What is the relationship between temperature and pressure in gases according to the Cambridge IGCSE Coordinated Science syllabus?

In the Cambridge IGCSE Coordinated Science syllabus, it is explained that the pressure of a gas is directly proportional to its temperature, provided the volume remains constant. This means that as the temperature of a gas increases, the kinetic energy of its molecules increases, leading to more frequent and forceful collisions with the walls of the container, thus increasing the pressure.

How does the concept of pressure changes apply to real-life scenarios in Thermal Physics?

Pressure changes are crucial in various real-life applications, such as in weather systems where changes in atmospheric pressure can indicate different weather conditions. In Thermal Physics, understanding pressure changes is also essential for designing pressure vessels and understanding how gases behave under different conditions, such as in car engines or refrigeration systems.

Why does pressure decrease with altitude, and how is this explained in the IGCSE Coordinated Science curriculum?

According to the IGCSE Coordinated Science curriculum, pressure decreases with altitude because the density of air molecules decreases as you move higher above sea level. This results in fewer air molecules exerting force on a given area, thus reducing pressure. This concept is important in understanding how altitude affects breathing and weather patterns.

What is Boyle's Law and how does it relate to pressure changes in gases?

Boyle's Law is a fundamental principle in Thermal Physics that states that the pressure of a gas is inversely proportional to its volume when temperature is held constant. This means that if the volume of a gas decreases, its pressure increases, and vice versa. This law is crucial for understanding how pressure changes occur in closed systems, such as syringes or hydraulic systems.

How do pressure changes affect boiling points, as discussed in the Cambridge IGCSE Coordinated Science course?

In the Cambridge IGCSE Coordinated Science course, it is taught that pressure changes can significantly affect boiling points. At higher altitudes, where atmospheric pressure is lower, the boiling point of water decreases, meaning it boils at a lower temperature. Conversely, increasing the pressure can raise the boiling point, which is why pressure cookers can cook food faster by increasing the pressure inside the pot.

Why is "Using Celsius instead of kelvin in gas law calculations" a common mistake in Pressure Changes?

Why it happens: Students forget to convert, especially when the temperature change is small. How to avoid it: Always add 273 to convert °C to K before substituting into any gas law. Celsius temperatures give wrong ratios.

Why is "Writing $p_{1}V_{1} = p_{2}V_{2}$ but then calculating $p_{2} = p_{1}V_{1} \times V_{2}$ (multiplying instead of dividing)" a common mistake in Pressure Changes?

Why it happens: Algebraic rearrangement error. How to avoid it: Rearrange carefully: p_2 = p_1V_1/V_2. If pressure increases, volume must decrease — check this sense before accepting your answer.

Why is "Forgetting to add atmospheric pressure to liquid pressure for total pressure" a common mistake in Pressure Changes?

Why it happens: p = h g gives only the pressure due to the liquid column. How to avoid it: Total pressure = liquid pressure + atmospheric pressure. Read the question: if it asks for pressure 'due to the water', use h g only.

Cambridge IGCSE Coordinated Science (0654)

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Short Notes - Pressure Changes

Detailed Study Notes

Detailed notes on Thermal Physics for Cambridge IGCSE Coordinated Science, covering key concepts, explanations, examples, and exam-focused revision points.

Pressure Changes in Gases — Cambridge IGCSE Coordinated Science 0654

Gas pressure, volume, and temperature are related by the gas laws. Cambridge tests Boyle's Law (P ∝ 1/V at constant T), the pressure law (P ∝ T at constant V), and qualitative understanding of Charles' Law.

At a glance

Boyle's Law: P₁V₁ = P₂V₂ (constant temperature). P ∝ 1/V.
Pressure Law: P/T = constant (constant volume). P ∝ T (in Kelvin).
Kelvin scale: T(K) = T(°C) + 273. Absolute zero = 0 K = −273°C.
Kinetic explanation: smaller volume → particles hit walls more frequently → higher pressure.
Higher temperature → particles move faster → harder and more frequent collisions → higher pressure.

