Chapter 9 Gases

9.3  The Ideal Gas Law

Learning Objectives

By the end of this section, you will be able to:

  • Use the ideal gas law, and related gas laws, to compute the values of various gas properties under specified conditions

The Ideal Gas Law

To this point, four separate laws have been discussed that relate pressure, volume, temperature, and the number of moles of the gas:

      • Boyle’s law    PV = constant at constant T and n
      • Amontons’s lawP / T = constant at constant V and n
      • Charles’s law   V / T = constant at constant P and n
      • Avogadro’s lawV / n = constant at constant P and T

Combining these four laws yields the ideal gas law, a relation between the pressure, volume, temperature, and number of moles of a gas:

PV=nRT
where P is the pressure of a gas, V is its volume, n is the number of moles of the gas, T is its temperature on the kelvin scale, and R is a constant called the ideal gas constant or the universal gas constant. The units used to express pressure, volume, and temperature will determine the proper form of the gas constant as required by dimensional analysis, the most commonly encountered values being 0.08206 L·atm/mol·K and 8.314 kPa·L/mol·K.

Gases whose properties of P, V, and T are accurately described by the ideal gas law (or the other gas laws) are said to exhibit ideal behavior or to approximate the traits of an ideal gas. An ideal gas is a hypothetical construct that may be used along with kinetic molecular theory to effectively explain the gas laws as will be described in a later module of this chapter. Although all the calculations presented in this module assume ideal behavior, this assumption is only reasonable for gases under conditions of relatively low pressure and high temperature. In the final module of this chapter, a modified gas law will be introduced that accounts for the non-ideal behavior observed for many gases at relatively high pressures and low temperatures.

The ideal gas equation contains five terms, the gas constant R and the variable properties P, V, n, and T. Specifying any four of these terms will permit use of the ideal gas law to calculate the fifth term as demonstrated in the following example exercises.

Example 9.9 – Using the Ideal Gas Law

Methane, CH4, is being considered for use as an alternative automotive fuel to replace gasoline. One gallon of gasoline could be replaced by 655 g of CH4. What is the volume of this much methane at 25°C and 745 torr?

 

Solution

We must rearrange PV = nRT to solve for V:

V=nRTP

If we choose to use R = 0.08206 L·atm/mol·K, then the amount must be in moles, temperature must be in kelvin, and pressure must be in atm.

Converting into the “right” units:

[latex]n = 6.66 \: \cancel {g \: CH_4} \; \frac {1 \: mol \: CH_4}{16.04 \: \cancel {g \: CH_4}} = 40.8 \: mol \: CH_4[/latex]
[latex]T = 25^\circ C + 273 = 298 \: K[/latex]
[latex]P = 745 \: \cancel {torr} \; \frac {1 \: atm}{760 \: \cancel {torr}} = 0.980 \: atm[/latex]
[latex][/latex]

[latex]V = \frac {nRT}{P} = \frac {(40.8 \: \cancel {mol})(0.08206 \: L \cdot \cancel {atm/mol \cdot K})(298 \: \cancel K)}{0.980 \: \cancel {atm}} = 1.02 \times 10^3 \: L[/latex]

It would require 1020 L (269 gal) of gaseous methane at about 1 atm of pressure to replace 1 gal of gasoline. It requires a large container to hold enough methane at 1 atm to replace several gallons of gasoline.

 

Check Your Learning

If the number of moles of an ideal gas are kept constant under two different sets of conditions, a useful mathematical relationship called the combined gas law is obtained:

P1V1T1=P2V2T2
using units of atm, L, and K. Both sets of conditions are equal to the product of nR (where n = the number of moles of the gas and R is the ideal gas law constant).

Example 9.10 – Using the Combined Gas Law

When filled with air, a typical scuba tank with a volume of 13.2 L has a pressure of 153 atm (Figure 9.16). If the water temperature is 27°C, how many liters of air will such a tank provide to a diver’s lungs at a depth of approximately 70 feet in the ocean where the pressure is 3.13 atm?

