Notes On Gaseous State
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 Crms2), where m = mass of one gas molecule, N = 1 mole or Avogadro’s No. = 6.02 x 1023, Crms = 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 Cx cm/s along the x direction. So that it’s momentum before collision on the wall of the vessel = mCx
While after collision on the face A it returns back with the same velocity
but in opposite direction. Momentum becomes = -mCx
Thus change in momentum in one collision = mCx – (-mCx) = 2mCx ----- eq. 1
No of collision per second ∝ Cx
No of collision per second ∝ 1 / 2l
No of collision per second = Cx/2l
=> As per eq. 1 rate of change in momentum, by the molecule in Cx/ 2l number of collision
= 2m Cx (Cx / 2l)
= m Cx 2 / l
Similarly, on face B rate of change
of momentum = m Cx 2/l
Along x-axis = m Cx 2/l + m Cx
2/l = 2 m Cx 2/l
Similarly along y-axis and
z-axis
Rate of change of momentum = 2m Cy
2/l & 2m CZ2/l
Total rate of change of momentum of the molecule
= 2 m Cx 2/l + 2m Cy 2/l + 2m CZ 2/l
=
2m/l (Cx 2 + Cy 2 + CZ
2)
=
(2m/l) C2
If C1 ,C2, C3, ………………….., Cn are the velocities of the gas molecules , then total rate of change of momentum of all the molecules
= (2m/l) (C12 + C22 + ……+ Cn2 )
= (2mN/l) x {(C12 + C22 + …+ Cn2 )/N}
= (2mN/l) x Crms2 = Force
Where Crms = Root mean square velocity
=> PV = 1/3 (mN Crms 2)
The above equation is the derived kinetic gas equation
Relationship between Kinetic Energy and Temperature:
Derivation of Gas Laws:
1. Boyles law: According to kinetic gas equation
PV = 1/3 mNc2
As it is known that K.E directly proportional to T
Hence ,
Boyle’s law derived
= 2/3 X 1/2 mNc2
PV = 2/3 X K.E …………. (2)
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.
According to kinetic gas equation
PV = 1/3 mNc2
Now applying this equation for gas 1
Hence from eq 1 and 2,
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,
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.
Where P1 and P2 are the partial pressure of the gas 1 an 2 respectively
PtotalV = 1/3 mNc2
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
=> P = P1+P2 …….(4)
Thus Dalton's law partial pressure explained.
=> C = rate of diffusion = (3RT/M)1/2
Maxwell Distribution of Molecular Velocities:
Maxwell Distribution Curve for Molecular Velocities:
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/m3/2} dE
Substituting for C2dC
and ½ mC2 in eq. 1, we have,
dN/N = 4π (m/2πkT)3/2 {(2E)1/2/m3/2}
dE exp (-E/kT)
dN/N = [(2E1/2) / {( π1/2).(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.
Average velocity (Cavg):
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.
The average distance traversed or covered by a molecule between two successive collisions with other molecules is called the mean free path.
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).
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.
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.
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:
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