Gas laws HL Papers 1& 2

B.3 — Gas Laws
HL Practice: Paper 2

15 extended-response questions • 1 lab investigation • Full expandable solutions

This original IB-style practice bank covers the ideal gas equation, gas-law relationships, kinetic theory, rms speed, and internal energy. The first seven questions preserve and develop the supplied examples. Questions involving heat, work, and thermodynamic paths connect B.3 to B.4; they are labelled accordingly.

Instructions, constants, and conventions

Show equations, working, units, and physical explanations. Attempt each question before opening its solution.
Use $R=8.31\,\mathrm{J\,mol^{-1}\,K^{-1}}$ , $k_B=1.38\times10^{-23}\,\mathrm{J\,K^{-1}}$ , and $N_A=6.02\times10^{23}\,\mathrm{mol^{-1}}$ .
Use absolute pressure, temperature in kelvin, volume in cubic metres, and molar mass in $\mathrm{kg\,mol^{-1}}$ . Here, $1\,\mathrm{L}=10^{-3}\,\mathrm{m^3}$ and $1\,\mathrm{mL}=10^{-6}\,\mathrm{m^3}$ .
Give final calculated answers to three significant figures unless uncertainty dictates otherwise. Keep intermediate values unrounded.
For the B.4 links, use $Q=\Delta U+W$ : $Q$ is heat supplied to the gas and $W$ is work done by the gas. Expansion gives positive $W$ ; compression gives negative $W$ .
Bracketed marks are suggested practice marks. This is a topic bank, not a complete official examination paper.

Key equations and their limits

$\displaystyle pV=nRT=Nk_BT,\qquad n=\frac{N}{N_A}$ $\displaystyle p=\frac13\rho v_{\mathrm{rms}}^2,\qquad v_{\mathrm{rms}}=\sqrt{\langle v^2\rangle}=\sqrt{\frac{3RT}{M}}$ $\displaystyle \langle E_{k,\mathrm{trans}}\rangle=\frac32k_BT$ For a monatomic ideal gas in the translational model: $\displaystyle U=\frac32nRT,\qquad\Delta U=\frac32nR\Delta T$ At constant pressure during a quasistatic change: $\displaystyle W=p\Delta V$ Here $N$ is particle number, $n$ is amount in moles, $M$ is molar mass, and $\rho$ is gas mass density. Rms speed depends on temperature and particle mass. Internal energy depends on temperature for a fixed amount and composition of ideal gas.

Extended-response questions

Question 1 — Ideal gas equation [5 marks]

(a) A container holds $0.25\,\mathrm{mol}$ of an ideal gas at $300\,\mathrm{K}$ in volume $4.0\times10^{-3}\,\mathrm{m^3}$ . Calculate its pressure. [2 marks]

(b) State one assumption of the ideal gas model. [1 mark]

(c) Explain how collisions with the walls produce pressure. [2 marks]

Show full solution and marking guidance

(a) [2 marks] $\displaystyle p=\frac{nRT}{V}=\frac{(0.25)(8.31)(300)}{4.0\times10^{-3}}\approx1.56\times10^5\,\mathrm{Pa}$

(b) [1 mark] Accept negligible particle volume compared with container volume; no intermolecular forces except during collisions; elastic collisions; or random motion.

(c) [2 marks] Particles change momentum when they collide with a wall. The rate of momentum transfer produces an average force; pressure is this force divided by wall area.

Question 2 — Boyle’s law and isothermal compression [6 marks]

(a) A fixed quantity of ideal gas has volume $2.5\times10^{-3}\,\mathrm{m^3}$ at pressure $1.2\times10^5\,\mathrm{Pa}$ . It is compressed isothermally to $1.5\times10^{-3}\,\mathrm{m^3}$ . Calculate its final pressure. [2 marks]

(b) Explain how the compression can be made approximately isothermal in practice. [2 marks]

(c) Explain the pressure increase using kinetic theory. [2 marks]

Show full solution and marking guidance

(a) [2 marks] $\displaystyle p_2=\frac{p_1V_1}{V_2}=\frac{(1.2\times10^5)(2.5\times10^{-3})}{1.5\times10^{-3}}=2.00\times10^5\,\mathrm{Pa}$

(b) [2 marks] Compress slowly while maintaining good thermal contact with a large constant-temperature reservoir. Energy transferred to the gas as work is transferred out as heat, so its internal energy and temperature remain constant.

