As a student, you are familiar with pressure. Work needs to be done, and there
is always a
limited time to do it. The less time there is, the more pressure you feel.
Gaseous pressure works
in much the same way. A force acts on a limited area to give pressure. If the
area shrinks (you
have less time), something has got to give: either the force shrinks (you take
on less work) or the
pressure rises.

Pressure is defined mathematically as force divided by the area over which the
force acts:

Pressure =

The SI unit of pressure is the Pascal. 1Pascal = 1 N / m^{2} = 1 kg m^{-1}s^{-2}. However, there will be times when you'll be
given
pressure in non-SI units. The table summarizes the most common pressure units
and their
conversion factors.

Unit

Relationship

Pascal (Pa)

1 = 1

Torr

atm

1 mm Hg at 0^{o} C

atm

atmosphere (atm)

101.3×10^{5}Pa

Bar

1×10^{5}Pa

1 Pound per Square Inch (psi)

0.0680atm

If you've ever checked your tires' pressure, you've probably encountered pounds
per square inch, or psi. The other units are less familiar: they arose because
gravity exerts a downward force on the atmosphere, which consequently exerts a
pressure on the Earth's surface and whatever else happens to be there. On a calm
day at sea level, the force gravity exerts is 1.01325×10^{5} N per 1 m^{2}. Since pressure = force / area, this gives a pressure of 1.01325×10^{5} Pa. One standard atmosphere (atm) is defined as exactly 1.01325×10^{5} Pa.

So how did we figure out that standard atmospheric pressure is 1.01325×10^{5} Pa in the first place? Atmospheric pressure was first measured with a
barometer. A barometer consists of a large dish and a long glass tube that
is sealed at one end. The tube and dish are filled with mercury (HG) or some
other liquid, and the tube is inverted into the dish. If all this is done
without any air entering the tube, a column of liquid will remain above the
dish.

When the tube full of mercury is inverted in the dish, the mercury level will
drop. It will continue
to drop until the pressure generated by the column's weight equals the
atmospheric pressure.
Since we know the column's height h, the density of mercury ρ, and the
acceleration due
to gravity g (9.81m s^{-2}), we can calculate the atmospheric
pressure P.

baroeq

P = ghρ

At 0 ^{o} Celsius, it turns out that the atmosphere can support a column of
mercury 760 mm tall (the unit mm Hg is thus equal to 1/760). 1 atm = 760 mm Hg
only at 0 ^{o} Celsius since the
ρ of Hg changes with temperature. It's a pain to recalculate ρ at
different
temperatures, but the units of Torr come to the rescue. 1 Torr = 1/760 atm at
all temperatures.
The Bar is related to the Torr, but is not quite as useful. 1 bar = 1×10^{5} Pascals.

Students are often given a non-Hg barometer or conditions where g does not
equal 9.8 m/s^{2}. Don't let these types of questions phase you; realize which
variables are
changing, convert everything down to SI units, and plug them all in to
@@Equation@@. There are examples in the problem section.