1) State the third law of thermodynamics. Give its limitations and importance.
The third law of thermodynamics states that:
“The entropy of a perfect crystal of each element and a compound is zero at
absolute zero.”
Limitations: If any disorder like impurity or imperfection is found in a
substance then the entropy of such crystal is non-zero at 0 K. For example: The
entropy of pure carbon dioxide and nitric oxide is zero at 0K. This shows that
there exists disorder in the arrangement of such molecules.
This law is applicable only to pure compounds. Thus we can say that, this law is
not applicable to glass which is a supercooled liquid. It is also not
applicable to amorphous substance and supercooled solutions.
Importance:
a. With the help of this law Thermodynamic properties can be calculated and
chemical affinity can be measured.
b. This law helps in explaining the behaviour of solids at very low
temperature.
Atomic structure
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2) Explain the laws of Thermodynamics.
a. Zeroth law: If any two systems are in thermal equilibrium with the third
system, then they are also in thermal equilibrium with each other.
b. First law: First law of thermodynamic states that energy can neither be
created nor be destroyed but it can only be converted from one form to another.
c. Second law: This law states that “all processes in nature tend to occur with
an increase in entropy and the direction of change always lead to the increase
in entropy.”
d. Third law: This law states that “The entropy of a perfect crystal of each
element and a compound is zero at absolute zero.”
3) Explain the following terms: Isolated system, Open system and closed system
and give example where ever possible.
a. Isolated system: A system that can neither exchange matter nor heat with the
surrounding is known as an isolated system. For example: Water placed in a
vessel that is closed as well as insulated.
b. Open system: A system that can exchange both matter and energy with the
surrounding is said to be an open system. For example: A reaction taking place
in an open vessel exchanges both energy and matter with the surrounding.
c. Closed system: A system that exchanges only energy and not matter with the
surrounding is said to be a closed system. For example: A reaction taking place
in a closed metallic vessel.
4) Give the relation between heat of reaction at constant pressure and that at
constant volume.
We know that
Qp = ΔH and qv = ΔE
At constant pressure
ΔH = ΔE +
PΔV
----------------------------i
Where ΔV is the change in volume, thus above equation can be written as
ΔH = ΔE + P (V2 – V1)
=ΔE + (PV2 – PV1)
---------------------------ii
Where V1 is the initial volume and V2 is the final volume
of the system
For ideal gases:
PV = nRT
So we have
PV1 = n1RT
PV2 = n2RT
Here n1 is the number of moles of the gaseous reactants and n2
is the number of moles of the gaseous products.
Substituting these in equation ii, we get
ΔH = ΔE + (n2RT – n1RT)
=ΔE + (n2 – n1) RT
ΔH = ΔE + Δng RT
Where Δng = n2 – n1 is the difference between
the number of moles of the gaseous products and those of the gaseous reactants.
Substituting the value of ΔE and ΔH the above equation becomes
qp = qv + Δng RT
The above equation gives us the relationship between heat of reaction at
constant pressure and that at constant volume.
5) Write a short note on Gibbs Free Energy and derive the equation for the
same.
This thermodynamic quantity states that the decrease in value during a process
is equal to the useful work done by the system. It is denoted by G and the
mathematical equation is:
G = H – TS
Where,
H = heat content
T = absolute temperature
S = entropy of the system
For isothermal process we have
G1 = H1 – TS1 for the initial state
G2 = H2 – TS2 for final stage
Therefore,
G2 – G1 = (H2 – H1) – T(S2
– S1)
Now,
ΔG = G2 – G1 is the change in Gibbs free energy
ΔH = H2 – H1 is the change in enthalpy of the system
ΔS = S2 – S1 Is the change in entropy of the system
Thus the above equation becomes:
ΔG = ΔH – TΔS is known as Gibbs-Helmoholtz equation.
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