What is Ideal Gas Law?
The ideal gas law, also known as the general gas equation is an equation of state of a hypothetical ideal gas. Although the ideal gas law has several limitations, it is a good approximation of the behaviour of many gases under many conditions. The ideal gas law was stated by Benoit Paul Émile Clapeyron in 1834 as a combination of the empirical Charles’s law, Boyle’s Law, Avogadro’s law, and GayLussac’s law.
The ideal gas law states that the product of the pressure and the volume of one gram molecule of an ideal gas is equal to the product of the absolute temperature of the gas and the universal gas constant.
The empirical form of ideal gas law is given by:
where, 
Ideal Gas Law Units
 When we use the gas constant R = 8.31 J/K.mol, then we have to plug in the pressure P in the units of pascals Pa, volume in the units of m^{3} and the temperature T in the units of kelvin K.
 When we use the gas constant R = 0.082 L.atm/K.mol then pressure should be in the units of atmospheres atm, volume in the units of litres L and the temperature T in the units of kelvin K.
For easy reference, the above information is summarized in the table as follows:


\(R=8.31\frac{J}{K.mol}\)  \(R=0.082\frac{L.atm}{K.mol}\) 
Pressure in pascals Pa 
Pressure in atmospheres atm 
Volume in m^{3} 
Volume in litres L 
The temperature in Kelvin K 
The temperature in Kelvin K 
Derivation of the Ideal Gas Law
The ideal gas law is derived from the observational work of Robert Boyle, GayLussac and Amedeo Avogadro. Combining their observation into a single expression we arrive at the Ideal gas equation which describes all the relationships simultaneously.
The three individual expressions are as follows:
Boyle’s Law
\(V\propto \frac{1}{P}\)
Charle’s Law
\(V\propto T\)
Avagadro’s Law
\(V\propto n\)
Combining these three expressions, we get
\(V\propto \frac{nT}{P}\)The above equation shows that volume is proportional to the number moles and the temperature while inversely proportional to the pressure.
This expression can be rewritten as follows:
\(V=\frac{RnT}{P}=\frac{nRT}{P}\)Multiplying both sides of the equation by P to clear off the fraction, we get
\(PV=nRT\)The above equation is known as the ideal gas equation.
Ideal Gas Solved Problems
1. What is the volume occupied by 2.34 grams of carbon dioxide gas at STP?
Solution:
To determine the volume, rearrange the ideal gas law as follows:
\(V=nRT/P\)
Substituting the values as follows, we get
\(V=[(2.34\,g/44\,g.mol^{1})(0.08206\,{L}\,atm\,mol^{1}\,K^{1})(273.0\,K)]/1.00\,atm\)
\(V=1.19\,L\)
2. Calculate the temperature at which 0.654 moles of neon gas occupies 12.30 litres at 1.95 atmospheres.
Solution:
To determine the temperature, rearrange the ideal gas equation as follows:
\(T=PV/nR\)
Substituting the value, we get
\(T=(1.95\,atm))(12.30\,L)/(0.654\,mol)(0.08206\,L\,atm\,mol^{1}K{1})\)
\(T=447\,K\)
Relationship between Pressure and Temperature
The graph here shows the relationship between temperature and pressure for different gases.
On extrapolating this graph, what we see is that irrespective of which gas we are graphing, the graphs always intercept the xaxis at a point we now call the absolute zero. This point represents the beginning of the Kelvin scale i.e Zero K. In the Celsius scale, 0 K is equivalent to 273.5 ^{o}C. This is the coldest temperature possible. As a molecule gets colder, it’s energy and consequently, its movement and vibrations decrease in amplitude. As we keep cooling it, at a point the atom will reach a state of minimum internal energy where the atom has lost almost all its energy and is stationary.
Frequently Asked Questions FAQs
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