# Physics Tutorial: Gas Pressure

In this Physics tutorial, you will learn:

• The composition of atmosphere
• What is air pressure?
• How to measure air pressure?
• What are the units used for measuring air pressure?
• Why mercury is used as a capillary liquid in barometers?
• How to measure a specific gas pressure?
• How to calculate the total pressure in water?
• How does the Atmospheric Pressure Vary with Altitude?
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9.4Gas Pressure

## Introduction

Try this simple experiment at home, if you can.

1. Take a newspaper and unfold it.
2. Place the unfolded newspaper on a table as shown in the figure.
3. Put a plastic ruler above the table but under the newspaper. Leave half of the ruler uncovered and outside the table.
4. Kick or hit the uncovered part of the ruler upwards very quickly. What happens? If you do the experiment according the instructions, i.e. if you kick the ruler very fast, it will break in two pieces. This is because air pressure above the newspaper prevents it to raise up immediately. As a result, the newspaper will block the part of the ruler under it. But if you kick the ruler slowly, you will give time to the air to enter under the newspaper and as a result, the air pressure is balanced in both sides of the newspaper. The ruler eventually will not break but it simply falls on the ground.

This experiment is an evidence of the air pressure strength, which people often underestimate. However, air pressure is an important part of our life. Therefore, in this tutorial we will discuss extensively about the pressure of air and other gases as well.

## Air as a Mixture of Gases

Atmosphere - as the layer that surrounds the Earth - is composed by a mixture of gases, which we call "air". Atmosphere is made of 78% nitrogen (N2), 21% oxygen (O2) and 1% are other gases where the most important is carbon dioxide (CO2). Air is a fluid, and as such, it possesses all properties a fluid has, where one of the most important is the pressure it exerts on objects in contact.

Given this, we will explain air pressure first and then we will deal with other specific gases.

## Air (Atmospheric) Pressure

The first scientist who was successful in his attempts to measure the air pressure was Evangelista Torricelli, an Italian physicist and mathematician and a student of Galileo. He invented an air pressure gauge called "mercury barometer" or simply "barometer" which consists of an open vessel with a closed vertical tube at it middle as shown in the figure. When the vessel is filled with mercury, its level is the same everywhere, even inside the tube (figure a), as this setup is a communicating vessel, as discussed in the previous tutorial "Liquid Pressure. Pascal's Principle ." This is because air exerts the same pressure on all parts of mercury surface.

If we remove the air from the upper part of the tube, mercury will start to raise up inside the tube, as there is no air above to hamper it from raising up. In this case, mercury acts as a capillary liquid, i.e. as a liquid that rises up in narrow tubes because of the change in pressure. This raise will continue until the liquid pressure caused by the mercury column will balance the air pressure aside (figure b). Therefore, since it is easy to measure the height h of mercury column and given that density of mercury is known (it is 13 600 kg/m3 or 13.6 g/cm3), we can calculate the air (atmospheric) pressure by working out the pressure of mercury column, i.e.

Pair = Pmercury column
= ρmercury × g × hmercury column

In normal atmospheric conditions (at sea level, in a sunny day at 150C), the mercury inside the column rises at 760 mm (76.0 cm or 0.760 m) above the neighbouring parts in contact with air. As a result, the atmospheric pressure in normal conditions is

Pair = 13 600 kg/m3 × 9.81 m/s2 × 0.760 m
≈ 101 325 Pa

Thus, we can write:

76 cm - Hg = 101 325 Pa

(Hg is the chemical symbol of mercury)

The above value can also be written as 1 atm (atmosphere). Thus, we have:

1 atm = 76 cm - Hg = 101 325 Pa

To facilitate calculations, another unit known as "bar" is often used. The conversion factor between bar and Pa is

1 bar = 100 000 Pa

Thus, we can say bar is slightly smaller than atm, as

1 bar = 100000/101325 atm = 0.9869 atm

On the other hand,

1 atm = 101325/100000
= 1.01325 bar
= 1013.25 mb

(mb means millibar)

High atmospheric pressure means the barometer shows values greater than 76 cm - Hg. Such a condition corresponds to cloudy skies and precipitations, as air charged with moisture is concentrated in the lower part of the sky. This increases the value of air pressure because an extra factor (moisture, i.e. water droplets) adds to the normal air pressure, giving a higher value in the barometer. On the other hand, low atmospheric pressure is associated with clear skies because the level of moisture in the air is very low.

