What is a Gaseous State?

It is well known that matter exists in different forms in our surroundings. There are five known states of matter, such as solids, gases, liquids, plasma and Bose-Einstein condensate. The last two are known newly in the recent days. Thus, the detailed forms of matter studied are solids, gases and liquids. The best example of a substance that is present in different states is water. It is solid ice, gaseous vapor or steam and liquid water depending on the temperature and pressure conditions. This is due to the difference in the intermolecular forces and distances. The occurrence of three different phases is due to the difference in the two major forces, the force which tends to tightly hold molecules i.e., forces of attraction and the disruptive forces obtained from the thermal energy of molecules.   

In a solid phase, the substance has a tight and rigid shape and structure as the molecules are said to be arranged in a tight regular manner, whereas in liquid phase mostly the substance takes the shape of the container as the molecules are arranged with little space and can move easily and in the gaseous phase the substance moves very freely without any shape as the molecules are very far from each other such as atmospheric air. Among these three phases, the studies dealing with gaseous phase is simplest compared to other states and is said as gaseous state. As mentioned, if the disruptive forces exceed the forces of attraction, then it leads the substance in gaseous state. 

Examples of some substances which exists in the gaseous state at normal conditions:

• Hydrogen, oxygen, nitrogen are pure gases that exist in the gaseous state.
• Argon is a noble gas or inert gas that exists in the gaseous state.
• Methane is a hydrocarbon gas that exists in the gaseous state.
• Carbon dioxide, nitrous oxide, ammonia are some other gases that exist in the gaseous state.

Characteristics Properties of Gases

The main characteristics of gases are: 

• Gases do not have a definite shape or volume and occupy the complete volume of the container. 
• Gases have great expansibility. They expand themselves to whatever extent that they fill the whole container in which they are present. 
• Gases are highly compressible. They can be compressed by applying pressure and can fill themselves in a very minimal space. 
• Gases exert pressure in all directions of the container. it can be can easily noticed in a balloon of gas. 
• Gases are highly diffusible. When a mixture of gases is present, each gas can diffuse easily within the other and form a homogeneous mixture.  

Kinetic Gas Theory

The theoretical explanation and mathematical derivations for all gas laws such as Boyle’s law, Charles law, Gay-Lussac law, Avogadro’s law etc., was postulated by Joule, Kronig, Clausius, Maxwell, Boltzmann. The hypotheses of the kinetic theory of gases are,

• All gases contain plenty of particles that are tiny with point masses termed molecules.  
• The gas molecules are very tiny and away from each other to the extent that the volume occupied by the gas molecules is insignificant compared to the complete volume of gases. 
• The gas molecules move randomly due to which collision between each other and also on the faces of the vessel or container occurs. This collision is elastic in nature i.e., no energy loss or exchange takes place. 
• The molecules are independent of each other i.e. there is no sort of force within the gas molecules. 
• The pressure of the gas obtained in the container is because of pressure exerted by the molecules on the container sides. 
• The kinetic energy of the gas molecules is proportionally related to the temperature at the absolute scale of the gas molecules. 

Based on these postulates, the kinetic gas equation is also derived and given as:

$$\text{PV = }\frac{\text{1}}{3}{\text{ m N C}}^{\text{2}}{}_{\text{RMS}}$$

where, P and V denotes pressure and volume of the gas, m denotes mass of the gas molecule, N for quantity of gas molecules in an integer and C for velocity taken by root mean square of gas molecules. The average of the squares of all molecular velocities such as C1, C2, C3 …. CN taken by square root gives root mean square velocity.

Effect of Pressure on Volume

The parameters pressure (P) and volume (V) at constant temperature (T), are related inversely to given mass (m) of gas. 

Mathematically, $\text{Vα }\frac{\text{1}}{\text{P}}$at constant T and m

Thus, at a constant temperature, $\text{PV}$is constant, which indicates Boyle’s law.

Effect of Temperature on Volume

The effect of temperature on volume for a gas is explained with Charlee’s law. Thus, according to Jacques Charles, by keeping the pressure constant, if the temperature at the absolute scale of gas is increased, its volume also increases and vice versa. 

