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 repulsion between them. 

6. The gas molecules move freely in all direction in straight lines. Their speed and direction change when these collide with one another or with the walls of the vessels. 

7. The average kinetic energy of the gas molecules is directly proportional to the absolute temperature of the gas. 

8. There is no effect of gravity on gas molecules.

Kinetic Gas Equation: A mathematical equation relating pressure, volume, mass and root mean square velocity, which is derived on the basis of kinetic molecular theory of gases is said to be kinetic gas equation . This equation is: 

PV = 1/3 (m N C_rms²), where m = mass of one gas molecule, N = 1 mole or Avogadro’s No. = 6.02 x 10²³, C_rms = root mean square velocity.  P and V are the pressure and volume respectively.

Derivation: Let us consider a cubical vessel of which each side is l cm. It contains one mole of gas which are moving in all the directions (with reference to x, y, z axis).

Consider a gas molecule inside the vessel which has mass ‘m’ moving with velocity C_x cm/s along the x direction. So that it’s momentum before collision on the wall of the vessel = mC_x 

  While after collision on the face A it returns back with the same velocity but in opposite direction. Momentum becomes = -mC_x

Thus change in momentum in one collision = mC_x – (-mC_x) = 2mC_x ----- eq. 1

No of collision per second ∝ C_x

No of collision per second ∝ 1 / 2l

No of collision per second = C_x/2l

=> As per eq. 1 rate of change in momentum, by the molecule in C_x/ 2l number of collision

= 2m C_x   (C_x / 2l)

= m C_x ² / l 

Similarly, on face B rate of change of momentum = m C_x ²/l

Along  x-axis = m C_x ²/l + m C_x ²/l  = 2 m C_x ²/l

Similarly along  y-axis  and  z-axis

Rate of change of momentum = 2m C_y ²/l   &   2m C_Z²/l

Total rate of change of momentum of the molecule 

=  2 m C_x ²/l + 2m C_y ²/l  + 2m C_Z ²/l

=  2m/l (C_x ² + C_y ² + C_Z ²)

 =  (2m/l) C²

    Where C² = C_x ² + C_y ² + C_Z ²
    If C₁ ,C₂, C₃, ………………….., C_n  are the velocities of the gas molecules , then total rate of change of momentum of all the molecules

= (2m/l) C₁² + (2m/l) C₂² + (2m/l) C_n²
= (2m/l) (C₁² + C₂² + ……+ C_n² )

Multiplying and dividing N
= (2mN/l)  x  {(C₁² + C₂² + …+ C_n² )/N}

= (2mN/l)  x C_rms² = Force

Where C_rms = Root mean square velocity

= {(C₁² + C₂² + …+ C_n² )/N} ^½ 

Area of cube = 6l²
Pressure (P) = F/A 

= {(2mN/l) x C_rms²} / 6l 

= 2mN C_rms²/6l³                     

= 2mN C_rms²/6V 

= 1/3 (mN C_rms²/V)  (Since l³= volume = V)  
=>  PV = 1/3 (mN C_rms ²)
The above equation is the derived kinetic gas equation

Relationship between Kinetic Energy and Temperature: 

According to kinetic gas equation  

PV = 1/3 mNc²

= 2/3  x 1/2 mNc²     

=   2/3 x ½  Mc²                                           

=>  PV = 2/3 K.E                                  

=> nRT = 2/3 K.E                                       

=> K.E = 3/2 n RT

For a given mass of gas the number of moles n is a constant.

Thus K.E directly proportional to T

Or    K.E   = K’T  (Where K' = proportionality constant)

Derivation of Gas Laws:

1. Boyles law: According to kinetic gas equation  

PV = 1/3 mNc²                                             

=  2/3  x 1/2 mNc²                                             

=   2/3 x ½  Mc²                                     

PV = 2/3 K.E       …………(2)
As it is known that    K.E directly proportional to T                

Or    K.E   = K’T
Hence ,        

PV = 2/3 K’T   

= 2/3  X Constant (Since T is kept constant in Boyle's law)                

PV = CONSTANT
Boyle’s law derived

2. Charles law : According to kinetic gas equation

PV = 1/3 mNc²
= 2/3 X 1/2 mNc² 

= 2/3  X 1/2 Mc²
PV = 2/3 X K.E                         …………. (2)

As it is known that K.E directly proportional to T
Hence ,  PV =2/3 K’T
Or V/T  = 2/3 K’/P
Or V/T = 2/3  X Constant (P is kept constant in Charle's law)
Or V/T = Constant
Charle’s law explained

3. Avogadro's Law : According to Avogadro’s law under the similar condition of temperature and pressure , equal volume of all the gases contain equal of molecules.

