IB Physics Topic 2.3: Gas laws | Notes & Flashcards | Aimnova

IB Physics — Gas laws

Topic 2.3 of IB Physics covers Gas laws, which is part of Unit 2: The particulate nature of matter. Students explore key concepts including Pressure, volume and temperature relationships, Ideal gas law, moles and Avogadro, Kinetic model of an ideal gas. A strong understanding of gas laws is essential for IB Physics exams and builds the foundation for connected topics across the syllabus.

Key concepts in Gas laws

Key Idea: This topic is about a gas's pressure P, volume V, temperature T and how much gas there is — and the handful of equations that tie them together. Hold one quantity fixed and the others are linked (the gas laws); count the particles and the ideal gas law PV = nRT = N kB T does it all at once; zoom in and the kinetic model explains why — pressure is particles hitting the walls and temperature is their average kinetic energy. It is examined on both papers — quick Paper 1A multiple-choice (compare two samples, P-against-1/V graphs, 'same T → same average KE') and longer Paper 2 structured questions (a before/after gas-law calculation, find moles or molecules, or explain the particle picture in words). One trap runs through the whole topic: temperature must be in kelvin.

📋 Key formulas

Every equation here is given in the data booklet — look for the booklet badge. There is nothing to memorise; the skill is choosing the right form and putting T in kelvin.

\frac{PV}{T} = \text{constant}

Combined gas law for a fixed amount of gas. Given. Used as a before/after pair: P₁V₁ ÷ T₁ = P₂V₂ ÷ T₂. T is the absolute (kelvin) temperature.

$P$: pressure of the gas (Pa)
$V$: volume of the gas (m³)
$T$: absolute temperature (K) — always in kelvin

PV = nRT = N k_B T

Ideal gas law. Given. Use the n-form (moles with R) OR the N-form (molecules with k_B) — never mix them. T in kelvin.

$P$: pressure (Pa)
$V$: volume (m³)
$n$: amount of gas, in moles (mol)
$R$: molar gas constant, 8.31 J K⁻¹ mol⁻¹ (given)
$N$: number of molecules (a plain count, no unit)
$k_B$: Boltzmann constant, 1.38 × 10⁻²³ J K⁻¹ (given)
$T$: absolute temperature (K — kelvin)

n = \frac{N}{N_A}

Converts between the amount in moles and the raw number of molecules. Given. N_A = 6.02 × 10²³ per mole, so N = n × N_A.

$n$: amount of gas, in moles (mol)
$N$: number of molecules (a plain count)
$N_A$: Avogadro constant, 6.02 × 10²³ mol⁻¹ (given)

\overline{E_k} = \tfrac{3}{2}k_B T

Average kinetic energy of ONE particle. Given. Depends only on T (in kelvin) — not on the gas or its mass.

$\overline{E_k}$: average kinetic energy of ONE particle (J)
$k_B$: Boltzmann constant, 1.38 × 10⁻²³ J K⁻¹ (given)
$T$: absolute temperature (K — kelvin)

⚖️ The three gas laws

Each gas law is the combined law PV ÷ T = constant with one quantity held fixed so it cancels. Boyle's needs no kelvin (no T in it); the other two must use kelvin.

Boyle's law P V = K rearranges to P = K × (1/V). So a graph of P against 1/V (at fixed temperature) is a straight line through the origin whose slope is K. Its SI unit is pressure × volume = Pa m³ = J (the joule). This is the classic Paper 1B data-handling task.

🔢 The three constants (don't mix them up)

R and kB describe the same gas, so the two forms PV = nRT and PV = NkBT must agree. Since N = n NA, that forces R = NA × kB. Check: 6.02 × 10²³ × 1.38 × 10⁻²³ = 8.31 ✓ — exactly the molar gas constant.

✏️ Worked exam-style questions

IB-style question — combined gas law (both P and T change)

A balloon holds 2.0 m³ of gas at a pressure of 100 kPa and a temperature of 27 °C. It is taken to a place where the pressure is 150 kPa and the temperature is 177 °C. Find the new volume of the gas.

