Gas Pressure and Pressure in Fluids
Scope note. On Combined Science you only need the first section, on gas pressure and the particle model. Boyle's law, pressure in fluids, atmospheric pressure, upthrust and hydraulics are Separate Physics only.
Gas Pressure and the Particle Model
Gases consist of particles that are in constant random motion. As these particles collide with the walls of their container, they exert a force on the surface. The total force per unit area is the gas pressure. The more frequent or forceful the collisions, the higher the pressure.
Increasing the temperature of a gas gives the particles more kinetic energy, so they move faster and collide with the walls more frequently and with greater force. If the volume of the container is fixed, this increases the pressure. This explains why aerosol cans carry warnings not to heat them — the pressure inside can build up enough to cause an explosion.
For a fixed mass of gas at constant temperature, the pressure and volume are inversely proportional: \(p \times V = \text{constant}\). This is known as Boyle’s Law.
If you compress a gas (reduce its volume), the particles have less space and hit the walls more often, so the pressure increases. If you expand the gas, the particles have more space, collide less often, and the pressure decreases. The relationship is: \(p_1 V_1 = p_2 V_2\), where subscripts 1 and 2 refer to the initial and final states.
A gas occupies 0.5 m³ at a pressure of 100 kPa. The gas is compressed to 0.2 m³ at constant temperature. What is the new pressure?
Using \(p_1 V_1 = p_2 V_2\):
$$p_2 = \frac{p_1 V_1}{V_2} = \frac{100 \times 0.5}{0.2} = 250 \text{ kPa}$$
Pressure in Fluids and Atmospheric Pressure
A fluid is any substance that can flow — both liquids and gases are fluids. Pressure acts in all directions within a fluid. The pressure at any point in a fluid depends on the depth, the density of the fluid and gravitational field strength.
The pressure due to a column of liquid is calculated using:
$$p = h \rho g$$
where \(p\) is pressure in pascals (Pa), \(h\) is the height of the liquid column in metres, \(\rho\) is the density of the liquid in kg/m³, and \(g\) is gravitational field strength (9.8 N/kg on Earth).
Pressure in a liquid increases with depth. A diver at 10 m experiences roughly double the atmospheric pressure — one atmosphere from the air above, plus one atmosphere-equivalent from the water.
Atmospheric pressure is caused by the weight of air molecules above a surface. At sea level, atmospheric pressure is approximately 101,325 Pa (about 101 kPa). As altitude increases, there are fewer air molecules above, so atmospheric pressure decreases. This is why air pressure is lower on top of a mountain than at sea level.
Because pressure increases with depth, the bottom of an object submerged in a fluid experiences a greater pressure than the top. This difference in pressure creates an upward force called upthrust (or buoyancy). If the upthrust equals the weight of the object, it floats. If the weight is greater than the upthrust, the object sinks.
An object floats when it displaces a volume of fluid whose weight equals its own weight. This is why a steel ship floats — its hollow shape displaces a large volume of water, creating enough upthrust to support its weight, even though steel is denser than water.
Pressure Differences and Everyday Examples
Pressure differences cause fluids to flow. When you drink through a straw, you reduce the air pressure inside the straw by expanding your lungs. The higher atmospheric pressure on the surface of the drink pushes the liquid up the straw. Similarly, suction cups work because removing air from inside the cup creates a lower pressure, and atmospheric pressure on the outside holds the cup in place.
Hydraulic systems use the fact that pressure is transmitted equally through an incompressible fluid. A small force applied to a small piston creates a pressure that acts equally on a larger piston, producing a larger force. The relationship is:
$$\frac{F_1}{A_1} = \frac{F_2}{A_2}$$
This principle is used in hydraulic brakes, car jacks and industrial presses.
Calculate the pressure at the bottom of a swimming pool that is 2.5 m deep. The density of water is 1000 kg/m³ and \(g = 9.8\) N/kg.
Using \(p = h \rho g\):
$$p = 2.5 \times 1000 \times 9.8 = 24{,}500 \text{ Pa} = 24.5 \text{ kPa}$$
This is the pressure from the water alone. The total pressure at the bottom would also include atmospheric pressure (approximately 101 kPa), giving a total of about 125.5 kPa.