1. THERMODYNAMICS Thermodynamics:
First Law of Thermodynamics:
energy can neither be created nor destroyed but
can be converted from one form to another.
Mathematical Form of First Law of
Thermodynamics:
Consider a system having energy E1 .Let q be the amount of
heat added to the system and w is the work done by the
system.
Total energy E2=E1+q+w
E2-E1=q+w
ΔE= q+w

2. THERMODYNAMICS If work done is the work of expansion. w= -PΔV
ΔE= q- PΔV
q= ΔE+ PΔV
For small change,
Enthalpy Or Heat Content:
Under condition of constant pressure
qp= ΔE+ PΔV……………i
δq=dE +PdV

3. THERMODYNAMICS Suppose when system absorbs q calorie of heat its
internal changes fromE1 to E2 and volume increases
from V1 toV2
ΔE= E2-E1………………ii
ΔV= V2-V1…………………iii
Then qp=( E2-E1)+ P(V2-V1)
qp= (E2+PV2)-(E1+PV1) …………..iv
But H=E+PV
qp=H2-H1

4. THERMODYNAMICS
qp= ΔH…………….v
Enthalpy change of a system is equal to the heat absorbed by
the system at constant pressure.
From equation (I) and (v) we get
The Enthalpy change during a process may also be defind as
the sum of the increase in internal energy of the system and
the pressure volume work done.
ΔH= ΔE+P ΔV

5. THERMODYNAMICS Heat Capacity: The amount of heat required to raise the
temperature of the system through 1 ˚C.
C=δq/dT ………….1
heat capacity are of two type:
1 Heat capacity at Constant Volume CV
According to first law of themodynamics
………………….2
…………………….3
δq=dE +PdV

6. THERMODYNAMICS dV=O…………………………4 then above equation becomes
When the volume is kept constant,
dV=O…………………………4
then above equation becomes
Thus heat capacity at constant volume is the rate change
of internal energy with temperature at constant
temperature
2 Heat capacity at Constant Pressure C p
When the pressure is kept constant during absorption of
heat, then equation (3) becomes
CP = (E/T)P + P(V/T)P………………..6
Cv= (E/T)v

7. THERMODYNAMICS H=E+PV differentiate w.r.t T at constant P
(dH/dT)P = (E/T)P + P(V/T)P. …………….7
Combining equation (6) and (7 )
Thus, CP = (H/T)p
Simply equation can be written as
CP=dH/dT

8. THERMODYNAMICS
Thus, heat capacity at constant pressure may be defined as
the rate of change enthalpy with temperature at
constant pressure
Relationship between CP and CV.
We know that μ > CV
CP-CV= dH/dT-dE/dT…………8
But H=E+PV
and PV=RT

9. THERMODYNAMICS H=E+RT Differentiate this equation w.r.t T, we get
dH/dT=dE/dT+R
dH/dT-dE/dT=R……………..9
combining equation (8) and (9) we get
CP-CV=R

10. THERMODYNAMICS Joule’s Law:
It Deals with the study of variation of internal
energy of a gas with volume at constant temperature.
The change in internal energy of an ideal gas with volume
at constant temperature is equal to zero.
Mathematically, (∂E/∂V)T =o
Proof: Taking E as a function of volume and temperature
i.e. E=f(V,T)
dE= (∂E/∂T)V dT + (∂E/∂V)T dV
But dE=0, dT=0 Hence,

11. THERMODYNAMICS (∂E/∂V)T dV=0 As dV≠ 0, It means (∂E/∂V)T =o
Which is Joule’s Law.
Joule-Thomson Effect(Adiabatic Expansion of Real Gas):
“ When a real gas is made to expand adiabatically from
region of high pressure to region of low pressure, a change
of temperature (generally cooling except for hydrogen and
helium) is produced”

12. THERMODYNAMICS Experimental Set –Up :
According to first law of thermodynamics
ΔE=q+w
Since the process is carried out adiabatically,
q=0
ΔE=w………….1

13. THERMODYNAMICS Work done on the system on the left side=P1V1
Work done by the system on the right side=-P2V2
Net work done by the system is
w= P1V1-P2V2………….2
Putting this value in equation (1). We get
ΔE = P1V1-P2V2
E2-E1= P1V1-P2V2
E2+P2V2 =E1+P1V1
H2=H1
H2-H1=0
ΔH=0
When the expansion of a gas takes place adiabatically through a porous
plug or a fine hole, the enthalpy of the system remain constant.