(c) [2 marks] At unchanged temperature, mean translational kinetic energy and rms speed remain unchanged. The smaller volume gives higher number density and more frequent impacts per unit wall area, increasing pressure.

Question 3 — Root-mean-square speed [6 marks]

(a) Calculate the rms speed of nitrogen molecules at $300\,\mathrm{K}$ . Use molar mass $M=28\times10^{-3}\,\mathrm{kg\,mol^{-1}}$ . [2 marks]

(b) Define rms speed and explain why it is not the mean velocity. [2 marks]

(c) Determine the rms speed at four times the original absolute temperature, for the same gas. [2 marks]

Show full solution and marking guidance

(a) [2 marks] $\displaystyle v_{\mathrm{rms}}=\sqrt{\frac{3RT}{M}}=\sqrt{\frac{3(8.31)(300)}{0.028}}\approx517\,\mathrm{m\,s^{-1}}$

(b) [2 marks] Rms speed is $\sqrt{\langle v^2\rangle}$ , the square root of the mean squared speed. In a gas at rest, the mean velocity is zero because randomly directed velocities cancel; rms speed is positive.

(c) [2 marks] For fixed molecular mass, $v_{\mathrm{rms}}\propto\sqrt T$ . Thus: $\displaystyle v_{\mathrm{rms,new}}=\sqrt4\,v_{\mathrm{rms,old}}\approx1.03\times10^3\,\mathrm{m\,s^{-1}}$

Question 4 — Internal energy of a monatomic ideal gas [5 marks]

(a) A monatomic ideal gas contains $0.40\,\mathrm{mol}$ at $350\,\mathrm{K}$ . Calculate its internal energy in the ideal translational model. [2 marks]

(b) Determine the internal energy if its absolute temperature doubles without changing the amount of gas. [1 mark]

(c) Explain why the same internal-energy expression should not automatically be used for a diatomic gas. [2 marks]

Show full solution and marking guidance

(a) [2 marks] $\displaystyle U=\frac32nRT=\frac32(0.40)(8.31)(350)\approx1.75\times10^3\,\mathrm{J}$

(b) [1 mark] For fixed $n$ , $U\propto T$ . Doubling $T$ doubles $U$ to approximately $3.49\times10^3\,\mathrm{J}$ .

(c) [2 marks] The monatomic model includes translational kinetic energy only. Diatomic molecules can also store rotational and, at sufficiently high temperatures, vibrational energy, so their total internal energy is not generally just the translational contribution.

Question 5 — Deriving pressure from molecular motion [8 marks]

(a) A gas occupies a cube of side $\ell$ . A particle of mass $m$ has velocity component $v_x$ perpendicular to one pair of walls. Show that its average contribution to the force on one wall is $mv_x^2/\ell$ , assuming elastic collisions. [3 marks]

(b) For $N$ particles, use random isotropic motion to derive $p=\frac13\rho v_{\mathrm{rms}}^2$ . [3 marks]

(c) Calculate pressure for density $1.20\,\mathrm{kg\,m^{-3}}$ and rms speed $500\,\mathrm{m\,s^{-1}}$ , and explain what the equation means physically. [2 marks]

Show full solution and marking guidance

(a) [3 marks] The magnitude of momentum transferred to the wall per collision is $2m|v_x|$ . The time between successive collisions with that same wall is $2\ell/|v_x|$ . Therefore: $\displaystyle F_{\mathrm{one\ particle}}=\frac{2m|v_x|}{2\ell/|v_x|}=\frac{mv_x^2}{\ell}$

(b) [3 marks] Add all particles’ force contributions and divide by wall area $\ell^2$ : $\displaystyle p=\frac{m}{\ell^3}\sum_{i=1}^{N}v_{x,i}^2=\frac{Nm}{V}\langle v_x^2\rangle$ Isotropy gives $\langle v_x^2\rangle=\langle v_y^2\rangle=\langle v_z^2\rangle=\langle v^2\rangle/3$ . Since $\rho=Nm/V$ and $v_{\mathrm{rms}}^2=\langle v^2\rangle$ : $\displaystyle p=\frac13\rho v_{\mathrm{rms}}^2$

(c) [2 marks] $\displaystyle p=\frac13(1.20)(500)^2=1.00\times10^5\,\mathrm{Pa}$ Pressure reflects momentum transfer from particle impacts. At fixed rms speed, greater density increases pressure; at fixed density, greater mean squared speed increases pressure.