Solution:

Convert both temperatures to kelvin first (°C + 273):
$T_1 = 27 + 273 = 300\ \text{K},\quad T_2 = 177 + 273 = 450\ \text{K}$
Start with the given combined gas law as a before/after pair:
$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$
Rearrange for V₂:
$V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1}$
Put in the numbers (V₁ = 2.0, P₁ = 100, P₂ = 150, T₁ = 300, T₂ = 450):
$V_2 = 2.0 \times \frac{100}{150} \times \frac{450}{300}$
Work it out — keep the unit:
$V_2 = 2.0 \times 0.667 \times 1.5 = 2.0\ \text{m}^{3}$

Final answer:

V₂ = 2.0 m³ — the volume is unchanged. The 1.5× pressure rise (which would shrink it) is exactly cancelled by the 1.5× kelvin-temperature rise (which would expand it).

IB-style question — pressure of a sealed rigid can (Gay-Lussac)

An aerosol can holds gas at 200 kPa at 27 °C. It is left in the sun and warms to 87 °C. The can is rigid (fixed volume). (a) Find the new pressure. (b) Find the percentage increase in pressure.

Solution:

(a) Convert both temperatures to kelvin:
$T_1 = 27 + 273 = 300\ \text{K},\quad T_2 = 87 + 273 = 360\ \text{K}$
(a) Volume is fixed, so the combined law reduces to Gay-Lussac's (V cancels):
$\frac{P_1}{T_1} = \frac{P_2}{T_2}\;\Rightarrow\; P_2 = P_1 \times \frac{T_2}{T_1}$
(a) Put in the numbers (P₁ = 200, T₁ = 300, T₂ = 360):
$P_2 = 200 \times \frac{360}{300} = 240\ \text{kPa}$
(b) Percentage change = change ÷ original × 100:
$\%\ \text{rise} = \frac{240 - 200}{200} \times 100 = 20\%$

Final answer:

(a) P₂ = 240 kPa. (b) a 20% rise. Note: 87 ÷ 27 = 3.2 is the WRONG ratio — pressure scales with the kelvin temperatures (360 ÷ 300 = 1.2), not the Celsius ones.

IB-style question — moles and molecules (ideal gas law)

A flask contains an ideal gas at a pressure of 2.0 × 10⁵ Pa in a volume of 3.0 × 10⁻⁴ m³ at a temperature of 27 °C. Taking kB = 1.38 × 10⁻²³ J K⁻¹ and NA = 6.02 × 10²³ mol⁻¹, find (a) the number of molecules N and (b) the amount of gas in moles.

Solution:

(a) Convert the temperature to kelvin: T = 27 + 273 = 300 K. Use the given law in its N-form, rearranged for N:
$PV = N k_B T \;\Rightarrow\; N = \frac{PV}{k_B T}$
(a) Put in the numbers (P = 2.0 × 10⁵, V = 3.0 × 10⁻⁴, T = 300):
$N = \frac{(2.0\times10^{5})(3.0\times10^{-4})}{(1.38\times10^{-23})(300)}$
(a) Work it out — N is a count, no unit:
$N = 1.4\times10^{22}$
(b) Convert to moles with the given bridge n = N ÷ NA:
$n = \frac{1.4\times10^{22}}{6.02\times10^{23}} = 2.4\times10^{-2}\ \text{mol}$

Final answer:

(a) N ≈ 1.4 × 10²² molecules. (b) n ≈ 2.4 × 10⁻² mol. Either form works — N uses kB, n uses R; here the N-form was quickest because the molecule count was asked for first.

IB-style question — average kinetic energy and comparing two gases

(a) A sample of neon is at 227 °C. With kB = 1.38 × 10⁻²³ J K⁻¹, find the average kinetic energy of one neon atom. (b) A sample of argon (heavier atoms) sits beside it at the same 227 °C. Compare the average kinetic energy of an argon atom with that of a neon atom.

