Pressure, Density & Matter

— Physics · AS Level (7408) · Fluids & States of Matter —

★ Exam Tip: Density is a ratio, not a rate — read it directly off a graph at one point (m₁/V₁), never from a gradient. This is a deliberately different word from "rate" (velocity, acceleration) and is directly tested.

1. Density

Density: the ratio of mass to volume of an object. Scalar, measured in kg m⁻³. ρ = m/V

ρ = m₁/V₁ — read the COORDINATE directly, no gradient needed

★ Trick — ratio vs rate: velocity/acceleration are rates ⟹ found from a graph's GRADIENT. Density is a ratio ⟹ found by reading one (m,V) point directly and dividing — a graph-reading trap tests exactly this distinction.

Measuring density

Measure mass with an electronic balance (simplest, always safe to write).
① Regular object: measure lengths, use the appropriate volume formula.
② Irregular object: use displacement — ΔV = volume of object (rise in a measuring cylinder's liquid level after submerging it).

2. Pressure & the Derivation of P=ρgh

Pressure: the ratio of force to cross-sectional area on which the force is applied. Scalar, measured in Pascals (Pa). P = F/A

P = F/A

Derivation of P = ρgh:

P = F/A = W/A = mg/A [weight of the liquid column above]
ρ=m/V ⟹ m=ρV=ρAh [V = base area × height]
P = (ρAh×g)/A ⟹ P = ρgh (the A's cancel)

★ ΔP = ρg Δh — for a CHANGE in depth

★ Exam Tip: h is always measured from the TOP (the free surface) downward — not from the base upward.

Structured — derive P=ρgh, container of liquidDerivation

Liquid density ρ fills a container to depth h, base area A. Derive P=ρgh from the definitions of pressure and density.

P = F/A = W/A = mg/Aρ=m/V ⟹ m=ρV=ρAhP = (ρAh×g)/A ⟹ P=ρgh — "proved"

3. Atmospheric Pressure, Gases & Liquids

★ Air pressure = 1 atm = 1.0125×10⁵ Pa | ★ 1 atm = 76 cm of mercury = 760 mm Hg

★ Gases exert equal pressure everywhere (uniform throughout a trapped gas).
★ Liquids exert equal pressure at the same height.

W-tube: equal-height points P have equal pressure (both sides in the same connected liquid)

Barometer: mercury column stands at 76cm, balanced by 1 atm on the reservoir

4. Matter — Solids, Liquids, Gases

Mass: two accepted definitions — (1) "the property of matter that resists changes in state of motion" (inertia, from F=ma), or (2) "the amount of matter in an object."

★ Trick — strong vs rigid: intermolecular forces that are STRONG cause a fixed volume. Forces that are RIGID cause a fixed shape. (Rigid objects can also exert a normal contact force — a rope can't, since it isn't rigid, only tension.)

State	Shape / Volume	Intermolecular forces	Particle behaviour
Solid	Fixed shape, fixed volume	Strong AND rigid	Closely packed, fixed positions, vibrate only
Liquid	Fixed volume, NO fixed shape	Strong, NOT rigid	Close together but can move from place to place ("flow")
Gas	No fixed volume, no fixed shape	Neither strong nor rigid	Move at high speed, large separation, compressible

★ Four exam-testable comparison facts:

Densities of solids and liquids are approximately the same (small numeric gap, e.g. ~0.03–0.05 on the g/cm³ scale).
Density of a gas is much lower than either.
Mean separation of molecules in solids and liquids is approximately the same (follows directly from fact 1).
Mean separation in a gas is very much larger.

Drawing tip: a liquid particle diagram should look almost as tightly packed as a solid's — only slightly more spread out to allow flow. Only the gas diagram should show particles far apart.

Structured — density define, water/vapour ratio, evidence for solid/liquid modelFull structured Q

(a) Define density → "Ratio of mass to volume of an object." [1]

Method — make the wanted quantity the SUBJECT first: V_vapour = mass/ρ_vapour and V_water = mass/ρ_water, THEN divide and cancel the equal masses — more reliable than combining the density ratio in your head first.

(i) V_v/V_w = (m/ρ_v)÷(m/ρ_w) = ρ_w/ρ_v = 1÷(1/1600) ⟹ 1600 (ii) Mean separation ratio: (l_v/l_w)³ = V_v/V_w (O-level "similar figures": volume ratio = length ratio cubed) ⟹ l_v/l_w = ∛1600 ⟹ ≈11.7

(d) Evidence for molecules in solids/liquids having approx. the same separation: "They have approximately the same densities." Evidence for strong & rigid forces in solids: strong → "Fixed volume of solids." rigid → "Fixed shape of solids."

Self-check (extra proof panel on the board): verify with ice vs water: ρ_ice=0.917, ρ_water=1 ⟹ V_ice/V_water=1/0.917=1.09 ⟹ l_ice/l_water=∛1.09≈1.03≈1 — confirms the general claim.

Sphere falling at terminal velocity through a liquidForces on a sphere

A small sphere falls with CONSTANT VELOCITY through a liquid. (i) Name the three forces. (ii) Give a word equation relating them.

Forces: Drag force (a.k.a. viscous force, in a liquid), Weight, Upthrust.

Word equation: W = F_D + U (terminal velocity ⟹ equilibrium ⟹ downward force = sum of upward forces)

Note: the follow-up "explain the energy changes" is covered properly in Work, Energy & Power — GPE converts to heat via work done against drag, while KE stays constant (speed is constant at terminal velocity).

5. Worked MCQ Practice

Density of a mixtureMCQ · two liquids mixed

A mass of liquid density ρ is thoroughly mixed with an equal mass of another liquid density 2ρ. No change in total volume occurs. What is the density of the mixture?