Question 6 — Adiabatic compression — link to B.4 [7 marks]

(a) Explain why compressing an ideal gas adiabatically can increase its temperature. State what happens to its internal energy. [3 marks]

(b) A fixed $0.200\,\mathrm{mol}$ monatomic ideal gas initially at $300\,\mathrm{K}$ receives $250\,\mathrm{J}$ of work during an adiabatic compression. Calculate the change in internal energy and the final temperature. [3 marks]

(c) Explain why adiabatic does not mean constant temperature. [1 mark]

Show full solution and marking guidance

(a) [3 marks] Adiabatic means $Q=0$ . Compression does work on the gas, so the work done by the gas is negative. From $Q=\Delta U+W$ , $\Delta U=-W>0$ . Increased internal energy raises temperature for a fixed ideal gas.

(b) [3 marks] Here $W=-250\,\mathrm{J}$ , so $\Delta U=+250\,\mathrm{J}$ . $\displaystyle \Delta T=\frac{\Delta U}{(3/2)nR}=\frac{250}{(3/2)(0.200)(8.31)}\approx100\,\mathrm{K}$ $\displaystyle T_f=300+100.281\approx400\,\mathrm{K}$

(c) [1 mark] Adiabatic excludes heat transfer, but work can still change internal energy and temperature. Isothermal specifically means temperature is constant.

Question 7 — Constant-pressure expansion — link to B.4 [7 marks]

(a) An ideal gas expands quasistatically at constant pressure $1.5\times10^5\,\mathrm{Pa}$ from $2.0\times10^{-3}\,\mathrm{m^3}$ to $5.0\times10^{-3}\,\mathrm{m^3}$ . Calculate the work done by the gas. [2 marks]

(b) State how this work is represented on a pressure–volume graph. [1 mark]

(c) If the gas is monatomic, calculate its change in internal energy without needing to know its amount. [2 marks]

(d) Calculate the heat supplied to the gas. [2 marks]

Show full solution and marking guidance

(a) [2 marks] $\displaystyle W=p\Delta V=(1.5\times10^5)(5.0-2.0)\times10^{-3}=450\,\mathrm{J}$ This is positive work done by the expanding gas.

(b) [1 mark] It is the rectangular area under the constant-pressure line between initial and final volumes. The vertical axis is $p$ and the horizontal axis is $V$ .

(c) [2 marks] For a fixed monatomic ideal gas, $U=\frac32nRT=\frac32pV$ . Hence: $\displaystyle \Delta U=\frac32p\Delta V=\frac32(450)=675\,\mathrm{J}$

(d) [2 marks] $\displaystyle Q=\Delta U+W=675+450=1125\,\mathrm{J}\approx1.13\times10^3\,\mathrm{J}$

Question 8 — Combined gas law and Celsius conversion [6 marks]

(a) A sealed sample has initial pressure $250\,\mathrm{kPa}$ , volume $1.80\,\mathrm{L}$ , and temperature $27.0\,^{\circ}\mathrm{C}$ . Its final pressure is $200\,\mathrm{kPa}$ and final temperature $87.0\,^{\circ}\mathrm{C}$ . Convert the temperatures and initial volume into SI units. [2 marks]

(b) Calculate its final volume, assuming ideal behaviour. [3 marks]

(c) Explain why a Celsius temperature ratio would give an incorrect result. [1 mark]

Show full solution and marking guidance

(a) [2 marks] $\displaystyle T_1=300.15\,\mathrm{K},\qquad T_2=360.15\,\mathrm{K},\qquad V_1=1.80\times10^{-3}\,\mathrm{m^3}$

(b) [3 marks] For fixed $n$ : $\displaystyle \frac{p_1V_1}{T_1}=\frac{p_2V_2}{T_2}$ $\displaystyle V_2=\frac{p_1V_1T_2}{p_2T_1}=\frac{(250\times10^3)(1.80\times10^{-3})(360.15)}{(200\times10^3)(300.15)}$ $\displaystyle V_2\approx2.70\times10^{-3}\,\mathrm{m^3}=2.70\,\mathrm{L}$

(c) [1 mark] The ideal gas law uses absolute temperature. Celsius has an offset from absolute zero, so ratios of Celsius temperatures do not represent the physical temperature ratios required.

