(b) Density and pressure

Learning objectives

5.3Density equationKnow and use density = mass ÷ volume, ρ = m/V.5.4Density practicalPractical: investigate density using direct measurements of mass and volume 5.5Pressure equationKnow and use pressure = force ÷ area, p = F/A.5.6Pressure in fluidsUnderstand how the pressure at a point in a gas or liquid at rest acts equally in all directions 5.7Pressure difference equationKnow and use pressure difference = height × density × gravitational field strength, p = hρg.

Density measures how much mass is packed into each unit of volume. A material with a greater density has more mass in the same volume.

\rho=\frac{m}{V}

$\rho$ is density, $m$ is mass and $V$ is volume. Useful rearrangements are $m=\rho V$ and $V=m/\rho$ . With mass in kilograms and volume in cubic metres, density is in kg/m³; with grams and cubic centimetres, it is in g/cm³.

Example: a solid has mass 540 g and volume 200 cm³. $\rho=540/200=2.70$ g/cm³. The mass and volume units already match, so no conversion is needed; the density unit follows from g divided by cm³.

Keep one consistent unit system. Dividing grams by metres cubed, or kilograms by centimetres cubed, gives a different numerical scale and an unsuitable mixed unit. Density identifies a material only when the sample is uniform and the measured value is sufficiently accurate.

To investigate density, measure the sample's mass and volume directly, then calculate $\rho=m/V$ . Choose the volume method to match the sample.

Sample	Mass measurement	Volume measurement
regular solid	zero a balance, then weigh the dry solid	measure the required dimensions and use the correct geometric volume formula
irregular solid	weigh it dry before putting it in water	record initial water volume $V_1$ , fully submerge the solid, record $V_2$ ; solid volume is $V_2-V_1$
liquid	tare an empty container, or subtract its mass from the filled mass	use a measuring cylinder on a level surface and read the bottom of the meniscus at eye level

For displacement, lower the object gently, remove trapped air and avoid losing water by splashing. Use a cylinder with a scale suited to the volume. Repeat measurements where practical, calculate density for each repeat and check any anomalous result before finding a mean.

Measure mass before immersing a solid, because water left on it increases the measured mass. The displaced volume equals the object's volume only when the object is completely submerged and no water is lost or extra air is trapped.

Pressure is the normal force acting on each unit of area. The same force produces greater pressure when it is concentrated on a smaller contact area.

p=\frac{F}{A}

$p$ is pressure in pascals (Pa), $F$ is the force normal to the surface in newtons (N), and $A$ is the contact area in square metres (m²). Rearrangements are $F=pA$ and $A=F/p$ . One pascal is one newton per square metre.

Example: a 600 N force acts normally over 0.030 m². $p=600/0.030=20\,000$ Pa. If the same force acted over twice the area, the pressure would halve.

Use force, not mass, and use the actual contact area. If mass is given, first calculate weight using $F=mg$ . Convert cm² to m² before substituting when the answer is required in pascals.

At one point in a gas or liquid at rest, pressure acts equally in all directions. A tiny surface placed at that point experiences a force perpendicular to the surface, whichever way the surface faces.

If pressure at the same point were greater in one direction, a small part of the fluid would have an unbalanced force and would start to move. The condition that the fluid is at rest therefore requires equal directional pressure at that point.

A small stationary air bubble under water is pushed inward from all sides. Likewise, identical holes at the same depth on opposite sides of a container, both opening to the same outside pressure, have the same pressure difference; water begins to leave with the same speed in opposite directions.

‘Equal in all directions’ refers to the same point. It does not mean pressure is equal everywhere in a fluid: in a liquid at rest, pressure changes with vertical depth. Pressure force is perpendicular to a surface, not along it.

In a fluid at rest, pressure increases with vertical depth because deeper points support the weight of more fluid above them. The pressure difference between two levels depends on their vertical separation, the fluid density and gravitational field strength.

p=h\rho g

$p$ is the pressure difference in pascals (Pa), $h$ is the vertical height difference in metres (m), $\rho$ is density in kg/m³, and $g$ is gravitational field strength in N/kg. A useful rearrangement is $h=p/(\rho g)$ .

Example: for water with $\rho=1000$ kg/m³ and $g=10$ N/kg, the pressure difference over 0.60 m is $p=0.60\times1000\times10=6000$ Pa. The deeper point has the greater pressure.

This relationship gives a pressure difference, not automatically the total pressure. Add the pressure at the upper surface when total pressure is required. Use vertical depth—not the sloping path—and convert centimetres and kilopascals before substituting in SI units.

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(b) Density and pressure

Learning objectives

Calculate density, mass and volume

Measure density directly

Calculate pressure from force and area

Pressure at a point acts in every direction

Calculate pressure difference with depth

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