What are ideal gas molecules in the context of Edexcel IGCSE Physics?

In Edexcel IGCSE Physics, ideal gas molecules refer to a theoretical model where gas molecules are considered as small, hard spheres that move in random, constant motion. They do not attract or repel each other, and collisions between them are perfectly elastic. This model helps simplify the study of gases by assuming no intermolecular forces and negligible volume of the molecules themselves.

How do ideal gas molecules differ from real gas molecules?

Ideal gas molecules differ from real gas molecules in that they do not experience any intermolecular forces and occupy no volume. Real gas molecules, on the other hand, do have intermolecular attractions and occupy a finite volume. These differences become significant at high pressures and low temperatures, where real gases deviate from ideal behavior.

Why is the concept of ideal gas molecules important in studying gases?

The concept of ideal gas molecules is important because it provides a simplified model that allows students to understand and predict the behavior of gases under various conditions. By using the ideal gas law, which relates pressure, volume, and temperature, students can solve problems and understand the fundamental principles of gas behavior without the complexities of real gas interactions.

What is the ideal gas law and how does it relate to ideal gas molecules?

The ideal gas law is a fundamental equation in physics that relates the pressure, volume, and temperature of an ideal gas. It is expressed as PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature in Kelvin. This law assumes the behavior of ideal gas molecules, allowing predictions of how changes in one variable affect the others.

Can you explain how temperature affects the movement of ideal gas molecules?

In the context of ideal gas molecules, temperature is directly related to the average kinetic energy of the molecules. As temperature increases, the kinetic energy of the molecules also increases, causing them to move faster. Conversely, a decrease in temperature results in slower molecular movement. This relationship is crucial for understanding how gases expand when heated and contract when cooled, as described by the kinetic theory of gases.

Why is "Using Celsius in $p/T$ = constant" a common mistake in Ideal Gas Molecules?

Why it happens: Reading the temperature from the question without converting. How to avoid it: ALWAYS convert °C to K before using gas laws. The relationship p T is only true when T is measured from ABSOLUTE ZERO (i.e. kelvin). Source: 4PH1 Examiner Reports 2022-2024 — single most common mark loss

Why is "Saying $p$ is directly proportional to $V$ (instead of inversely)" a common mistake in Ideal Gas Molecules?

Why it happens: Confusing with the constant-V law. How to avoid it: Compress a gas → volume falls AND pressure RISES → inverse relationship. Picture pushing a syringe to half the volume: pressure doubles.

Why is "Inconsistent volume units (mixing cm³ and m³) in Boyle's law" a common mistake in Ideal Gas Molecules?

Why it happens: Question gives one in cm³ and the other in m³. How to avoid it: Use the SAME unit on both sides. Units cancel in Boyle's law — but only if they're consistent.

Why is "Forgetting to mention CHANGE of momentum when explaining gas pressure" a common mistake in Ideal Gas Molecules?

Why it happens: Skipping straight to 'particles hit the walls'. How to avoid it: Edexcel mark scheme expects: random motion → collisions → CHANGE OF MOMENTUM → force = rate of change of momentum → force/area = pressure. State each link. Source: 4PH1/1P Examiner Reports 2022-2024

Why is "Drawing the wrong shape for p-V and p-T graphs" a common mistake in Ideal Gas Molecules?

Why it happens: Mixing up Boyle's and Gay-Lussac's graphical signatures. How to avoid it: At constant T: p-V graph is a HYPERBOLA (curve falling away from both axes). At constant V: p-T graph is a STRAIGHT LINE through the origin (if T in kelvin), or through −273 °C on the temperature axis (if T in °C).

Solids, Liquids and GasesPearson Edexcel IGCSE Physics 4PH1 Higher

Ideal Gas Molecules

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Ideal Gas Molecules — Pearson Edexcel International GCSE Physics 4PH1 Study Notes (2026 onwards, Spec Issue 4)

How random molecular motion produces gas pressure; absolute zero −273 °C; the Kelvin scale; temperature ∝ average KE; the qualitative p-V (constant T) and p-T (constant V) graphs; Gay-Lussac's law $p_1/T_1 = p_2/T_2$ ; Boyle's law $p_1V_1 = p_2V_2$ .

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4PH1 syllabus (2026 onwards).

