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PHYSICS
Atmospheric pressure
Air possesses weight as a result of the Earth’s gravity. The sensation of weight exerted on anything it comes into
contact with is called
atmospheric pressure
.
Theweight of the air is determined by theweight of 1
m³
of dry air at a temperature of 20
°C
and at a pressure of
760
mmHg
(760
mm
of mercury column), and is equivalent to 1,03
Kg
.
One of the characteristics of atmospheric pressure is that it varies according to the altitude. Themaximum value is
at sea level (zero altitude) and decreases as the altitude increases. This is due to theweight of the air on the lower
layers of the atmosphere being greater than theweight on the upper layers. At sea level the atmospheric pressure
is an average of 760
mmHg
and decreases by about 1
cm
column of mercury per 100
m
of elevation.
At an altitude of 1000
m
the column has an average height of 66
cm
. For high-altitudes, pressure decreases less
rapidly.
The value of the atmospheric pressure can be measured by reproducing the experiment devised by the Italian
physicist Evangelista Torricelli (1608-1647), a pupil of Galileo Galilei (1564-1642).
Figure 2
Pos. 1
: take a glass tube closed at one endwith length1
m
andwith an internal diameter of about 12
mm
(about
1
cm²
), and fill it completely with mercury. Hold a finger over the open end of the tube and invert it. Place the
inverted tube inabasinalso containingmercury, taking carenot to release the endof the tubeuntil it is immersed.
When the finger is removed,mercurywill flow from the tube into thebowl, the level ofmercury retained in the tube
can bemeasured from the surface of themercury in the bowl to a height of 76
cm
.
Themercury in the tube has not flowed entirely into the basin because of the atmospheric pressure acting on the
free surface of themercury in the basin.We can say that the atmospheric pressure is equal to the pressure exerted
by a column of mercury 76
cm
high.
With this informationwe can calculate theweight of the air.
Calculate the volume of the column:
V
column
=
Areaof thebase * height
1 * 76 =76
cm
3
V
column
=
76
cm
3
With the knowledge that the specific weight
P
s
of mercury is 0,01359
Kg/cm³
, we calculate the mass of the
mercury column:
m
column
=
V
column
* P
s
76 * 0,01359 =1,03
Kg
m
column
=
1,03
Kg
Figure 2
Pos. 2
: in order to balance themercury columnwemust create an opposing Force, usingweight for example.
Theweight Force
F
p
corresponds to:
F
p
= 9,81
[N ⁄ Kg]
* 1,03
[Kg]
=10,1
N
F
p
=
10,1
N
Figure 2
Pos. 3
: starting from the result of the previous experiment we calculate the height reached in a column of water
instead of mercury:
V
column
=
Area of base * height
1
[cm
2
]
* x
[cm]
= x
[cm
3
]
m
column
=
V
column
* P
s
x
[cm
3
]
* m
vol.
[Kg ⁄ cm
3
]
=1,03
Kg
m
column
=
1,03
Kg
Knowing that the density
m
vol.
of water is 0,001
Kg/cm³
and that the mass of the mercury column is 1,03
Kg
calculate the height of thewater column
x
:
1,03
[kg]
= 1
[cm
2
] * x [cm] * m
vol.
[Kg cm
3
]
1,03
[kg]
=1
[cm
2
] * x [cm] *
0,001
[Kg ⁄ cm
3
]
x
=
1,03
[kg]
x=
1,03
[kg]
x
=1030
cm
1
[cm
2
] *
0,001
[Kg ⁄ cm
3
]
0,001
[Kg ⁄ cm]
1
13
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