Question 9 — Gauge pressure and a heated tyre [6 marks]

(a) A sealed tyre has gauge pressure $2.20\times10^5\,\mathrm{Pa}$ at $293\,\mathrm{K}$ . Atmospheric pressure is $1.00\times10^5\,\mathrm{Pa}$ . Calculate the initial absolute gas pressure. [1 mark]

(b) The gas warms to $333\,\mathrm{K}$ . Assume tyre volume and atmospheric pressure remain unchanged and no gas escapes. Calculate the final absolute pressure and gauge pressure. [3 marks]

(c) Explain why the gauge pressure cannot be used directly in the gas-law temperature ratio. State one limitation of the calculation. [2 marks]

Show full solution and marking guidance

(a) [1 mark] $\displaystyle p_{\mathrm{abs},1}=p_{\mathrm{gauge},1}+p_{\mathrm{atm}}=3.20\times10^5\,\mathrm{Pa}$

(b) [3 marks] At fixed $n,V$ , $p_{\mathrm{abs}}/T$ is constant: $\displaystyle p_{\mathrm{abs},2}=(3.20\times10^5)\frac{333}{293}\approx3.64\times10^5\,\mathrm{Pa}$ $\displaystyle p_{\mathrm{gauge},2}=p_{\mathrm{abs},2}-1.00\times10^5\approx2.64\times10^5\,\mathrm{Pa}$

(c) [2 marks] Gauge pressure excludes atmospheric pressure, whereas the gas equation requires the total pressure due to molecular impacts. A real tyre can expand, develop non-uniform gas temperature, or leak; any one valid limitation earns credit.

Question 10 — Amount, particle number, and density [8 marks]

(a) An ideal nitrogen sample has pressure $1.01\times10^5\,\mathrm{Pa}$ , volume $2.00\times10^{-3}\,\mathrm{m^3}$ , and temperature $290\,\mathrm{K}$ . Calculate its amount in moles. [2 marks]

(b) Calculate its number of molecules using $N_A=6.02\times10^{23}\,\mathrm{mol^{-1}}$ . [2 marks]

(c) Calculate its mass density using $M=0.0280\,\mathrm{kg\,mol^{-1}}$ . [2 marks]

(d) Derive $\rho=pM/(RT)$ and explain why density increases with pressure at fixed temperature and composition. [2 marks]

Show full solution and marking guidance

(a) [2 marks] $\displaystyle n=\frac{pV}{RT}=\frac{(1.01\times10^5)(2.00\times10^{-3})}{(8.31)(290)}\approx0.0838\,\mathrm{mol}$

(b) [2 marks] $\displaystyle N=nN_A\approx(0.0838209)(6.02\times10^{23})=5.05\times10^{22}$

(c) [2 marks] $\displaystyle m_{\mathrm{sample}}=nM\approx2.35\times10^{-3}\,\mathrm{kg}$ $\displaystyle \rho=\frac{nM}{V}\approx1.17\,\mathrm{kg\,m^{-3}}$

(d) [2 marks] Substitute $n=m_{\mathrm{sample}}/M=\rho V/M$ into $pV=nRT$ : $\displaystyle pV=\frac{\rho V}{M}RT\quad\Rightarrow\quad\rho=\frac{pM}{RT}$ At fixed $T,M$ , higher pressure requires more mass per unit volume.

