For chemists, there is no single equation that conveys the behavior of every gas. Still, the basic properties gases do share in common are combined to produce the simplistic equation of state. We will use this equation to determine the ideal gas law molecular weight.
Deriving the Ideal Gas LawWhen we compress a gas, its volume shrinks. That is, the pressure is proportional to the inverse of the volume.
P ∝ 1/VThis has been known for hundreds of years as Boyle’s Law.
In addition, as temperature increases, gas volume increases (think hot-air balloon).
T ∝ VThis, too, has long been known and has been assigned a name. It is Charles’ Law.
Now obviously, the more gas there is, the greater its volume will be. Gas is often listed in terms of moles*, n.
n ∝ VObviously, there is no name for this relationship.
We now combine above. Since we have not been discussing equalities, we wind up with a “proportion” equation, not an “equals” equation. We need to introduce a constant to get around this. We call it the Ideal Gas constant, R¹. Its value is 0.082 liter-atmospheres per mole-degrees Kelvin. This allows us to state the Ideal Gas Law in exact mathematical form,
PV = nRT
Simple Ideal Gas Law ApplicationThe ideal gas law equation is useful to describe the physical characteristics of simple non-interactive gases.
The number of moles, n, equals the weight of gas present divided by the molar mass (atomic or molecular weight²). For example: at standard temperatures and pressures (STP), one mole of a gas is maintained at 273 degrees Kelvin, at a pressure of one atmosphere. What is the volume this gas occupies?
Rewriting our formula, we get,
V = nRT/PThis works out to approximately 22.414 liters under these conditions.
Ideal Gas Law Molecular WeightThe number of moles, n, can be converted into the mass divided by the molar mass (essentially, the molecular weight), m/M. Then,
PV = (m/M)(RT)Rearranging, the above equation becomes,
M = (m/PV)(RT)Plugging in the correct values, we should be able to determine the ideal gas law molecular weight of a gas.
The Units We ChooseAt standard temperatures and pressures (STP), for instance, the pressure is one atmosphere. The volume is 22.414 liter. The temperature is 273° Kelvin. Under these conditions, the mass of the gas equals its molar mass. To demonstrate this, we evaluate 4.000 grams of helium under these conditions. We calculate its molar mass,
M = [4.000/(1)(22.414)][(0.082)(273)] = 3.995 grams.The actual M (atomic weight) for helium is 4.003 grams.
Other and varied complications are listed in the references for those desiring to pursue the ideal gas law further.
The Ideal Gas Law in Terms of DensityFor the purpose of showing how other forms of the ideal gas law can be derived, consider the following:
PV = m/MRT
m/V = RT/MP
δ = RT/MPAnother (if strange) way of viewing things is the concentration, C, of a gas in space—the number of moles of gas in an otherwise empty volume. The ideal gas law is then written,
n/V = P/RT or
C = P/RT
When the Ideal Gas Law is InsufficientResults from application of the ideal gas law will never match reality, since no gas is an “ideal gas.” An ideal gas requires no interaction of its component atoms or molecules. The size of an “ideal atom” is zero. Clearly this is not reality either. Nevertheless, the law is a good one. It is not a bad approximation for non-reactive gases that consist of very tiny particles, as is the case for helium. The ideal gas law molecular weight determination is an excellent instructional aid for beginning students.
¹ The value of R is closely associated with Avogadro’s Number. See references. ² A simple gas may consist of atoms, such as He, or small molecules, such as H₂.References:
- University of Waterloo: The Ideal Gas Law
- AUS-e-TUTE: Chemistry Tutorial – Ideal Gas Law
- YouTube: Find the Density of a Gas (Ideal Gas Law)