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Notes On Gaseous State (BSc and Integrated Standard For all Concerned Entrance Examination)

 Notes On Gaseous State      This page provides you all details about the gaseous state of BSc standard. Concerned problems will be solved at the end of each concept. Continue till the end to find the easiest explanation of every concept you need in this regard . Postulates of Kinetic Theory of Gases: 1. All gas consist of a very large number of minute particles, called molecules.  2. The gas molecules are extremely small in size and are separated by large distance. The actual volume of the gas molecules is thus negligible as compared to the total volume occupied by the gas.  3. The pressure exerted by the gas is due to the bombardment of the molecules on the walls of the vessel.  4. The gas molecules collide with one another and also with the walls of the vessels. These collisions are perfectly elastic and there is no loss of energy during these collisions.  5. The distance between the gas molecules are very large. Thus, there is no effective force of attraction or r

Some extra knowledge on chemical thermodynamics

 Know more about chemical thermodynamics

In this page you will find some extra knowledge on chemical thermodynamics. If you are a true learner you should be here. Find your required concept by scrolling down. The list includes:

1. Transfer of thermal and mechanical energy

2. Interesting example on Joule's Law

3. Exact and inexact differential with respect to state and path function.

4. Molecular interpretation of internal energy, relation with degrees of freedom and equipartition principle.

1. Transfer of thermal and mechanical energy: By molecular interpretation, heating is the transfer of energy that makes use of the disorderly molecular motion in the surroundings. This disorderly motion of molecules is called thermal motion. If the system heats its surroundings, molecules of the system enhance the thermal motion of the molecules in the surroundings. But transfer of mechanical energy (as work) from the system to the surroundings, increases the orderly motion of the molecules of the surroundings. Later on the topic you will come to know about entropy and disorderly (random) motion and the relation of it with heat.

2. Interesting example on Joule's Law: When water falls form a height towards the bottom, its potential energy is converted into heat and thus temperature is raised slightly. (For understanding this knowledge of heat capacity and specific heat capacity is required.)

    If "m g of water at a temperature T1 falls from a height 'h', and if it’s final temperature is T2, 

then its potential energy = m . g. h.

Energy and work are almost equivalent, Thus P.E = Energy

Hence m . g. h = w = JQ = J (m . s . ΔT) = J m . s . (T2 - T1) 

where s = specific heat

Knowing the value of all other terms we can calculate final temperature.

3. Exact and inexact differential with respect to state and path function: If we look into the following two integrals, we find the exact and inexactness of the differentials.


The differential dU is an exact differential but not the dq. Exact differentials can be integrated between the appropriate limits. But this can not be done with an inexact differential.

State functions are exact differentials.

To know details about the mathematical background of exact differential Click here

4. Molecular interpretation of internal energy: The total energy stored inside a system is called the internal energy. This in turn is equal to the sum total of the kinetic energy (Due to motion of particles) and potential energy (due to position or placement of the atoms in the molecules).

For Ideal gases, since there is no attractive force between molecules, the positions of the molecules with respect to each other is immaterial. Thus potential energy of the ideal gas is considered to be zero. For this reason the total internal energy of an ideal gas is equal to the kinetic energy. K.E = U

Now let us consider monatomic molecules like inert gases (He, Ne, Ar, etc.). Any monoatomic molecule has equal probability in moving linearly along any one of the three perpendicular axes (X, Y and X axes).


This is called the translational motion. Total energy of a monatomic molecule can be written equal to the sum total of the kinetic energies along these there axes as:

Etotal = ( ½)mvx2 + ( ½)mvy2  + ( ½)mvz2

Thus three different terms (called squared term, e.g., ( ½)mvx2) are needed to express the total kinetic energy of the molecule and we consider each (squared) term as a degrees of freedom.

The number of independent (squared) term in the expression of kinetic energy of the molecule is called the degrees of freedom.

Thus monoatomic molecules have 3 degrees of freedom (Dof or simply f).

