Cambridge IGCSE0625

Pressure

Physics 0625 Chapter Notes

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Pressure
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In this chapter

  1. 1What is Pressure?
  2. 2Pressure Calculations
  3. 3Pressure in Liquids
  4. 4Pressure and Volume of Gases: Boyle's Law

1. What is Pressure?

Pressure is defined as the force acting normally (at right angles) per unit area. It tells us how concentrated a force is. A large force spread over a large area can result in low pressure, while a small force concentrated on a tiny area can create very high pressure. The formula for pressure is p = F/A, where p is pressure, F is the force applied perpendicular to the surface, and A is the area over which the force is distributed.

p = F/A

Key term

Pressure: The force acting normally per unit area.

Examiner insight

Examiners expect students to clearly state the formula for pressure and understand the inverse relationship between pressure and area when the force is constant.

Common pitfall

Forgetting that the force must be perpendicular to the area for the formula p=F/A to be directly applied.

Fun fact

High heels exert more pressure on the ground than an elephant's foot, due to the tiny area of the heel!

Worked example 13 marks

A force of 200 N acts on an area of 4 m². What pressure is produced?

  1. 1

    Identify the given values: Force (F) = 200 N, Area (A) = 4 m².

  2. 2

    Recall the formula for pressure: p = F/A.

  3. 3

    Substitute the values into the formula: p = 200 N / 4 m².

  4. 4

    Calculate the pressure: p = 50 Pa.

Recap

  • Pressure is force per unit area.
  • The formula for pressure is p = F/A.
  • Pressure is measured in Pascals (Pa), which is equivalent to N/m².
  • A smaller area for a given force results in higher pressure.
  • A larger area for a given force results in lower pressure.

Quick check

  1. State the SI unit for pressure.1 mark
  2. If a force of 50 N is applied over an area of 0.5 m², what is the pressure?2 marks

2. Pressure Calculations

The pressure formula p = F/A can be rearranged to find any of the three variables if the other two are known. To find force, we use F = p × A. To find area, we use A = F/p. It is crucial to ensure all units are consistent; typically, force in Newtons (N), area in square metres (m²), and pressure in Pascals (Pa). Kilopascals (kPa) are often used for larger pressures, where 1 kPa = 1000 Pa.

F = p × A

A = F/p

Key term

Pascal (Pa): The SI unit of pressure, defined as one Newton per square metre (1 N/m²).

Examiner insight

Candidates often lose marks by not converting units (e.g., cm² to m², kPa to Pa) before calculations. Always check and convert to SI units.

Common pitfall

Incorrectly rearranging the pressure formula, leading to errors in calculating force or area.

Fun fact

The pressure exerted by a sharp knife can be millions of Pascals, allowing it to cut through materials easily.

Worked example 13 marks

The wind pressure on a wall is 100 Pa. If the wall has an area of 6 m², what is the force on it?

  1. 1

    Identify the given values: Pressure(p) = 100 Pa, Area (A) = 6 m².

  2. 2

    Rearrange the pressure formula to find force: F = p × A.

  3. 3

    Substitute the values: F = 100 Pa × 6 m².

  4. 4

    Calculate the force: F = 600 N.

Worked example 25 marks

A concrete block has a mass of 2600 kg. If the block measures 0.5 m by 1.0 m by 2.0 m, what is the maximum pressure it can exert when resting on the ground? (Assume g = 10 N/kg).

  1. 1

    Calculate the weight (force) of the block: F = mass × g = 2600 kg × 10 N/kg = 26000 N.

  2. 2

    To exert maximum pressure, the block must rest on its smallest area. The possible areas are: 0.5 m × 1.0 m = 0.5 m², 0.5 m × 2.0 m = 1.0 m², 1.0 m × 2.0 m = 2.0 m².

  3. 3

    The smallest area is 0.5 m².

  4. 4

    Calculate the maximum pressure: p = F/A = 26000 N / 0.5 m².

  5. 5

    p = 52000 Pa or 52 kPa.

Recap

  • The formula p = F/A can be rearranged to F = p × A or A = F/p.
  • Always convert units to SI (Newtons, square metres, Pascals) before calculation.
  • Maximum pressure occurs when a given force acts on the smallest possible area.
  • Minimum pressure occurs when a given force acts on the largest possible area.
  • Weight is a force, calculated as mass × gravitational field strength (F = mg).

