Physics: Foundations of Physical Science

MGU Physics · Preparatory Volume

Chapter 35: Density and Pressure

35.1 What a Fluid Is

A fluid is any substance that can flow and deform continuously under the application of a shear stress (a force applied parallel to a surface). Both liquids and gases are fluids. Solids, by contrast, resist shear deformation — they can be pushed without flowing. The defining property of a fluid is the inability to sustain a static shear stress: apply a sideways force to a layer of fluid and it will flow, no matter how small the force. This distinguishes water from ice (a solid that resists shear), and air from glass (which behaves like an extremely viscous fluid over geological timescales, despite seeming solid on human timescales). Fluids do resist compression and tension (normal stresses), which is why you cannot push your hand through concrete even though concrete technically flows over centuries. But for all practical purposes of this chapter and the next two, fluids are treated as substances that transmit pressure forces in all directions and deform without limit when sheared.

35.2 Liquids vs. Gases

Liquids and gases share the property of being fluids, but they differ in important ways Physics: Foundations of Physical Science Chapter 35 — Density and Pressure that arise from their microscopic structure. A liquid has a definite volume but no definite shape. Its molecules are close together and interact through intermolecular forces, but they can slide past each other (unlike a solid). Liquids are nearly incompressible: changing the pressure by a factor of two changes the volume by much less than 1%. This makes liquids excellent for hydraulic systems, where pressure must be transmitted with minimal energy storage in compression. A gas has neither definite volume nor definite shape. Its molecules are widely separated and interact very weakly. Gases are highly compressible — halving the volume roughly doubles the pressure (Boyle’s Law, Chapter 32). Gases fill any container entirely. These properties make gas behaviour strongly dependent on pressure and temperature in ways that liquid behaviour is not. Property Liquid Gas Definite volume? Yes No — fills container Definite shape? No — takes container shape No — fills container Compressible? Negligibly (treat as incompressible) Yes, significantly Density High (hundreds to thousands kg/m³) Low (fraction to few kg/m³) Molecular spacing Close together Far apart Free surface? Yes (liquid–gas interface) No free surface

Table 35.1

35.3 Density

Density (ρ, rho) is the mass per unit volume of a substance. It is an intensive property

— it does not depend on how much material is present, only on what the material is and its thermodynamic state (temperature, pressure).

ρ = m / V

Physics: Foundations of Physical Science Chapter 35 — Density and Pressure Density measures how tightly mass is packed. Dense materials (lead, steel, gold) have large amounts of mass in small volumes. Less dense materials (wood, foam, air) have small amounts of mass in larger volumes. Water has a density of almost exactly 1000 kg/m³ (or 1 g/cm³) at 4°C — this is not a coincidence but was built into the original definition of the kilogram. Substance

ρ (kg/m³)

ρ (g/cm³)

Notes Osmium (densest element) 22 590 22.59 Densest naturally occurring element Gold 19 300 19.3 Heavy precious metal Lead 11 340 11.34 Dense industrial metal Iron / steel 7 870 7.87 Common structural metal Aluminium 2 700 2.70 Lightweight structural metal Glass 2 500 2.50 Approximate; varies with type Seawater 1 025 1.025 Slightly denser than fresh water Water (4°C) 1 000 1.000 Reference density for specific gravity Ice (0°C) 0.917 Less dense than liquid water Ethanol 0.789 Common organic solvent Air (20°C, 1 atm) 1.204 0.001204 Much less dense than liquids Helium (20°C, atm) 0.164 0.000164 Lightest noble gas

Table 35.2

35.4 Calculating Density

Density calculations involve three possible unknowns: density, mass, and volume. Given any two, the third is determined by ρ = m/V, m = ρV, or V = m/ρ.

Step 1: Identify: What is known (two of ρ, m, V) and what is asked (the third).

Physics: Foundations of Physical Science Chapter 35 — Density and Pressure

Step 2: Units: If density is in kg/m³, mass must be in kg and volume in m³. If density

in g/cm³, mass in g and volume in cm³.

Step 3: Compute: Apply the formula. Note: 1 mL = 1 cm³; 1 L = 1000 cm³ = 10 ³ m³.

A solid gold bar measures 25 cm × 8 cm × 4 cm.

Volume = 800 cm³ = 800 × 10 ⁶ m³. Mass = ρV = 19 300 × 800 × 10 ⁶ = 15.44 kg

15.4 kg.