While expressing the DOF of monoatomic molecule we have not considered the rotational motion. This is because, when a monoatomic molecule rotates around its own axis, the radius of rotation is negligibly small. Or in simple words there is no change in the position of the atom when it rotates around its own axis; hence it has no rotational degrees of freedom.

{{{{  From the physics point of view, the moment of inertia = I = mr2 , this moment of inertia is very small due to small radius for monoatomic molecule. The rotational kinetic energy =( ½) I w2 . Thus rotational kinetic energy is also negligibly small as  ‘’ I ‘’ is very small. Thus monoatomic molecule has no rotational degrees of freedom. }}}}

A monoatomic molecule has no vibrational motion. Accordingly it has no vibrational degrees of freedom. You will get more idea about the concept of vibration in case of diatomic molecule. A monoatomic molecule can independently move along three axes only.

Thus DOF can otherwise be defined as the number of independent ways in which a molecule of gas can move. The term ‘move’ refers to linear, rotational and vibrational motion of the molecule.

Now consider diatomic molecules like O2 , N2, H2, etc. Any molecule can have three translational degrees of freedom as already discussed.

Now if we will discuss about rotational degrees of freedom. Then we can see the molecule changes its position when rotates thorough any two of the perpendicular axes except the one passing through both the nucleus (Called molecular axis). Thus a diatomic molecule has two rtaional degrees of freedom.


A diatomic molecule can vibrate along its bond, and hence it has one vibrational degrees of freedom.

But it should be noted that vibrational degrees of freedom becomes active at very high temperature around 3000 K. Hence at room temperature we don’t consider the vibrational degrees of freedom.

Thus the total degrees of freedom of a diatomic molecule = 3 tanslational DOF + 2 rotational degrees of freedom => f = 5

In a similar way we can consider for triatomic molecule also.

For any molecule having N number of atoms the total number of degrees of freedom = 3N

For a linear molecule, there should be 3 tanslational DOF + 2 rotational degrees of freedom. Hence the vibrational degrees of freedom = 3N – 5

For a nonlinear molecule, there should be 3 tanslational DOF + 3 rotational degrees of freedom. Hence the vibrational degrees of freedom = 3N - 6

Law of equipartition principle: In thermal equilibrium, the total energy of a molecule is shared equally in all DOFs.

In other words each term in the kinetic energy expression are equal or energy in each DOF is same.

It can be shown that the average kinetic energy associated with each degrees of freedom of a molecule is (1/2) kT. Where k is the Boltzmann constant and T is in Kelvin.  K = R / N , where R = universal gas constant, N = Avogadro’s number

k = 1.380649 × 10-23 m2 kg s-2 K-1

Now for a monoatomic molecule, there are three degrees of freedom (f =3). Thus the total kinetic energy of a molecule = internal energy = U = (1/2) kT + (1/2) kT + (1/2) kT = (3/2) kT = (f / 2) kT

Energy of one mole of monoatomic molecules = N (3/2) kT = (3/2) RT = (f / 2) RT.  Here we can look back to the kinetic gas equation from which derived, the kinetic energy of one mole of gas =  (3/2) RT

Energy of n mole of monoatomic molecules = n (f / 2) RT 

Similarly for one diatomic molecules The internal energy = U = (f / 2) kT = (5 /2) kT  (Since at room temperature a diatomic molecule has 5 degrees of freedom)

Problem: Find kinetic energy of 1 molecule of Ne at 300 k.

Solution: Ne is a monoatomic molecule having f = 3.

K.E = (f / 2) k T = (3/2) k T = (3/2) 1.380649 × 10-23 . 300  

 Problem: Find K.E of 3 moles of N2 at 300 K.

Solution: N2  is a diatomic molecule having f = 5

I.E =K.E = U = (f / 2) R T = (5 /2) 8.314 x 300 

Again  since Internal energy of n mole of gas = U = n (f / 2) RT

Change in internal energy = U2 – U1 = ΔU = n (f / 2) R ΔT

It should be noted that, for isothermal process, temperature is kept constant. Thus ΔT=0 and hence ΔU=0


To be continued .....

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