### Example 1

The barometer in a weather station shows 74 cm - Hg.

1. What kind of weather is at the weather station location?
2. What is the value of atmospheric pressure in Pa and millibar (mb)?

### Solution 1

1. The value shown by the barometer (74 cm - Hg) is smaller than the value of normal atmospheric pressure (76 cm - Hg). This means there is a good weather (clear sky) in the given location.
2. We use the cross-production rule to convert the value in the desired units. Thus,
76 cm - Hg = 101 325 Pa
74 cm - Hg = x Pa

We have

x = 101 325 × 74/76
= 98 659 Pa

The same procedure is used to convert the atmospheric pressure from Pa to millibar. Thus,

100 000 Pa = 1 bar
98 659 Pa = x bar

We have

x = 98 659 × 1/100 000 = 0.98659 bar
= 986.59 mb

## Why is Mercury Used as Capillary Liquid in Barometers?

Mercury is a rare and expensive material. Also, it is very poisonous if consumed accidentally. Then, why it is used in barometers (and other pressure gauges we will discuss later)?

We will give answer to this question through the following exercise.

### Example 2

Calculate the liquid's height inside the barometer column in normal atmospheric conditions if we use water as a capillary liquid instead of mercury. Density of water is 1000 kg/m3.

### Solution 2

We know that the value of normal atmospheric pressure is 101 325 Pa. Thus, from the formula of liquid pressure, we can write

Patm = ρwater × g × hwater
hwater = Patm/ρwater × g
= 101 325/1000 × 9.81
= 10.33 m

Now, it is clear why mercury is used in barometers despite its shortcomings. If we used another liquid such as water instead of mercury, we would need a 11 + meter high device. Such barometer would be inappropriate to carry and use in indoor conditions.

## Gas Pressure

We cannot use standard barometers to measure pressure of a specific gas, as we would need a closed environment in which there is only this gas. Therefore, we use a combination of U-shaped tube and barometer as an equipment to measure gas pressure if the value of atmospheric pressure is known. This equipment is called "open-tube manometer" or "pressure gauge" which consists in a closed deposit at one side, in which there is the gas whose pressure needs to be measured, and an open end at the other side of a U-shaped tube, which is in contact with the air (atmospheric pressure). The separation between these two gases (air and the specific gas to be measured) is ensured by a liquid, such as water, mercury etc., as shown in the figure. 1) In the case shown above, air pressure is equal to the gas pressure as both gases push equally the capillary liquid at bottom of the U-shaped tube. As a result, the liquid level is the same in both sides of the tube.

If the capillary liquid inside the manometer is mercury, we can write

Pgas = Patm (in cm - Hg)

2) When gas pressure is greater than atmospheric pressure, the gas pushes the capillary liquid more than air. As a result, there will be a disparity in the liquid level in both sides of the U-shaped tube, as shown below. The abovementioned disparity is represented mathematically by the height h, similarly as in the U-shaped tube used to calculate the density of an unknown liquid discussed in the Physics tutorial Physics Tutorial: Liquid Pressure. Pascal's Principle

Mathematically we have:

Pgas = Patm + h (cm - Hg)

3) When air pressure is greater than gas pressure, the air will push the capillary liquid more than the gas. As a result, we will obtain the following figure. Mathematically we have:

Pgas = Patm-h (cm - Hg)

### Example 3

Calculate the gas pressure (in cm - Hg) in the three examples shown below if the actual atmospheric pressure is 74 cm - Hg. ### Solution 3

I. In this case, Patm > Pgas and h = 8 cm - Hg. Thus, given that Patm = 74 cm - Hg, we obtain for the gas pressure:

Pgas = Patm - h (cm - Hg)
= 74 cm - Hg - 8 cm - Hg
= 66 cm - Hg

II. In this case, Patm < Pgas and h = 10 cm - Hg. Thus, given that Patm = 74 cm - Hg, we obtain for the gas pressure:

Pgas = Patm + h (cm - Hg)
= 74 cm - Hg + 10 cm - Hg
= 84 cm - Hg

III. In this case, Patm = Pgas as mercury level is the same in both sides of the tube. Therefore, since Patm = 74 cm - Hg, we obtain the same value for the gas pressure, i.e.

Pgas = Patm = 74 cm - Hg

Remark! The atmospheric pressure in normal conditions is often referred as P0 instead of Patm.

