Cambridge IGCSE0972

Pressure

Physics 0972 Chapter Notes

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Pressure
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In this chapter

  1. 1What is Pressure?
  2. 2Pressure from Solid Objects
  3. 3Pressure in Liquids
  4. 4Atmospheric Pressure and Manometers
  5. 5Gas Pressure and Boyle's Law

1. What is Pressure?

Pressure is a measure of how concentrated a force is. It's defined as the force acting perpendicularly on a unit area of a surface. Imagine pushing a drawing pin into a board. You apply the same force to both ends, but only the sharp end goes in. This is because the tiny area of the pin's point concentrates the force, creating a very high pressure. The formula that links pressure (p), force (F), and area (A) is p = F/A. The standard unit for pressure is the Pascal (Pa), where one Pascal is equal to a force of one Newton acting over an area of one square metre (1 Pa = 1 N/m²). Because the Pascal is a small unit, we often use kilopascals (kPa), where 1 kPa = 1000 Pa.

p = F / A

F = p × A

A = F / p

Key term

Pressure: The force exerted per unit area on a surface, with the force acting perpendicular to the surface.

Examiner insight

Examiners award marks for correctly stating the formula p = F/A and showing the substitution of values. Always write down the formula first.

Common pitfall

Confusing force and pressure. They are not the same. A huge force can create a low pressure if spread over a large enough area, like a hovercraft.

Fun fact

A ballerina standing on the tip of one pointe shoe can exert more pressure on the floor (over 2000 kPa) than an elephant standing on one foot (around 250 kPa).

Worked example 12 marks

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

  1. 1

    Step 1: Identify the known values. Force (F) = 200 N, Area (A) = 4 m².

  2. 2

    Step 2: State the formula for pressure. p = F / A.

  3. 3

    Step 3: Substitute the values into the formula. p = 200 N / 4 m².

  4. 4

    Step 4: Calculate the result. p = 50 N/m² = 50 Pa.

Worked example 22 marks

The pressure of the wind on a large window is 150 Pa. If the window has an area of 5 m², what is the total force acting on it?

  1. 1

    Step 1: Identify the known values. Pressure(p) = 150 Pa, Area (A) = 5 m².

  2. 2

    Step 2: State the rearranged formula for force. F = p × A.

  3. 3

    Step 3: Substitute the values into the formula. F = 150 Pa × 5 m².

  4. 4

    Step 4: Calculate the result. F = 750 N.

Recap

  • Pressure is defined as force per unit area (p = F/A).
  • The standard unit of pressure is the Pascal (Pa), which is equivalent to N/m².
  • A small area concentrates force, resulting in high pressure.
  • A large area spreads out a force, resulting in low pressure.
  • The force in the pressure equation must be perpendicular to the area.

Quick check

  1. State the formula linking pressure, force, and area.1 mark
  2. If a force of 50 N acts over an area of 10 m², what is the pressure?1 mark

2. Pressure from Solid Objects

When a solid object rests on a surface, it exerts a pressure due to its weight. The force (F) in the pressure equation is the object's weight (W), which you can calculate using W = mg, where 'm' is the mass in kg and 'g' is the gravitational field strength (usually taken as 10 N/kg in IGCSE). A rectangular block can exert different pressures depending on which face it rests on. To get the maximum pressure, you must place it on its smallest face, as this concentrates the force. To get the minimum pressure, you place it on its largest face, spreading the force out.

p = F / A

W = m × g

Key term

Weight (W): The force of gravity acting on an object's mass, calculated as mass multiplied by the gravitational field strength (W = mg).

Examiner insight

For maximum/minimum pressure questions, examiners look for the correct calculation of weight, the identification of the smallest and largest areas, and two separate pressure calculations.

Common pitfall

Forgetting to convert the object's mass in kg into a weight in Newtons before using the pressure formula. Remember, weight is the force.

Worked example 15 marks

A concrete block has a mass of 120 kg and measures 0.2 m by 0.5 m by 1.0 m. Calculate the maximum and minimum pressure it can exert on the ground. (Take g = 10 N/kg).