Mathematically we have to prove that,      N₁ = N₂                
According to kinetic gas equation
PV = 1/3 mNc²             

= 2/3  x 1/3 mNc²
Now applying this equation for gas 1            

P₁V₁  = 2/3  X 1/2 m₁N₁c²₁   ….(1)

Similarly for gas 2            

P₂V₂  = 2/3  X 1/2 m₂N₂C₂²   ….(2)

Since for both the gases , P and V are remaining same , so that                                          

P₁V₁ = P₂V₂
Hence from eq 1 and 2,  

2/3  X 1/2 m₁N₁c²₁= 2/3  X 1/2 m₂N₂C₂²    ……….(3)
Again for both the gases temperature remains constant
     (K.E)₁     =    (K.E)₂
Hence    ½ M₁c₁²   =  ½ M₂c₂²   ……..(4)
Now dividing equation (3) by (4)  we get, N₁ = N₂   
Avogadro's Law derived. 
4. Dalton's Law of Partial Pressure: According to Dalton’s , under the conditions of gases which don’t react chemically are taken in a container then pressure of the gas mixture becomes equal to that of sum of the partial pressure.

 Mathematically we have to prove that,   P = P_1 + P₂           …… (1)
Where    P₁   and    P₂  are the partial pressure of the gas 1 an 2 respectively

According to kinetic gas equation
P_totalV = 1/3 mNc²    

=  2/3  x 1/2 mNc²                                                

=   2/3 x ½  Mc                                             

 =  2/3  (K.E)_total                       

Or  3/2 P_totalV =  (K.E)_total   ..... eq. 2       
Now applying the above equation (2) into gas 1 and 2
We have
(K.E)₁   = 3/2 P₁V
 And     (K.E)₂   = 3/2 P₂V                        ………(3)
Since (K.E)_total   = (K.E)₁ + (K.E)₂
 => 3/2 PV   = 3/2 P₁V   + 3/2 P₂V
=> 3/2 PV   = 3/2 V  (P₁+P₂)      
=> P =   P₁+P₂    …….(4)      
Thus Dalton's law partial pressure explained.              
 5. Grahm's Law of Diffusion: According to kinetic gas equation,   

PV = 1/3 mNc²   = 1/3 Mc²

    Here c is the velocity of the gas which can be considered to be the rate of diffusion of the gas.
=> c²  = 3PV/M = 3RT/M (Since, PV = RT for one mole of gas)
=> C = rate of diffusion = (3RT/M)^1/2

    Thus rate of diffusion of a gas is inversely proportional to the square root of its molecular mass. Graham's law of diffusion is thus is proved.

Maxwell Distribution of Molecular Velocities:

   According to kinetic theory of gases, molecules of a gas inside a vessel move with different velocities. Also, as the molecules collide redistribution of molecular velocities takes  place. But at a particular instant, a fraction of total number of molecules will have same particular velocity.

    Maxwell studied the distribution of molecular velocities applying laws of probability. He then deduced an empirical equation relating velocities with the fraction of molecules moving with those velocities.

    f(c) = dNc/N = 4π (m/2πkT)^3/2 C² exp (-mC²/2kT)    .... eq. 1

    Where dNc = Number of molecules moving with velocities between C and C+dC

Maxwell Distribution Curve for Molecular Velocities:

    When we plot a graph taking fractions of molecules versus concerned velocities of these fractions, we obtain a graph known as Maxwell distribution curve for molecular velocities at increasing temperatures T1 < T2 < T3.

    The following observations were made:

1. No molecules are at rest (moving with zero velocity).

2. As the velocity increases from zero, the fractions of molecules moving with those velocities also increases.

3. This increasing trend continues till it reaches a maximum and then start decreasing.

4. The velocity corresponding to the largest fraction of the molecules is called the most probable velocity. Molecules find this velocity easy to move with. its value at a given temperature depends on the volume of the gas.