What you’ll learn

Mapped to the Cambridge IGCSE 0654 syllabus (2025-2027).

P3.2a — Recall and use Boyle's Law (P₁V₁ = P₂V₂).
P3.2b — Recall and use the Pressure Law (P/T = constant).
P3.2c — Convert between Celsius and Kelvin temperature scales.
P3.2d — Explain gas pressure changes in terms of particle motion.

Boyle's Law — Pressure and Volume

At constant temperature, doubling pressure halves volume. PV = constant.

Boyle's Law:

P₁V₁ = P₂V₂ (constant temperature, fixed amount of gas) P ∝ 1/V

P = pressure (Pa), V = volume (m³ or cm³)

Graphs:

P vs V: hyperbola (P decreases as V increases)
P vs 1/V: straight line through origin (directly proportional)

Kinetic explanation:

Smaller volume → particles travel shorter distance between collisions with walls
More frequent collisions per second → higher pressure

Example: A gas has P₁ = 100 kPa and V₁ = 3.0 L. Volume is compressed to V₂ = 1.5 L. Find P₂.

P₂ = P₁V₁/V₂ = 100 × 3.0 / 1.5 = 200 kPa

Applications:

Bicycle pump: compressing air in pump increases pressure → forces air into tyre
Breathing: diaphragm lowers → lung volume increases → pressure drops → air rushes in

Boyle's Law: P₁V₁ = P₂V₂ at constant T. Halving volume → doubling pressure.
More particles per unit volume → more wall collisions → higher pressure.
P vs V: hyperbola. P vs 1/V: straight line through origin.

The Pressure Law and Temperature

At constant volume, pressure is proportional to absolute (Kelvin) temperature.

Kelvin (absolute) temperature scale:

T (K) = T (°C) + 273

Absolute zero (0 K = −273°C): the temperature at which all particle motion ceases; minimum possible temperature
0°C = 273 K; 100°C = 373 K; −10°C = 263 K

Pressure Law (Gay-Lussac's Law):

P/T = constant (constant volume, fixed amount of gas) P₁/T₁ = P₂/T₂

Temperature MUST be in Kelvin

Kinetic explanation of pressure law:

Higher temperature → particles move faster (higher KE)
Faster particles → harder and more frequent collisions with walls
→ Higher pressure

Graph: P vs T(K): straight line through the origin (directly proportional)

Example: A gas at 300 K has pressure 150 kPa. What is the pressure at 450 K (constant volume)?

P₂ = P₁ × T₂/T₁ = 150 × 450/300 = 225 kPa

Charles' Law (qualitative):

V ∝ T at constant pressure

Heating gas at constant pressure → volume increases
Example: hot air balloon inflates because heated air expands

T(K) = T(°C) + 273. Always use Kelvin in gas law calculations.
Pressure law: P₁/T₁ = P₂/T₂ at constant volume.
Higher T → faster particles → more/harder wall collisions → higher pressure.

Quick recap

Boyle's Law: P₁V₁ = P₂V₂ (constant T). Pressure × volume = constant.
Pressure Law: P₁/T₁ = P₂/T₂ (constant V). Must use Kelvin.
Absolute zero: 0 K = −273°C. T(K) = T(°C) + 273.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Boyle's Law — P₁V₁ = P₂V₂ (constant temperature); P ∝ 1/V.
Pressure Law — P₁/T₁ = P₂/T₂ (constant volume); must use Kelvin.
Absolute zero — 0 K = −273°C; minimum possible temperature; no particle motion.
T(K) = T(°C) + 273 — always convert to Kelvin for gas calculations.

How it’s examined

Paper 4: 'A sealed gas has volume 2.0 L at pressure 100 kPa. Calculate the pressure when volume is reduced to 0.8 L at the same temperature' (2 marks — P₂ = 100 × 2.0/0.8 = 250 kPa). 'A gas at 27°C has pressure 200 kPa. Calculate the pressure at 127°C (constant volume)' (3 marks — T₁ = 300 K; T₂ = 400 K; P₂ = 200 × 400/300 = 267 kPa). 'Explain using particle theory why increasing temperature at constant volume increases pressure' (3 marks).