Mathematically, $\text{V}\alpha \text{ T }$ at constant P and m

$\text{V = KT}$

So, ${\text{V}}_{\text{1}}{\text{T}}_{\text{1}}{\text{ = V}}_{\text{2}}{\text{T}}_{\text{2}}$

The absolute temperature is a scale of temperature in which $–273º\text{C}$is equal to zero and this scale of temperature is termed as Kelvin scale. Thus, the degree Celsius and Kelvin scale can be interchanged by increasing the given value of Celsius by $273$. This was identified by Lord Kelvin.

Gay-Lussac’s Law

Similar to the temperature and volume relationship, by keeping the volume constant, the gas pressure increases as temperature increases and vice versa. Mathematically, $\text{Pα T }$at constant V and m $$\text{P = KT}$$

So, ${\text{P}}_{\text{1}}{\text{T}}_{\text{1}}{\text{ = P}}_{\text{2}}{\text{T}}_{\text{2}}$

Avogadro’s Law  

The volume of a gas (V) and the number of molecules (N) present in it is given by Avogadro’s law which predicts that at unchanged gas temperature and pressure, an equal volume of gases contains an equal number of molecules in it. Mathematically, $\text{Vα N}$ at constant P and T

But, at particular temperature and pressure parameters, the quantity of molecules available in a mass of gas is proportionally related to the mole number (n). 

So, $\text{Nα n}$

$\text{Vα n}$at constant P and T

So, ${\text{V}}_{\text{1}}{\text{n}}_{\text{1}}{\text{ = V}}_{\text{2}}{\text{n}}_{\text{2}}$

At standard temperature ($273\text{ K}$) and pressure (1 atmosphere), i.e STP conditions volume is said to be $22.4\text{ L}$, the number of molecules in a mole of any atoms, ions or molecules is defined as the Avogadro’s number i.e.$6.023\times {10}^{23}$.

Ideal Gas Law 

Based on the laws of gases, all the equations are combined to derive a relationship between the parameters pressure, volume, absolute temperature for a given quantity of gas at various conditions.   

$\text{Vα }\frac{\text{1}}{\text{P}}$a particular T – Boyle’s law

$\text{V}\alpha \text{ T }$ a particular P – Charles law

$\text{Vα n}$ a particular T and P – Avogadro’s law

By combining, it will be $\text{PV α nT }$

$\text{PV = nRT}$

R is termed as gas constant or molar gas constant. The above equation is termed an ideal gas Law and it obeys well to ideal behavior gases. The value of R is taken based on the unit of other parameters.  

$\begin{array}{l}\text{R = 0}{\text{.0821 L atm K}}^{\text{-1}}{\text{ mol}}^{\text{-1}}\hfill \\ \text{ = 8}{\text{.314 × 10}}^{\text{7}}{\text{ erg K}}^{\text{-1}}{\text{ mol}}^{\text{-1}}\hfill \\ \text{= 8}{\text{.314 J K}}^{\text{-1}}{\text{ mol}}^{\text{-1}}\hfill \\ \text{ = 1}{\text{.987 cal K}}^{\text{-1}}{\text{ mol}}^{\text{-1}}\hfill \end{array}$

Solved Problems  

Calculate the pressure of a gas of 5 mol present in a closed container with 224 L at absolute temperature? 

Solution:

Given: $\text{no: of moles }\left(\text{n}\right)=5\text{ mol}$

$\text{Volume }\left(\text{V}\right)=224\text{ L}$

$\text{Temperature }\left(\text{T}\right)\text{ = 273 K }$(Absolute temperature)

$\begin{array}{l}\text{Pressure }\left(\text{P}\right)\text{ = ? }\hfill \\ \text{ R = 0}{\text{.0821 L atm K}}^{\text{-1}}{\text{ mol}}^{\text{-1}}\hfill \end{array}$

Substitute the given parameters in ideal gas equation:  

$\begin{array}{l}\text{PV = nRT}\hfill \\ \text{ P = nRT / V }\hfill \\ \text{ = 5 }\times \text{ 0}\text{.0821 }\times \text{ 273 / 224 }\hfill \\ \text{ = 0}\text{.5 atm }\hfill \end{array}$

Thus, the pressure of the gas is found to be $0.5\text{ atm}$.

Context and Applications   

This topic is significant in the professional exams for both undergraduate and graduate courses, especially for 

• B.Sc. in Chemistry 
• M.Sc. in Chemistry