2.1 Pressure in Liquids
Liquid pressure is the pressure exerted by a liquid on an object immersed in it or on the walls of its container, arising from the weight of the liquid column above. In everyday life, liquid pressure explains why the base of a dam must be constructed thicker than its top, why household water tanks are elevated, and how medical intravenous drips function.
Learning Objectives
- Derive the liquid pressure formula \(P = h\rho g\) from \(P = \frac{F}{A}\) and \(\rho = \frac{m}{V}\).
- Identify the factors affecting pressure in liquids: depth, density and gravitational acceleration.
- State that liquid pressure does not depend on surface area, volume or container shape.
- Explain that pressure acts in all directions and points at the same level in a static liquid have the same pressure.
- Calculate liquid pressure and actual pressure using \(\text{Actual pressure} = h\rho g + P_{\text{atm}}\).
- Compare the densities of two immiscible liquids using a U-tube based on \(h_1\rho_1 = h_2\rho_2\).
- Explain daily applications of liquid pressure, including water tanks, IV drips, dam construction and the siphon system.
Factors Affecting Liquid Pressure and U-Tube Equilibrium
See how the liquid pressure formula is derived, how depth and density affect manometer readings, and how two immiscible liquids balance in a U-tube.
Liquid Column and Pressure Formula
A liquid column of height \(h\), base area \(A\) and density \(\rho\) has volume \(V = Ah\), mass \(m = Ah\rho\) and weight \(W = Ah\rho g\). Therefore, pressure at the base is:
Liquid pressure
\(P = h\rho g\)\(P\) = liquid pressure, \(h\) = depth, \(\rho\) = density, \(g\) = gravitational acceleration.
U-Tube with Immiscible Liquids
At the common horizontal level containing points A and B, pressure is equal:
U-tube equilibrium
\(h_1\rho_1 = h_2\rho_2\)Measure \(h_1\) and \(h_2\) to compare the densities of two immiscible liquids.
Short Explanation
What is Liquid Pressure?
Liquid pressure is the pressure exerted by a liquid on an object immersed in it or on the walls of its container, caused by the weight of the liquid column above.
Derivation of the Formula
For a liquid column of height \(h\), base area \(A\) and density \(\rho\): \(V = Ah\), \(m = Ah\rho\), \(W = Ah\rho g\). Therefore, \(P = \frac{W}{A} = h\rho g\).
Factors Affecting Liquid Pressure
Liquid pressure depends directly on depth \(h\), density \(\rho\), and gravitational acceleration \(g\).
Actual Pressure
When the liquid surface is open to the atmosphere, the total pressure experienced by an immersed object is \(\text{Actual pressure} = h\rho g + P_{\text{atm}}\).
| Factor | Change in Factor | Effect on Liquid Pressure |
|---|---|---|
| Depth, \(h\) | Greater depth | Higher liquid pressure |
| Density, \(\rho\) | Higher density | Higher liquid pressure |
| Gravitational acceleration, \(g\) | Greater \(g\) | Higher liquid pressure |
| Container shape / surface area / volume | Any change | No effect on liquid pressure at a given depth |
Try to Answer First
Answer in your mind, then press “Check Answer”.
A fish is swimming at a depth of \(24\ \mathrm{m}\) in a freshwater lake. If the density of water is \(1\,000\ \mathrm{kg\,m^{-3}}\) and gravitational acceleration is \(9.81\ \mathrm{m\,s^{-2}}\), what is the water pressure experienced by the fish?
Check Answer
A diver dives to a depth of \(35\ \mathrm{m}\) in sea water of density \(1\,060\ \mathrm{kg\,m^{-3}}\). Given \(g = 9.81\ \mathrm{m\,s^{-2}}\) and atmospheric pressure is \(100\ \mathrm{kPa}\), calculate the actual pressure acting on the diver's body.
Check Answer
In a U-tube containing water of density \(1\,000\ \mathrm{kg\,m^{-3}}\) and olive oil, the height of the water column above the boundary level is \(15.0\ \mathrm{cm}\), while the height of the olive oil column is \(16.5\ \mathrm{cm}\). Find the density of the olive oil.
Check Answer
Common Mistakes
- Forgetting to add atmospheric pressure \(P_{\text{atm}}\) when a question asks for the actual pressure rather than just the liquid pressure.
- Using liquid depth in centimetres instead of converting to standard S.I. units in metres when using \(P = h\rho g\).
- Mixing up units when adding \(h\rho g\) in pascals to \(P_{\text{atm}}\) given in kilopascals without converting both to the same unit first.
- Assuming container width, base surface area, or total liquid volume affects liquid pressure at a specific depth.
Concept Misunderstandings
A wider container filled with water exerts higher water pressure at the bottom than a narrow tube of water at the exact same depth.
Liquid pressure depends strictly on depth \(h\), density \(\rho\), and \(g\). Base area, volume, and container shape have no effect on liquid pressure.
Pressure in a liquid acts only downwards onto the seabed or container base.
Liquid pressure acts in all directions equally at any given depth, which is why water shoots equally from holes on all sides at the same depth.
In a siphon, liquid moves because gravity pulls on the upper intake tube.
Liquid flows out at the lower exit because of the height difference, which creates a lower pressure region at the tube bend; external atmospheric pressure pushes water into the higher intake.
Summary
- Liquid pressure is governed by the formula \(P = h\rho g\).
- The S.I. unit of pressure is pascal, \(\mathrm{Pa}\), equivalent to \(1\ \mathrm{N\,m^{-2}}\).
- Liquid pressure increases with depth \(h\), liquid density \(\rho\), and gravitational acceleration \(g\).
- Liquid pressure is independent of container shape, surface area, and total liquid volume.
- In static liquids, points at the same horizontal level experience the same pressure in all directions.
- Total pressure experienced by an immersed object is \(\text{Actual pressure} = h\rho g + P_{\text{atm}}\).
- For two immiscible liquids in a U-tube, equilibrium at a common level satisfies \(h_1\rho_1 = h_2\rho_2\).
- Practical applications include dam base thickening, elevated water storage, IV drips, and siphons.
Short Activity
Objective: Identify the factors affecting liquid pressure, actual pressure, and applications of liquid pressure.
A. Objective Quiz
1 Which of the following factors affects the pressure at a specific point in a static liquid?
2 Why is the wall of a hydroelectric dam constructed significantly thicker at the base than at the top?
B. Fill in the Blanks
3 The formula to calculate the actual pressure experienced by an object submerged in a liquid open to the atmosphere is Actual pressure = hρg + .
4 The S.I. unit for pressure is the , which is equivalent to 1 N m⁻².
C. Matching / Drag and Drop
Match each liquid pressure phenomenon or application to its correct scientific principle or function. If using a phone, tap the answer first, then tap the matching box.
