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Physics - Pressure (Liquids and Gases)

Grade 9IGCSE

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

🔑Concepts

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Pressure in a fluid (liquid or gas) at a point acts equally in all directions.

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The pressure in a liquid is caused by the weight of the liquid above and increases with depth (hh) and density (ρ\rho).

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Pressure in a liquid is calculated using the formula P=hhogP = h ho g, where gg is the gravitational field strength (typically 9.8 N/kg9.8 \text{ N/kg} or 10 N/kg10 \text{ N/kg} for IGCSE).

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Atmospheric pressure is exerted by the air around us; at sea level, it is approximately 1.01×105 Pa1.01 \times 10^5 \text{ Pa} or 760 mmHg760 \text{ mmHg}.

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A barometer measures atmospheric pressure, while a manometer measures the pressure difference between two gases using a liquid column.

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

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Total pressure at a depth in a liquid is the sum of the liquid pressure and the atmospheric pressure: Ptotal=Patmosphere+hhogP_{total} = P_{atmosphere} + h ho g.

📐Formulae

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

P=hρgP = h \rho g

P1V1=P2V2P_1 V_1 = P_2 V_2

ΔP=ρgΔh\Delta P = \rho g \Delta h

💡Examples

Problem 1:

Calculate the pressure exerted by water at the bottom of a swimming pool that is 3 m3 \text{ m} deep. (Density of water ρ=1000 kg/m3\rho = 1000 \text{ kg/m}^3, g=10 m/s2g = 10 \text{ m/s}^2)

Solution:

P=hρg=3 m×1000 kg/m3×10 m/s2=30,000 PaP = h \rho g = 3 \text{ m} \times 1000 \text{ kg/m}^3 \times 10 \text{ m/s}^2 = 30,000 \text{ Pa}

Explanation:

We use the liquid pressure formula P=hρgP = h \rho g. By substituting the height (3 m3 \text{ m}), density (1000 kg/m31000 \text{ kg/m}^3), and gravity (10 m/s210 \text{ m/s}^2), we find the pressure to be 30 kPa30 \text{ kPa}.

Problem 2:

A gas cylinder contains 0.04 m30.04 \text{ m}^3 of nitrogen at a pressure of 2.0×105 Pa2.0 \times 10^5 \text{ Pa}. If the gas is allowed to expand into a larger container with a total volume of 0.10 m30.10 \text{ m}^3 at constant temperature, what is the new pressure?

Solution:

P2=P1V1V2=(2.0×105 Pa)×(0.04 m3)0.10 m3=80,000 PaP_2 = \frac{P_1 V_1}{V_2} = \frac{(2.0 \times 10^5 \text{ Pa}) \times (0.04 \text{ m}^3)}{0.10 \text{ m}^3} = 80,000 \text{ Pa}

Explanation:

According to Boyle's Law (P1V1=P2V2P_1 V_1 = P_2 V_2), as volume increases, pressure must decrease. Rearranging to solve for P2P_2 gives 8.0×104 Pa8.0 \times 10^4 \text{ Pa} or 80 kPa80 \text{ kPa}.

Problem 3:

A manometer is connected to a gas supply. The mercury level in the arm open to the atmosphere is 15 cm15 \text{ cm} higher than the arm connected to the gas. If atmospheric pressure is 76 cmHg76 \text{ cmHg}, what is the gas pressure in cmHg\text{cmHg}?

Solution:

Pgas=Patm+h=76 cmHg+15 cmHg=91 cmHgP_{gas} = P_{atm} + h = 76 \text{ cmHg} + 15 \text{ cmHg} = 91 \text{ cmHg}

Explanation:

In a manometer, if the liquid is higher on the atmospheric side, the gas pressure is greater than atmospheric pressure. We add the height difference (hh) to the atmospheric pressure (PatmP_{atm}).