{273}$ (or use of any read-off point with its kelvin temperature)(1)
$P=192\ \mathrm{kPa}$ (accept 190 to 194 kPa); award [0] for this mark if Celsius temperatures are used in the ratio(1)
(c) Pressure is doubled at every temperature / the gradient is doubled (as $P=\dfrac{nR}{V}T$)(1)
The line still meets the temperature axis at the same point (absolute zero, $\approx-273\ ^\circ\mathrm{C}$)(1)
Answer: (a) about $-273\ ^\circ$C (b) 192 kPa (c) twice the gradient, same intercept at absolute zero
Examiner tip: Extrapolate to zero pressure, not to the edge of the grid, and always use kelvin in P/T = constant.

6Hard6 marks

At ground level, air has a pressure of $1.01\times10^{5}\ \mathrm{Pa}$ and a density of $1.20\ \mathrm{kg\,m^{-3}}$. Treat air as an ideal gas with a mean molar mass of $0.029\ \mathrm{kg\,mol^{-1}}$. $R=8.31\ \mathrm{J\,mol^{-1}\,K^{-1}}$.

(a) Show that the root mean square speed of the air molecules is about $500\ \mathrm{m\,s^{-1}}$. [2]

(b) Determine the temperature of the air. [2]

(c) Air is mainly nitrogen (molar mass $0.028\ \mathrm{kg\,mol^{-1}}$) and oxygen (molar mass $0.032\ \mathrm{kg\,mol^{-1}}$). Calculate $\dfrac{\text{r.m.s. speed of nitrogen molecules}}{\text{r.m.s. speed of oxygen molecules}}$ in the air. [2]

(a) $P=\tfrac13\rho\langle c^2\rangle\Rightarrow\langle c^2\rangle=\dfrac{3P}{\rho}=\dfrac{3\times1.01\times10^5}{1.20}\ (=2.53\times10^5\ \mathrm{m^2\,s^{-2}})$(1)
$c_{\mathrm{rms}}=\sqrt{2.53\times10^5}=502\ \mathrm{m\,s^{-1}}$ (must be given to at least 3 s.f. or with working shown, as this is a 'show that')(1)
(b) $\tfrac12m\langle c^2\rangle=\tfrac32k_BT$ with $m=M/N_A$, giving $T=\dfrac{M\langle c^2\rangle}{3R}$ (or $PV=nRT$ with $\dfrac{n}{V}=\dfrac{\rho}{M}$: $T=\dfrac{PM}{\rho R}$)(1)
$T=\dfrac{0.029\times2.53\times10^5}{3\times8.31}=294\ \mathrm{K}$ (accept 290 to 295 K; accept $21\ ^\circ\mathrm{C}$)(1)
(c) Same temperature so same mean kinetic energy: $\tfrac12m_{N}\langle c^2\rangle_N=\tfrac12m_O\langle c^2\rangle_O$, so ratio $=\sqrt{\dfrac{M_O}{M_N}}=\sqrt{\dfrac{0.032}{0.028}}$(1)
$=1.07$ (award [1 max] for 0.935 (inverted) or 1.14 (no square root))(1)
Answer: (a) 502 m s$^{-1}$ (b) 294 K (c) 1.07
Examiner tip: Equal temperature means equal mean kinetic energy, not equal speed: the lighter molecule is faster by the square root of the mass ratio.

7Standard6 marks

A sealed syringe contains a fixed mass of an ideal gas.

Explain, using the kinetic model of an ideal gas, how the gas exerts a pressure on the walls of the syringe, and why the pressure increases when

the gas is compressed slowly to a smaller volume at constant temperature
the gas is heated at constant volume.

[6]

Molecules are in continuous random motion and collide with the walls(1)
In each (elastic) collision the momentum of a molecule changes (reverses its perpendicular component, $\Delta p=2mu$)(1)
By Newton's second law the wall exerts a force on the molecule equal to the rate of change of momentum; by Newton's third law the molecule exerts an equal and opposite force on the wall(1)
Pressure is the total (average) force from very many collisions divided by the area of the wall(1)
Compression at constant $T$: mean speed / mean kinetic energy of molecules is unchanged, so momentum change per collision is unchanged(1)
but molecules have less distance to travel between walls / more molecules per unit volume, so collisions with each unit area of wall are more frequent, giving a greater rate of change of momentum and a greater pressure(1)
Heating at constant volume: mean kinetic energy (and mean speed) increases since $\bar E_k=\tfrac32k_BT$(1)
so the momentum change per collision is greater and collisions are more frequent, so the rate of change of momentum (force per unit area) increases(1)