Question 11 — Comparing molecular speeds and energies [7 marks]

(a) Helium and oxygen are at $300\,\mathrm{K}$ . Their molar masses are $0.00400\,\mathrm{kg\,mol^{-1}}$ and $0.0320\,\mathrm{kg\,mol^{-1}}$ . Calculate both rms speeds and the ratio helium divided by oxygen. [3 marks]

(b) Calculate the mean translational kinetic energy per particle for each gas. [2 marks]

(c) Explain why equal temperatures do not imply equal rms speeds, and why equal rms speeds for these gases would not imply equal temperatures. [2 marks]

Show full solution and marking guidance

(a) [3 marks] $\displaystyle v_{\mathrm{rms,He}}=\sqrt{\frac{3(8.31)(300)}{0.00400}}\approx1.37\times10^3\,\mathrm{m\,s^{-1}}$ $\displaystyle v_{\mathrm{rms,O_2}}=\sqrt{\frac{3(8.31)(300)}{0.0320}}\approx483\,\mathrm{m\,s^{-1}}$ $\displaystyle \frac{v_{\mathrm{rms,He}}}{v_{\mathrm{rms,O_2}}}=\sqrt{\frac{0.0320}{0.00400}}=\sqrt8\approx2.83$

(b) [2 marks] For each gas: $\displaystyle \langle E_k\rangle=\frac32k_BT=\frac32(1.38\times10^{-23})(300)=6.21\times10^{-21}\,\mathrm{J}$ This is mean translational kinetic energy, not the full molecular energy including rotation.

(c) [2 marks] At equal temperature, mean translational kinetic energies are equal, so lighter particles have higher rms speed. Since $T=Mv_{\mathrm{rms}}^2/(3R)$ , equal rms speed would require different temperatures when molar masses differ.

Question 12 — Interpreting pressure–volume data [8 marks]

(a) A fixed sample of $0.120\,\mathrm{mol}$ has the idealised measurements below. Calculate $pV$ for each state and identify the process. [2 marks]

Data for Question 12
Volume / m³	Pressure / Pa
1.00 × 10⁻³	3.00 × 10⁵
1.50 × 10⁻³	2.00 × 10⁵
2.00 × 10⁻³	1.50 × 10⁵
3.00 × 10⁻³	1.00 × 10⁵

(b) Describe the graph of $p$ against $1/V$ , including its gradient and the gradient’s units. [2 marks]

(c) Use the gradient to calculate the gas temperature. [2 marks]

(d) Explain why a pressure–volume graph alone does not prove that no gas escaped during measurement. [2 marks]

Show full solution and marking guidance

(a) [2 marks] Each product is $300\,\mathrm{Pa\,m^3}=300\,\mathrm{J}$ . For a fixed ideal gas, constant $pV=nRT$ implies constant temperature: the process is isothermal.

(b) [2 marks] The graph is a straight line through the origin because $p=(nRT)(1/V)$ . Its gradient is $nRT=300\,\mathrm{Pa\,m^3}$ , equivalent to joules.

(c) [2 marks] $\displaystyle T=\frac{300}{(0.120)(8.31)}\approx301\,\mathrm{K}$

(d) [2 marks] The measured relationship determines the product $nT$ . If gas escaped while temperature changed, $nT$ could remain constant. Independent checks of temperature and sealing are needed to justify the constant- $n$ interpretation.

Question 13 — Two paths between the same states — link to B.4 [10 marks]

(a) A fixed monatomic ideal gas starts at A with $p_A=1.00\times10^5\,\mathrm{Pa}$ , $V_A=2.00\times10^{-3}\,\mathrm{m^3}$ , and $T_A=300\,\mathrm{K}$ . It ends at B with $p_B=2.00\times10^5\,\mathrm{Pa}$ and $V_B=4.00\times10^{-3}\,\mathrm{m^3}$ . Calculate $T_B$ . [2 marks]

(b) Path 1 first doubles pressure at constant volume, then doubles volume at constant pressure. Path 2 first doubles volume at the initial pressure, then doubles pressure at constant volume. Calculate the work done by the gas along each path. [2 marks]

(c) Calculate the change in internal energy for each path. [2 marks]

(d) Calculate the heat supplied along each path. [2 marks]

(e) Explain why the internal-energy changes agree but the work and heat transfers differ. [2 marks]

Show full solution and marking guidance

(a) [2 marks] $\displaystyle \frac{T_B}{T_A}=\frac{p_BV_B}{p_AV_A}=4\quad\Rightarrow\quad T_B=1200\,\mathrm{K}$

