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:
The SI unit of pressure is the Pascal. 1Pascal = 1 N / m2 = 1 kg m-1s-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.
|1 mm Hg at 0o C||
|1 Pound per Square Inch (psi)||
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 1.01325×105 N per 1 m2. Since pressure = force / area, this gives a pressure of 1.01325×105 Pa. One standard atmosphere (atm) is defined as exactly 1.01325×105 Pa.
So how did we figure out that standard atmospheric pressure is 1.01325×105 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.
|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 = 1×105 Pascals.
Students are often given a non-Hg barometer or conditions where g does not equal 9.8 m/s2. 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.
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