A 250 mL bottle contains 198 g of

ethanol. Volume = 250 cm³. ρ = 198/250 = 0.792 g/cm³ = 792 kg/m³. (Table value:

789 kg/m³ — close, consistent with real ethanol.) For irregular objects, volume can be found by water displacement: submerge the object in a graduated cylinder and measure the volume of water displaced. This technique was famously used by Archimedes to check whether a crown was pure gold.

35.5 Relative Density and Specific Gravity

Specific gravity (SG), also called relative density, is the ratio of the density of a substance to the density of a reference substance, usually water at 4°C:

SG = ρ_substance / ρ_water = ρ_substance / 1000

(where ρ is in kg/m³)

Specific gravity is convenient because its numerical value in the metric system equals the density in g/cm³ and makes comparisons

immediate: SG = 0.92 means the

substance is 92% as dense as water (ice floats); SG = 7.87 means iron is 7.87 times

denser than water (iron sinks). No unit conversion is needed. Specific gravity is measured by a hydrometer: a weighted float that sinks deeper in denser liquids. Hydrometers are used to check battery acid concentration, the sugar content of wine musts (brix measurement), antifreeze concentration, and the salinity of aquaria.

35.6 Pressure

Pressure is the force exerted perpendicular to a surface per unit area of that surface. It is one of the most important quantities in fluid mechanics and connects the macroscopic behaviour of fluids directly to the forces they exert. Physics: Foundations of Physical Science Chapter 35 — Density and Pressure

P = F / A

The same force applied over a smaller area produces greater pressure. This is why a stiletto heel sinks into soft ground while a flat boot does not: the same body weight (force) is concentrated into a much smaller area by the heel. Why a sharp knife cuts more easily than a dull one: the same force acts over a much smaller edge area, producing very high pressure. Why snowshoes allow walking on snow: the body weight is spread over a large area, keeping the pressure below the snow’s yield strength.

area

0.5 cm² = 0.5 × 10 ⁴ m²). P = mg/A = (60 × 9.81)/(0.5 × 10 ⁴) = 588/0.00005 =

11.8 MPa. This is comparable to the yield strength of some soft materials — hence the

damage stiletto heels cause to wooden floors.

35.8 Fluid Pressure and Depth

In a static fluid (one not in motion), pressure increases with depth because the weight of the fluid above presses down on the fluid below. The pressure at any depth h below

the free surface of a fluid of density ρ is:

P = P₀ + ρgh

₀ This equation has several key consequences. First, pressure depends only on depth, not on the shape of the container or the horizontal position. A point 5 m below the surface of a narrow well and 5 m below the surface of the ocean both have the same pressure (if the same fluid and same surface pressure). This is a non-obvious result that puzzled natural philosophers until Pascal systematised it. Second, pressure increases linearly with depth. Every additional metre of water adds

ρgh = 1000 × 9.81 × 1

9810 Pa 0.097 atm 1 psi of pressure. At 10 m depth, water pressure is approximately 1 atm, so the total absolute pressure is about 2 atm (atmospheric plus water). At 100 m, about 11 atm. At the deepest point of the ocean (Challenger Deep, 11 km), the pressure is about 1100 atm.

Find the absolute pressure at the bottom of a 4.0 m

deep swimming pool. P₀ = 101 325 Pa. ρ = 1000 kg/m³. P = 101 325 + 1000 × 9.81 ×

4.0 = 101 325 + 39 240 = 140 565 Pa

140.6 kPa

1.39 atm.

Physics: Foundations of Physical Science Chapter 35 — Density and Pressure

Pressure at a given depth is the same in all directions — up, down, and sideways. This follows from Pascal’s principle: a fluid cannot sustain a shear stress, so the pressure at any point must be isotropic (the same in all directions). A submerged object therefore experiences pressure pushing on it from all sides.

35.9 Atmospheric Pressure

Atmospheric pressure is the pressure exerted by the weight of the column of air above a given point in the atmosphere. At sea level, standard atmospheric pressure P_atm =

101 325 Pa = 101.325 kPa

1 atm 760 mmHg

14.7 psi.

Atmospheric pressure decreases with altitude because the column of air above is shorter and less dense. At 5500 m altitude (approximately the height of the summit of Kilimanjaro), atmospheric pressure is roughly half of sea-level value — each breath contains half as many oxygen molecules as at sea level. At 10 000 m (typical cruising altitude for aircraft), pressure is about 26 kPa, which is why aircraft cabins are pressurised. Atmospheric pressure varies with weather: high-pressure systems (anticyclones) have slightly above-average P_atm; low-pressure systems (cyclones, storms) have below- average P_atm. A falling barometric pressure reading signals approaching unsettled weather. The variation is typically 1–2% around the standard value, significant for aviation and precision measurement. The weight of the entire atmosphere acts on every surface. A horizontal flat surface of area 1 m² has about 101 325 N 10 tonnes of air weight pressing down on it at sea level. This force is balanced by the atmospheric pressure acting upward on the underside of any real surface, so we do not feel it as a net force. When pressure is removed from one side (as in a vacuum seal or suction cup), the net force becomes apparent.