Level	Marks	What it looks like
Level 3	5–6	Explains pressure fully in terms of momentum change at the walls and the resulting force per unit area, using Newton's laws; correctly distinguishes the two cases: compression changes collision frequency only, heating changes both momentum change per collision and collision frequency. Logical and uses correct terminology.
Level 2	3–4	Links collisions with the walls to a force and to pressure, and gives a valid explanation for at least one of the two changes; the other may be incomplete, or the role of momentum is stated without clear linking.
Level 1	1–2	Isolated relevant statements, e.g. molecules hit the walls, or molecules move faster when heated, with little or no link to momentum or force.
Level 0	0	No relevant content.

Answer: Pressure = rate of change of momentum of molecules colliding with the walls per unit area; compression raises collision frequency only; heating raises both collision frequency and momentum change per collision.
Examiner tip: State that heating changes both the momentum change per collision and the collision rate; compression at constant temperature changes only the rate.

8Hard6 marks

Flask A has a volume of $2.0\times10^{-3}\ \mathrm{m^3}$ and contains an ideal gas at a pressure of $300\ \mathrm{kPa}$ and a temperature of $300\ \mathrm{K}$. Flask B has a volume of $3.0\times10^{-3}\ \mathrm{m^3}$ and contains the same gas at $100\ \mathrm{kPa}$ and $400\ \mathrm{K}$. The flasks are joined by a narrow tube of negligible volume, closed by a tap. $R=8.31\ \mathrm{J\,mol^{-1}\,K^{-1}}$, $k_B=1.38\times10^{-23}\ \mathrm{J\,K^{-1}}$, $N_A=6.02\times10^{23}\ \mathrm{mol^{-1}}$.

The tap is opened and the whole system is then brought to a temperature of $350\ \mathrm{K}$.

(a) Determine the final pressure of the gas. [4]

(b) Calculate the number of molecules that move from flask A to flask B. [2]

(a) $n_A=\dfrac{3.00\times10^5\times2.0\times10^{-3}}{8.31\times300}=0.241\ \mathrm{mol}$(1)
$n_B=\dfrac{1.00\times10^5\times3.0\times10^{-3}}{8.31\times400}=0.0903\ \mathrm{mol}$(1)
Total amount is conserved: $n=n_A+n_B=0.331\ \mathrm{mol}$ in a total volume of $5.0\times10^{-3}\ \mathrm{m^3}$(1)
$P=\dfrac{0.331\times8.31\times350}{5.0\times10^{-3}}=1.9\times10^{5}\ \mathrm{Pa}$ (accept 192 to 193 kPa); award [1 max] for (a) for an answer that averages the two pressures (200 kPa or 180 kPa)(1)
(b) Same $P$ and $T$ throughout so amount in B is proportional to volume: $n_B'=\dfrac{3.0}{5.0}\times0.331=0.199\ \mathrm{mol}$ (or $n_B'=\dfrac{PV_B}{RT}$)(1)
Moved $=(0.199-0.0903)\times6.02\times10^{23}=6.5\times10^{22}$ molecules (accept $6.4$ to $6.6\times10^{22}$; ECF from (a))(1)
Answer: (a) $1.9\times10^5$ Pa (b) $6.5\times10^{22}$
Examiner tip: Pressures cannot be averaged: conserve the number of moles, then apply PV = nRT to the combined volume.

9Standard6 marks

A helium-filled balloon has a volume of $8.0\times10^{-3}\ \mathrm{m^3}$. The helium is at a pressure of $1.05\times10^{5}\ \mathrm{Pa}$ and a temperature of $293\ \mathrm{K}$. Treat helium as a monatomic ideal gas. $R=8.31\ \mathrm{J\,mol^{-1}\,K^{-1}}$.

(a) Show that the internal energy of the helium is about $1.3\ \mathrm{kJ}$. [2]

(b) The balloon is carried into a room at $308\ \mathrm{K}$. The pressure of the helium remains constant.