Sources: Cambridge IGCSE Coordinated Sciences 0654 syllabus 2025-2027 (P3); 0654 Examiner Reports 2022-2024. Last reviewed 2026-05-14.

Take this whole topic with you

Step-by-step worked examples — Pressure Changes

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

1Boyle's law calculation
Core• Boyle's law, pV constant
Question
A gas occupies $2.0\,\text{L}$ at a pressure of $100\,\text{kPa}$ at constant temperature. The gas is compressed to $0.5\,\text{L}$ . Find the new pressure.
Step-by-step solution
1. Step 1
  Boyle's law: $p_{1}V_{1} = p_{2}V_{2}$ at constant temperature.
  $100 \times 2.0 = p_{2} \times 0.5$
2. Step 2
  Solve for $p_2$ .
  $p_{2} = \dfrac{100 \times 2.0}{0.5} = 400\,\text{kPa}$
Answer
$p_{2} = 400\,\text{kPa}$
Examiner tip
Units cancel as long as both pressures are in the same unit and both volumes are in the same unit. No conversion needed if consistent.
2Pressure in a liquid column
Extended• pressure in liquids, p=hρg
Question
A diver is $25\,\text{m}$ below the surface of the sea. The density of sea water is $1025\,\text{kg/m}^{3}$ . Take $g = 10\,\text{N/kg}$ . Calculate the pressure due to the water at this depth.
Step-by-step solution
1. Step 1
  Use $p = h\rho g$ .
  $p = 25 \times 1025 \times 10 = 256\,250\,\text{Pa} \approx 256\,\text{kPa}$
Answer
$p \approx 256\,\text{kPa}$ (pressure due to the water column only).
Examiner tip
Total pressure on the diver = water pressure + atmospheric pressure ( $\approx 101\,\text{kPa}$ ). Read the question carefully for what it asks.
3Combined gas law — changing all three variables
Challenge• combined gas law, pV/T
Question
A gas at $27\,°\text{C}$ and $150\,\text{kPa}$ has a volume of $3.0\,\text{m}^{3}$ . It is heated to $127\,°\text{C}$ and compressed to $2.0\,\text{m}^{3}$ . Find the new pressure.
Step-by-step solution
1. Step 1
  Convert temperatures to kelvin: $T(\text{K}) = T(°\text{C}) + 273$ .
  $T_{1} = 27 + 273 = 300\,\text{K}; \quad T_{2} = 127 + 273 = 400\,\text{K}$
2. Step 2
  Apply the combined gas law.
  $\dfrac{p_{1}V_{1}}{T_{1}} = \dfrac{p_{2}V_{2}}{T_{2}} \implies p_{2} = \dfrac{p_{1}V_{1}T_{2}}{T_{1}V_{2}}$
3. Step 3
  Substitute values.
  $p_{2} = \dfrac{150 \times 3.0 \times 400}{300 \times 2.0} = \dfrac{180\,000}{600} = 300\,\text{kPa}$
Answer
$p_{2} = 300\,\text{kPa}$
Examiner tip
ALWAYS convert Celsius to kelvin before using any gas law. $T = 0\,°\text{C}$ is NOT the same as $0\,\text{K}$ .

Key Formulae — Pressure Changes

The formulae you need to memorise for pressure changes on the Cambridge IGCSE 0654 paper, with every variable defined in plain English and a note on when to use it.