5.15 — Understand the existence of random motion of molecules in a gas and that this motion produces pressure on the walls of a container.
5.16 — Understand that there is an absolute zero of temperature which is −273 °C.
5.17 — Describe the Kelvin scale of temperature and convert between Kelvin and degree Celsius: T/K = θ/°C + 273.
5.18 — Understand that an increase in temperature results in an increase in the average speed of gas molecules.
5.19 — Understand that the Kelvin temperature of a gas is proportional to the average kinetic energy of its molecules.
5.20 — Describe qualitatively the pressure-volume relationship at constant temperature, and the pressure-temperature relationship at constant volume.
5.21 — Know and use the relationship between the pressure and Kelvin temperature of a fixed mass of gas at constant volume: $p_1/T_1 = p_2/T_2$ .
5.22 — Know and use the relationship between the pressure and volume of a fixed mass of gas at constant temperature: $p_1V_1 = p_2V_2$ .

Pressure from molecular motion (spec 5.15)

Collisions with walls produce pressure.

The picture. A gas in a container consists of vast numbers of molecules moving randomly in all directions at a range of speeds.

How pressure arises.

A molecule hits a wall and bounces back.
The molecule's MOMENTUM has changed (typically reversed in direction).
Newton's 2nd law: force = rate of change of momentum. The wall exerts a force on the molecule during the collision.
Newton's 3rd law: the molecule exerts an EQUAL AND OPPOSITE force on the wall.
Many molecules colliding many times per second → a continuous AVERAGE force on the wall.
Total force ÷ wall area = PRESSURE.

Why pressure is the same in all directions (spec 5.6 / 5.15). The randomness of molecular motion means molecules hit each part of the wall equally on average → pressure at a point inside the gas is the same in all directions.

Random molecular motion — each wall collision exerts a tiny outward force; together they make gas pressure.

How pressure changes when you change the conditions.

More gas in the SAME volume. More molecules → more collisions per second → higher pressure.
Smaller volume, same gas. Same collision rate per unit area is achieved with fewer molecules per wall section, BUT walls are closer → much higher collision rate. Pressure RISES.
Higher temperature, same volume. Higher T → faster molecules → harder collisions AND more frequent collisions → pressure RISES.

Random motion + collisions = pressure.
Force per area = pressure.
Pressure rises with: more gas, less volume, or higher T.

See the full worked example for ideal gas molecules →

Absolute zero and the Kelvin scale (spec 5.16, 5.17)

Lowest possible T = −273 °C = 0 K.

Absolute zero. As a gas cools, its molecules slow down. Extrapolating the p-T graph back to where pressure would be zero gives a temperature of about $-273$ °C — ABSOLUTE ZERO (spec 5.16). At this temperature, molecules have minimum kinetic energy and cannot be cooled further.

Kelvin scale (spec 5.17). An absolute temperature scale that starts at 0 K. The size of a degree is the same as on the Celsius scale.

$T / \text{K} = \theta / °\text{C} + 273$

Description	Celsius	Kelvin
Absolute zero	−273 °C	0 K
Water freezes	0 °C	273 K
Room temperature	20 °C	293 K
Body temperature	37 °C	310 K
Water boils	100 °C	373 K

Why Kelvin matters for gas laws. The relationships $p \propto T$ (constant V) and $V \propto T$ (constant p) are only true when $T$ is measured FROM absolute zero (i.e. in kelvin). Using Celsius produces a nonsense result — e.g. doubling temperature from 1 °C to 2 °C should barely change the pressure, but Celsius would suggest a factor of 2.

Mark-scheme tip. Convert TO kelvin AT THE START of any gas-law calculation. Then convert BACK to Celsius at the end if the question asks for the answer in °C.

Absolute zero = $-273$ °C = 0 K.
T/K = θ/°C + 273.
ALWAYS use kelvin in gas-law formulas.

See the full worked example for ideal gas molecules →

Temperature ↔ KE (spec 5.18, 5.19)

Kelvin T ∝ average KE.

Spec 5.18. Raising the temperature of a gas raises the AVERAGE SPEED of its molecules.

Spec 5.19. The Kelvin temperature of a gas is DIRECTLY PROPORTIONAL to the average kinetic energy of its molecules.

$\langle KE \rangle \propto T_{\text{(in K)}}$

Consequences.

Doubling T (in K) doubles the average KE → multiplies the average speed by $\sqrt{2}$ .
A gas at 600 K has 2× the average KE of the same gas at 300 K.
At absolute zero (0 K), average KE = 0 — molecules would be (in this classical model) stationary.