35.10 Gauge Pressure vs. Absolute Pressure

Absolute pressure is pressure measured from zero (a perfect vacuum). It is always non- negative. Gauge pressure is pressure measured relative to atmospheric pressure: P_gauge = P_abs P_atm. Gauge pressure can be positive (pressure above atmospheric, as in an inflated tyre) or negative (pressure below atmospheric, as in a partial vacuum or a suction cup). Gauge pressure is zero when the absolute pressure equals atmospheric. Physics: Foundations of Physical Science Chapter 35 — Density and Pressure

P_abs = P_gauge + P_atm

Most everyday pressure gauges read gauge pressure: a flat tyre reads 0 psi gauge, not

14.7 psi absolute. A tyre inflated to “32 psi” has gauge pressure 32 psi and absolute

pressure 32 + 14.7 = 46.7 psi. Always clarify which type of pressure is being used;

using gauge where absolute is required (or vice versa) causes errors in gas law and fluid pressure calculations.

Gas law calculations (Chapters 32–34) require absolute pressure. Fluid

depth-pressure calculations using P = P₀ + ρgh require absolute pressure at the

surface (P₀ = P_atm for an open container). Most real-world gauges read gauge

pressure. Always convert before applying formulas.

35.11 Pascal's Principle

Pascal's Principle states: a pressure change applied to a confined fluid is transmitted undiminished to every part of the fluid and to the walls of the container.

This follows directly from the hydrostatic pressure equation P = P₀ + ρgh. If P₀ changes

(for example, by applying a force to a piston), then P changes by the same amount at

every point in the fluid (since ρgh is unchanged for any fixed depth h). The pressure

change propagates instantaneously throughout the entire fluid volume. Pascal's principle makes fluids ideal for transmitting forces over distances and around corners without mechanical linkages. It is the operating principle behind every hydraulic system: brakes, lifts, presses, and jacks.

A pump increases the pressure at one end of a hydraulic line by 500 kPa. The pressure at a cylinder 10 m away (same elevation) increases by exactly 500 kPa. The 10 m distance does not reduce the transmitted pressure change.

35.12 Hydraulic Systems

A hydraulic system transmits force from one piston to another through an incompressible fluid. The key insight is that pressure is the same throughout the fluid (Pascal’s principle), but the force exerted by the fluid depends on the area of the piston. A small force applied to a small piston can be amplified into a much larger force at a large piston.

P = F₁/A₁ = F₂/A₂

⇒

F₂ = F₁ × A₂/A₁

Physics: Foundations of Physical Science Chapter 35 — Density and Pressure The mechanical advantage is A₂/A₁. If the output piston is 10 times larger in area than the input piston, the output force is 10 times the input force. But energy is conserved:

the output piston moves a shorter distance. Specifically, A₁d₁ = A₂d₂ (fluid volume is

conserved), so d₂ = d₁ A₁/A₂. Force is amplified by A₂/A₁; distance is reduced by the

same factor. Work done in = work done out (in the ideal case).

A car lift has a small input piston (area 5.0 cm²) and a large output piston (area 200 cm²). A mechanic pushes the input piston with 150 N.

Output force: F₂ = 150 × 200/5.0 = 6000 N. This is enough to lift a car of mass

6000/9.81 612 kg. The mechanical advantage is 200/5 = 40. Hydraulic systems power car brakes (master cylinder amplifies foot force to disc/drum pressure), aircraft control surfaces (pilots move controls against minimal resistance; hydraulics drive the actual surface), construction machinery (excavators, bulldozers), and industrial presses (forging, stamping, injection moulding). The combination of Pascal’s principle and incompressibility makes hydraulic force transmission essentially lossless at practical engineering speeds.