(b) [2 marks] Constant-volume segments do no boundary work. $\displaystyle W_1=(2.00\times10^5)(2.00\times10^{-3})=400\,\mathrm{J}$ $\displaystyle W_2=(1.00\times10^5)(2.00\times10^{-3})=200\,\mathrm{J}$

(c) [2 marks] For both paths: $\displaystyle \Delta U=\frac32(p_BV_B-p_AV_A)=\frac32(800-200)=900\,\mathrm{J}$

(d) [2 marks] $\displaystyle Q_1=\Delta U+W_1=1300\,\mathrm{J}$ $\displaystyle Q_2=\Delta U+W_2=1100\,\mathrm{J}$

(e) [2 marks] Internal energy is a state function, so its change depends only on the initial and final states. Work depends on the pressure during the volume change, or the area under the path on a pressure–volume graph. The heat transfers differ accordingly through energy conservation.

Question 14 — Real gases and the ideal approximation [7 marks]

(a) One mole of a hypothetical gas at $300\,\mathrm{K}$ has the three states in the table below. Calculate $Z=pV/(nRT)$ for each state. [3 marks]

Hypothetical real-gas data for Question 14
State	Pressure / Pa	Volume / m³
1	1.00 × 10⁵	2.50 × 10⁻²
2	5.00 × 10⁶	4.00 × 10⁻⁴
3	2.00 × 10⁷	1.60 × 10⁻⁴

(b) Identify the state closest to ideal behaviour and explain why the ideal approximation is often better at low pressure and high temperature. [2 marks]

(c) Suggest a molecular explanation for a value below one at intermediate pressure and above one at very high pressure. [2 marks]

Show full solution and marking guidance

(a) [3 marks] $\displaystyle Z_1=\frac{(1.00\times10^5)(2.50\times10^{-2})}{(1.00)(8.31)(300)}\approx1.00$ $\displaystyle Z_2=\frac{(5.00\times10^6)(4.00\times10^{-4})}{(1.00)(8.31)(300)}\approx0.802$ $\displaystyle Z_3=\frac{(2.00\times10^7)(1.60\times10^{-4})}{(1.00)(8.31)(300)}\approx1.28$

(b) [2 marks] State 1 is closest to ideal because its Z value is nearest one. Low pressure generally means large separation and a negligible particle-volume fraction; high temperature makes attractive interaction energies less significant relative to kinetic energy.

(c) [2 marks] Attractions can reduce pressure below the ideal prediction, giving Z below one. At very high density, finite particle size and short-range repulsion can increase pressure above the ideal prediction, giving Z above one. These are qualitative explanations, not a complete real-gas equation.

Question 15 — Uncertainty and a changing amount of gas [8 marks]

(a) An ideal gas has $p=(2.00\pm0.04)\times10^5\,\mathrm{Pa}$ , $V=(3.00\pm0.06)\times10^{-3}\,\mathrm{m^3}$ , and $T=(300\pm1)\,\mathrm{K}$ . Calculate its amount in moles. [2 marks]

(b) Estimate the maximum percentage uncertainty in the amount by adding fractional contributions. Treat the gas constant as exact. [3 marks]

(c) State the amount with an absolute uncertainty. [1 mark]

(d) In a separate change, volume remains constant, $10\%$ of the molecules escape, and absolute temperature rises by $5\%$ . Calculate the final-to-initial pressure ratio and explain why pressure does not increase. [2 marks]

Show full solution and marking guidance

(a) [2 marks] $\displaystyle n=\frac{pV}{RT}=\frac{(2.00\times10^5)(3.00\times10^{-3})}{(8.31)(300)}\approx0.241\,\mathrm{mol}$

(b) [3 marks] $\displaystyle \frac{\delta n}{n}\approx\frac{0.04}{2.00}+\frac{0.06}{3.00}+\frac{1}{300}=0.0433$ Maximum percentage uncertainty is $4.33\%$ .