Distribution of speeds. At any temperature, molecules have a RANGE of speeds (some fast, some slow). Increasing T shifts the distribution to higher speeds and broadens it. (Not required at 4PH1 level but the average speed concept matters.)

Linking to evaporation. At a temperature well below boiling, the FAST tail of the speed distribution can escape from the surface of a liquid → evaporation. Evaporation REMOVES the fastest molecules → the remaining liquid's average KE drops → it cools.

$\langle KE \rangle \propto T$ (kelvin).
Doubling T doubles average KE.
Molecules have a range of speeds at any T.

Qualitative p-V and p-T graphs (spec 5.20)

Hyperbola and straight line.

Pressure vs Volume at constant temperature (Boyle's-law shape).

Plot $p$ on y-axis, $V$ on x-axis.
Shape: HYPERBOLA — falls away from both axes.
Smaller V → higher p; if you double V, p halves.
Alternative plot: $p$ vs $1/V$ → STRAIGHT LINE through origin (proves inverse proportionality).

Pressure vs Temperature at constant volume (Gay-Lussac shape).

Plot $p$ on y-axis, $T$ on x-axis.
Shape: STRAIGHT LINE.
If $T$ in KELVIN: line passes through ORIGIN (0, 0).
If $T$ in °C: line passes through the temperature axis at $-273$ °C (i.e. extrapolating to $p = 0$ gives absolute zero).

Volume vs Temperature at constant pressure (Charles's law — not in 4PH1 but worth knowing).

Same shape as p-T: straight line through 0 K (or through $-273$ °C).

Edexcel asks for qualitative graphs. Memorise the SHAPES and the intercepts. You don't need to plot specific numerical values unless the question gives them.

Gas-law graph shapes — a p-V hyperbola at constant T, and a straight p-T line through the origin (T in kelvin).

p-V at constant T: hyperbola.
p vs 1/V: straight line through origin.
p-T at constant V: straight line through origin (if T in K).
Extrapolating p-T (in °C) gives intercept at $-273$ °C.

Constant-volume law (spec 5.21)

$p_1/T_1 = p_2/T_2$ — Kelvin only.

Formula.

$\dfrac{p_1}{T_1} = \dfrac{p_2}{T_2}$

For a fixed mass of gas at CONSTANT VOLUME.

$p$ in Pa (or any consistent pressure unit; units cancel).
$T$ MUST be in KELVIN.

Worked example. A sealed gas cylinder is at 300 K and $1.5 \times 10^{5}$ Pa. If it's heated to 600 K, what's the new pressure?

$p_2 = p_1 \times \dfrac{T_2}{T_1} = 1.5 \times 10^{5} \times \dfrac{600}{300} = 3.0 \times 10^{5}\,\text{Pa}$

Doubling kelvin temperature doubles the pressure.

Real-world examples.

Tyre pressure on a hot day. A car tyre at 25 °C (298 K) inflated to 200 kPa might reach 35 °C (308 K) → pressure rises to $200 \times 308/298 \approx 207$ kPa. Larger temperature swings on race tracks can be significant.
Aerosol cans warning. 'Do not heat above 50 °C' — heated gas → higher pressure → can may burst.
Pressure cooker. Heating a sealed pot of water to 120 °C (393 K) raises the internal pressure to about 1.4 atmospheres → water can be hotter than 100 °C without boiling → food cooks faster.

Microscopic explanation. Higher T → faster molecules → harder and more frequent collisions with walls → higher pressure (spec 5.15).

$p_1/T_1 = p_2/T_2$ — T in KELVIN.
Doubling T (K) doubles pressure.
Aerosol cans and tyre pressure are everyday examples.

See the full worked example for ideal gas molecules →

Boyle's law (spec 5.22)

$p_1 V_1 = p_2 V_2$ — same T.

Formula.

$p_1 V_1 = p_2 V_2$

For a fixed mass of gas at CONSTANT TEMPERATURE.

$p$ in Pa (any consistent pressure unit).
$V$ in m³ (or any consistent volume unit — units cancel in this ratio).

Worked example. A gas in a cylinder has $p_1 = 100$ kPa and $V_1 = 500$ cm³. Slowly compress it to $V_2 = 200$ cm³ at the same temperature. What's the new pressure?

$p_2 = p_1 \times \dfrac{V_1}{V_2} = 100 \times \dfrac{500}{200} = 250\,\text{kPa}$

Roughly halving the volume more than doubled the pressure.

Boyle's law — same molecules in a smaller volume hit the walls more often, so pressure rises.

