Important Thermodynamics Formulas for JEE and NEET

 Page 1


  
    
     
  
       
THERMODYNAMICS
Thermodynamic processes :
1. Isothermal process : T = constant
dT = 0
?T = 0
2. Isochoric process : V = constant
dV = 0
?V = 0
3. Isobaric process : P = constant
dP = 0
?P = 0
4. Adiabatic process : q = 0
or heat exchange with the surrounding = 0(zero)
IUPAC Sign convention about Heat and Work :
Work done on the system  = Positive
Work done by the system  = Negative
1
st
 Law of Thermodynamics
?U = (U
2 
? U
1
) = q + w
Law of equipartion of energy :
U = 
2
f
nRT (only for ideal gas)
?E  = 
2
f
 nR (?T)
where f = degrees of freedom for that gas. (Translational + Rotational)
f = 3 for monoatomic
  = 5 for diatomic or linear polyatmic
  = 6 for non - linear polyatmic
Page 2


  
    
     
  
       
THERMODYNAMICS
Thermodynamic processes :
1. Isothermal process : T = constant
dT = 0
?T = 0
2. Isochoric process : V = constant
dV = 0
?V = 0
3. Isobaric process : P = constant
dP = 0
?P = 0
4. Adiabatic process : q = 0
or heat exchange with the surrounding = 0(zero)
IUPAC Sign convention about Heat and Work :
Work done on the system  = Positive
Work done by the system  = Negative
1
st
 Law of Thermodynamics
?U = (U
2 
? U
1
) = q + w
Law of equipartion of energy :
U = 
2
f
nRT (only for ideal gas)
?E  = 
2
f
 nR (?T)
where f = degrees of freedom for that gas. (Translational + Rotational)
f = 3 for monoatomic
  = 5 for diatomic or linear polyatmic
  = 6 for non - linear polyatmic
  
Calculation of heat (q) :
Total heat capacity :
C
T
 = 
dT
dq
T
q
?
?
?
= J/ºC
Molar heat capacity :
C = 
ndT
dq
T n
q
?
?
?
= J mole
?1 
K
?1
C
P
 = 
1 ?
R
?
?
C
V
 = 
1 ?
R
?
Specific heat capacity (s) :
S = 
mdT
dq
T m
q
?
?
?
= J gm
?1
 K
?1
WORK DONE  (w)  :
Isothermal Reversible expansion/compression of an ideal gas :
W = ? nRT ln (V
f
/V
i
)
Reversible and irreversible isochoric processes.
Since dV = 0
So dW = ? P
ext
 . dV = 0.
Reversible isobaric process :
W = P (V
f
 ? V
i
)
Adiabatic reversible expansion :
?
1
2 2
V T
? ?
= 
1
1 1
V T
? ?
Reversible  Work :
W = 
1
V P V P
1 1 2 2
? ?
?
 = 
1
) T T ( nR
1 2
? ?
?
Irreversible  Work :
W = 
1
V P V P
1 1 2 2
? ?
?
 = 
1
) T T ( nR
1 2
? ?
?
= nC
v
 (T
2
 ? T
1
) = ? P
ext
 (V
2
 ? V
1
)
and use 
2
2 2
1
1 1
T
V P
T
V P
?
Free expansion?Always going to be irrerversible and since P
ext
 = 0
so dW = ? P
ext 
. dV = 0
If no. heat is supplied q  = 0
then ?E = 0 so ?T = 0.
Page 3


  
    
     
  
       
THERMODYNAMICS
Thermodynamic processes :
1. Isothermal process : T = constant
dT = 0
?T = 0
2. Isochoric process : V = constant
dV = 0
?V = 0
3. Isobaric process : P = constant
dP = 0
?P = 0
4. Adiabatic process : q = 0
or heat exchange with the surrounding = 0(zero)
IUPAC Sign convention about Heat and Work :
Work done on the system  = Positive
Work done by the system  = Negative
1
st
 Law of Thermodynamics
?U = (U
2 
? U
1
) = q + w
Law of equipartion of energy :
U = 
2
f
nRT (only for ideal gas)
?E  = 
2
f
 nR (?T)
where f = degrees of freedom for that gas. (Translational + Rotational)
f = 3 for monoatomic
  = 5 for diatomic or linear polyatmic
  = 6 for non - linear polyatmic
  