(c) [1 mark] $\displaystyle \delta n\approx(0.240674)(0.0433)\approx0.010\,\mathrm{mol}$ $\displaystyle n=(0.241\pm0.010)\,\mathrm{mol}$

(d) [2 marks] At fixed volume, $p\propto NT$ : $\displaystyle \frac{p_2}{p_1}=(0.900)(1.05)=0.945$ Pressure decreases by $5.50\%$ . The reduction in particle number outweighs the temperature increase; the fixed-amount gas-law ratio is not applicable.

Lab investigation — Boyle’s law with dead volume

Investigation 1 — Gas syringe and pressure sensor [15 marks]

A sealed gas syringe is connected to an absolute-pressure sensor by tubing. The syringe scale gives volume $V_s$ , but the tubing and sensor contain an unknown constant dead volume $V_d$ . The total gas volume is $V_s+V_d$ . The idealised equilibrium data below are taken at constant temperature $298\,\mathrm{K}$ with no leaks. Treat the tabulated values as exact for the calculations.

Equilibrium absolute-pressure measurements
Indicated syringe volume / mL	Absolute pressure / kPa
5.00	400
7.00	300
11.0	200
15.0	150
23.0	100

(a) Describe a method to test Boyle’s law. State the independent and dependent variables, controls, and how approximately isothermal conditions are achieved. [3 marks]

(b) Calculate the products of pressure and indicated syringe volume for the first and last readings. Explain why they differ even though the sample may obey Boyle’s law. [2 marks]

(c) Let $K=nRT$ . Derive a linear equation for a graph of $V_s$ against $1/p$ . State the meaning and units of its gradient and intercept. [3 marks]

(d) Use the readings at indicated volumes 11.0 mL and 23.0 mL to calculate the dead volume and K. [3 marks]

(e) Calculate the amount of gas in the sealed syringe-and-tubing system. [2 marks]

(f) Predict the effect of taking a pressure reading immediately after rapid compression. Explain how the procedure should address this. [2 marks]

Show investigation solution and evaluation

(a) [3 marks] Connect a gas syringe to an absolute-pressure sensor through short sealed tubing. Change the indicated syringe volume, the independent variable, and record the equilibrium absolute pressure, the dependent variable. Maintain the same sealed quantity of gas, room temperature, and tubing. Compress in small slow steps, wait for the reading and temperature to stabilise, and repeat during expansion to check consistency. Check leaks and sensor calibration, include tubing volume in the analysis, and remain within the syringe and sensor pressure ratings.

(b) [2 marks] First reading: $(4.00\times10^5)(5.00\times10^{-6})=2.00\,\mathrm{Pa\,m^3}$ . Last reading: $(1.00\times10^5)(23.0\times10^{-6})=2.30\,\mathrm{Pa\,m^3}$ . These products exclude the constant dead volume. Boyle’s law applies to the total gas volume $V_s+V_d$ , not just the syringe scale reading.

(c) [3 marks] $\displaystyle p(V_s+V_d)=K\quad\Rightarrow\quad V_s=K\left(\frac1p\right)-V_d$ With $V_s$ in $\mathrm{m^3}$ and $1/p$ in $\mathrm{Pa^{-1}}$ , the gradient is $K$ in $\mathrm{Pa\,m^3}$ , and the vertical intercept is $-V_d$ in $\mathrm{m^3}$ .

(d) [3 marks] $\displaystyle 2.00\times10^5(11.0\times10^{-6}+V_d)=1.00\times10^5(23.0\times10^{-6}+V_d)$ $\displaystyle V_d=1.00\times10^{-6}\,\mathrm{m^3}=1.00\,\mathrm{mL}$ $\displaystyle K=(1.00\times10^5)(24.0\times10^{-6})=2.40\,\mathrm{Pa\,m^3}$

(e) [2 marks] $\displaystyle n=\frac{K}{RT}=\frac{2.40}{(8.31)(298)}\approx9.69\times10^{-4}\,\mathrm{mol}$

(f) [2 marks] Rapid compression can temporarily warm the gas, giving a pressure above the isothermal value for that total volume. Wait for the gas to return to room temperature before recording; slower compression and good thermal contact reduce the transient. An immediate high reading need not indicate failure of the ideal gas equation.