Why constant T matters. Compressing a gas quickly typically heats it (because work is done on it by the piston) → temperature rises → pressure rises further than Boyle's law alone would predict. Boyle's law strictly applies only when the compression is SLOW and the gas is kept at a fixed temperature (isothermal change).

Real-world example: scuba diving.

At the surface, air in a diver's lungs is at $\sim 1$ atmosphere.
At 10 m depth, water pressure adds another atmosphere → total 2 atm.
If the diver inhales at depth and then surfaces while holding their breath, the air in their lungs DOUBLES in volume → potentially fatal lung rupture.
This is why divers are trained to EXHALE continuously while ascending.

Microscopic explanation. Halving the volume halves the space between molecules → twice as many collisions per second on each unit area of wall → pressure doubles. (Same number of molecules, same speed → same average KE, same T.)

$p_1 V_1 = p_2 V_2$ — constant T.
Inverse proportion: half the volume → double the pressure.
Volume units cancel — use consistent units.
Slow compression only — otherwise T changes.

See the full worked example for ideal gas molecules →

How it’s examined

Ideal-gas content appears on every 4PH1 paper — usually 8-12 marks across Paper 1 and Paper 2. Highest-frequency items: Celsius-Kelvin conversions (1-2 marks); Boyle's-law calculations (4 marks); Gay-Lussac calculations (4-5 marks including K conversion); explanations of gas pressure in terms of molecular motion (4-5 marks); qualitative p-V and p-T graph sketches (2-3 marks).

Sources: Pearson Edexcel International GCSE Physics 4PH1 specification — Issue 4, September 2024 (valid for 2026 onwards); Pearson Edexcel 4PH1 Examiner Reports 2022-2024; 4PH1/1P May/Jun 2024 question paper and mark scheme; 4PH1/1P January 2024 question paper and mark scheme. Last reviewed 2026-05-12.

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Step-by-step worked examples — Ideal Gas Molecules

Step-by-step solutions to past-paper-style questions on ideal gas molecules, written exactly the way a tutor would explain them at the board.