Boyle's law
$p_{1}V_{1} = p_{2}V_{2} \quad (\text{constant } T)$
$p$
pressure (Pa or kPa)
$V$
volume (m³ or L)
When to use
When temperature is constant and pressure or volume changes.
Pressure-temperature law
$\dfrac{p_{1}}{T_{1}} = \dfrac{p_{2}}{T_{2}} \quad (\text{constant } V)$
$p$
pressure (Pa)
$T$
temperature (K)
When to use
When volume is constant and pressure and temperature change.
Combined gas law
$\dfrac{p_{1}V_{1}}{T_{1}} = \dfrac{p_{2}V_{2}}{T_{2}}$
$p$
pressure (Pa)
$V$
volume (m³)
$T$
temperature (K)
When to use
When all three gas variables change simultaneously.
Pressure in a liquid
$p = h\rho g$
$h$
depth below the surface (m)
$\rho$
density of liquid (kg/m³)
$g$
gravitational field strength (N/kg)
When to use
Finding pressure at a given depth in a static liquid.

Key Definitions and Keywords — Pressure Changes

Definitions to memorise and the exact keywords mark schemes credit for pressure changes answers — sharpened from recent examiner reports for the 2026 0654 sitting.

Boyle's law
Examiner keyword
At constant temperature, the pressure of a fixed mass of gas is inversely proportional to its volume: $pV = \text{constant}$ .
Absolute zero
Examiner keyword
The lowest possible temperature, $0\,\text{K}$ ( $-273\,°\text{C}$ ), at which particles have minimum internal energy. Gas pressure and volume are theoretically zero at absolute zero.
Kelvin (temperature scale)
Examiner keyword
The absolute temperature scale. $T(\text{K}) = T(°\text{C}) + 273$ . Must be used in all gas law calculations.
Pascal (Pa)
The SI unit of pressure. $1\,\text{Pa} = 1\,\text{N/m}^{2}$ .

Common Mistakes and Misconceptions — Pressure Changes

The traps other students keep falling into on pressure changes questions — taken from recent Cambridge IGCSE 0654 examiner reports and mark schemes — and how to avoid them.

✕Using Celsius instead of kelvin in gas law calculations
Why it happens
Students forget to convert, especially when the temperature change is small.
How to avoid it
Always add 273 to convert °C to K before substituting into any gas law. Celsius temperatures give wrong ratios.
✕Writing $p_{1}V_{1} = p_{2}V_{2}$ but then calculating $p_{2} = p_{1}V_{1} \times V_{2}$ (multiplying instead of dividing)
Why it happens
Algebraic rearrangement error.
How to avoid it
Rearrange carefully: $p_{2} = p_{1}V_{1}/V_{2}$ . If pressure increases, volume must decrease — check this sense before accepting your answer.
✕Forgetting to add atmospheric pressure to liquid pressure for total pressure
Why it happens
$p = h\rho g$ gives only the pressure due to the liquid column.
How to avoid it
Total pressure = liquid pressure + atmospheric pressure. Read the question: if it asks for pressure 'due to the water', use $h\rho g$ only.

Pressure Changes — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Cambridge IGCSE Coordinated Science (0654)

Pressure Changes

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Short Notes - Pressure Changes

Detailed Study Notes

Detailed notes on Thermal Physics for Cambridge IGCSE Coordinated Science, covering key concepts, explanations, examples, and exam-focused revision points.

Pressure Changes in Gases — Cambridge IGCSE Coordinated Science 0654

At a glance

Boyle's Law: P₁V₁ = P₂V₂ (constant temperature). P ∝ 1/V.
Pressure Law: P/T = constant (constant volume). P ∝ T (in Kelvin).
Kelvin scale: T(K) = T(°C) + 273. Absolute zero = 0 K = −273°C.
Kinetic explanation: smaller volume → particles hit walls more frequently → higher pressure.
Higher temperature → particles move faster → harder and more frequent collisions → higher pressure.

What you’ll learn

Mapped to the Cambridge IGCSE 0654 syllabus (2025-2027).

P3.2a — Recall and use Boyle's Law (P₁V₁ = P₂V₂).
P3.2b — Recall and use the Pressure Law (P/T = constant).
P3.2c — Convert between Celsius and Kelvin temperature scales.
P3.2d — Explain gas pressure changes in terms of particle motion.