14. THERMODYNAMICS Joule Thomson Coefficient: μ= (∂T/∂P)
Temperature change in degrees produced by a drop of one
atmospheric pressure when the gas expands under conditions of
constant enthalpy.
1.For cooling , μ will be positive
2. For heating, μ will be negative
3. If μ=0, the gas gets neither heated up nor cooled on
adiabatically expansion
Inversion Temperature:
The temperature at a particular pressure at
which μ=0 i.e. the gas gets neither cooled nor heated up on a
adiabatic expansion and below which μ is positive and above
which μ is negative.

15. THERMODYNAMICS thermodynamic quantity:
Relationship between μ and other
thermodynamic quantity:
H= f(T,P)……………1
dH= (∂H/∂T)P dT + (∂H/∂P)T dP…………2
In Joule Thomson Effect, H remain Constant i.e
dH=0
Putting this value in above equation, we get
(∂H/∂T)P dT + (∂H/∂P)T dP=0………3
Dividing above equation throughout by dP and
keeping enthalpy constant.
(∂H/∂T)P (dT/dP)H + (∂H/∂P)T=0……………..4

16. THERMODYNAMICS But (∂H/∂T)P = CP, heat capacity at constant pressure,
and (∂T/∂P)P= μ, Joule Thomson coefficient
Putting these value in equation (4)
(∂H/∂P)T + μ CP=0
μ= -1/CP. (∂H/∂P)T
Joule Thomson Coefficient for an ideal gas:
μ=-1/CP. (∂H/∂P)T ……………..1
Putting H=E+PV in above equation
μ=-1/CP. (∂E+PV/∂P)T…………..2
=- 1/CP.[ (∂E/ ∂ P)T+ ∂ (PV)/ ∂ P)T]

17. THERMODYNAMICS =- 1/CP.[ (∂E/ ∂V. ∂V/ ∂P )T+ ∂ (PV)/ ∂ P)T]………….3
But for an ideal gas ,(∂E/ ∂ V)T=0
Therefore [ (∂E/ ∂V. ∂V/ ∂P )T=0………….4
Also for an ideal gas , PV=constant , at constant
temperature.
∂(PV)/ ∂ P)T=0…………….5
Substituting the values of equation (4) and (5) in (3)
We get
μ=0

18. THERMODYNAMICS
Isothermal Reversible Expansion of Ideal Gas: (Calculation
of w, q, ΔE and ΔH)
An isothermal process is conducted in such a manner so that
temperature remains constant during the entire operation.
Consider an ideal gas contained in cylinder fitted with
frictional piston, and placed in a thermostat. Let the
pressure of gas which is undergoing isothermal expansion by
reversible process be P. Let the external pressure be reduced
by a very small quantity dP. The gas expand in the cylinder
until the internal pressure and external pressure become
equal. Suppose the volume dV changes when gas expands.
The work done δw by the gas in reversible process is given by

19. THERMODYNAMICS δw=-(P-dP) dV………………i = -PdV+dP dV…………..ii
Product dP.dV being small can be neglected. Equation i
become
δw= -P dV…………iii
Total work done w When volume changes from V1 toV2 is
given by v2
w= …………..iv v1
For n moles PV= n RT
P=n RT/V, Substitute this value in equation (iv)