Quick check

  1. What force is produced if a pressure of 2 kPa acts on an area of 0.2 m²?3 marks
  2. Explain why a tractor has large tyres.2 marks

3. Pressure in Liquids

Unlike solids, liquids exert pressure in all directions. The pressure at a point within a liquid depends on the depth, the density of the liquid, and the gravitational field strength. This is known as hydrostatic pressure. Consider a column of liquid of height h and cross-sectional area A. The volume of this column is V = A × h. The mass of the liquid is m = density × volume = ρ × A × h. The weight (force) of this liquid column is F = m × g = ρ × A × h × g. Since pressure is F/A, the pressure at depth h is p = (ρ × A × h ×g) / A, which simplifies to p = ρgh. This formula shows that pressure in a liquid increases with depth and density, but is independent of the shape or total volume of the liquid.

p = ρgh

Key term

Hydrostatic Pressure: The pressure exerted by a fluid at equilibrium at a given point within the fluid, due to the force of gravity.

Examiner insight

Students must be able to derive p = ρgh from p = F/A and F = mg, showing understanding of how density, gravity, and depth contribute to liquid pressure.

Common pitfall

Confusing the total pressure at a depth with the pressure due to the liquid only. Total pressure often includes atmospheric pressure.

Fun fact

The deepest part of the ocean, the Mariana Trench, has a pressure of over 1000 atmospheres, equivalent to 11,000,000 Pa!

Worked example 13 marks

If the density of water is 1000 kg/m³, what is the pressure due to the water at the bottom of a swimming pool 2 m deep? (Assume g = 10 N/kg).

  1. 1

    Identify the given values: Density (ρ) = 1000 kg/m³, Depth(h) = 2 m, Gravitational field strength(g) = 10 N/kg.

  2. 2

    Recall the formula for pressure in a liquid: p = ρgh.

  3. 3

    Substitute the values: p = 1000 kg/m³ × 10 N/kg × 2 m.

  4. 4

    Calculate the pressure: p = 20000 Pa.

Worked example 23 marks

A rectangular storage tank 4 m long by 3 m wide is filled with paraffin to a depth of 2 m. The density of paraffin is 800 kg/m³. Calculate the pressure at the bottom of the tank due to the paraffin. (Assume g = 10 N/kg).

  1. 1

    Identify the given values: Density (ρ) = 800 kg/m³, Depth(h) = 2 m, Gravitational field strength(g) = 10 N/kg. (Note: tank dimensions are irrelevant for pressure calculation at a specific depth).

  2. 2

    Recall the formula for pressure in a liquid: p = ρgh.

  3. 3

    Substitute the values: p = 800 kg/m³ × 10 N/kg × 2 m.

  4. 4

    Calculate the pressure: p = 16000 Pa.

Recap

  • Pressure in a liquid is given by p = ρgh.
  • ρ is the density of the liquid, g is gravitational field strength, h is the depth.
  • Liquid pressure increases with depth.
  • Liquid pressure increases with liquid density.
  • Liquid pressure acts equally in all directions at a given depth.
  • The shape or base area of the container does not affect the pressure at a specific depth.

Quick check

  1. How does the pressure at a point in a liquid change if the depth is doubled?1 mark
  2. What is the pressure due to a column of water 5 m deep? (Density of water = 1000 kg/m³, g = 10 N/kg).2 marks

4. Pressure and Volume of Gases: Boyle's Law

Boyle's Law describes the relationship between the pressure and volume of a fixed mass of gas at a constant temperature. It states that the pressure of a gas is inversely proportional to its volume. This means that if the volume of a gas is halved, its pressure doubles, and vice versa. Mathematically, this relationship can be expressed as p ∝ 1/V, or pV = constant. For changes between two states, this leads to the formula p₁V₁ = p₂V₂, where p₁ and V₁ are the initial pressure and volume, and p₂ and V₂ are the final pressure and volume. This law is fundamental to understanding gas behaviour in various applications, from scuba diving to internal combustion engines.

p₁V₁ = p₂V₂

Key term

Boyle's Law: For a fixed mass of gas at constant temperature, the pressure is inversely proportional to the volume.

Examiner insight

When applying Boyle's Law, ensure the conditions (fixed mass, constant temperature) are stated or implied. Also, consistent units for pressure and volume on both sides of the equation are essential, though they don't have to be SI units as long as they are consistent.