  1. 1

    Step 1: Calculate the weight of the block. Weight (Force) = mass × g = 120 kg × 10 N/kg = 1200 N.

  2. 2

    Step 2: For maximum pressure, find the smallest area. Smallest area = 0.2 m × 0.5 m = 0.1 m².

  3. 3

    Step 3: Calculate the maximum pressure. p_max = F / A_min = 1200 N / 0.1 m² = 12000 Pa.

  4. 4

    Step 4: For minimum pressure, find the largest area. Largest area = 0.5 m × 1.0 m = 0.5 m².

  5. 5

    Step 5: Calculate the minimum pressure. p_min = F / A_max = 1200 N / 0.5 m² = 2400 Pa.

Recap

  • The force exerted by a resting object is its weight (W = mg).
  • Maximum pressure occurs when the object rests on its smallest area.
  • Minimum pressure occurs when the object rests on its largest area.
  • Always convert mass (kg) to weight (N) before calculating pressure.
  • Ensure all length measurements are in metres (m) before calculating area (m²).

Quick check

  1. An object has a mass of 40 kg. What is its weight? (g = 10 N/kg)1 mark
  2. To get the highest pressure from a brick, should you stand it on its end, its side, or lay it flat?1 mark

3. Pressure in Liquids

Pressure in a liquid is caused by the weight of the liquid column above a certain point. This means that pressure increases as you go deeper. The pressure at a certain depth in a liquid does not depend on the shape of the container, only on the depth, the density of the liquid, and the gravitational field strength. This is described by the formula p = ρgh, where 'p' is the pressure, 'ρ' (rho) is the density of the liquid (in kg/m³), 'g' is the gravitational field strength (in N/kg), and 'h' is the vertical depth (in m). A key property is that at any given depth, the pressure acts equally in all directions.

p = ρ × g × h

Key term

Density (ρ): The mass per unit volume of a substance, usually measured in kilograms per cubic metre (kg/m³).

Common pitfall

Using the total volume or surface area of a liquid in the pressure calculation instead of the vertical depth 'h'. The formula p=ρgh only cares about how deep you are.

Fun fact

The immense pressure at great ocean depths is why deep-sea submersibles need incredibly thick hulls. The pressure at the bottom of the Mariana Trench is over 1,000 times atmospheric pressure!

Worked example 13 marks

A submarine is at a depth of 250 m in the sea. Calculate the pressure exerted on the submarine by the water. (Density of seawater = 1030 kg/m³, g = 10 N/kg).

  1. 1

    Step 1: Identify the known values. Depth(h) = 250 m, Density (ρ) = 1030 kg/m³, g = 10 N/kg.

  2. 2

    Step 2: State the formula for pressure in a liquid. p = ρgh.

  3. 3

    Step 3: Substitute the values into the formula. p = 1030 kg/m³ × 10 N/kg × 250 m.

  4. 4

    Step 4: Calculate the result. p = 2,575,000 Pa or 2575 kPa.

Worked example 23 marks

A rectangular water tank has a base measuring 3 m by 2 m and is filled to a depth of 1.5 m. Calculate the pressure at the bottom of the tank. (Density of water = 1000 kg/m³, g = 10 N/kg).

  1. 1

    Step 1: Identify the relevant values for the pressure calculation. Depth(h) = 1.5 m, Density (ρ) = 1000 kg/m³, g = 10 N/kg. Note that the base dimensions are not needed for this calculation.

  2. 2

    Step 2: State the formula. p = ρgh.

  3. 3

    Step 3: Substitute the values. p = 1000 kg/m³ × 10 N/kg × 1.5 m.

  4. 4

    Step 4: Calculate the pressure. p = 15,000 Pa or 15 kPa.

Recap

  • Pressure in a liquid increases with depth.
  • The formula for liquid pressure is p = ρgh.
  • Pressure at a given depth is the same, regardless of the container's shape.
  • Pressure in a liquid acts equally in all directions.
  • Ensure depth is in metres (m) and density is in kg/m³ for calculations in Pascals.