(i) Calculate the new volume of the balloon. [1]

(ii) Calculate the increase in the internal energy of the helium. [2]

(a) $U=\tfrac32nRT$ and $nRT=PV$ used, so $U=\tfrac32PV$ (or $n=0.345\ \mathrm{mol}$ found)(1)
$U=1.5\times1.05\times10^5\times8.0\times10^{-3}=1260\ \mathrm{J}$ (must see answer to at least 3 s.f. or full substitution)(1)
(b)(i) $V_2=8.0\times10^{-3}\times\dfrac{308}{293}=8.4\times10^{-3}\ \mathrm{m^3}$(1)
(ii) $n=\dfrac{PV}{RT}=0.345\ \mathrm{mol}$ and $\Delta U=\tfrac32nR\Delta T$ with $\Delta T=15\ \mathrm{K}$(1)
$\Delta U=1.5\times0.345\times8.31\times15=65\ \mathrm{J}$ (accept 64 to 66 J; allow $1324-1260$ approach)(1)
(c) There are no intermolecular forces (except during collisions), so the particles have no (intermolecular) potential energy; do not accept 'no collisions'(1)
Answer: (a) 1260 J (b)(i) $8.4\times10^{-3}\ \mathrm{m^3}$ (ii) 65 J
Examiner tip: Use ΔU = (3/2)nRΔT with the change in temperature, not (3/2)nRT at the final temperature.

10Stretch7 marks

A nitrogen molecule of mass $4.65\times10^{-26}\ \mathrm{kg}$ moves inside a cubical box of side $0.10\ \mathrm{m}$. Its velocity component perpendicular to one face, X, of the box is $500\ \mathrm{m\,s^{-1}}$. Assume all collisions are elastic and ignore collisions with other molecules. $R=8.31\ \mathrm{J\,mol^{-1}\,K^{-1}}$, $k_B=1.38\times10^{-23}\ \mathrm{J\,K^{-1}}$, $N_A=6.02\times10^{23}\ \mathrm{mol^{-1}}$.

(a) Calculate the magnitude of the change in momentum of the molecule each time it collides with face X. [1]

(b) Calculate the average force exerted on face X by this molecule. [2]

(c) The box is filled with nitrogen at $300\ \mathrm{K}$ and $1.0\times10^{5}\ \mathrm{Pa}$. The molecules move in random directions with a range of speeds.

(i) Explain why the pressure is given by $P=\dfrac{1}{3}\dfrac{Nm\langle c^2\rangle}{V}$ rather than by $\dfrac{Nm\langle c^2\rangle}{V}$. [2]

(ii) Calculate the root mean square speed of the molecules. [2]

(a) $\Delta p=2mu=2\times4.65\times10^{-26}\times500=4.65\times10^{-23}\ \mathrm{N\,s}$ (accept $4.7\times10^{-23}$; award [0] for $2.3\times10^{-23}$)(1)
(b) Time between successive collisions with face X: $\Delta t=\dfrac{2L}{u}=\dfrac{0.20}{500}=4.0\times10^{-4}\ \mathrm{s}$(1)
$F=\dfrac{\Delta p}{\Delta t}=\dfrac{4.65\times10^{-23}}{4.0\times10^{-4}}=1.2\times10^{-19}\ \mathrm{N}$ (accept $1.16\times10^{-19}$; ECF from (a))(1)
(c)(i) Only the velocity component perpendicular to a face contributes to the momentum change at that face: $c^2=u_x^2+u_y^2+u_z^2$(1)
Motion is random so the mean squares of the three components are equal: $\langle u_x^2\rangle=\tfrac13\langle c^2\rangle$(1)
(ii) $\tfrac12m\langle c^2\rangle=\tfrac32k_BT$ so $c_{\mathrm{rms}}=\sqrt{\dfrac{3k_BT}{m}}=\sqrt{\dfrac{3\times1.38\times10^{-23}\times300}{4.65\times10^{-26}}}$ (or via $N=\dfrac{PV}{k_BT}=2.4\times10^{22}$ and $P=\tfrac13\dfrac{Nm\langle c^2\rangle}{V}$)(1)
$c_{\mathrm{rms}}=517\ \mathrm{m\,s^{-1}}$ (accept 510 to 520 $\mathrm{m\,s^{-1}}$)(1)
Answer: (a) $4.65\times10^{-23}$ N s (b) $1.2\times10^{-19}$ N (c)(ii) 517 m s$^{-1}$
Examiner tip: The molecule returns to the same face after travelling 2L, not L, and its momentum change is 2mu because it rebounds.

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