Density and Pressure
Example 1
Example 2
Example 3
Example 4
Example 5
Example 6
Example 7


Density and Pressure in Fluids


All fluids

— can flow. Some like pitch flow slowly.
— assume the dimensions of their containers.
— cannot withstand a shearing stress.
—exert a force in the direction perpendicular to its surface.
— do not have orderly structure of constinuent atoms.
— are liquids and gases.


— density ρ = [Δm/Δv] .
— we assume that a fluid sample is large relative to atomic dimensions and thus with uniform density) so density becomes ρ = [m/v] where m and V are the mass and volume of the sample.
— is scalar.
— unit is kgm-3
— density of gases varies considerably with pressure, but the density of liquids and solids do not because gases are readily compressible but liquids and solids are not.


— pressure of uniform force on flat area.
— p = F/A where F is the magnitude of the normal force on area A.
— Pressure is a scalar.
— Unit of pressure is the Newton per square meter Nm-2, called the pascal (Pa).
— Normal atmospherica pressure = 1.01 X 105 Pa.

Fluids at Rest

— pressure increase with depth. Divers know that pressure increases with depth in water.
— Pressure decreases with altitude in the atmosphere.
— In fluids at rest pressure in fluids is called hydrostatic pressure.

Fig 1

Absolute and Gauge Pressure

Consider a cylinder of water bounded on the top by y1 and the bottom by y2 in Fig 1.

— The forces on the cylinder of water are:

— F1 on the top surface.
— F2 on the bottom surface.
— mg is the weight of the cylinder (caused by gravity).

— The pressures on the cylinder of water are:

— P1 on the top surface.
— P2 on the bottom surface.

— The dimensions of the cylinder:

— y1 - y2 is the height
— A is the area of the top or bottom surface.

Note: The water is at rest (no translation or rotation) so it is in static equilibrum.
ΣF = 0
F2 + (-F1) + (-mg) = 0
F2 = F1 + mg --- Eq1

By definition F2 = P1 A and F2 = P2 A
So Eq 1 becomes P2A = P1A + mg --- Eq2
Mass of water in cylinder = ρV where V is the volume of the cylinder
Mass of water in cylinder = ρ [A (y1 - y2)]
So Eq2 becomes P2A = P1A + ρg [A (y1 - y2)]
P2 = P1 + ρg (y1 - y2)
This becomes the famous equation P= ρgh where h is the depth of the liquid if y1 becomes 0 . This is called gauge pressure as it is the pressure difference between the suface and the point as measured by a gauge.
If the pressure at the surface is p0 then the absolute pressure is p0 + ρgh
Gauge Pressure = ρgh
Absolute Pressure = p0 + ρgh

Divers experience p0 + ρgh
Implication: Pressure at a point in a fluid in static equilibrium depends on depth at that point and is the same on a particular horizontal level.


The Mercury Barometer

— long glass tube is filled with mercury and inverted with its open end in a dish of mercury.
— space above the mercury column contains only mercury vapor.
— pressure of mercury vapor is so small at ordinary temperatures that it can be neglected.

The weight of the mercury column is supported by atmospheric pressure so:

Po = ρgh

Note: The vertical height h of the mercury column

— does not depend on the cross-sectional area of the vertical tube.
— depends on the value of g at the location of the barometer
— on the density of mercury, which varies with temperature
— corrections are made for g and temperature.


The Open-Tube Manometer

— a U-tube containing a liquid, with one end of the tube connected to the gas supply and the other end open to the atmosphere.
— measures gauge pressure of a gas.
— Pg = ρgh
— gauge pressure can be positive or negative e.g. positive in a tyre andnegative in a vacuum cleaner input. A good example is pressure in your lungs which is positive when exhaling and negative when inhaling.


Pascal’s principle

If an incompressible fluid experiences a change in the pressure, then that pressure is transmitted undiminished to every portion of the fluid and to the walls of its container.








Concept by Kishore Lal. Programmed by Kishore Lal... Copyright © 2015 Kishore Lal. All rights reserved.