Quick check

  1. What three quantities determine the pressure at a point within a liquid?1 mark
  2. Does the pressure at the bottom of a wide lake differ from the pressure at the same depth in a narrow pipe, assuming both contain fresh water?1 mark

4. Atmospheric Pressure and Manometers

The Earth is surrounded by a layer of air called the atmosphere. This air has weight, and it exerts a pressure on everything within it, known as atmospheric pressure. At sea level, this pressure is approximately 101,300 Pa (or 101.3 kPa). To measure pressure differences, we use a device called a manometer. A simple U-shaped manometer contains a liquid (like water or mercury) and is connected to a gas supply on one side and left open to the atmosphere on the other. The difference in the liquid levels(h) tells you the 'gauge pressure' – how much more (or less) pressure the gas has compared to the atmosphere. The actual pressure of the gas, called the absolute pressure, is found by adding the gauge pressure to the atmospheric pressure: P_absolute = P_atmosphere + P_gauge (where P_gauge = ρgh).

P_absolute = P_atmosphere + P_gauge

P_gauge = ρgh

Key term

Manometer: An instrument for measuring the pressure of a gas, typically by observing the difference in height of a column of liquid in a U-tube.

Examiner insight

Examiners look for the correct conversion of the height difference into metres before using it in the p=ρgh equation. Marks are often lost for simple unit conversion errors.

Common pitfall

Forgetting to add atmospheric pressure to the calculated gauge pressure to find the absolute pressure of the gas.

Worked example 14 marks

A manometer containing mercury is used to measure the pressure of a gas supply. The mercury level on the gas side is 50 mm lower than the level open to the atmosphere. If atmospheric pressure is 101,000 Pa, what is the absolute pressure of the gas in Pa? (Density of mercury = 13600 kg/m³, g = 10 N/kg)

  1. 1

    Step 1: The gas pressure is higher than atmospheric pressure because it has pushed the liquid down. Find the height difference in metres. h = 50 mm = 0.05 m.

  2. 2

    Step 2: Calculate the gauge pressure using p = ρgh. P_gauge = 13600 kg/m³ × 10 N/kg × 0.05 m = 6800 Pa.

  3. 3

    Step 3: Calculate the absolute pressure. P_absolute = P_atmosphere + P_gauge.

  4. 4

    Step 4: Substitute values. P_absolute = 101,000 Pa + 6800 Pa = 107,800 Pa.

Recap

  • Atmospheric pressure is the pressure exerted by the weight of the air in the atmosphere.
  • A manometer measures gauge pressure, which is the pressure difference relative to the atmosphere.
  • Absolute pressure is the total pressure, found by adding atmospheric pressure to the gauge pressure.
  • If the liquid level is lower on the gas side, the gas pressure is higher than atmospheric.
  • If the liquid level is higher on the gas side, the gas pressure is lower than atmospheric.

Quick check

  1. What is the approximate value of standard atmospheric pressure in Pascals?1 mark
  2. A manometer shows the liquid level is higher on the side connected to a gas. Is the gas pressure greater or less than atmospheric pressure?1 mark

5. Gas Pressure and Boyle's Law

For a fixed amount of gas kept at a constant temperature, its pressure and volume are inversely proportional. This means if you decrease the volume of the container, the gas particles will be more crowded, collide more often with the walls, and thus exert a higher pressure. Conversely, increasing the volume gives the particles more space, leading to fewer collisions and lower pressure. This relationship is known as Boyle's Law. It can be expressed mathematically as p₁V₁ = p₂V₂, where p₁ and V₁ are the initial pressure and volume, and p₂ and V₂ are the final pressure and volume.

p₁V₁ = p₂V₂ (at constant temperature)

p ∝ 1/V

Key term

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

Examiner insight

Examiners reward clear working that shows the identification of p₁, V₁, p₂, and V₂ before substituting them into the correct formula.

Common pitfall

Applying Boyle's Law in situations where the temperature changes or where gas is allowed to enter or leave the container.

Worked example 13 marks

A gas is trapped in a syringe with a volume of 50 cm³ at a pressure of 100 kPa. The plunger is pushed in, reducing the volume to 20 cm³. What is the new pressure of the gas, assuming the temperature remains constant?