Boyle's Law — Pressure and Volume

At constant temperature, doubling pressure halves volume. PV = constant.

Boyle's Law:

P₁V₁ = P₂V₂ (constant temperature, fixed amount of gas) P ∝ 1/V

P = pressure (Pa), V = volume (m³ or cm³)

Graphs:

P vs V: hyperbola (P decreases as V increases)
P vs 1/V: straight line through origin (directly proportional)

Kinetic explanation:

Smaller volume → particles travel shorter distance between collisions with walls
More frequent collisions per second → higher pressure

Example: A gas has P₁ = 100 kPa and V₁ = 3.0 L. Volume is compressed to V₂ = 1.5 L. Find P₂.

P₂ = P₁V₁/V₂ = 100 × 3.0 / 1.5 = 200 kPa

Applications:

Bicycle pump: compressing air in pump increases pressure → forces air into tyre
Breathing: diaphragm lowers → lung volume increases → pressure drops → air rushes in

Boyle's Law: P₁V₁ = P₂V₂ at constant T. Halving volume → doubling pressure.
More particles per unit volume → more wall collisions → higher pressure.
P vs V: hyperbola. P vs 1/V: straight line through origin.

The Pressure Law and Temperature

At constant volume, pressure is proportional to absolute (Kelvin) temperature.

Kelvin (absolute) temperature scale:

T (K) = T (°C) + 273

Absolute zero (0 K = −273°C): the temperature at which all particle motion ceases; minimum possible temperature
0°C = 273 K; 100°C = 373 K; −10°C = 263 K

Pressure Law (Gay-Lussac's Law):

P/T = constant (constant volume, fixed amount of gas) P₁/T₁ = P₂/T₂

Temperature MUST be in Kelvin

Kinetic explanation of pressure law:

Higher temperature → particles move faster (higher KE)
Faster particles → harder and more frequent collisions with walls
→ Higher pressure

Graph: P vs T(K): straight line through the origin (directly proportional)

Example: A gas at 300 K has pressure 150 kPa. What is the pressure at 450 K (constant volume)?

P₂ = P₁ × T₂/T₁ = 150 × 450/300 = 225 kPa

Charles' Law (qualitative):

V ∝ T at constant pressure

Heating gas at constant pressure → volume increases
Example: hot air balloon inflates because heated air expands

T(K) = T(°C) + 273. Always use Kelvin in gas law calculations.
Pressure law: P₁/T₁ = P₂/T₂ at constant volume.
Higher T → faster particles → more/harder wall collisions → higher pressure.

Quick recap

Boyle's Law: P₁V₁ = P₂V₂ (constant T). Pressure × volume = constant.
Pressure Law: P₁/T₁ = P₂/T₂ (constant V). Must use Kelvin.
Absolute zero: 0 K = −273°C. T(K) = T(°C) + 273.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Boyle's Law — P₁V₁ = P₂V₂ (constant temperature); P ∝ 1/V.
Pressure Law — P₁/T₁ = P₂/T₂ (constant volume); must use Kelvin.
Absolute zero — 0 K = −273°C; minimum possible temperature; no particle motion.
T(K) = T(°C) + 273 — always convert to Kelvin for gas calculations.

How it’s examined

Sources: Cambridge IGCSE Coordinated Sciences 0654 syllabus 2025-2027 (P3); 0654 Examiner Reports 2022-2024. Last reviewed 2026-05-14.

