# 6.2: Measurable Quantities of Gases

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Learning Objectives
• Identify and define the four principle measurable quantities of gases.
• Represent the four principle measurable quantities of gases as variables.
• Identify the units in which the four principle measurable quantities of gases are expressed.
• Apply a conversion factor to change a pressure measurement into a corresponding value in a different unit.

Chapter 1 presented the five principle measurable quantities of chemistry.  However, while these fundamental measurements are generally applicable to a wide variety of chemicals, two of these quantities, length and time, cannot be utilized to measure chemicals that exist in the gaseous state of matter.  Furthermore, as a result of the the characteristics of gases that were described in the previous section, the units in which the remaining quantities, mass, volume, and temperature, are expressed must be adapted before being applied in gas-related calculations.  Finally, the pressure of a gas qualifies as a fundamental measurement, due to the impact that this force has on the constituent particles in and, therefore, the volume of, gaseous chemicals.

## Volume

Volume, V, is defined as the amount of space that is occupied by an object.  The metric unit for measuring volume is the liter (L), and the cubic meter (m3) is the primary unit that is used in the SI system.  While both of these units are appropriate for expressing the volumetric measurements of gases, the conversions that are required to change liters into alternative units, such as milliliters, are much more straight-forward than the process of converting cubic meters into other units, such as cubic centimeters or cubic inches.  Therefore, the volumes of gaseous chemicals are primarily expressed in liters (L) and milliliters (mL).

## Temperature

Temperature, T, is defined as the measure of how hot an object is. The U.S., metric, and SI units for measuring temperatures are degrees Fahrenheit (°F), degrees Celsius (°C), and Kelvin (K), respectively.  As will be discussed in greater detail in the following sections, the measurable quantities of gases can be related to one another using equations that involve multiplication and division.  Therefore, if a negative temperature value were to be utilized in one of these calculations, the resultant solution, which could correspond to a volumetric measurement, must also be negative.  However, because a volumetric quantity cannot be negative, by definition, negative numerical values cannot be incorporated into gas-related calculations.  Therefore, because both the Fahrenheit and Celsius scales allow for the existence of negative values, the temperatures of gases cannot be measured using these units.  In contrast, the Kelvin scale is based on a quantity called absolute zero, which is defined as the temperature at which all motion stops.  Since absolute zero, which is assigned a numerical value of 0 Kelvin, is the lowest possible temperature that can be theoretically achieved, the Kelvin scale does not allow for negative temperatures.  Therefore, incorporating temperatures that are recorded in Kelvin into the gas-related equations that will be developed will result in positive solutions, and, as a result, the temperatures of gaseous substances must be expressed in Kelvin (K).

## Amount

Mass, m, is defined as the quantity of substance that is contained in an object.  However, because the volume of a gaseous chemical is largely comprised of the "empty" spaces between its particles, the number of gaseous particles that are present in a given container is small, relative to the quantity of solid or liquid particles that would occupy the same amount of space.  As a result, the masses of gaseous substances that are recorded using standard units, such as grams or kilograms, have very small numerical values.  Because incorporating these measurements into the gas-related equations that will be developed would result in the calculation of correspondingly-small solutions, the amount of gas that is present in a container is not quantified by its mass.  Instead, the amountn, of a gaseous chemical is expressed in moles (mol).

## Pressure

As stated above, because the volume of a gaseous chemical is largely comprised of the "empty" spaces between its particles, the application of pressure causes those particles to shift their positions to occupy the "empty" spaces, which effectively decreases, or compresses, the overall volume of the gas by increasing the physical proximity of its particles.  The ability to forcibly alter the volume of a chemical through an application of pressure (P), which is defined as a force exerted over an area, is unique to gaseous substances, which qualifies this type of measurement as fundamental to the study of gases.  While pressure can be exerted on a gaseous substance, chemists are primarily interested in measuring the pressure that is exerted by gaseous particles as they collide with the surfaces of the container in which they are held.

The U.S. unit that is utilized to report pressure measurements is psi, which is an abbreviation for "pounds per square inch."  This unit, which can be symbolized as lb/in2, is the only pressure unit that consists of both a numerator and a denominator, as indicated verbally by the word "per" and symbolically by the "/".  Additionally, because U.S. units are not commonly utilized by those individuals who reside outside of the United States, very few scientific pressure measurements are recorded in psi.  Therefore, this unit will not be discussed further in this text.

