Recall and use the equation for the change in pressure beneath the surface of a liquid Δp = ρ g Δh

1.8 Pressure

Learning Objective

Recall and use the hydrostatic‑pressure equation

\(\displaystyle \Delta p = \rho\,g\,\Delta h\)

Quick‑Recall Box – Pressure Units & Conversions

Unit	Symbol	Equivalent
Pascal	Pa	1 Pa = 1 N m⁻²
Kilopascal	kPa	1 kPa = 10³ Pa
Atmosphere	atm	1 atm = 101 325 Pa ≈ 1.01 × 10⁵ Pa
Bar	bar	1 bar = 10⁵ Pa

Fundamental Definition

• Pressure: \(p = \dfrac{F}{A}\) – force applied per unit area.
• SI unit: pascal (Pa = N m⁻²).

Quantitative Everyday Example

A 500 N person standing on a shoe sole of area \(0.20\ \text{m}^2\) exerts

\[ p = \frac{F}{A}= \frac{500\ \text{N}}{0.20\ \text{m}^2}= 2.5\times10^{3}\ \text{Pa}=2.5\ \text{kPa}. \]

This illustrates how pressure increases when the same force acts over a smaller area (e.g. a high‑heeled shoe).

Atmospheric & Absolute Pressure

• Atmospheric pressure at sea level: \(p_{\text{atm}} = 1.0\ \text{atm}=101\,325\ \text{Pa}\).
• Gauge pressure (\(\Delta p\)): pressure above atmospheric pressure.
• Absolute pressure (\(p_{\text{abs}}\)): total pressure, \(p_{\text{abs}} = p_{\text{atm}} + \Delta p\).

Hydrostatic (Fluid) Pressure

• Pressure in a liquid increases **linearly** with depth because \(\Delta p\) is directly proportional to \(\Delta h\) (the hydrostatic formula).
• This linear increase explains why the base of a dam must be thicker than the top – the water depth (and therefore pressure) is greatest at the bottom.

Derivation of \(\Delta p = \rho g \Delta h\)

Consider a horizontal slab of fluid at depth \(h\) with thickness \(\Delta h\) and cross‑sectional area \(A\).

1. Mass of the slab: \(m = \rho A \Delta h\).
2. Weight of the slab: \(W = mg = \rho A \Delta h\,g\).
3. Pressure difference between the bottom and the top of the slab: \[ \Delta p = \frac{W}{A}= \frac{\rho A \Delta h\,g}{A}= \rho g \Delta h. \]

Notation Table (Consistent with the Syllabus)

Symbol	Meaning	SI Unit
\(\Delta p\)	Gauge (change) pressure	Pa (N m⁻²)
\(p_{\text{abs}}\)	Absolute pressure	Pa (or atm, bar)
\(\rho\)	Density of the liquid	kg m⁻³
g	Acceleration due to gravity	9.81 m s⁻² (≈10 m s⁻² for quick work)
\(\Delta h\)	Depth difference (vertical distance)	m
p_{\text{atm}}	Atmospheric pressure at sea level	Pa (or atm)

Typical Densities of Common Liquids

Liquid	Density \(\rho\) (kg m⁻³)
Water (4 °C)	1000
Sea water	1025
Vegetable oil	≈ 920
Mercury	13 600

Worked Example

Problem: Calculate the pressure increase at a depth of \(5.0\ \text{m}\) in fresh water.

Given: \(\rho = 1000\ \text{kg m}^{-3}\), \(g = 9.81\ \text{m s}^{-2}\), \(\Delta h = 5.0\ \text{m}\).

Solution:

\[ \Delta p = \rho g \Delta h = (1000)(9.81)(5.0)=49\,050\ \text{Pa}\approx4.9\times10^{4}\ \text{Pa}. \]

Gauge pressure in atmospheres:

\[ \frac{\Delta p}{p_{\text{atm}}}= \frac{49\,050}{101\,325}\approx0.48\ \text{atm}. \]

Absolute pressure at that depth:

\[ p_{\text{abs}} = p_{\text{atm}} + \Delta p \approx 1.48\ \text{atm}. \]

Practice Questions

1. What is the pressure increase at a depth of \(2.0\ \text{m}\) in oil (\(\rho = 920\ \text{kg m}^{-3}\))? Use \(g = 10\ \text{m s}^{-2}\) for a quick calculation.
2. A diver is \(12\ \text{m}\) below the surface of sea water. Calculate the absolute pressure on the diver’s suit if atmospheric pressure is \(1.0\ \text{atm}\). (Take \(\rho_{\text{sea}} = 1025\ \text{kg m}^{-3}\), \(g = 9.8\ \text{m s}^{-2}\).)
3. Explain, using the hydrostatic‑pressure equation, why a dam must be built thicker at its base than at its top.

Model Answers

1. \(\Delta p = \rho g \Delta h = (920)(10)(2.0)=18\,400\ \text{Pa}=0.182\ \text{kPa}\).
2. \(\Delta p = (1025)(9.8)(12)=1.21\times10^{5}\ \text{Pa}=1.20\ \text{atm}\).
   \(p_{\text{abs}} = 1.0\ \text{atm} + 1.20\ \text{atm}=2.20\ \text{atm}\).
3. The pressure on a vertical wall is \(\Delta p = \rho g h\). At the base (\(h\) large) the pressure is greatest, so a larger resisting area (thicker wall) is required. Near the surface (\(h\) small) the pressure is low, allowing a thinner wall.

Common Mistakes to Avoid

• Forgetting to convert depth to metres (e.g., using cm or mm directly).
• Using the wrong value of \(g\); remember the exam permits \(g = 10\ \text{m s}^{-2}\) for quick work, but the more accurate value is \(9.81\ \text{m s}^{-2}\).
• Mixing gauge and absolute pressure – always add \(p_{\text{atm}}\) when an absolute value is required.
• Omitting units or mixing units (Pa with atm, kPa, bar).
• Neglecting the linear relationship \(\Delta p \propto \Delta h\) when explaining real‑world applications (e.g., dams, submarine hulls).

Diagram Suggestion

Summary Checklist

• Define pressure: \(p = F/A\); know that 1 Pa = 1 N m⁻².
• Recall the hydrostatic formula \(\Delta p = \rho g \Delta h\) and its linear dependence on depth.
• Identify each symbol, its meaning, and SI unit (see notation table).
• Distinguish gauge pressure (\(\Delta p\)) from absolute pressure (\(p_{\text{abs}} = p_{\text{atm}} + \Delta p\)).
• Convert between Pa, kPa, atm, and bar when required.
• Apply the formula to everyday situations (e.g., dam design, diving, hydraulic lifts).
• Check calculations: correct \(g\) value, depth in metres, and inclusion of units.