4. After most probable velocity, the fraction of molecules moving with higher velocities start decreasing.

5. Fractions of molecules moving with very low and very high velocities are very small.

6. As the temperature increases (T1 < T2 < T3), it is observed that the entire distribution curve shifts to the right. It indicates that, with the rise in temperature, the most probable velocity increases and so also the fractions of molecules moving with higher velocities. This is evident from the equation provided by Maxwell. The factor exp (-MC²/2RT) of eq. 1 indicates the same. The exponent has a negative sign and the temperature T is in the denominator. Thus, as the temperature increases the dN/N (fraction of molecules) increases for higher velocities.

7. As the molecular mass of a gas decreases, the distribution curve flattens and shifts rightward (following similar logic as above). Thus decreases in molecular mass increases the fraction of molecules with higher velocities.

You can also find other notes on our website such as atomic structure, periodic table chemical bonding,  ionic equilibrium etc.

Maxwell distribution of molecular kinetic energy:

Kinetic energy = E = ½ mC² (when velocity is C)

C² = 2E/m or C = (2E/m)^1/2

Differentiating both sides,

2CdC = (2/m)dE

CdC = dE/m

C²dC = C (dE/m) = (2E/m)^1/2 (dE/m) = {(2E)^1/2/m^3/2} dE

Substituting for C²dC and  ½ mC²  in eq. 1, we have,

dN/N = 4π (m/2πkT)^3/2 {(2E)^1/2/m^3/2} dE exp (-E/kT)

dN/N = [(2E^1/2) / {( π^1/2).(kT)^3/2] exp (-E/kT) dE     ........... eq. 2

Eq. 2 is called the Maxwell distribution of molecular kinetic energy and gives the fraction of molecules having kinetic energies in the range E and E+dE.

Most Probable Velocity (α or M.P.V or C_mpv):

    As has been previously mentioned the velocity with which most of the molecules move in a container at a particular temperature is called the most probable velocity.

    We can see from the Maxwell distribution curve that the most probable velocity is the peak of the curve where the value of the derivative must be zero. Accordingly the derivative of dNc with respect to dc (from eq. 1) must be equal to zero.

Information: The above most probable velocity pe jir molecule is (2kT/M)^1/2 but we replace k by R throughout the derivation we find the most probable velocity as (2RT/M)^1/2 per mole.

Average velocity (C_avg):
   The average velocity or mean speed is the speed of all the particles divided by the total number of particles. The number of particles moving within velocities C and C+dc is given by dNc. Thus when we multiply dNc with C and integrate the product from zero to infinity it gives the sum total of the speeds of all of the particles. Then by dividing total particles we can have the mean and thus average velocity is obtained.

Root Mean Square Velocity:

       This is obtained when we calculate for the root of the mean of the square of the velocities of all the particles. In a similar way, by integrating the product of C² and dNc within a limit 0 to infinity we can find the sum total of the squares of all the velocities of all of the particles. Then by dividing total particles we can have the mean, and the root of this mean gives the root mean square velocity.

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Collision Parameters:

Collision Diameter: At the moment when two molecules collide, that becomes their closest approach. The distance between the centres of two molecules (particles) when they are at their closest approach is called the collision diameter (d).

    Collision Number (Z1): The number of collisions made by a single molecule  with other molecules per second per unit volume of the gas is called the collision number.

    Consider a molecule moving with average speed Cavg with respect to the other molecules hypothetically imagined to be at rest. Every second the molecule moves, the following cylindrical shape of length Cavg (distance covered by one molecule in one second) and radius d will be made. Whenever the centre of another molecule comes within this cylinder, a collision occurs.

    Volume of the cylinder = πd²C_avg   (since V = πr²h) .......... eq. 1

  Number density of molecules inside the cylinder is defined as the number of molecules per unit volume = n = N/V

    The actual fact is that all the molecules are in motion. Hence the collision number should be determined by the velocity of the molecules relative to each other.

    Now since the collision is equally possible at all angles (0 to 180 degree), the average angle of collision may be taken as 90 degree.

    Thus, the average relative velocity of two molecules = Crel

           = (C_avg + C_avg)^1/2 = 2^1/2 C_avg

    By replacing Cavg by Crel in equation 1, we have,

    Volume of the cylinder, V = πd²Crel

    Thus collision number = Z1 = number of molecules per unit volume in the cylinder X  volume of the collision cylinder = n X πd²Crel = n πd² 2^1/2 Cavg  ........... eq 2 

Collision Frequency (Z11): The total number of collisions made by the molecules per second per unit volume of the gas is called the  collision frequency.