The most common scientific unit for measuring pressure is the atmosphere (atm), which is defined as the pressure that is exerted by the gases that are present in the atmosphere at sea level.  Because pressure decreases as altitude increases, changing the elevation of a gas impacts the extent to which the corresponding force can causes gaseous particles to shift their positions within a defined volume.  Atmospheric pressure is measured using a piece of equipment called a barometer, which was invented by a seventeenth-century Italian scientist named Evangelista Torricelli.  As shown below in Figure $$\PageIndex{1}$$, a barometer is constructed by inverting a tube in a pool of mercury.  Torricelli selected mercury for this application because this element is the only known metal that exists in the liquid state of matter at room temperature.  Furthermore, due to its high surface tension, mercury particles are resistant to separation at the molecular level and, therefore, will adhere to one another when disturbed by an outside force, such as pressure.  As a result, mercury has a convex, or inverted meniscus, meaning that this substance curves upward, instead of downward, when contained within a tube.  Because the barometric tube is sealed at one end, an application of pressure, such as atmospheric pressure, causes the mercury to be pulled into the column until the downward pressure of mercury in the tube is exactly equal to the force that is being applied.  At sea level, the mercury level in a barometer will rise to an height of exactly 760 millimeters of mercury (mmHg).  In honor of Torricelli's invention of the barometer, this atmospheric pressure measurement has also been equated to 760 torr (Torr), and, therefore, 1 millimeter of mercury is equivalent to 1 torr. Figure $$\PageIndex{1}$$: Mercury barometer.

However, because these units are all dependent on the elevation at which the gas is being measured, none of these units are appropriately-qualified to be designated as an SI unit.  Instead, the N/m2, or "Newton per square meter," which is a unit that is derived directly from the mathematical definition of pressure, was established as the SI unit for measuring pressure.  This quantity is more commonly-known as the pascal (Pa), which was named in honor of Blaise Pascal, a French mathematician who used barometers extensively in his studies of fluids.  Scientists defined atmospheric pressure to have a numerical value of 101,325 pascals.

Since all of the pressure units that were defined above can be equated to atmospheric pressure, these units can be related in a single equality string, as shown below.  Finally, any two of the values that are present within this equality can be rewritten in the form of a conversion factor that can, in turn, be utilized to change a pressure measurement to a corresponding value in a different unit.

1 atmosphere (atm) = 760 millimeters of mercury (mmHg) = 760 torr (Torr) = 101,325 pascals (Pa)

Exercise $$\PageIndex{1}$$

Hurricane Wilma, which was a Category 5 storm with maximum reported wind speeds of 185 miles per hour, is the most intense tropical cyclone that has occurred in the Atlantic basin.  However, the pressure within the eye of this storm, which is the relatively-calm region at the center of a hurricane, was recorded at 0.827 atmospheres, which is the lowest central pressure on record for any Atlantic hurricane.  Using conversion factors based on the equality string that is shown above, convert this pressure to

1. millimeters of mercury,
2. torr, and
3. pascals.
In order to bring about the desired unit transformation, a conversion factor must be developed using the first and second quantities in the equality string that is shown above.  Applying the correct number of significant figures to the calculated quantity results in the final answer that is shown below.

$${0.827 \; \cancel{\rm{atm}}} \times \dfrac{760 \; \rm{mmHg}}{1 \; \cancel{\rm{atm}}}$$ = $${\text {628.52}}$$ $${\rm{mmHg}}$$ ≈ $${\text {629}}$$ $${\rm{mmHg}}$$

In order to bring about the desired unit transformation, a conversion factor must be developed using the first and third quantities in the equality string that is shown above.  Applying the correct number of significant figures to the calculated quantity results in the final answer that is shown below.

$${0.827 \; \cancel{\rm{atm}}} \times \dfrac{760 \; \rm{torr}}{1 \; \cancel{\rm{atm}}}$$ = $${\text {628.52}}$$ $${\rm{torr}}$$ ≈ $${\text {629}}$$ $${\rm{torr}}$$

Alternatively, this unit transformation could have been accomplished by relating the answer to Part (a) of this Exercise to torr through the application of a conversion factor using the second and third quantities in the equality string that is shown above.
$${0.827 \; \cancel{\rm{atm}}} \times \dfrac{101,325 \; \rm{Pa}}{1 \; \cancel{\rm{atm}}}$$ = $${\text {83,795.775}}$$ $${\rm{Pa}}$$ ≈ $${\text {83,800}}$$ $${\rm{Pa}}$$