In previous lesson, we have discussed liquid pressure, where we discussed the formula of liquid pressure, characteristics of liquid pressure and some applications of liquid pressure.

In gas pressure, we are going to explain how gas pressure is produced by using the kinetic theory of gas. After that, we will shift to atmospheric pressure where we will discuss the characteristics of atmospheric pressure, proofs or phenomena involving atmospheric pressure, how to measure atmospheric pressure and some applications of atmospheric pressure.

Let’s begin with kinetic theory of gas. We use kinetic theory of gas to explain how gas pressure is produced.

There are many assumptions in kinetic theory of gas. However, in SPM, we only focus on 2 assumptions, namely:

- The gas consists of very small particles, all with non-zero mass.
- These particles are at constant and random movements; hence constantly collide with the walls of the container.

Now, let’s use these 2 assumptions to explain how gas pressure is produced in a container

Gases consists of very small particles which are at constant and random movements. The particles constantly collide with the walls of the container. When the particles collide with the wall of the container and bounce back, they experience a change in momentum. The momentum change exerts a force on the wall and hence produces a pressure on the wall of the container. This is how the gas pressure is produced.

In SPM, you need to memorise this explanation because in exam you may be asked to explain how the gas pressure is produced by using the kinetic theory of gas. We have just discussed kinetic theory of gas. Let’s continue with atmospheric pressure.

Atmospheric pressure is the pressure exerted on the surface of any objects by the atmospheric gas. It is caused by the weight of the atmospheric gases above the earth surface. In SPM, you need to know some experiments that prove the existence of atmospheric pressure, including the crushing can experiment, the Magdeburg hemisphere and experiment where water covered with a cardboard in an upside down glass does not flow out.

We have just discussed the proofs of existence of atmospheric pressure. In this slide, we are going to discuss how to measure atmospheric pressure. Atmospheric pressure can be measure by Aneroid Barometer, Fortin Barometer and mercury barometer. In SPM, most of the questions related to measuring atmospheric pressure are questions regarding the mercury barometer. Therefore, the focus of our discussion is on the mercury barometer.

The diagram shows the illustration of a mercury barometer. A mercury barometer is also called a simple barometer. The green colour substance represents the mercury and there is vacuum at the upper end of the tube.

From my teaching experience, I know many students facing problem in finding atmospheric pressure by using a mercury barometer. Finding atmospheric pressure by using mercury barometer is actually very simple. If you follow this explanation step by step, you will find that it is just piece of cake.

Now, in order to learn how to use a mercury barometer to find atmospheric pressure, firstly, we must know how to find the pressure “in the mercury” in unit of cmHg if the column above the mercury is vacuum.

Second, we need to know how to find the pressure “in the mercury” in unit of cmHg if the column above the mercury is not vacuum.

Third, we need to know how to convert the unit of pressure from cmHg to Pascal.

Forth, we need to know some important notes regarding the gas pressure and liquid pressure. Last but not least, we learn how to find the atmospheric pressure by using the mercury barometer.

Next, we will learn how to find the pressure in the mercury in the unit of cmHg.

First of all, we need to know that the pressure of the vacuum is 0. At point A, the surface of the mercury is exposed to the vacuum, hence its pressure is equal to the pressure of the vacuum, which is 0.

The pressure of any point in the mercury is equal to the depth of the point from the vacuum. For example, point B is 24cm below the surface. Therefore, the pressure at B is 24 cmHg. How about point C? Point C is 75cm below the surface. The pressure is 75 cmHg. Lastly, point D. Point D is 83cm below the surface, therefore the pressure is 83cmHg.

If some gas is trapped in the space inside the tube, the pressure inside the tube is not 0 anymore. This will affect the pressure inside the mercury. The pressure inside the mercury will be the gas pressure + the mercury pressure.

For example, if the pressure of the gas is 20 cmHg, then the pressure at point B will become 20 + 24 = 44 cmHg. How about point C? Pressure at point C will become 20 + 75 = 95 cmHg. Lastly, point D. Pressure at point D is 103 cmHg.

So far, we have learned that the pressure (in the unit of cmHg) in the mercury of a mercury barometer is equal to the depth of the mercury from the vacuum. Therefore the pressure at B, C and D are 24, 75 and 83 cmHg respectively. However, sometime, we are asked to find the pressure in the unit of Pascal (Pa), but not cmHg. How do we convert the unit in cmHg to Pa?