1Celsius - Kelvin conversion (3 marks, Paper 1)
Core• Adapted from 4PH1/1P May/Jun 2024• Kelvin, spec-5.16, spec-5.17, Paper 1
Question
Convert the following temperatures: (a) 27 °C to kelvin; (b) 400 K to °C; (c) absolute zero from kelvin to Celsius. (3 marks)
Step-by-step solution
1. Step 1
  (a) Conversion (spec 5.17). $T/\text{K} = \theta/°\text{C} + 273$ . $\therefore T = 27 + 273 = 300$ K.
2. Step 2
  (b) Reverse. $\theta = T - 273 = 400 - 273 = 127$ °C.
3. Step 3
  (c) Absolute zero (spec 5.16). $T = 0$ K → $\theta = 0 - 273 = -273$ °C.
Answer
(a) 300 K. (b) 127 °C. (c) Absolute zero = −273 °C = 0 K.
Examiner tip
Edexcel accepts 273 OR the more accurate 273.15. ALWAYS convert to kelvin before applying $p/T =$ constant or in any T-dependent gas law. Most common slip: forgetting to convert before using Gay-Lussac's law.
2Boyle's law calculation (4 marks, Paper 1)
Core• Adapted from 4PH1/1P January 2024• Boyle's law, spec-5.22, Paper 1
Question
A gas in a sealed cylinder has a pressure of 2.0 × 10⁵ Pa and a volume of 300 cm³. The piston is slowly pushed in so that the volume becomes 150 cm³. The temperature remains constant. Calculate the new pressure. (4 marks)
Step-by-step solution
1. Step 1
  Identify the conditions. Temperature constant → Boyle's law (spec 5.22).
  $p_1 V_1 = p_2 V_2$
2. Step 2
  Substitute (keeping consistent volume units).
  $(2.0 \times 10^{5}) \times 300 = p_2 \times 150$
3. Step 3
  Rearrange for $p_2$ .
  $p_2 = \dfrac{(2.0 \times 10^{5}) \times 300}{150}$
4. Step 4
  Evaluate.
  $p_2 = 4.0 \times 10^{5}\ \text{Pa}$
Answer
New pressure = $4.0 \times 10^{5}$ Pa.
Examiner tip
Edexcel: 1 mark for the law, 1 substitution, 1 rearrangement, 1 final answer with unit. Halving the volume DOUBLES the pressure (inverse proportion). Note: volume units cancel out in this calculation, so cm³ or m³ works — but be CONSISTENT.
3Gay-Lussac (constant-V) law (5 marks, Paper 1)
Extended• Adapted from 4PH1/1P May/Jun 2024• constant volume, Kelvin, spec-5.21, Paper 1
Question
A rigid metal can contains air at 300 K and a pressure of $1.0 \times 10^{5}$ Pa. The can is heated until the pressure reaches $1.4 \times 10^{5}$ Pa. Calculate the temperature of the air at this new pressure, in °C. (5 marks)
Step-by-step solution
1. Step 1
  Identify the conditions. Volume constant (rigid can) → use spec 5.21.
  $\dfrac{p_1}{T_1} = \dfrac{p_2}{T_2}$
2. Step 2
  Substitute.
  $\dfrac{1.0 \times 10^{5}}{300} = \dfrac{1.4 \times 10^{5}}{T_2}$
3. Step 3
  Rearrange for $T_2$ .
  $T_2 = 300 \times \dfrac{1.4 \times 10^{5}}{1.0 \times 10^{5}}$
4. Step 4
  Evaluate.
  $T_2 = 300 \times 1.4 = 420\ \text{K}$
5. Step 5
  Convert to °C. $\theta = T - 273 = 420 - 273 = 147$ °C.
Answer
$T_2 = 420$ K = 147 °C.
Examiner tip
CRITICAL: temperatures in $p/T$ MUST be in KELVIN. Examiner reports flag this as the most common mistake. Convert °C → K BEFORE using the formula. The final answer in °C is just a unit re-write at the end.
4Explaining gas pressure (5 marks, Paper 1)
Core• Adapted from 4PH1/1P January 2024• gas pressure, molecular motion, spec-5.15, Paper 1
Question
Explain, in terms of molecular motion, how a gas exerts a pressure on the walls of its container. State how the pressure changes when the gas is heated at constant volume. (5 marks)
Step-by-step solution
1. Step 1
  Random motion (spec 5.15). Gas molecules move RANDOMLY in all directions with a range of speeds.
2. Step 2
  Collisions. Molecules collide with the walls of the container. At each collision, the molecule's momentum CHANGES (reverses direction).
3. Step 3
  Force from collisions. Newton's 2nd law: rate of change of momentum = force. The wall exerts a force on the molecule; by Newton's 3rd law, the molecule exerts an equal and opposite force on the wall.
4. Step 4
  Pressure. Total force from all the collisions divided by the wall area = pressure.
5. Step 5
  Heated at constant volume. Higher temperature → particles have HIGHER average KE → faster on average. They hit the walls MORE OFTEN and with GREATER force per collision → pressure INCREASES. This matches the gas law $p \propto T$ (spec 5.21, in kelvin).
Answer
Random molecular motion; collisions with walls reverse momentum → force on walls → pressure. Heating → faster molecules → more frequent and harder collisions → higher pressure.
Examiner tip
Edexcel mark scheme rewards: (1) random motion, (2) collisions with walls, (3) change of momentum, (4) force × area = pressure, (5) heating → faster → more frequent/harder collisions. State each link explicitly.

Key Formulae — Ideal Gas Molecules

The formulae you need to memorise for ideal gas molecules on the Pearson Edexcel IGCSE 4PH1 paper, with every variable defined in plain English and a note on when to use it.

Celsius - Kelvin conversion (spec 5.17)
$T/\text{K} = \theta/°\text{C} + 273$
When to use
$T$ in kelvin, $\theta$ in degrees Celsius. Always convert to kelvin BEFORE using any gas-law formula that has T in it.
Gay-Lussac's law — constant volume (spec 5.21)
$\dfrac{p_1}{T_1} = \dfrac{p_2}{T_2}$
When to use
Both papers. Pressure proportional to absolute temperature at constant volume. T in KELVIN. Used for sealed rigid containers (gas can, tyre on a hot day).
Boyle's law — constant temperature (spec 5.22)
$p_1 V_1 = p_2 V_2$
When to use
Both papers. Pressure inversely proportional to volume at constant temperature. Volume units cancel; use any consistent unit. Used for slow compression / expansion in a cylinder with piston.

Key Definitions and Keywords — Ideal Gas Molecules

Definitions to memorise and the exact keywords mark schemes credit for ideal gas molecules answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IGCSE 4PH1 sitting.