Take this whole topic with you

Step-by-step worked examples — Pressure Changes

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

1Boyle's law calculation
Core• Boyle's law, pV constant
Question
A gas occupies $2.0\,\text{L}$ at a pressure of $100\,\text{kPa}$ at constant temperature. The gas is compressed to $0.5\,\text{L}$ . Find the new pressure.
Step-by-step solution
1. Step 1
  Boyle's law: $p_{1}V_{1} = p_{2}V_{2}$ at constant temperature.
  $100 \times 2.0 = p_{2} \times 0.5$
2. Step 2
  Solve for $p_2$ .
  $p_{2} = \dfrac{100 \times 2.0}{0.5} = 400\,\text{kPa}$
Answer
$p_{2} = 400\,\text{kPa}$
Examiner tip
Units cancel as long as both pressures are in the same unit and both volumes are in the same unit. No conversion needed if consistent.
2Pressure in a liquid column
Extended• pressure in liquids, p=hρg
Question
A diver is $25\,\text{m}$ below the surface of the sea. The density of sea water is $1025\,\text{kg/m}^{3}$ . Take $g = 10\,\text{N/kg}$ . Calculate the pressure due to the water at this depth.
Step-by-step solution
1. Step 1
  Use $p = h\rho g$ .
  $p = 25 \times 1025 \times 10 = 256\,250\,\text{Pa} \approx 256\,\text{kPa}$
Answer
$p \approx 256\,\text{kPa}$ (pressure due to the water column only).
Examiner tip
Total pressure on the diver = water pressure + atmospheric pressure ( $\approx 101\,\text{kPa}$ ). Read the question carefully for what it asks.
3Combined gas law — changing all three variables
Challenge• combined gas law, pV/T
Question
A gas at $27\,°\text{C}$ and $150\,\text{kPa}$ has a volume of $3.0\,\text{m}^{3}$ . It is heated to $127\,°\text{C}$ and compressed to $2.0\,\text{m}^{3}$ . Find the new pressure.
Step-by-step solution
1. Step 1
  Convert temperatures to kelvin: $T(\text{K}) = T(°\text{C}) + 273$ .
  $T_{1} = 27 + 273 = 300\,\text{K}; \quad T_{2} = 127 + 273 = 400\,\text{K}$
2. Step 2
  Apply the combined gas law.
  $\dfrac{p_{1}V_{1}}{T_{1}} = \dfrac{p_{2}V_{2}}{T_{2}} \implies p_{2} = \dfrac{p_{1}V_{1}T_{2}}{T_{1}V_{2}}$
3. Step 3
  Substitute values.
  $p_{2} = \dfrac{150 \times 3.0 \times 400}{300 \times 2.0} = \dfrac{180\,000}{600} = 300\,\text{kPa}$
Answer
$p_{2} = 300\,\text{kPa}$
Examiner tip
ALWAYS convert Celsius to kelvin before using any gas law. $T = 0\,°\text{C}$ is NOT the same as $0\,\text{K}$ .

Key Formulae — Pressure Changes

The formulae you need to memorise for pressure changes on the Cambridge IGCSE 0654 paper, with every variable defined in plain English and a note on when to use it.

Boyle's law
$p_{1}V_{1} = p_{2}V_{2} \quad (\text{constant } T)$
$p$
pressure (Pa or kPa)
$V$
volume (m³ or L)
When to use
When temperature is constant and pressure or volume changes.
Pressure-temperature law
$\dfrac{p_{1}}{T_{1}} = \dfrac{p_{2}}{T_{2}} \quad (\text{constant } V)$
$p$
pressure (Pa)
$T$
temperature (K)
When to use
When volume is constant and pressure and temperature change.
Combined gas law
$\dfrac{p_{1}V_{1}}{T_{1}} = \dfrac{p_{2}V_{2}}{T_{2}}$
$p$
pressure (Pa)
$V$
volume (m³)
$T$
temperature (K)
When to use
When all three gas variables change simultaneously.
Pressure in a liquid
$p = h\rho g$
$h$
depth below the surface (m)
$\rho$
density of liquid (kg/m³)
$g$
gravitational field strength (N/kg)
When to use
Finding pressure at a given depth in a static liquid.

Key Definitions and Keywords — Pressure Changes

Definitions to memorise and the exact keywords mark schemes credit for pressure changes answers — sharpened from recent examiner reports for the 2026 0654 sitting.

