Some extra knowledge on chemical thermodynamics
Know more about chemical thermodynamics
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.
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).
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
To be continued .....