35.13 Common Mistakes in Density and Pressure Problems

•

Confusing mass and weight. Density = mass/volume, not weight/volume. The

weight of a fluid does appear in the pressure formula P = ρgh (as ρg = weight

per unit volume), but density itself is a mass property. • Using gauge pressure when absolute pressure is required. The hydrostatic

formula P = P₀ + ρgh gives absolute pressure. If P₀ is gauge (zero for open

containers), you must add P_atm to get absolute. Failing to include P_atm leads to underestimating pressure by about 101 kPa. • Forgetting to convert units consistently. If density is given in g/cm³ and depth in

metres, convert before applying ρgh. Either convert ρ to kg/m³ and use g in

m/s², or convert h to cm and use g in cm/s² (inconvenient). Easiest: always use SI units (kg, m, Pa). • Thinking pressure depends on the shape of the container (the hydrostatic paradox). The pressure at a given depth depends only on the fluid density, the depth, and the surface pressure — not on whether the container is wide, narrow, tilted, or U-shaped. A narrow tube and a wide tank containing the same fluid have the same pressure at the same depth. Physics: Foundations of Physical Science Chapter 35 — Density and Pressure • Misapplying hydraulic MA: confusing area ratio with force ratio. The force ratio

equals the area ratio (F₂/F₁ = A₂/A₁), but the distance ratio is the inverse (d₂/d₁

= A₁/A₂). Both must be stated consistently or the energy conservation check fails. • Using diameter instead of area. Hydraulic problems often give piston diameters.

Area = πr² = π(d/2)². Using diameter directly instead of computing area first

gives the wrong force ratio.

pressure (P_atm for an open container) or a gauge pressure that must be corrected? This is the most common source of error in hydrostatic pressure problems.

Chapter Summary

Density (ρ = m/V) measures mass per unit volume. Pressure (P = F/A) measures force

per unit area. Both are foundational to understanding all fluid behaviour. •

Density SI unit: kg/m³. Specific gravity = ρ/ρ_water (dimensionless; equals

density in g/cm³). •

Pressure SI unit: pascal (Pa = N/m²). Acts equally in all directions at a point in a

static fluid. •

Hydrostatic pressure: P = P₀ + ρgh. Increases linearly with depth; independent of

container shape. • Atmospheric pressure 101 325 Pa at sea level; decreases with altitude. •

Absolute = gauge + atmospheric. Gas laws and depth-pressure formulas require

absolute pressure. • Pascal's principle: pressure change in a confined fluid is transmitted undiminished throughout. •

Hydraulic systems: F₂ = F₁(A₂/A₁). Force amplified by area ratio; distance reduced

by same factor; work conserved.

Key Terms

fluid: Any substance that flows and deforms

density (ρ): ρ = m/V; mass per unit volume; SI unit kg/m³; an intensive, material-

Physics: Foundations of Physical Science

specific gravity (SG): SG = ρ/ρ_water; dimensionless ratio of substance density to

pressure (P): P = F/A; perpendicular force per unit area; SI unit pascal (Pa = N/m²);

hydrostatic pressure: P = P₀ + ρgh; the pressure at depth h in a static fluid of density

atmospheric pressure (P_atm): Pressure exerted by the weight of the atmosphere;

gauge pressure: Pressure measured relative to atmospheric pressure; P_gauge = P_abs

absolute pressure: Pressure measured from zero (perfect vacuum); P_abs = P_gauge +

Pascal's principle

hydraulic mechanical advantage: The force amplification ratio in a hydraulic system:

hydrostatic paradox: The counterintuitive result that fluid pressure at a given depth

Concept Review Questions

Answer in complete sentences. No calculations required unless stated.

Explain why a sharp knife cuts more easily than a dull knife using the concept of

pressure. What changes between a sharp and a dull knife when the same force is applied?

Water pressure at a depth of 5 m is the same in a narrow pipe and in a wide

lake. Explain why container shape does not affect pressure at a given depth.

Explain Pascal's principle in your own words and give one everyday example of it

being used in an engineering system.

A hydraulic press has a force advantage of 50 (output force = 50 × input force).

What is the disadvantage? Why does this not violate conservation of energy?

Atmospheric pressure is about 101 kPa at sea level but only 26 kPa at 10 km

altitude. Explain why the pressure decreases with altitude. Physics: Foundations of Physical Science Chapter 35 — Density and Pressure

Ice has density 917 kg/m³ and water has density 1000 kg/m³. What fraction of a

floating iceberg is below the waterline? (Use the floating equilibrium condition from Chapter 36, but reason it out from densities here.)

Distinguish absolute pressure from gauge pressure. Give an example where using

gauge pressure instead of absolute pressure would lead to a serious error.

Why are liquids more useful than gases for hydraulic systems? What property of

liquids makes hydraulic force transmission nearly lossless?