ρ=m/V₁ ⟹ V₁=m/ρ. 2ρ=m/V₂ ⟹ V₂=V₁/2 ρ_f = 2m/(V₁+V₁/2) = 2m/[(3V₁)/2] = 4m/(3V₁) = 4/3 ρ

Answer: A (4/3 ρ)

Maximum straw lengthMCQ · ΔP=ρgΔh

A child drinks liquid density ρ through a vertical straw. Atmospheric pressure p₀; the child can lower the pressure at the top of the straw by 10%. Maximum straw length?

ΔP = ρgΔh: (100p₀/100)−(90p₀/100) = ρgΔh Δh = (10p₀/100)/(ρg) ⟹ p₀/(10ρg)

Answer: A

Flask heated, U-tube liquid risesMCQ · pressure increase

Flask (air, atmospheric pressure) connected to a U-tube of liquid density ρ. Heated gently; liquid on the right rises by h. What is the increase in pressure of the heated air?

Trick: if one side rises by h, the other side FALLS by h (liquid is conserved) — so the total height difference between the two columns changes by 2h, not h.

P = ρg(2h) ⟹ P=2ρgh

Answer: D

Tapered vessels P and QMCQ · shape is a distractor

Vessels P (tapers outward) and Q (tapers inward), sides at 45°, same liquid depth h. Liquid in P is twice as dense as in Q. Ratio (pressure on base of P)/(pressure on base of Q)?

Common mistake: P=ρgh depends ONLY on density and depth — NOT on the container's shape, taper, or cross-sectional area. The 45° taper is a pure distractor.

P_P/P_Q = 2ρgh/ρgh = 2/1

Answer: A

Gas trapped by water in an angled pipeMCQ · vertical height only

Pipe closed at one end, gas trapped by a water column. Vertical height difference to the open end = 5.0m (slanted section = 7.0m). Atmospheric pressure 1.0×10⁵ Pa, ρ_water=1000 kg/m³, g=10 m/s². Gas pressure?

Note: the 7.0m slanted length is a distractor — pressure depends on VERTICAL depth only, not distance along a slanted path.

P = (1×10⁵) + (1000×10×5) ⟹ 1.5×10⁵ Pa

Answer: C

Two bulbs X, Y with mercury threadsCalculation · possible h₁, h₂

Bulbs X (16000 Pa) and Y (8000 Pa) connected via a tube with two mercury threads h₁, h₂. Density of mercury 13600 kg/m³. Which pair of h₁, h₂ is possible?

From X: P=16000−ρgh₁. From Y: P=8000−ρgh₂ Equate: 8000 = ρg(h₁−h₂) ⟹ h₁−h₂ = 8000/(13600×9.81) = 0.06m = 6cm

Answer: D (h₁=18.0, h₂=12.0 — difference exactly 6cm)

W-shaped tube, trapped air (x, y, P)Calculation · possible values

W-tube, mercury open to atmosphere both ends, air at pressure P trapped between. Distances x, y shown; 1 atm = 760mm Hg; a fixed 50mm offset given. Which values of x, y, P are possible?

Left: P=760+x. Right: P=760+(50−y)=810−y Equate: 760+x=810−y ⟹ x+y=50 Check each option against x+y=50 AND the resulting P from both sides matching

Answer: B (x=20, y=30 — sum=50, both sides give P=780)

✓ Quick Recap — Full Chapter Cheat-Sheet

Density is a RATIO — read directly off a graph (m₁/V₁), never from a gradient (unlike velocity/acceleration, which ARE rates).
ρ = m/V (kg m⁻³) | P = F/A (Pa) | P = ρgh (h measured from the TOP/free surface)
★ ΔP = ρgΔh — always prefer this for any question giving TWO pressure/height readings: it automatically cancels atmospheric pressure and fixes the "wrong reference height" trap in one move.
1 atm = 1.0125×10⁵ Pa = 76 cm Hg = 760 mm Hg. Gases: equal pressure everywhere. Liquids: equal pressure at equal heights.
Strong intermolecular forces → fixed volume. Rigid intermolecular forces → fixed shape.
Solids/liquids: similar density AND similar mean molecular separation. Gases: much lower density, much larger separation.
Terminal velocity in a fluid: W = F_D + U (drag/viscous force + upthrust balance weight).
Container shape/taper never affects P=ρgh — only density and vertical depth matter.

Two immiscible liquids — density of liquid M from a p-x graphStructured Q — the ΔP=ρgΔh trap

Container: liquid L, then liquid M added on top (immiscible), total depth 0.17m. Graph gives pressure p vs height x above the base — two segments, "liquid M" (upper) and "liquid L" (lower).

(i) Atmospheric pressure = value at the very top (x=0.17, lowest pressure on the graph) = 9.10×10⁴ Pa

★ CRITICAL Exam Trap: do NOT pick one point on liquid M's segment and plug into plain P=ρgh — that height would be measured from the wrong reference (the base, not the top of M) AND would still include atmospheric pressure. Instead use ΔP=ρgΔh between the TWO ENDPOINTS of liquid M's own segment — this cancels atmospheric pressure and fixes the reference-height automatically.

Using liquid M's segment endpoints (x=0.10 to x=0.17): ΔP=(9.15−9.10)×10⁴; Δh=0.17−0.10 ρ = ΔP/(gΔh) = (9.15−9.10)×10⁴ / [9.81×(0.17−0.10)] ⟹ ρ≈728 kg m⁻³

General rule: whenever you're taking a pressure CHANGE (two readings, two heights), reach for ΔP=ρgΔh rather than the bare P=ρgh formula — it's simpler and self-corrects two common mistakes at once.