Solution:

(a) Convert the temperature to kelvin:
$T = 227 + 273 = 500\ \text{K}$
(a) Use the given formula:
$\overline{E_k} = \tfrac{3}{2}k_B T = \tfrac{3}{2}(1.38\times10^{-23})(500)$
(a) Work it out — keep the unit:
$\overline{E_k} = 1.0\times10^{-20}\ \text{J}$
(b) Average kinetic energy depends only on temperature — same T means the same Ēₖ. The heavier argon atoms simply move more slowly to carry that same energy.

Final answer:

(a) Ēₖ ≈ 1.0 × 10⁻²⁰ J per atom. (b) Identical average kinetic energy — temperature alone sets it; mass changes only the speed, not the energy.

🧠 Quick self-check

Tap each card to reveal the answer.

🎯 Exam tips

Exam Tips

Temperature ALWAYS in kelvin (K = °C + 273) before any gas-law or kinetic formula. Scaling on the Celsius numbers is the classic trap — pressure and volume scale with the kelvin temperatures.
For a before/after change, write the combined law P₁V₁ ÷ T₁ = P₂V₂ ÷ T₂. Whatever is held fixed cancels: fixed T → Boyle (P₁V₁ = P₂V₂); fixed P → Charles; fixed V → Gay-Lussac.
Ideal gas law: use n with R (8.31), or N with kB (1.38 × 10⁻²³) — never mix the two forms in one equation.
To compare two samples, write PV = NkT (or nRT) for each and DIVIDE one by the other — any equal quantity (P, V or T) cancels, leaving a clean ratio.
Convert moles ↔ molecules with n = N ÷ NA (NA = 6.02 × 10²³). To get N from n you multiply: N = n × NA.
A graph of P against 1/V (fixed temperature) is a straight line through the origin; its slope is the Boyle constant K, with SI unit Pa m³ = J.
Average kinetic energy Ēₖ = (3/2)kB T depends ONLY on the kelvin temperature — same T means the same average KE for any gas; heavier particles just move slower.
Pressure comes from particles hitting the walls; compressing a gas quickly does work on it, raising the particles' average KE — so a higher temperature and faster particles.

IB Physics — Gas laws

Key concepts in Gas laws

Key Idea: This topic is about a gas's pressure P, volume V, temperature T and how much gas there is — and the handful of equations that tie them together. Hold one quantity fixed and the others are linked (the gas laws); count the particles and the ideal gas law PV = nRT = N kB T does it all at once; zoom in and the kinetic model explains why — pressure is particles hitting the walls and temperature is their average kinetic energy. It is examined on both papers — quick Paper 1A multiple-choice (compare two samples, P-against-1/V graphs, 'same T → same average KE') and longer Paper 2 structured questions (a before/after gas-law calculation, find moles or molecules, or explain the particle picture in words). One trap runs through the whole topic: temperature must be in kelvin.

📋 Key formulas

Every equation here is given in the data booklet — look for the booklet badge. There is nothing to memorise; the skill is choosing the right form and putting T in kelvin.

\frac{PV}{T} = \text{constant}

Combined gas law for a fixed amount of gas. Given. Used as a before/after pair: P₁V₁ ÷ T₁ = P₂V₂ ÷ T₂. T is the absolute (kelvin) temperature.

$P$: pressure of the gas (Pa)
$V$: volume of the gas (m³)
$T$: absolute temperature (K) — always in kelvin

PV = nRT = N k_B T

Ideal gas law. Given. Use the n-form (moles with R) OR the N-form (molecules with k_B) — never mix them. T in kelvin.

$P$: pressure (Pa)
$V$: volume (m³)
$n$: amount of gas, in moles (mol)
$R$: molar gas constant, 8.31 J K⁻¹ mol⁻¹ (given)
$N$: number of molecules (a plain count, no unit)
$k_B$: Boltzmann constant, 1.38 × 10⁻²³ J K⁻¹ (given)
$T$: absolute temperature (K — kelvin)

n = \frac{N}{N_A}

Converts between the amount in moles and the raw number of molecules. Given. N_A = 6.02 × 10²³ per mole, so N = n × N_A.