20. THERMODYNAMICS V2 V2 w= =n RT V1 V1 On integrating,
w= -n RT ln V2/V1 ……………v
w= n RT log V2/V1………..vi
Since P1V1=P2V2i.e, V2/V1=P1/P2 Equation( vi)
become
w=-n RT ln P1/P2
w= n RT log P1/P2……………..vii

21. THERMODYNAMICS Expression for ΔE: For ideal gas, energy depends upon
temperature only. Since Δ T=o for isothermal process. Hence
Expression for q:
Δ E=q +w
For isothermal Process, Δ E=o
Therefore q=-w
It shows that work is done at the expense of heat absorbed
q=2.303 n RT log V2/V1…………..viii
q= n RT log P1/P2……………..ix
Δ E=o

22. THERMODYNAMICS Expression for ΔH: ΔH= Δ(E+PV)= ΔE+ Δ(PV)
= ΔE + Δ(n RT)= ΔE +n R(ΔT)
= ΔE+n R((o)
ΔH =o

23. THERMODYNAMICS Adiabatic Reversible Expansion of Ideal Gas:
In adiabatic process δq=0
According to first law of thermodynamics
dE=δq+w…………..i
We get w=δq
If work involved is the work of expansion only, then if dV is small increase in volume and P is pressure of gas. Then
w= -P dV……………..ii
Putting this value in equation i,we get
dE= -P dV…………iii
For an ideal gas
Cv=dE /dT or dE=Cv dT…………….iv

24. THERMODYNAMICS Calculation of ΔH: We know that For finite Changes,
ΔE =Cv ΔT…………….v
Calculation of ΔH:
We know that
H=E+PV
ΔH = ΔE + Δ(PV)
= ΔE + Δ(RT)
=ΔE+R ΔT
= Cv ΔT+ R ΔT
=(Cv+R) ΔT
ΔH =Cp ΔT……………….vi

25. THERMODYNAMICS PV=RT or P=RT/V Comparing equation iii and iv we get
-P dV=Cv dT……….vii
Relationship between temperature and volume:
PV=RT or P=RT/V
Putting this value in equation vii , we get
-Cv dT= RT dV/V
Cv DT = -RT dV/V
or Cv dT/T = -RdV/V……………..viii
Let volume of gas change from V1 to V2 and temperature
change from T1 to T2.
Assuming Cv to be independent of temperature and integrating
equation viii between limits V1,V2 and T1and T2, we get

26. THERMODYNAMICS T2 V2 Cv = - T1 V1 or Cv ln T2/T1 = R ln V1/V2……….ix
We know that Cp-Cv =R
Putting this value in equation ix, we get
Cv ln T2/T1 = (Cp-Cv) ln V1/V2
Dividing throught by Cv,
ln T2/T1 = (Cp/Cv- 1) ln V1/V2
Putting Cp/Cv= γ , ratio of two heat capacities

27. THERMODYNAMICS Relation between temperature and pressure:
ln T2/T1= (γ-1) ln V1/V2
Taking natural log on both side,
T2/T1= (V1/V2) (γ-1) ………………x
T2V2 (γ-1) = T1V1 (γ-1)
Relation between temperature and pressure:
P1V1=RT1 and P2V2=RT2
From above equation, we get
V1/V2 = P2T1/P1T2
Putting this value in equation x, we get
T2/T1 = (P2T1/P1T2) (γ-1)
After solving this we get
(T2/T1) = (P1/P2) 1- γ …………………….xi
Taking γ th root on both side
T V γ-1 = Constant
TP 1- γ/ γ = constant

28. THERMODYNAMICS From Equation ix and x , we get
(V1/V2) (γ-1) = (P1/P2) 1- γ/ γ …………..xi
(V2/V1)1- γ = (P1/P2) 1- γ/ γ
V2/V1 =( P1/P2 )1/ γ
(V2/V1) γ = P1/P2
P1V1 γ = P2V2 γ
PV γ = constant