  1. 1

    Step 1: Identify the initial and final states. p₁ = 100 kPa, V₁ = 50 cm³, V₂ = 20 cm³, p₂ = ?

  2. 2

    Step 2: State Boyle's Law. p₁V₁ = p₂V₂.

  3. 3

    Step 3: Rearrange the formula to solve for p₂. p₂ = (p₁V₁) / V₂.

  4. 4

    Step 4: Substitute the values. p₂ = (100 kPa × 50 cm³) / 20 cm³.

  5. 5

    Step 5: Calculate the result. p₂ = 5000 / 20 = 250 kPa. Note that units for volume can be left as cm³ as long as they are consistent on both sides.

Worked example 23 marks

An air bubble has a volume of 2 cm³ at the bottom of a lake where the pressure is 400,000 Pa. It rises to the surface where the pressure is 100,000 Pa. What is its volume at the surface, assuming no temperature change?

  1. 1

    Step 1: Identify the initial and final states. p₁ = 400,000 Pa, V₁ = 2 cm³, p₂ = 100,000 Pa, V₂ = ?

  2. 2

    Step 2: State Boyle's Law. p₁V₁ = p₂V₂.

  3. 3

    Step 3: Rearrange the formula to solve for V₂. V₂ = (p₁V₁) / p₂.

  4. 4

    Step 4: Substitute the values. V₂ = (400,000 Pa × 2 cm³) / 100,000 Pa.

  5. 5

    Step 5: Calculate the result. V₂ = 800,000 / 100,000 = 8 cm³.

Recap

  • Boyle's Law applies to a fixed mass of gas at a constant temperature.
  • Pressure and volume are inversely proportional; if one doubles, the other halves.
  • The formula for calculations is p₁V₁ = p₂V₂.
  • Ensure you use consistent units for pressure (e.g., both in kPa) and volume (e.g., both in cm³) on both sides of the equation.
  • The law works because compressing a gas increases the frequency of particle collisions with the container walls.

Quick check

  1. A sealed syringe of gas has its volume tripled at constant temperature. What happens to the pressure inside?1 mark
  2. What two conditions must be met for Boyle's Law to be applicable?2 marks

End-of-chapter exercise

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

  1. A force of 450 N is applied to an area of 0.03 m². Calculate the pressure exerted in Pascals.2 marks
  2. Explain, using the principles of pressure, why a sharp knife cuts better than a blunt one.2 marks
  3. A rectangular block of wood has a mass of 60 kg and measures 1.0 m by 0.5 m by 0.4 m. Calculate the maximum pressure it can exert when resting on the ground. Take g = 10 N/kg.4 marks
  4. Calculate the pressure at a depth of 12 m in a freshwater lake. The density of fresh water is 1000 kg/m³ and g = 10 N/kg.2 marks
  5. A gas is stored in a cylinder with a volume of 0.5 m³ at a pressure of 200 kPa. The gas is transferred to a 2.5 m³ container. Assuming the temperature is constant, what will be the new pressure?3 marks
  6. A manometer connected to a gas tap shows the water level is 40 cm higher on the side open to the atmosphere. If atmospheric pressure is 101 kPa, calculate the absolute pressure of the gas. The density of water is 1000 kg/m³ and g = 10 N/kg.4 marks
  7. A vehicle of weight 120,000 N is supported by four tyres. If the pressure in each tyre is 300,000 Pa, calculate the contact area of a single tyre with the ground, assuming the weight is distributed equally.4 marks
  8. An air bubble with a volume of 0.5 cm³ is released from a diver at a depth where the total pressure is 3.5 atm. It rises to the surface where the pressure is 1 atm. Calculate its volume just as it reaches the surface, assuming the temperature is constant.3 marks
  9. A rectangular swimming pool has a base of 10 m by 4 m and is filled with water to a depth of 2.5 m. Calculate (a) the weight of the water in the pool, and (b) the pressure on the bottom of the pool. (Density of water = 1000 kg/m³, g = 10 N/kg).5 marks
  10. Explain why tractors used on soft farmland have very large, wide tyres.2 marks

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