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Chemistry LibreTexts

13.4: Pressure Units and Conversions

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  • There are several benefits to maintaining the proper air pressure in a car tire. The ride is smoother and safer than with lowered pressure. The car gets better gas mileage and the tires don't wear out as fast. The recommended pressure for that model of car (usually somewhere between \(32\)-\(35 \: \text{psi}\) is usually listed in the owner's manual or stamped somewhere inside the door. The pressure on the tire is the maximum pressure for that tire, not the recommended one. Tire pressure is best measured when the tire is cold, since driving the car for a while will heat up the air in the tire and increase the pressure.

    Pressure Units and Conversions

    A barometer measures gas pressure by the height of the column of mercury. One unit of gas pressure is the millimeter of mercury \(\left( \text{mm} \: \ce{Hg} \right)\). An equivalent unit to the \(\text{mm} \: \ce{Hg}\) is called the \(\text{torr}\), in honor of the inventor of the barometer, Evangelista Torricelli. The pascal \(\left( \text{Pa} \right)\) is the standard unit of pressure. A pascal is a very small amount of pressure, so the most useful unit for everyday gas pressures is the kilopascal \(\left( \text{kPa} \right)\). A kilopascal is equal to 1000 pascals. Another commonly used unit of pressure is the atmosphere \(\left( \text{atm} \right)\). Standard atmospheric pressure is called \(1 \: \text{atm}\) of pressure and is equal to \(760 \: \text{mm} \: \ce{Hg}\) and \(101.3 \: \text{kPa}\). Atmospheric pressure is also often stated as pounds per square inch \(\left( \text{psi} \right)\). The atmospheric pressure at sea level is \(14.7 \: \text{psi}\).

    \[1 \: \text{atm} = 760 \: \text{mm} \: \ce{Hg} = 760 \: \text{torr} = 101.3 \: \text{kPa} = 14.7 \: \text{psi}\]

    It is important to be able to convert between different units of pressure. To do so, we will use the equivalent standard pressures shown above.

    Example \(\PageIndex{1}\)

    The atmospheric pressure in a mountainous location is measured to be \(613 \: \text{mm} \: \ce{Hg}\). What is this pressure in \(\text{atm}\) and in \(\text{kPa}\)?


    Step 1: List the known quantities and plan the problem.


    • Given: \(613 \: \text{mm} \: \ce{Hg}\)
    • \(1 \: \text{atm} = 760 \: \text{mm} \: \ce{Hg}\)
    • \(101.3 \: \text{kPa} = 760 \: \text{mm} \: \ce{Hg}\)


    • Pressure \(= ? \: \text{atm}\)
    • Pressure \(= ? \: \text{kPa}\)

    Use conversion factors from the equivalent pressure units to convert from \(\text{mm} \: \ce{Hg}\) to \(\text{atm}\) and from \(\text{mm} \: \ce{Hg}\) to \(\text{kPa}\).

    Step 2: Solve.

    \[613 \: \text{mm} \: \ce{Hg} \times \frac{1 \: \text{atm}}{760 \: \text{mm} \: \ce{Hg}} = 0.807 \: \text{atm}\]

    \[613 \: \text{mm} \: \ce{Hg} \times \frac{101.3 \: \text{kPa}}{760 \: \text{mm} \: \ce{Hg}} = 81.7 \: \text{kPa}\]

    Step 3: Think about your result.

    The air pressure is about \(80\%\) that of standard atmospheric pressure at sea level. For significant figure purposes, the standard pressure of \(760 \: \text{mm} \: \ce{Hg}\) has three significant figures.


    • Calculations are described for converting between different pressure units.

    Contributors and Attributions

    • CK-12 Foundation by Sharon Bewick, Richard Parsons, Therese Forsythe, Shonna Robinson, and Jean Dupon.