Exam tips and common pitfalls

Gas-law pressure must be absolute. Add atmospheric pressure to a gauge reading before using $pV=nRT$ .
Use kelvin for temperature ratios; Celsius differences can equal kelvin differences, but Celsius ratios cannot replace kelvin ratios.
Distinguish particle mass from molar mass and particle number from amount in moles.
For the same gas, rms speed varies as $\sqrt T$ ; at the same temperature, lighter particles have greater rms speed.
Mean translational kinetic energy per particle is set by temperature. Total internal energy also depends on the amount and molecular degrees of freedom.
Use $U=\frac32nRT$ for a monatomic ideal gas, not automatically for every gas.
Isothermal means constant temperature; adiabatic means no heat transfer. They are different conditions.
Keep the work sign convention consistent. Work depends on the path; internal energy is a state function.
For gas experiments, check leaks, temperature equilibration, sensor calibration, and dead volume.

Faith connection — Sustained creation and faithful learning

Hebrews 1:1–3 presents Jesus as the One through whom God made the universe and who sustains all things. In 1 John 1:9, we are invited to confess our sins and receive God’s forgiveness; Revelation 22:17 offers an invitation to receive the water of life. Together, these passages point us toward God’s sustaining power, mercy, and invitation through Jesus.

The study of gases helps us investigate the physical world with care and curiosity. Let your learning also cultivate honesty: record measurements faithfully, acknowledge uncertainty, and treat others with Christ’s love. Reflect on how integrity in scientific work can express your faith.

Further reading: OpenStax — Pressure, temperature, and rms speed.

B.3 — Gas Laws
HL Practice: Paper 2

Instructions, constants, and conventions

Key equations and their limits

Extended-response questions

Question 1 — Ideal gas equation [5 marks]

Question 2 — Boyle’s law and isothermal compression [6 marks]

Question 3 — Root-mean-square speed [6 marks]

Question 4 — Internal energy of a monatomic ideal gas [5 marks]

Question 5 — Deriving pressure from molecular motion [8 marks]

Question 6 — Adiabatic compression — link to B.4 [7 marks]

Question 7 — Constant-pressure expansion — link to B.4 [7 marks]

Question 8 — Combined gas law and Celsius conversion [6 marks]

Question 9 — Gauge pressure and a heated tyre [6 marks]

Question 10 — Amount, particle number, and density [8 marks]

Question 11 — Comparing molecular speeds and energies [7 marks]

Question 12 — Interpreting pressure–volume data [8 marks]

Question 13 — Two paths between the same states — link to B.4 [10 marks]

Question 14 — Real gases and the ideal approximation [7 marks]

Question 15 — Uncertainty and a changing amount of gas [8 marks]

Lab investigation — Boyle’s law with dead volume

Investigation 1 — Gas syringe and pressure sensor [15 marks]

Exam tips and common pitfalls

Faith connection — Sustained creation and faithful learning

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Gas laws HL Papers 1& 2

B.3 — Gas LawsHL Practice: Paper 2

Instructions, constants, and conventions

Key equations and their limits

Extended-response questions

Question 1 — Ideal gas equation [5 marks]

Question 2 — Boyle’s law and isothermal compression [6 marks]

Question 3 — Root-mean-square speed [6 marks]

Question 4 — Internal energy of a monatomic ideal gas [5 marks]

Question 5 — Deriving pressure from molecular motion [8 marks]

Question 6 — Adiabatic compression — link to B.4 [7 marks]

Question 7 — Constant-pressure expansion — link to B.4 [7 marks]

Question 8 — Combined gas law and Celsius conversion [6 marks]

Question 9 — Gauge pressure and a heated tyre [6 marks]

Question 10 — Amount, particle number, and density [8 marks]

Question 11 — Comparing molecular speeds and energies [7 marks]

Question 12 — Interpreting pressure–volume data [8 marks]

Question 13 — Two paths between the same states — link to B.4 [10 marks]

Question 14 — Real gases and the ideal approximation [7 marks]

Question 15 — Uncertainty and a changing amount of gas [8 marks]

Lab investigation — Boyle’s law with dead volume

Investigation 1 — Gas syringe and pressure sensor [15 marks]

Exam tips and common pitfalls

Faith connection — Sustained creation and faithful learning

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B.3 — Gas Laws
HL Practice: Paper 2