$n$: amount of gas, in moles (mol)
$N$: number of molecules (a plain count)
$N_A$: Avogadro constant, 6.02 × 10²³ mol⁻¹ (given)

\overline{E_k} = \tfrac{3}{2}k_B T

Average kinetic energy of ONE particle. Given. Depends only on T (in kelvin) — not on the gas or its mass.

$\overline{E_k}$: average kinetic energy of ONE particle (J)
$k_B$: Boltzmann constant, 1.38 × 10⁻²³ J K⁻¹ (given)
$T$: absolute temperature (K — kelvin)

⚖️ The three gas laws

Each gas law is the combined law PV ÷ T = constant with one quantity held fixed so it cancels. Boyle's needs no kelvin (no T in it); the other two must use kelvin.

Boyle's law P V = K rearranges to P = K × (1/V). So a graph of P against 1/V (at fixed temperature) is a straight line through the origin whose slope is K. Its SI unit is pressure × volume = Pa m³ = J (the joule). This is the classic Paper 1B data-handling task.

🔢 The three constants (don't mix them up)

R and kB describe the same gas, so the two forms PV = nRT and PV = NkBT must agree. Since N = n NA, that forces R = NA × kB. Check: 6.02 × 10²³ × 1.38 × 10⁻²³ = 8.31 ✓ — exactly the molar gas constant.

✏️ Worked exam-style questions

IB-style question — combined gas law (both P and T change)

Solution:

Convert both temperatures to kelvin first (°C + 273):
$T_1 = 27 + 273 = 300\ \text{K},\quad T_2 = 177 + 273 = 450\ \text{K}$
Start with the given combined gas law as a before/after pair:
$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$
Rearrange for V₂:
$V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1}$
Put in the numbers (V₁ = 2.0, P₁ = 100, P₂ = 150, T₁ = 300, T₂ = 450):
$V_2 = 2.0 \times \frac{100}{150} \times \frac{450}{300}$
Work it out — keep the unit:
$V_2 = 2.0 \times 0.667 \times 1.5 = 2.0\ \text{m}^{3}$

Final answer:

V₂ = 2.0 m³ — the volume is unchanged. The 1.5× pressure rise (which would shrink it) is exactly cancelled by the 1.5× kelvin-temperature rise (which would expand it).

IB-style question — pressure of a sealed rigid can (Gay-Lussac)

An aerosol can holds gas at 200 kPa at 27 °C. It is left in the sun and warms to 87 °C. The can is rigid (fixed volume). (a) Find the new pressure. (b) Find the percentage increase in pressure.

Solution:

(a) Convert both temperatures to kelvin:
$T_1 = 27 + 273 = 300\ \text{K},\quad T_2 = 87 + 273 = 360\ \text{K}$
(a) Volume is fixed, so the combined law reduces to Gay-Lussac's (V cancels):
$\frac{P_1}{T_1} = \frac{P_2}{T_2}\;\Rightarrow\; P_2 = P_1 \times \frac{T_2}{T_1}$
(a) Put in the numbers (P₁ = 200, T₁ = 300, T₂ = 360):
$P_2 = 200 \times \frac{360}{300} = 240\ \text{kPa}$
(b) Percentage change = change ÷ original × 100:
$\%\ \text{rise} = \frac{240 - 200}{200} \times 100 = 20\%$

Final answer:

(a) P₂ = 240 kPa. (b) a 20% rise. Note: 87 ÷ 27 = 3.2 is the WRONG ratio — pressure scales with the kelvin temperatures (360 ÷ 300 = 1.2), not the Celsius ones.

IB-style question — moles and molecules (ideal gas law)

Solution:

(a) Convert the temperature to kelvin: T = 27 + 273 = 300 K. Use the given law in its N-form, rearranged for N:
$PV = N k_B T \;\Rightarrow\; N = \frac{PV}{k_B T}$
(a) Put in the numbers (P = 2.0 × 10⁵, V = 3.0 × 10⁻⁴, T = 300):
$N = \frac{(2.0\times10^{5})(3.0\times10^{-4})}{(1.38\times10^{-23})(300)}$
(a) Work it out — N is a count, no unit:
$N = 1.4\times10^{22}$
(b) Convert to moles with the given bridge n = N ÷ NA:
$n = \frac{1.4\times10^{22}}{6.02\times10^{23}} = 2.4\times10^{-2}\ \text{mol}$

Final answer:

(a) N ≈ 1.4 × 10²² molecules. (b) n ≈ 2.4 × 10⁻² mol. Either form works — N uses kB, n uses R; here the N-form was quickest because the molecule count was asked for first.

