All ideal gases conform, in behavior, to a particular relationship between pressure, volume, moles, and temperature as dictated by the ideal gas law.

In this equation, R is the ideal gas constant. Rearranging the equation allows any one of the variables to be calculated as long as the other three are known. 

For example, what is the volume of one mole of an ideal gas under standard temperature and pressure conditions? Abbreviated as STP, these conditions are 0 °C or 273 K and 1 atm. 

Rearranging the equation and substituting in the values for n (1 mole), temperature (273 K), pressure (1 atm), and the ideal gas constant (0.08206 L·atm/mol·K), one mole of an ideal gas occupies a volume of 22.4 liters. This is the molar volume at STP, which is also a good approximation for many common gases.

At higher temperatures and lower pressures, the gas expands and its molar volume is larger than it is at standard conditions. At lower temperatures and higher pressures, the molar volume is smaller.

Another useful quantity of a gas is its density. Recall that the number of moles, n, is equal to the mass of the gas divided by its molar mass. Substituting this relationship into the ideal gas equation, and then rearranging, yields an expression for mass over volume or density. 

From this equation, the density of a gas is directly proportional to its molar mass. This is why helium balloons float away when released outside. The molar mass, and thus density, of helium, is much less than that of air, which is primarily nitrogen and oxygen. 

Also, notice that density and temperature are inversely related. This is observed when piloting a hot air balloon. Turning on the burner heats up the air molecules within the balloon and they move faster. 

The pressure in the balloon increases, but the balloon is designed so that some of the air escapes. This makes the air in the balloon less dense than the surrounding air. Because of this difference in density, the balloon ascends.

Conversely, turning off the burner and opening the vent, allows the warm to escape. As the balloon contracts, outside air enters, increasing the density in the balloon to that of the surroundings. Then, because of the weight in the basket, the balloon descends.

The equation, when rearranged, also allows us to calculate the molar mass of an unknown gas.

Suppose an unknown gas with a mass of 12.5 grams occupies a volume of 6.08 liters and exerts a pressure of 1.2 atm at 40.0 °C.

The density of the gas is known from the given mass and volume. Then, the temperature in degree Celsius is converted to units of kelvin and substituted into the equation along with the values for pressure and the gas constant.

Solving for M yields a molar mass of 44 g/mol. Therefore, carbon dioxide is the unknown gas.