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Motion, Forces and Energy - Pressure

Grade 12IGCSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Pressure is defined as the force acting per unit area: P=FAP = \frac{F}{A}. The SISI unit is the Pascal (PaPa), where 1 Pa=1 N/m21\ Pa = 1\ N/m^2.

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In solids, a smaller contact area results in a higher pressure for the same applied force. Conversely, a larger area reduces pressure.

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Pressure in a fluid (liquids and gases) at rest acts equally in all directions.

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Fluid pressure increases with depth because of the weight of the fluid above. It is calculated using P=hρgP = h \rho g, where hh is depth, ho ho is density, and gg is the gravitational field strength (9.81 N/kg9.81\ N/kg).

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Atmospheric pressure at sea level is approximately 1.01×105 Pa1.01 \times 10^5\ Pa or 760 mmHg760\ mmHg. This is measured using a mercury barometer.

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A manometer measures the pressure difference between a gas and the atmosphere. The pressure of the gas is given by Pgas=Patm±ΔhρgP_{gas} = P_{atm} \pm \Delta h \rho g.

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Pascal's Principle: Pressure applied to an enclosed incompressible fluid is transmitted undiminished to every portion of the fluid and to the walls of the container.

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Boyle's Law for gases: For a fixed mass of gas at a constant temperature, the pressure is inversely proportional to the volume: P∝1VP \propto \frac{1}{V} or P1V1=P2V2P_1 V_1 = P_2 V_2.

📐Formulae

P=FAP = \frac{F}{A}

ΔP=hρg\Delta P = h \rho g

Ptotal=Patm+hρgP_{total} = P_{atm} + h \rho g

P1V1=P2V2P_1 V_1 = P_2 V_2

F1A1=F2A2\frac{F_1}{A_1} = \frac{F_2}{A_2}

💡Examples

Problem 1:

Calculate the pressure exerted by a block of wood with a weight of 600 N600\ N and a base area of 0.5 m20.5\ m^2.

Solution:

P=FA=600 N0.5 m2=1200 PaP = \frac{F}{A} = \frac{600\ N}{0.5\ m^2} = 1200\ Pa

Explanation:

The force is the weight of the block. By dividing the weight by the contact area, we obtain the pressure in Pascals.

Problem 2:

A diver is at a depth of 20 m20\ m in seawater. If the density of seawater is 1025 kg/m31025\ kg/m^3 and g=9.8 m/s2g = 9.8\ m/s^2, calculate the pressure due to the water alone.

Solution:

P=hρg=20 m×1025 kg/m3×9.8 m/s2=200,900 PaP = h \rho g = 20\ m \times 1025\ kg/m^3 \times 9.8\ m/s^2 = 200,900\ Pa

Explanation:

The pressure in a liquid depends only on the depth, the density of the liquid, and the gravitational field strength.

Problem 3:

A gas occupies 2.0 m32.0\ m^3 at a pressure of 100 kPa100\ kPa. If the volume is compressed to 0.5 m30.5\ m^3 at a constant temperature, what is the new pressure?

Solution:

P1V1=P2V2⇒100 kPa×2.0 m3=P2×0.5 m3⇒P2=2000.5=400 kPaP_1 V_1 = P_2 V_2 \Rightarrow 100\ kPa \times 2.0\ m^3 = P_2 \times 0.5\ m^3 \Rightarrow P_2 = \frac{200}{0.5} = 400\ kPa

Explanation:

According to Boyle's Law, when the volume is reduced by a factor of 4, the pressure must increase by a factor of 4, provided the temperature remains constant.

Pressure - Revision Notes & Key Formulas | IGCSE Grade 12 Physics