TOPIC 1B: ENERGY and PRESSURE

An aerosol can contains a liquid. Every time you press the button, a fine spray of liquid jets out. But what makes it come out of the bottle?

The liquid inside is under pressure. That means it is **pushing outwards in all directions**. This is why it is dangerous to puncture an aerosol bottle as the pressure will force the liquid out of the hole at a high speed.

**Figure 1. An aerosol spray
**

Fizzy drinks contain a gas that produces a pressure. The pressure pushes outwards on all sides of the container. An easy test to identify a bottle of sparkling fizzy water compared to a flat, still water bottle is to squeeze the sides. The sparkling water bottle will be hard to compress, because of the pressure pushing outwards, as shown in figure 2.

**Figure 2. Pressure in a fizzy drink bottle
**

In figure 2, the fizzy water produces a pressure in all directions. This pushes on the sides, and a 'push' is a force. As can be seen in the image, there is a force of 4 N acting on two small 1cm^{2} sections of the sides of the bottle. In total, there is a force of 8 N on these two sections, and a much large force on the whole bottle's surface. However, we say that the pressure is producing** 4 newtons per centimetre squared.** (4 N/cm^{2}). Pressure as a quantity is simply the **force acting on a unit area**, either a cm^{2} for small surfaces or a m^{2} for larger areas.

Any fluid can move freely, so the pressure in a fluid is the same at all points and is distributed evenly. if you push the sides of a bottle of water, you are increasing the pressure under your fingers, but this increase spreads throughout the liquid and is the same at all points.

To calculate pressure we use the following formula:

(N/cmpressure ^{2} or N/m^{2}) |
= |
force (N) |

area (cm^{2} or m^{2}) |

p = |
F |

A |

**Notes:**

- If we use m
^{2}for the surface area, then the pressure will be in units of N/m^{2}. This is the international unit for pressure, and is also called a**pascal (Pa)**. So 4 pascals pressure is the same as 4 N/m^{2}.

**Example:**

A steel block rests on a desk. The weight of the block pushing down on the desk is 40.0 N. The area touching the desk is 95.0 cm^{3}. Calculate the pressure acting on the desk.

**Answer:**

We know that:

p = |
F |
and therefore:- | P = | 40.0 |

A |
96.0 |

This gives a pressure of 0.417 N/cm^{2} to 3 sig. figs.

Try the practice questions below, to check you understand how to use this formula:

**Questions:**

1. A syringe is used to squeeze a liquid down a tube. A force of 5N is applied to an area of 0.8 cm^{2}. Calculate the pressure on the liquid.

We know that

p = | F |

A |

p = | 5 |

0.8 |

2. A classroom stool has a weight of 160 N. Each of the 4 stool legs produces a pressure of 8.0 N/cm^{2} on the floor.

Calculate the area of each stool leg that is in contact with the floor.

Each stool takes ¼ of the weight. This means the force on each leg is ¼ of 160 N, or 40 N each.

We know that:

p = | F |

A |

A = |
F | and therefore | A = |
40 |

p | 8 |

**A**** = ****5.0 cm ^{2}**
(The pressure was given in N/

When putting in a fence in a field, a farmer must drive thick wooden stakes into the ground. What shape should the end of the stake be to make it easy to push into the soil?

The answer is that the stake should have a sharp point on the end, and this seems like common sense. However, the reason the sharp point is used is that the surface area of the end is very small. When the stake is hammered downwards, a **force is applied** on a **very small surface area**, and this makes
a **large pressure**. The large pressure applied makes it much easier to drive the stake into the ground.

**Figure 3: A fence post driven into hard soil **

A tractor, on the other hand, has very large wheels with a large surface area. Even though the tractor is very heavy and the force on the ground is large, the surface area of the wheels in contact with the ground is also large so that the pressure is small. This prevents the wheels sinking into the soft, muddy ground and getting stuck. Note that you should always state the equation (shown above) to prove this relationship between area and pressure is true in exam questions.