We can convert the unit in cmHg to Pa by using the equation of pressure, P = hrg. For example, at point B, the depth, h = 24cm or 0.24m. The density of mercury, r = 13,600 kgm^{-3} (this will be given in the question) and the gravitational field strength, g = 10 Nkg^{-1}. Substitute these values into the equation. Pressure at B equals to 32,640 Pa. At C, the depth is 0.75m, density and gravitational field strength is still the same. Substitute these values into the equation; the pressure at C is 102,000 Pa. The depth at D is 0.83m. Substitute this into the equation. Pressure at D is 112,880 Pa. This is how we convert the unit of pressure from cmHg to Pascal.

Sometime, the density may be given in g/cm^{3}. You need to convert it to kg/m^{3} before finding the value of the pressure.

In SPM, usually, the density of an object is given in g/cm^{3} or kg/m^{3}. Sometime, you will need to convert these units from one to another. You should have learned how to convert the unit of density in Chapter 1, introduction to physics.

However, we can convert these 2 units of density faster and easier by using a factor of conversion.

In chapter 1, you should have learned that 1g/cm^{3} is equal to 1000kg/m^{3}. Therefore, we can convert g/cm^{3} to kg/m^{3} by multiplying a factor of 1000. For example, 1 g/cm^{3 }= 1000kg/m^{3}; 2 g/cm^{3 }= 2000kg/m^{3}; 3 g/cm^{3 }= 3000kg/m^{3}; 2.5 g/cm^{3 }= 2500kg/m^{3}…….

In opposite, we can convert kg/m^{3} to g/cm^{3} by dividing 1000. For example, 3000kg/m^{3} = 3g/cm^{3}; 1500kg/m^{3} = 1.5g/cm^{3}; 800kg/m^{3} = 0.8g/cm^{3}; 200kg/m^{3} = 0.2g/cm^{3}……

You are advised to memorise this factor of conversion because conversion of the unit of density is very common in Physics and Chemistry.

Before we learned how to find the atmospheric pressure by using a mercury barometer, there are a few things that we need to know.

- The difference of gas pressure at different height can be ignored if the difference of the height is not too big. For example, we can assume that the gas pressure at point D is equal to the gas pressure at point E even though point D is higher than point E. This is also true for the gas inside the container. The pressure at point A will be equal to the pressure at point F.
- The pressure on the surface of a liquid is equal to the pressure of the gas it exposed. For example, in the diagram, the pressure of the gas inside the container is 200,000Pa. Surface A is exposed to the gas inside the container. Therefore, the pressure at point A is also 200,000 Pa. The atmospheric pressure is 100,000Pa. Surface B exposes to the atmosphere, hence the pressure at point B is equal to the atmospheric pressure, which is 100,000 Pa.
- Pressure of a given liquid is the same at the same level. For example, A and C are at the same level. Therefore pressure at C is equal to pressure at A, which is 200,000Pa

These are the 3 important points that you need to know when using a mercury barometer to find atmospheric pressure. So, how do we determine the atmospheric pressure from a mercury barometer?

Well, we have just learned that the pressure on the surface of a liquid is equal to the gas pressure of the gas it exposed. Point A exposes to the atmosphere, hence the pressure at A is equal to the atmospheric presssure. Point B is at the same level as A, therefore, pressure at B is equal to pressure at A, which is also equal to the atmospheric pressure. We write, P_{A} = atmospheric pressure. We demote atmospheric pressure by using the symbol P_{atm}.

Now, let’s look at point B. Point B are at the same level as point A. Therefore, their pressure should be the same. Write this as P_{A} = P_{B} = the atmospheric pressure. Now, our problem is, how much is the pressure at point B? Point B is 76cm below the vacuum. Therefore, the pressure at point B is 76 cmHg.

Since the pressure at point B is also equal to the atmospheric pressure, hence we can deduce that the atmospheric pressure is equal to 76 cmHg. Yes! This is how we find the atmospheric pressure by using a mercury barometer. The atmospheric pressure is equal to the height of the mercury column in the tube.

We can convert the unit from cmHg to Pascal by using the formula P=hρg. In this case, h = 0.76m. The unit of h must be in meter. The density of mercury, ρ = 13,600 kg/m^{3}. This will be given in the question. g is 10 N/kg. Plug in all these numbers into the equation. The atmospheric pressure is 103360 Pa. We have just finished our discussion on mercury barometer. Now we are going to continue with Aneroid barometer and Fortin barometer.