    Each molecule in unit volume collides with the rest molecules present in that particular volume. Thus the collision frequency is expected to be n times the collision number. But since a collision involves two molecules, the number of collision for one molecule is 1/2. In this regard, the collision frequency = Z11 = 1/2 X n X Collision number = 1/2 X n X Z1 = 1/2 n n πd² 2^1/2 Cavg = (1/2^1/2) n²  πd²  Cavg .......... eq 3

The Z11 here represents the collision between molecules of same gas.

If we write Z12, it represents collision of molecule of type 1 with that of type 2. This time the collision frequency will depend on the number densities of both molecules ρ1 and ρ2. The square of number density in eq 2 must be replaced by the product of number densities of gas 1 and 2.

Z12 = (1/2^1/2) ρ1 ρ2  πd²  Cavg ..... eq 4

We know PV = n RT = n NA K T = NKT ........ eq 5 (Since R = NK = Avogadro's number X Boltzmann constant and n X NA = N = total number of molecules)

From eq 5,

=> N/V = n = ρ = P/KT   ........ eq 6

Now we can express the collision number and collision frequency of gases in terms of pressure and Boltzmann constant as (refer eq 2),

Z1 = (P/KT) πd² 2^1/2 C_{avg  ........ eq 7}

Z11 = (1/2^1/2) (P/KT)² πd²  Cavg .......... eq 

Mean Free Path (λ): 
    The average distance traversed or covered by a molecule between two successive collisions with other molecules is called the mean free path.

    The value of mean free path is not the same for all the molecules as the molecules are not equally spaced and also the velocity and direction of the molecules change as they collide. Hence an average distance is considered. If d1, d2, d3, ...... dn are the distances covered by a molecule in successive collisions then the mean free path is expressed as,

    λ = (d1 + d2 + d3 + ....... + dn) / n

    Since Z1 is the number of collision made by a molecule per second with other molecules, therefore its inverse (Z1^-1) should give the average time between collision. The average distance covered by molecule (mean free path) in this interval is = Cavg . Z1^-1

Thus λ = Cavg/ Z1 = Cavg / n πd² 2^1/2 C_{avg =} 1 / n π  d² 2^1/2

Since n = N/V

=>  λ = V / N π  d² 2^1/2

Degrees of freedom:

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:

E_total = ( ½)mv_x² + ( ½)mv_y²  + ( ½)mv_z²

Thus three different terms (called squared term, e.g., ( ½)mv_x²) 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

Behaviour of real gases:

    The kinetic theory gases remind us that:

1. Actual volume of all gas molecules is negligible as compared to the volume of the container in which they are kept.

2. There is no force of interaction between gas molecules.

    In reality there are no gases for which the above two points are valid. The hypothetical gases for which the above two points are valid and which obey the gas laws under all conditions of temperature and pressure are called the Ideal Gasses.

     Real gases like the gases of the atmosphere (CO2, H2 etc) certainly occupy some volume of the container and have either attractive or repulsive forces between gas molecules. These gases do not obey the gas laws strictly under all set of temperature and pressure. This is the reason why real gases deviate from the ideal behaviour.

Compressibility factor and demonstration of deviation of real gases from ideal behaviour:

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Expected Long Questions from BSc (First Semester Physical)

Gaseous State

1. What are the two faulty statements of kinetic theory of gasses that cant be applied to real gasses?

Derive an expression for the relation between pressure, volume, mass of gas and root mean square velocity.

2. Derive the value of most probable velocity from Maxwell distribution of molecular velocity equation.

Show a graph for the position of most probable  velocity , root mean square velocity  and average velocity.

3. Derive a relationship between collision number , and collisiion frequency for a homonuclear bimolecular collision.

4. Discuss degrees of freedom.

5. What is equipartition principle?

6.How can you show the deviation of real gasses from ideal behaviour?

7. Derive the Vander Waal's equation of state.

8. How can you calculate the Vander Waal's constant if Tc and Pc are known?

What is law of corresponding principle?

Ionic Equilibrium

1. What is Buffer? What is the mechanism of Basic buffer? Derive Henderson's equation for basic buffer.

2. What is hydrolysis of salt? Discuss and derive for the pH of salt (of strong acid and weak base) hydrolysis.