THERMODYNAMICSTHERMODYNAMICS

Department:METALLURGY & MATERIAL ENGINEERING

Veer surendra sai university of technologyBurla,odisha-768018

TOPIC :TOPIC :Plotting of different parameters

entropy, enthalpy, Gibbs free energy, heat capacity, and slope calculation.

entropy

Entropy-Temperature Graph

Internal energy vs entropy

dU=TdS-PdV At constant volume, dV=0 dU=TdS dU/dS=T so slope=(dU/dS)V=T

ENTHALPY

Pressure Vs Enthalpy dH=TdS+VdP At constant entropy, dS=0 dH=VdP dp/dH= 1/V

so, slope =(dP/dH)s=1/V

FREE ENERGY

∆∆GGoo = ∆H = ∆Hoo - T∆S - T∆Soo

Gibbs Gibbs free energyfree energy change = change =total energy change for system total energy change for system - energy lost in disordering the system- energy lost in disordering the system

If reaction isIf reaction is•• exothermic (negative ∆ Hexothermic (negative ∆ Hoo) ) (energy dispersed)(energy dispersed)•• and entropy increases (positive ∆Sand entropy increases (positive ∆Soo) )

(matter dispersed)(matter dispersed)•• thenthen ∆G ∆Goo must bemust be NEGATIVE NEGATIVE• reaction is spontaneous (and product-

favored).

∆∆GGoo = ∆H = ∆Hoo - T∆S - T∆Soo

Gibbs free energy change =Gibbs free energy change =total energy change for system total energy change for system - energy lost in disordering the system- energy lost in disordering the system

If reaction isIf reaction is•• endothermic (positive ∆Hendothermic (positive ∆Hoo))•• and entropy decreases (negative ∆Sand entropy decreases (negative ∆Soo))•• then then ∆G∆Goo must be must be POSITIVEPOSITIVE•• reaction is reaction is not spontaneousnot spontaneous (and is (and is reactant-reactant-

favoredfavored).).

Gibbs free energy change with reaction

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)

Gibbs energy vs Pressure dG=-SdT+VdP At constant

temperature, dT=0 so, dG=VdP dG/dP=V so slope=(dG/dP)T =

V

Helmholtz energy vs Temperature

dA=-SdT-PdV At constant volume, dV=0 (dA/dT)v = -S Slope = (dA/dT)v = -

S

Heat Capacity, C

Tq

turein tempera increaseabsorbedheat 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.

Specific heat capacity: The energy (joules) required to raise the temperature of 1 gram of substance by 1C

Unit: J g-1K-1 or J g-1C-1

Molar heat capacity: The energy (joules) required to raise the temperature of 1 mol of substance by 1C

Unit: J mol-1 K-1 or J mol-1C-1

mCCs

nCCm

Heat capacity Vs Temperature

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

G=H-TSdG=dH-TdS-SdTdG=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.

Superscript, S and L denote the solid and liquid phases

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 differencebetween the enthalpy of these two phases is equal to the latent heat of fusion L.

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.

THANK YOU

Submitted by

Sidhant Barik – 15010841

Soumya Ranjan Nayak- 15010842

Soumya Ranjan Sahoo- 15010843