Boyle's law
Examiner keyword
At constant temperature, the pressure of a fixed mass of gas is inversely proportional to its volume: $pV = \text{constant}$ .
Absolute zero
Examiner keyword
The lowest possible temperature, $0\,\text{K}$ ( $-273\,°\text{C}$ ), at which particles have minimum internal energy. Gas pressure and volume are theoretically zero at absolute zero.
Kelvin (temperature scale)
Examiner keyword
The absolute temperature scale. $T(\text{K}) = T(°\text{C}) + 273$ . Must be used in all gas law calculations.
Pascal (Pa)
The SI unit of pressure. $1\,\text{Pa} = 1\,\text{N/m}^{2}$ .

Common Mistakes and Misconceptions — Pressure Changes

The traps other students keep falling into on pressure changes questions — taken from recent Cambridge IGCSE 0654 examiner reports and mark schemes — and how to avoid them.

✕Using Celsius instead of kelvin in gas law calculations
Why it happens
Students forget to convert, especially when the temperature change is small.
How to avoid it
Always add 273 to convert °C to K before substituting into any gas law. Celsius temperatures give wrong ratios.
✕Writing $p_{1}V_{1} = p_{2}V_{2}$ but then calculating $p_{2} = p_{1}V_{1} \times V_{2}$ (multiplying instead of dividing)
Why it happens
Algebraic rearrangement error.
How to avoid it
Rearrange carefully: $p_{2} = p_{1}V_{1}/V_{2}$ . If pressure increases, volume must decrease — check this sense before accepting your answer.
✕Forgetting to add atmospheric pressure to liquid pressure for total pressure
Why it happens
$p = h\rho g$ gives only the pressure due to the liquid column.
How to avoid it
Total pressure = liquid pressure + atmospheric pressure. Read the question: if it asks for pressure 'due to the water', use $h\rho g$ only.

Pressure Changes — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Pressure Changes

Short Notes - Pressure Changes

Detailed Study Notes

At a glance

What you’ll learn

Boyle's Law — Pressure and Volume

The Pressure Law and Temperature

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Boyle's law calculation

2Pressure in a liquid column

3Combined gas law — changing all three variables

Boyle's law

Pressure-temperature law

Combined gas law

Pressure in a liquid

Boyle's law

Absolute zero

Kelvin (temperature scale)

Pascal (Pa)

✕Using Celsius instead of kelvin in gas law calculations

✕Writing p1V1=p2V2p_{1}V_{1} = p_{2}V_{2}p1​V1​=p2​V2​ but then calculating p2=p1V1×V2p_{2} = p_{1}V_{1} \times V_{2}p2​=p1​V1​×V2​ (multiplying instead of dividing)

✕Forgetting to add atmospheric pressure to liquid pressure for total pressure

Pressure Changes — frequently asked questions

Pressure Changes

Short Notes - Pressure Changes

Detailed Study Notes

At a glance

What you’ll learn

Boyle's Law — Pressure and Volume

The Pressure Law and Temperature

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Boyle's law calculation

2Pressure in a liquid column

3Combined gas law — changing all three variables

Boyle's law

Pressure-temperature law

Combined gas law

Pressure in a liquid

Boyle's law

Absolute zero

Kelvin (temperature scale)

Pascal (Pa)

✕Using Celsius instead of kelvin in gas law calculations

✕Writing p1V1=p2V2p_{1}V_{1} = p_{2}V_{2}p1​V1​=p2​V2​ but then calculating p2=p1V1×V2p_{2} = p_{1}V_{1} \times V_{2}p2​=p1​V1​×V2​ (multiplying instead of dividing)

✕Forgetting to add atmospheric pressure to liquid pressure for total pressure

Pressure Changes — frequently asked questions

✕Writing $p_{1}V_{1} = p_{2}V_{2}$ but then calculating $p_{2} = p_{1}V_{1} \times V_{2}$ (multiplying instead of dividing)

✕Writing $p_{1}V_{1} = p_{2}V_{2}$ but then calculating $p_{2} = p_{1}V_{1} \times V_{2}$ (multiplying instead of dividing)