IB-style question — average kinetic energy and comparing two gases

Solution:

(a) Convert the temperature to kelvin:
$T = 227 + 273 = 500\ \text{K}$
(a) Use the given formula:
$\overline{E_k} = \tfrac{3}{2}k_B T = \tfrac{3}{2}(1.38\times10^{-23})(500)$
(a) Work it out — keep the unit:
$\overline{E_k} = 1.0\times10^{-20}\ \text{J}$
(b) Average kinetic energy depends only on temperature — same T means the same Ēₖ. The heavier argon atoms simply move more slowly to carry that same energy.

Final answer:

(a) Ēₖ ≈ 1.0 × 10⁻²⁰ J per atom. (b) Identical average kinetic energy — temperature alone sets it; mass changes only the speed, not the energy.

🧠 Quick self-check

Tap each card to reveal the answer.

🎯 Exam tips

Exam Tips

Temperature ALWAYS in kelvin (K = °C + 273) before any gas-law or kinetic formula. Scaling on the Celsius numbers is the classic trap — pressure and volume scale with the kelvin temperatures.
For a before/after change, write the combined law P₁V₁ ÷ T₁ = P₂V₂ ÷ T₂. Whatever is held fixed cancels: fixed T → Boyle (P₁V₁ = P₂V₂); fixed P → Charles; fixed V → Gay-Lussac.
Ideal gas law: use n with R (8.31), or N with kB (1.38 × 10⁻²³) — never mix the two forms in one equation.
To compare two samples, write PV = NkT (or nRT) for each and DIVIDE one by the other — any equal quantity (P, V or T) cancels, leaving a clean ratio.
Convert moles ↔ molecules with n = N ÷ NA (NA = 6.02 × 10²³). To get N from n you multiply: N = n × NA.
A graph of P against 1/V (fixed temperature) is a straight line through the origin; its slope is the Boyle constant K, with SI unit Pa m³ = J.
Average kinetic energy Ēₖ = (3/2)kB T depends ONLY on the kelvin temperature — same T means the same average KE for any gas; heavier particles just move slower.
Pressure comes from particles hitting the walls; compressing a gas quickly does work on it, raising the particles' average KE — so a higher temperature and faster particles.

IB Physics — Gas laws

Key concepts in Gas laws

📋 Key formulas

⚖️ The three gas laws

🔢 The three constants (don't mix them up)

✏️ Worked exam-style questions

IB-style question — combined gas law (both P and T change)

IB-style question — pressure of a sealed rigid can (Gay-Lussac)

IB-style question — moles and molecules (ideal gas law)

IB-style question — average kinetic energy and comparing two gases

🧠 Quick self-check

🎯 Exam tips

Exam Tips

What you'll learn in Topic 2.3

Study resources — 2.3 Gas laws

Pressure, volume and temperature relationships

Ideal gas law, moles and Avogadro

Kinetic model of an ideal gas

Ready to study Gas laws?

Ready to practice?

IB Physics — Gas laws

Key concepts in Gas laws

📋 Key formulas

⚖️ The three gas laws

🔢 The three constants (don't mix them up)

✏️ Worked exam-style questions

IB-style question — combined gas law (both P and T change)

IB-style question — pressure of a sealed rigid can (Gay-Lussac)

IB-style question — moles and molecules (ideal gas law)

IB-style question — average kinetic energy and comparing two gases

🧠 Quick self-check

🎯 Exam tips

Exam Tips

What you'll learn in Topic 2.3

Study resources — 2.3 Gas laws

Pressure, volume and temperature relationships

Ideal gas law, moles and Avogadro

Kinetic model of an ideal gas

Ready to study Gas laws?

Ready to practice?