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Figure 9.16  Scuba divers use compressed air to breathe while underwater. (credit: modification of work by Mark Goodchild)

Solution

Letting 1 represent the air in the scuba tank and 2 represent the air in the lungs, and noting that body temperature (the temperature the air will be in the lungs) is 37°C, we have:

P1V1T1=P2V2T2
(153atm)(13.2L)300K=(3.13atm)V2310K

Solving for V2:

V2=(153atm)(13.2L)(310K)(300K)(3.13atm)=667L

(Note: Be advised that this particular example is one in which the assumption of ideal gas behavior is not very reasonable, since it involves gases at relatively high pressures and low temperatures. Despite this limitation, the calculated volume can be viewed as a good “ballpark” estimate.)

 

Check Your Learning

 

Chemistry in Everyday Life

The Interdependence between Ocean Depth and Pressure in Scuba Diving

Whether scuba diving at the Great Barrier Reef in Australia (shown in Figure 9.17) or in the Caribbean, divers must understand how pressure affects a number of issues related to their comfort and safety.

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Figure 9.17  Scuba divers, whether at the Great Barrier Reef or in the Caribbean, must be aware of buoyancy, pressure equalization, and the amount of time they spend underwater, to avoid the risks associated with pressurized gases in the body. (credit: Kyle Taylor)

Pressure increases with ocean depth, and the pressure changes most rapidly as divers reach the surface. The pressure a diver experiences is the sum of all pressures above the diver (from the water and the air). Most pressure measurements are given in units of atmospheres, expressed as “atmospheres absolute” or ATA in the diving community: Every 33 feet of salt water represents 1 ATA of pressure in addition to 1 ATA of pressure from the atmosphere at sea level. As a diver descends, the increase in pressure causes the body’s air pockets in the ears and lungs to compress; on the ascent, the decrease in pressure causes these air pockets to expand, potentially rupturing eardrums or bursting the lungs. Divers must therefore undergo equalization by adding air to body airspaces on the descent by breathing normally and adding air to the mask by breathing out of the nose or adding air to the ears and sinuses by equalization techniques; the corollary is also true on ascent, divers must release air from the body to maintain equalization. Buoyancy, or the ability to control whether a diver sinks or floats, is controlled by the buoyancy compensator (BCD). If a diver is ascending, the air in their BCD expands because of lower pressure according to Boyle’s law (decreasing the pressure of gases increases the volume). The expanding air increases the buoyancy of the diver, and they begin to ascend. The diver must vent air from the BCD or risk an uncontrolled ascent that could rupture the lungs. In descending, the increased pressure causes the air in the BCD to compress and the diver sinks much more quickly; the diver must add air to the BCD or risk an uncontrolled descent, facing much higher pressures near the ocean floor. The pressure also impacts how long a diver can stay underwater before ascending. The deeper a diver dives, the more compressed the air that is breathed because of increased pressure: If a diver dives 33 feet, the pressure is 2 ATA and the air would be compressed to one-half of its original volume. The diver uses up available air twice as fast as at the surface.

 

Standard Conditions of Temperature and Pressure

We have seen that the volume of a given quantity of gas and the number of molecules (moles) in a given volume of gas vary with changes in pressure and temperature. Chemists sometimes make comparisons against a standard temperature and pressure (STP) for reporting properties of gases: 273.15 K and 1 atm (101.325 kPa)[1]. At STP, one mole of an ideal gas has a volume of about 22.4 L—this is referred to as the standard molar volume (Figure 9.18).

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Figure 9.18  Regardless of its chemical identity, one mole of gas behaving ideally occupies a volume of ~22.4 L at STP.

  1. The IUPAC definition of standard pressure was changed from 1 atm to 1 bar (100 kPa) in 1982, but the prior definition remains in use by many literature resources and will be used in this text.

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