Random molecular motion (spec 5.15)
Examiner keyword
Gas molecules move continuously in random directions at a range of speeds. Collisions with the container walls produce a force on the walls — gas pressure.
Absolute zero (spec 5.16)
Examiner keyword
The lowest temperature physically possible: −273 °C = 0 K. At this temperature the particles have minimum kinetic energy (zero in the classical picture).
Kelvin scale (spec 5.17)
Examiner keyword
An absolute temperature scale that starts at 0 K (= −273 °C). 1 K increment = 1 °C increment. Conversion: T/K = θ/°C + 273. Used in all gas-law calculations.
Kelvin temperature ∝ average KE (spec 5.19)
Examiner keyword
The absolute (Kelvin) temperature of a gas is proportional to the average kinetic energy of its molecules. Doubling T (in K) doubles the average KE.
Boyle's law (spec 5.22)
Examiner keyword
At constant temperature, the pressure of a fixed mass of gas is INVERSELY proportional to its volume. $p_1 V_1 = p_2 V_2$ .
Pressure law (spec 5.21)
Examiner keyword
At constant volume, the pressure of a fixed mass of gas is DIRECTLY proportional to its absolute (Kelvin) temperature. $p_1/T_1 = p_2/T_2$ .

Common Mistakes and Misconceptions — Ideal Gas Molecules

The traps other students keep falling into on ideal gas molecules questions — taken from recent Pearson Edexcel IGCSE 4PH1 examiner reports and mark schemes — and how to avoid them.

✕Using Celsius in $p/T$ = constant
4PH1 Examiner Reports 2022-2024 — single most common mark loss
Why it happens
Reading the temperature from the question without converting.
How to avoid it
ALWAYS convert °C to K before using gas laws. The relationship $p \propto T$ is only true when $T$ is measured from ABSOLUTE ZERO (i.e. kelvin).
✕Saying $p$ is directly proportional to $V$ (instead of inversely)
Why it happens
Confusing with the constant-V law.
How to avoid it
Compress a gas → volume falls AND pressure RISES → inverse relationship. Picture pushing a syringe to half the volume: pressure doubles.
✕Inconsistent volume units (mixing cm³ and m³) in Boyle's law
Why it happens
Question gives one in cm³ and the other in m³.
How to avoid it
Use the SAME unit on both sides. Units cancel in Boyle's law — but only if they're consistent.
✕Forgetting to mention CHANGE of momentum when explaining gas pressure
4PH1/1P Examiner Reports 2022-2024
Why it happens
Skipping straight to 'particles hit the walls'.
How to avoid it
Edexcel mark scheme expects: random motion → collisions → CHANGE OF MOMENTUM → force = rate of change of momentum → force/area = pressure. State each link.
✕Drawing the wrong shape for p-V and p-T graphs
Why it happens
Mixing up Boyle's and Gay-Lussac's graphical signatures.
How to avoid it
At constant T: p-V graph is a HYPERBOLA (curve falling away from both axes). At constant V: p-T graph is a STRAIGHT LINE through the origin (if T in kelvin), or through −273 °C on the temperature axis (if T in °C).

Practice questions

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Ideal Gas Molecules — frequently asked questions

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Ideal Gas Molecules

Short Study Notes in the form of Slides

Short Study Notes — Ideal Gas Molecules

Detailed Notes

What you’ll learn

How it’s examined

1Celsius - Kelvin conversion (3 marks, Paper 1)

2Boyle's law calculation (4 marks, Paper 1)

3Gay-Lussac (constant-V) law (5 marks, Paper 1)

4Explaining gas pressure (5 marks, Paper 1)

Celsius - Kelvin conversion (spec 5.17)

Gay-Lussac's law — constant volume (spec 5.21)

Boyle's law — constant temperature (spec 5.22)

Random molecular motion (spec 5.15)

Absolute zero (spec 5.16)

Kelvin scale (spec 5.17)

Kelvin temperature ∝ average KE (spec 5.19)

Boyle's law (spec 5.22)

Pressure law (spec 5.21)

✕Using Celsius in p/Tp/Tp/T = constant

✕Saying ppp is directly proportional to VVV (instead of inversely)

✕Inconsistent volume units (mixing cm³ and m³) in Boyle's law

✕Forgetting to mention CHANGE of momentum when explaining gas pressure

✕Drawing the wrong shape for p-V and p-T graphs

Practice questions

Past paper style quiz

Assess your understanding

Ideal Gas Molecules — frequently asked questions

✕Using Celsius in $p/T$ = constant

✕Saying $p$ is directly proportional to $V$ (instead of inversely)