## Total Pressure in Water

As discussed in the previous tutorial Physics Tutorial: Liquid Pressure. Pascal's Principle, water pressure is calculate by the equation

Pwater = ρwater × g × h

where h is the depth from the water surface.

However, this is only a part of the total pressure exerted on an object immersed in water, i.e. the pressure exerted only by water (we called it "gauge water pressure"). Since water is within the atmosphere layer, we must add the value of air pressure to the gauge water pressure to obtain the total pressure on objects immersed in water. Mathematically, we have:

Ptot = Pwater + Patm

### Example 4

What is the total pressure exerted on an object located at the deepest point of oceans, at Challenger Deep in the Mariana Trench (Pacific Ocean) if the depth in that place is 10 929 m? Take the density of Pacific Ocean in deep sectors equal to 1050 kg/m3 and the air pressure above water in normal values.

### Solution 4

The total pressure exerted on objects immersed in the given location, is

Ptot = Pwater + Patm

where

Pwater = ρwater × g × h
= 1050 × 9.81 × 10929 Pa
= 112 574 165 Pa

Adding this value to the normal atmospheric pressure (P0 = 101 325 Pa), we obtain for the total pressure at the deepest point of oceans:

Ptot = 112 574 165 Pa + 101 325 Pa
= 112 675 490 Pa

Such a big value (about 112.6 Million Pa) makes impossible any animal life in that depth. Furthermore, it is a complete darkness in such depths as light is entirely absorbed by the water.

## How does the Atmospheric Pressure Vary with Altitude?

The pressure at any level in the atmosphere represents the total weight of the air above a unit area. At higher altitudes, there are fewer air molecules above a given surface than on a similar surface at lower levels. For example, there are fewer molecules at 50 km above the Earth surface than at 10 km of altitude.

The following figure shows the air pressure values in some altitudes. Atmospheric pressure decreases with the increase in altitude. Since most of the atmosphere's molecules are near the earth's surface because of the attracting force of gravity, air pressure decreases rapidly at first (from sea level to h = 12 km), then more slowly at higher altitudes (from h = 12 km to h = 50 km) as air becomes very rare. Thus, since more than half of air molecules are in the first 5.5 km of atmosphere, air pressure halves at that altitude, i.e. Pair (5.5 km) = 500 mb = 1/2 P0. (This altitude is not shown in the figure in order to not make it confusing.) Such a decrease continues, albeit at slower rate, until it becomes 1 mb or 1/1000 P0 at h = 50 km.

There is a mathematical formula, which expresses the relationship between air pressure and altitude. Below is shown a simplified version of it.

P(h) = P0 × (1 - 2.25577 × 10-5 × h)5.25588

where P0 = 101 325 Pa is the standard atmospheric pressure in normal conditions and h is the altitude in metres. The other numbers are values obtained from operations with various constants.

### Example 5

What is the value of air pressure in the boundaries of ozone layer? Most of ozone lies from 10 km to 16 km above the sea level. Take the atmospheric conditions at sea level as normal (standard).

### Solution 5

Let's write the values of altitude and air pressure at sea level in standard notation. We have h1 = 10 km = 104 m and h2 = 16 km = 1.6 × 104 m. Also, P0 = 101 325 Pa = 1.01325 × 105 Pa.

Since

P(h) = P0 × (1 - 2.25577 × 10-5 × h)5.25588

we obtain for P1 (P at 10 km above sea level):

P(10 km) = 1.01325 × 105 × (1 - 2.25577 × 10-5 × 104 )5.25588
= 1.01325 × 105 × (1 - 2.25577 × 10-1 )5.25588
= 1.01325 × 105 × (1 - 0.225577)5.25588
= 1.01325 × 105 × (0.774423)5.25588
= 1.01325 × 105 × 0.2609
= 0.26436 × 105 Pa
= 26 436 Pa

and for P2 (at 16 km above the sea level):

P(16 km) = 1.01325 × 105 × (1 - 2.25577 × 10-5 × 1.6 × 104 )5.25588
= 1.01325 × 105 × (1 - 3.609232 × 10-1 )5.25588
= 1.01325 × 105 × 0.6395.25588
= 1.01325 × 105 × 0.095
= 0.09626 × 105 Pa
= 9626 Pa

As you see, these values are very small when compared to standard atmospheric pressure. This means the air density is very low in such altitudes as air pressure is produced when air molecules hit a given surface by a certain force, i.e. it is proportional to the number of air molecules available.

## Whats next?

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