## The Ideal Gas Law

Contents

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 law: \(\cfrac{P}{T}\) = constant at constant
*V*and*n* - Charles’s law: \(\cfrac{V}{T}\) = constant at constant
*P*and*n* - Avogadro’s law: \(\cfrac{V}{n}\) = 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^{–1} K^{–1} and 8.314 kPa L mol^{–1} K^{–1}.

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 topic in this tutorial.

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 tutorial, 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

### Using the Ideal Gas Law

Methane, CH_{4}, is being considered for use as an alternative automotive fuel to replace gasoline. One gallon of gasoline could be replaced by 655 g of CH_{4}. What is the volume of this much methane at 25 °C and 745 torr?

### Solution

We must rearrange *PV* = *nRT* to solve for *V*: \(V=\phantom{\rule{0.2em}{0ex}}\cfrac{nRT}{P}\)

If we choose to use *R* = 0.08206 L atm mol^{–1} K^{–1}, then the amount must be in moles, temperature must be in kelvin, and pressure must be in atm.

Converting into the “right” units:

\(n=6\phantom{\rule{0.2em}{0ex}}55\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{g}\phantom{\rule{0.2em}{0ex}}{\text{CH}}_{4}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\cfrac{1\phantom{\rule{0.2em}{0ex}}\text{mol}}{16.043\phantom{\rule{0.2em}{0ex}}{\require{cancel}\cancel{\text{g CH}}}_{4}}\phantom{\rule{0.2em}{0ex}}=40.8\phantom{\rule{0.2em}{0ex}}\text{mol}\)

\(T=25\phantom{\rule{0.2em}{0ex}}\text{°C}+273=298\phantom{\rule{0.2em}{0ex}}\text{K}\)

\(P=745\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{torr}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\cfrac{1\phantom{\rule{0.2em}{0ex}}\text{atm}}{760\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{torr}}}\phantom{\rule{0.2em}{0ex}}=0.980\phantom{\rule{0.2em}{0ex}}\text{atm}\)

\(V=\phantom{\rule{0.2em}{0ex}}\cfrac{nRT}{P}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{\left(40.8\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{mol}}\right)\left(0.08206\phantom{\rule{0.2em}{0ex}}\text{L}\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{{\text{atm mol}}^{–1}\phantom{\rule{0.2em}{0ex}}{\text{K}}^{\text{–1}}}\right)\left(298\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{K}}\right)}{0.980\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{atm}}}\phantom{\rule{0.2em}{0ex}}=1.02\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\text{L}\)

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.

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: \(\cfrac{{P}_{1}{V}_{1}}{{T}_{1}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{{P}_{2}{V}_{2}}{{T}_{2}}\) using units of atm, L, and K. Both sets of conditions are equal to the product of *n*\(×\)*R* (where *n* = the number of moles of the gas and *R* is the ideal gas law constant).

## Example

### 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 (see the figure below). 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?

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:

\(\cfrac{{P}_{1}{V}_{1}}{{T}_{1}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{{P}_{2}{V}_{2}}{{T}_{2}}\phantom{\rule{0.4em}{0ex}}⟶\phantom{\rule{0.4em}{0ex}}\cfrac{\left(153\phantom{\rule{0.2em}{0ex}}\text{atm}\right)\left(13.2\phantom{\rule{0.2em}{0ex}}\text{L}\right)}{\left(300\phantom{\rule{0.2em}{0ex}}\text{K}\right)}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{\left(3.13\phantom{\rule{0.2em}{0ex}}\text{atm}\right)\left({V}_{2}\right)}{\left(310\phantom{\rule{0.2em}{0ex}}\text{K}\right)}\)

Solving for *V*_{2}:

\({V}_{2}=\phantom{\rule{0.2em}{0ex}}\cfrac{\left(153\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{atm}}\right)\left(13.2\phantom{\rule{0.2em}{0ex}}\text{L}\right)\left(310\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{K}}\right)}{\left(300\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{K}}\right)\left(3.13\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{atm}}\right)}\phantom{\rule{0.2em}{0ex}}=667\phantom{\rule{0.2em}{0ex}}\text{L}\)

(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.)

### The Interdependence between Ocean Depth and Pressure in Scuba Diving

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

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 his 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 she or he begins 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.