Chapter 4: Atmospheric pressure

Question 1

Atmospheric pressure is

Answer 1

the weight of a column of air exerted on a surface of a unit area by the activity of the molecules composing the fluid (air).

Question 2

The pressure of the atmosphere at a point on the earth’s surface is therefore equivalent to

Answer 2

the weight of the whole column of air standing on unit area at that point.

Question 3

if one considers a point at some height
above the surface, it seen that the pressure at that height is equal to

Answer 3

the weight of the air above

Question 4

A pressure of 1000 mb (100 kPa) results from the weight exerted by

Answer 4

10,000 kg of air overlying one square meter of surface, accelerated by
gravity of 10 m sec^-2
.

Question 5

The huge pressure does not crush us because

Answer 5

it is exactly balanced by outward pressure from the inside of our bodies.

Question 6

Ears popping due to change in altitude are the result of

Answer 6

the pressure difference between the inside of our heads and the surrounding air.

Question 7

Pressure decreases with altitude, due to

Answer 7

the reduction of the mass of overlying air with height.

Question 8

Pressure is 700 mb at

Answer 8

3,000 m

Question 9

pressure is 500 mb at

Answer 9

5500m

Question 10

pressure is 300 mb at

Answer 10

10,000 m

Question 11

In SI units the pressure is measured by

Answer 11

Pascals.

Question 12

One Pascal =

Answer 12

one Newton per square meter [1 Pa = 1 N/m²]

Question 13

In meteorology the unit of pressure, which is a

Answer 13

force per unit area, is Pascal

Question 14

In meteorology the unit of
pressure, which is a force per unit area, is Pascal. This unit is very small so
that the pressure is usually expressed by

Answer 14

millibars (mb) or hectoPascal s
(hPa).

Question 15

The relation between mb, hPa and Pascal is the following:

Answer 15

1 mb= 1 hPa = 100 Pa

Question 16

The relation between mb, hPa and Pascal is the following:
1 mb= 1 hPa = 100 Pa.
Also the pressure is measured in

Answer 16

millimeters and inches of mercury (1 in = 254 mm)

Question 17

……………………. is used to measure atmospheric pressure

Answer 17

simple barometer

Question 18

A simple
barometer, which is called

Answer 18

Torricelli barometer or simply a mercury barometer

Question 19

Torricelli correctly described the atmosphere as

Answer 19

a vast ocean of air that exerts a pressure on us and all things about us.

Question 20

the weight of the mercury in the column equaled

Answer 20

the weight of that extended from the ground to the top of the atmosphere

Question 21

The length of mercury column varies with

Answer 21

temperature.

Question 22

Therefore,
Meteorologists determine the length of mercury column under

Answer 22

standard conditions.

Question 23

Under standard conditions the temperature is that of melting ice

Answer 23

0 c

Question 24

Under standard conditions the temperature is that of melting ice (0cand the gravitational acceleration is

Answer 24

9.80665 m/s²

Question 25

Under standard conditions the temperature is that of melting ice (0oC) and the gravitational acceleration is 9.80665 m/s2. On the sea level the pressure under standard conditions is equal

Answer 25

760 mm (1013.25 mb)

Question 26

air is also measured by

Answer 26

aneroid barometer

Question 27

aneroid means

Answer 27

without liquid

Question 28

Instead of having a mercury column held up by air pressure, the aneroid uses a

Answer 28

partially evacuated metal chamber. A chamber, being very sensitive to variations in air pressure, changes shape, compressing as the pressure increasing and expands as pressure decreases.

Question 29

An advantage of the aneroi d barometer

Answer 29

it can be easily connected to a recording mechanism

Question 30

An advantage of the aneroi d barometer is that it can be easily connected to a recording mechanism. The resulting is a new instrument, which is called the

Answer 30

barograph.

Question 31

The barograph

Answer 31

records the pressure continuously with passage of time

Question 32

In the atmosphere the amount of pressure is determined by the effect of two factors

Answer 32

temperature and density

Question 33

A temperature variations of the air affect the

Answer 33

density of the air (the mass of the air per unit volume)

Question 34

As temperature decreases the air molecules are

Answer 34

closely packed (greater density)

Question 35

As temperature decreases the air molecules are closely packed (greater density), so the air pressure is

Answer 35

high.

Question 36

The pressure at any given altitude in the atmosphere depends mainly at the

Answer 36

weight of the air directly above that point

Question 37

With ascending through the atmosphere, the air becomes

Answer 37

less dense because the lesser amount (weight) of the air above, the result is a decrease in pressure

Question 38

The rate at which the pressure decreases with height is

Answer 38

not a constant

Question 39

The rate at which the pressure decreases with height is not a constant. The rate is greatest near

Answer 39

the earth’s surface and with increasing altitude the rate of pressure decreasing becomes less and less

Question 40

The rate is greatest near the earth’s surface and with increasing altitude the rate of pressure decreasing becomes less and less (this is because

Answer 40

the amount of gases decreases rapidly in the lower troposphere where in the first 5 km concentrated 50% of the whole amount of gases, the pressure at this height is only half its value near the earth’s surface).

Question 41

The atmospheric pressure is reduced approximately one-half for each

Answer 41

5-kilometer increase in altitude.

Question 42

Mean sea level (MSL)

Answer 42

Since pressure varies with altitude, we cannot readily compare station pressures between stations at different altitudes. To make them comparable, we must adjust them to some common level.

Question 43

To compute sea level pressure, it is necessary to determine the

Answer 43

station pressure. We then add to this value the weight of air, which would occupy a column of unit area of cross-section between the station and MSL

Question 44

to express the pressure at a station of height 3 km into the sea level pressure, (see figure 3.3), we can use the basic principle of

Answer 44

sea level pressure reduction

Question 45

for 3 km the pressure will decrease

Answer 45

300 mb

Question 46

If the rate of pressure decreasing with height is 10 mb for each 100 pressure is measured

Answer 46

700 mb at this height

Question 47

for 3 km the pressure will decrease 300 mb. If the rate of pressure decreasing with height is 10 mb for each 100 pressure is measured 700 mb at this height, then the sea level pressure for this station is

Answer 47

700 + 300 or 1000 mb.

Question 48

To meet the need of weather forecaster, a network of observing stations has been built up, which covers the greatest part of the earths surface. The density of the network depends on different factors such as th

Answer 48

difficulties in climate and terrain, including about 70% of the earths surface covered by oceans, the distribution of population, etc.

Question 49

In studying the weather, it is necessary to obtain observations from

Answer 49

very large meteorological stations simultaneously and at regular intervals.

Question 50

Isobars.

Answer 50

lines of equal pressure

Question 51

Low

Answer 51

a center of pressure surrounded on all sides by higher pressure; also called a cyclone. Cyclonic curvature is the curvature of isobars to the left when you stand with lower pressure to your left

Question 52

High

Answer 52

a center of pressure surrounded on all sides by lower pressure; also called an anticyclone. Anticyclone curvature is the curvature of isobars to the right when you stand with lower pressure to the left.

Question 53

Trough

Answer 53

an elongated area of low pressure with the lowest pressure along a line marking maximum cyclonic curvature

Question 54

Ridge

Answer 54

an elongated area of high pressure with the highest pressure along a line marking maximum anticyclonic curvature.

Question 55

Col

Answer 55

the neutral area between two highs and two lows. It is also is the intersection of a trough and a ridge.