Common pitfall

Forgetting the 'constant temperature' condition, or assuming it applies to liquids as well as gases.

Fun fact

Boyle's Law was one of the first quantitative laws in chemistry and physics, discovered by Robert Boyle in the 17th century using a J-shaped tube filled with mercury.

Worked example 14 marks

An air bubble has a volume of 2 cm³ when released at a depth of 20 m in water. What will its volume be when it reaches the surface? Assume that the temperature does not change and that atmospheric pressure is equivalent to the pressure from a column of water 10 m deep.

  1. 1

    Determine initial pressure (p₁): Atmospheric pressure is 10 m water equivalent. Pressure due to 20 m water depth is 20 m water equivalent. So, p₁ = 10 m + 20 m = 30 m of water (or 3 atm).

  2. 2

    Determine final pressure (p₂): At the surface, the pressure is atmospheric pressure, so p₂ = 10 m of water (or 1 atm).

  3. 3

    Identify initial volume (V₁): V₁ = 2 cm³.

  4. 4

    State Boyle's Law: p₁V₁ = p₂V₂.

  5. 5

    Substitute the values: (30 m water) × 2 cm³ = (10 m water) × V₂.

  6. 6

    Solve for V₂: V₂ = (30 × 2) / 10 = 6 cm³.

Worked example 23 marks

A fixed mass of gas occupies a volume of 12 m³ at a pressure of 100 kPa. If its volume is compressed to 6 m³ at the same temperature, what is the new pressure?

  1. 1

    Identify initial conditions: p₁ = 100 kPa, V₁ = 12 m³.

  2. 2

    Identify final conditions: V₂ = 6 m³, p₂ is unknown.

  3. 3

    Apply Boyle's Law: p₁V₁ = p₂V₂.

  4. 4

    Substitute values: 100 kPa × 12 m³ = p₂ × 6 m³.

  5. 5

    Solve for p₂: p₂ = (100 × 12) / 6 = 1200 / 6 = 200 kPa.

Recap

  • Boyle's Law relates pressure and volume for a fixed mass of gas at constant temperature.
  • Pressure and volume are inversely proportional: p ∝ 1/V.
  • The formula p₁V₁ = p₂V₂ is used for calculations involving changes in gas pressure and volume.
  • If volume decreases, pressure increases proportionally, and vice versa.
  • Units for pressure and volume must be consistent on both sides of the equation.

Quick check

  1. State the conditions under which Boyle's Law applies.2 marks
  2. If the pressure of a gas doubles, what happens to its volume, assuming constant temperature and fixed mass?1 mark

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. A force of 300 N is applied uniformly over a surface area of 0.5 m². Calculate the pressure exerted on the surface.2 marks
  2. A pressure of 250 Pa is exerted on a window pane with an area of 1.2 m². Calculate the total force acting on the window.2 marks
  3. A person weighing 600 N stands on one foot. If the area of their shoe in contact with the ground is 0.015 m², calculate the pressure exerted on the ground.2 marks
  4. A rectangular block of mass 50 kg measures 0.2 m by 0.5 m by 1.0 m. Calculate the maximum pressure it can exert when resting on a surface. (Assume g = 10 N/kg).4 marks
  5. Explain why a person wearing snowshoes can walk on soft snow without sinking, while a person wearing ordinary boots would sink.3 marks
  6. A submarine is submerged to a depth of 50 m in seawater. If the density of seawater is 1020 kg/m³, calculate the pressure due to the seawater at this depth. (Assume g = 10 N/kg).3 marks
  7. A U-tube contains two immiscible liquids: oil on one side and water on the other. If the oil column is 15 cm high and has a density of 800 kg/m³, what height of water (density 1000 kg/m³) would exert the same pressure?4 marks
  8. Describe two characteristics of pressure in liquids.2 marks
  9. A fixed mass of gas has a volume of 400 cm³ at a pressure of 120 kPa. If the temperature remains constant, what will be the new pressure if the volume is reduced to 250 cm³?3 marks
  10. An air bubble released from a deep-sea vent has a volume of 5 cm³ at a pressure of 300 kPa. When it rises to a point where the pressure is 150 kPa, what will be its new volume, assuming constant temperature?3 marks

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