This document discusses various thermodynamic concepts including entropy, enthalpy, Gibbs free energy, heat capacity, and their relationships with temperature. It provides equations relating these concepts and explains how to calculate slopes from plots of these variables. Specifically, it states that the slope of a temperature vs entropy graph equals the system's temperature. It also discusses how phase changes occur based on the relative free energy of different phases and how entropy affects phase stability at different temperatures.

1. THERMODYNAMICSTHERMODYNAMICSTHERMODYNAMICSTHERMODYNAMICS
Department:
METALLURGY & MATERIAL ENGINEERING
Veer surendra sai university of technology
Burla,odisha-768018
Department:
METALLURGY & MATERIAL ENGINEERING
Veer surendra sai university of technology
Burla,odisha-768018

2. TOPIC :TOPIC :
Plotting of different parameters entropy,
enthalpy, Gibbs free energy, heat capacity,
and slope calculation.

3. entropy

4. Entropy-Temperature Graph

5. Internal energy vs entropy
dU=TdS-PdV
At constant volume,
dV=0
dU=TdS
dU/dS=T
so slope=(dU/dS)V=T

6. ENTHALPY

7. Pressure Vs Enthalpy
dH=TdS+VdP
At constant entropy,
dS=0
dH=VdP
dp/dH= 1/V
so, slope =(dP/dH)s=1/V

8. FREE ENERGY

9. ∆∆GGoo
= ∆H= ∆Hoo
- T∆S- T∆Soo
GibbsGibbs free energyfree energy change =change =
total energy change for systemtotal energy change for system
- energy lost in disordering the system- energy lost in disordering the system
If reaction isIf reaction is
•• exothermic (negative H∆exothermic (negative H∆ oo
)) (energy dispersed)(energy dispersed)
•• and entropy increases (positive S∆and entropy increases (positive S∆ oo
))
(matter dispersed)(matter dispersed)
•• thenthen G∆G∆ oo
must bemust be NEGATIVENEGATIVE
• reaction is spontaneous (and product-
favored).

10. ∆∆GGoo
= ∆H= ∆Hoo
- T∆S- T∆Soo
Gibbs free energy change =Gibbs free energy change =
total energy change for systemtotal energy change for system
- energy lost in disordering the system- energy lost in disordering the system
If reaction isIf reaction is
•• endothermic (positive H∆endothermic (positive H∆ oo
))
•• and entropy decreases (negative S∆and entropy decreases (negative S∆ oo
))
•• thenthen G∆G∆ oo
must bemust be POSITIVEPOSITIVE
•• reaction isreaction is not spontaneousnot spontaneous (and is(and is reactant-reactant-
favoredfavored).).

11. Gibbs free energy change with reaction

12. Gibbs energy versus Temperature
As we know that
G=H-TS
dG=dH-TdS (1)
Compairing equation 1
with the general equation of
striaght line i.e. Y=mX+C
then
Y=dG ,X=T, m= -dS and
C=dH
so, slope =(dG/T)=-dS (2)

13. Gibbs energy vs Pressure
dG=-SdT+VdP
At constant temperature,
dT=0
so, dG=VdP
dG/dP=V
so slope=(dG/dP)T = V

14. Helmholtz energy vs Temperature
dA=-SdT-PdV
At constant volume,
dV=0
(dA/dT)v = -S
Slope = (dA/dT)v = -S

15. Heat Capacity, C
T
q
turein temperaincrease
absorbedheat
C
∆
==
“C” is an extensive property; so a large object has a larger
heat capacity than a small object made of the same material.
Using the Equation:
Looking at the figures on the
left, it can be seen that the
temperature change is constant,
but the heat absorbed by the
larger object is greater.
This results in a larger heat
capacity for the larger object
because more heat is absorbed.

16. Specific heat capacity: The energy (joules) required to
raise the temperature of 1 gram of substance by 1°C
Unit: J g-1
K-1
or J g-1
°C-1
Molar heat capacity: The energy (joules) required to
raise the temperature of 1 mol of substance by 1°C
Unit: J mol-1
K-1
or J mol-1
°C-1
m
C
Cs =
n
C
Cm =

17. Heat capacity Vs Temperature

20. Variation of G,H and S with T
Typical variation of
thermodynamic parameters are
shown in the figure.
From the definition, we know that
the slope of the enthalpy at any
temperature is equal to the specific
heat at that temperature.
As ,Cp=H/T
dH/dT=Cp

21. G=H-TS
dG=dH-TdS-SdT
dG=dE+PdV+VdP-TdS-SdT since H=E+PV
dG=dQ-PdV+PdV+VdP-TdS-SdT since dE=dQ-PdV
dG=TdS+VdP-TdS-SdT since dQ/T=dS
dG=VdP-SdT
So at a constant pressure the slope of the free
energy curve(dG/dT) is –S.

23. Superscript, S and L denote the solid and liquid
phases

24. Free energy for liquid phase changes more
drastically compared to the free energy of solid.
Because entropy of the liquid phase is always
higher than the solid phase,which is the slope
of the free energy.
 At melting point, free energy for both the
phases are the same and the difference
between the enthalpy of these two phases is
equal to the latent heat of fusion L.

25. On the other hand, at higher temperature range, phase having
higher entropy will be stable since in this range “TS” term will
dominate. That is why liquid phase is stable at high
temperature range.
It must be apparent that one particular phase at certain
temperature range will be stable, if the free energy, G is lower
of that phase than the other phase.
 At low temperature range, one particular phase will be stable,
which has low enthalpy, since “TS” term will not dominate.
That is why solid phase is stable at low tempareture range.

26. THANK YOU
 Submitted by
Sidhant Barik– 15010841
Soumya Ranjan Nayak- 15010842
Soumya Ranjan Sahoo- 15010843