Density Problems

A solid copper sphere has radius 3.0 cm. Find its mass. (ρ_Cu = 8960 kg/m³; V =

(4/3)πr³)

A 2.0 L bottle contains 1640 g of a liquid. Find its density in kg/m³ and in g/cm³.

What liquid might this be?

A steel I-beam (density 7870 kg/m³) has a cross-sectional area of 8000 mm² and

a length of 6.0 m. Find its mass.

An object with mass 450 g is submerged in water and displaces 380 cm³. Find its

density. Does it float or sink in water?

An alloy is 60% aluminium (ρ = 2700 kg/m³) and 40% copper (ρ = 8960 kg/m³)

by volume. Find the density of the alloy.

Pressure Problems

A concrete dam holds water to a depth of 40 m. Find the gauge pressure and

absolute pressure at the base of the dam.

A 90 kg man stands on snowshoes with total contact area 0.12 m². Find the

pressure exerted on the snow. Compare to the same man standing on boots (contact area 0.024 m²). Why do snowshoes work?

A submarine operates at a depth of 300 m in seawater (ρ = 1025 kg/m³). Find

the absolute pressure outside the hull in Pa, kPa, and atm.

A mercury manometer shows a height difference of 25 cm on one side. Find the

gauge pressure in Pa. (ρ_Hg = 13 600 kg/m³)

Physics: Foundations of Physical Science Chapter 35 — Density and Pressure

A tyre gauge reads 220 kPa. Find the absolute pressure inside the tyre. (P_atm =

101.3 kPa)

Fluid Pressure Exercises

A U-tube manometer contains water on one side and an unknown oil on the

other, separated by a valve. When the valve is opened, the water side rises 3.0 cm and the oil side drops. If the oil column is 4.2 cm higher than the water column at equilibrium, find the density of the oil. (Equal pressures at the bottom:

ρ_oil g h_oil = ρ_water g h_water; be careful with the geometry.)

A diver at 25 m depth in fresh water. (a) Find gauge pressure. (b) Find absolute

pressure. (c) If the diver moves to 25 m depth in the ocean (ρ = 1025 kg/m³),

how does the pressure change?

An aluminium pressure vessel (capable of withstanding 5.0 MPa gauge) contains

water at 20 m depth in the ocean. What is the gauge pressure inside if the vessel is open to the seawater at 20 m depth? Is the vessel adequate?

Two connected tanks of different cross-sections are filled with the same liquid.

Tank A has cross-sectional area 0.10 m² and liquid depth 0.50 m; Tank B has area

0.30 m². If both tanks are at the same level and connected at the bottom, and

the total volume of liquid is constant, find the liquid depth in Tank B. (Hint: pressure at the connection point must be equal.) Hydraulic System Applications

A hydraulic car lift has an input piston of diameter 2.0 cm and an output piston

of diameter 18 cm. (a) Find the area of each piston. (b) A force of 200 N is applied to the input. Find the output force. (c) What mass car can be lifted?

A hydraulic brake system has a master cylinder (bore diameter 22 mm) and four

wheel cylinders (bore diameter 38 mm each). The driver applies 500 N to the brake pedal, which produces 1500 N force on the master cylinder piston. (a) Find the pressure in the brake fluid. (b) Find the force at each wheel cylinder.

A hydraulic press compresses a metal part. The input piston (area 4.0 cm²)

moves 120 mm. The output piston area is 600 cm². (a) Find the volume of fluid transferred. (b) How far does the output piston move? (c) If 200 N is applied to Physics: Foundations of Physical Science Chapter 35 — Density and Pressure input, find the output force and the output work. Compare to the input work (F × d). Is energy conserved?

A hydraulic system is used to lift a 2000 kg car using a pump that generates 15

MPa of pressure. What minimum output piston area is needed?

Chapter 35 Checkpoint

Before moving on to Chapter 36, confirm you can do each of the following. I can… See Section Define fluid and explain how liquids differ from gases 35.1–35.2

Define density and calculate ρ, m, or V given the other two

35.3–35.4 Define and calculate specific gravity 35.5 Define pressure and calculate P, F, or A given the other two 35.6–35.7

Apply P = P₀ + ρgh to find pressure at depth in a static fluid

35.8 State the value of standard atmospheric pressure in Pa, kPa, and atm 35.9 Convert between gauge pressure and absolute pressure 35.10 State Pascal's principle and explain how it applies to confined fluids 35.11

Analyse hydraulic systems using F₂ = F₁(A₂/A₁) and energy

conservation 35.12 Identify and avoid common mistakes in density and pressure problems 35.13