You should be aware of some other applications where large or small surface areas make a small or large pressure:

**Large area, therefore small pressure:**

- Snow shoes
- Caterpillar tracks (e.g. on a tank or bulldozer)
- Skis
- Elephant / camel feet

**Small area, therefore large pressure: **

- Drawing pins
- Needles used for injections
- tent pegs
- Stiletto heels on shoes

A fluid is a liquid or a gas, containing particles that are free to move. When under gravity, fluids are pulled downwards. The ocean and the Earth's atmosphere are both pulled down towards the centre of the Earth, giving the fluid a weight. (see section 1.5). If you are a diver, you will know that this weight of water makes the pressure increase as you dive deeper into the ocean.

This can also be seen in the water in this bottle that is leaking. (Figure 4). The jet at the bottom of the bottle is much stronger than one at the top, making the water move faster and further away from the bottle. This is because the **pressure is larger **at the bottom of the bottle, due to the weight of water above pushing downwards.

**Figure 4: A leaking bottle**

The pressure under 10 m of water is much larger than the pressure under 10 m of air because water is a much denser fluid than air.

The pressure in a fluid depends on three factors: The depth or height of fluid, the density of the fluid, and the gravitational field strength pulling the fluid downwards. This can be seen in this formula for the pressure in a fluid:

**pressure difference= density **x** gravitational field strength **x** height**

**Δp** = **ρ ****g** **Δh**

[Pa] = [kg/m^{3}] x [N/kg] x [m]

Notes:

- It is easy to confuse the
(rho) symbols when using this formula.**'p'**and the**'ρ'** - The pressure is usually given in pascals as the examples are often on a larger scale than earlier questions.
- A change on depth or height results in a change in pressure, so
and**Δp**are used in the formula in this exam.**Δh** - If the question involves a column of water or other liquid on Earth, then the pressure difference calculated is the pressure increase compared to the pressure at the surface of the liquid.

**Example:**

A diver is 30 m under the ocean. The density of salt water is 1030 kg/m^{3}. Atmospheric pressure at the surface is 100 kPa.

Calculate:

(i) The additional pressure caused by diving under water.

(ii) The total pressure acting on the diver at this point.

**Answers:**

(i) We know that g = 9.8 N/kg. Using the formula above:

**Δp** = **ρ****g****Δh**

So p = 1030 x 9.8 x 30 = 303 000 N/m^{2 } or **303 kilopascals. **(kPa)

(ii) Using the answer in kPa from above, we can simply add the atmospheric pressure, as both the atmosphere and the water are pushing down on the diver.

p = 303 + 100 = **403 kPa **

*Note: Divers know that each 10 m of depth in water is equivalent to approximately one more 'atmosphere' of pressure. *

The Earth's atmosphere is a thick layer of a mixture of gases. This produces a pressure just like a deep pool of water, and is actually quite high. At sea level, the atmospheric pressure is about 100 000 Pa (100 kPa), although this varies with the weather. Sometimes, scientists talk about a pressure of '1 atmosphere' ( defined as 101325 Pa). We never notice this pressure as we have an equal pressure inside us pushing outwards.

When diving in a swimming pool, we are under atmospheric pressure at the surface, and under an additional pressure due to the water as we swim downwards.

Here are some typical examples of the kind of questions you will meet:

Questions:

3. A whale is diving for food in sea water of density 1050 kg/m^{3}.

- a) Calculate the water pressure on the whale at a depth of 500 m.
- b) The whale's eye has an area of 0.0015 m
^{2}. Calculate the inward force acting on the eye at this pressure.

a) We know that Δp = *ρ* g Δh.

The value of 'g' on Earth is taken as 9.8 N/kg.

Therefore:

p = 1050 x 9.8 x 500

**p = 5 145 000 Pa or 5.15 MPa (to 3 sig.figs.)**

b) We know that:

p = |
F |

A |

So F = p x A,

F = 5 150 000 x 0.0015

**F = 7 725 N **

4. The maximum pressure increase for safe recreational diving in water is given as 400 kPa. In a fresh water lake of density 1000 kg/m^{3}, how deep can a diver go so that this limit is not exceeded?

We know that Δp = *ρ* g Δh.

So 400 000 = 1000 x 9.8 x h

400 000 = 9 800 x h

So:

**h= 40.8 m **

So 400 000 = 1000 x 9.8 x h

400 000 = 9 800 x h

So:

h = |
400 000 |

9 800 |

**Now test your understanding using these quick, 10 minute questions on this topic from Grade Gorilla:**