Calculation of heat (q) :
Total heat capacity :
C
T
 = 
dT
dq
T
q
?
?
?
= J/ºC
Molar heat capacity :
C = 
ndT
dq
T n
q
?
?
?
= J mole
?1 
K
?1
C
P
 = 
1 ?
R
?
?
C
V
 = 
1 ?
R
?
Specific heat capacity (s) :
S = 
mdT
dq
T m
q
?
?
?
= J gm
?1
 K
?1
WORK DONE  (w)  :
Isothermal Reversible expansion/compression of an ideal gas :
W = ? nRT ln (V
f
/V
i
)
Reversible and irreversible isochoric processes.
Since dV = 0
So dW = ? P
ext
 . dV = 0.
Reversible isobaric process :
W = P (V
f
 ? V
i
)
Adiabatic reversible expansion :
?
1
2 2
V T
? ?
= 
1
1 1
V T
? ?
Reversible  Work :
W = 
1
V P V P
1 1 2 2
? ?
?
 = 
1
) T T ( nR
1 2
? ?
?
Irreversible  Work :
W = 
1
V P V P
1 1 2 2
? ?
?
 = 
1
) T T ( nR
1 2
? ?
?
= nC
v
 (T
2
 ? T
1
) = ? P
ext
 (V
2
 ? V
1
)
and use 
2
2 2
1
1 1
T
V P
T
V P
?
Free expansion?Always going to be irrerversible and since P
ext
 = 0
so dW = ? P
ext 
. dV = 0
If no. heat is supplied q  = 0
then ?E = 0 so ?T = 0.
  
Application of Ist Law :
?U = ?Q + ?W ? ?W = ?P ?V
? ?U = ?Q ?P?V
Constant volume process
Heat given at constant volume = change in internal energy
? du = (dq)
v
du = nC
v
dT
C
v 
 =
dT
du
.
n
1
 = 
2
f
R
Constant pressure process :
H ? Enthalpy (state function and extensive property)
H = U + PV
? C
p
 ? C
v
 = R (only for ideal gas)
Second Law Of Thermodynamics :
?S 
universe
 = ?S 
system
  + ??S 
surrounding
  > 0 for a spontaneous process.
Entropy  (S) :
?S
system
 = 
?
B
A
rev
T
dq
Entropy calculation for an ideal gas undergoing a process :
State A
irr
S
irr
?
?? ?
State B
P
1
, V
1
, T
1
P
2
, V
2
, T
2
?S
system 
 = nc
v
 ln 
1
2
T
T
+ nR ln 
1
2
V
V
(only for an ideal gas)
Third Law Of Thermodynamics :
The entropy of perfect crystals of all pure elements & compounds is zero
at the absolute zero of temperature.
Gibb?s free energy ?G) :  (State function and an extensive property)
G
system
 =  H 
system
 ? TS 
system
Criteria of spontaneity :
(i) If ?G 
system 
 is (?ve) < 0 ? process is spontaneous
(ii) If ?G
 system
 is > 0 ? process is non spontaneous
(iii) If ?G
 system
    = 0 ? system is at equilibrium.
Page 4


  
    
     
  
       
THERMODYNAMICS
Thermodynamic processes :
1. Isothermal process : T = constant
dT = 0
?T = 0
2. Isochoric process : V = constant
dV = 0
?V = 0
3. Isobaric process : P = constant
dP = 0
?P = 0
4. Adiabatic process : q = 0
or heat exchange with the surrounding = 0(zero)
IUPAC Sign convention about Heat and Work :
Work done on the system  = Positive
Work done by the system  = Negative
1
st
 Law of Thermodynamics
?U = (U
2 
? U
1
) = q + w
Law of equipartion of energy :
U = 
2
f
nRT (only for ideal gas)
?E  = 
2
f
 nR (?T)
where f = degrees of freedom for that gas. (Translational + Rotational)
f = 3 for monoatomic
  = 5 for diatomic or linear polyatmic
  = 6 for non - linear polyatmic
  
Calculation of heat (q) :
Total heat capacity :
C
T
 = 
dT
dq
T
q
?
?
?
= J/ºC
Molar heat capacity :
C = 
ndT
dq
T n
q
?
?
?
= J mole
?1 
K
?1
C
P
 = 
1 ?
R
?
?
C
V
 = 
1 ?
R
?
Specific heat capacity (s) :
S = 
mdT
dq
T m
q
?
?
?
= J gm
?1
 K
?1
WORK DONE  (w)  :
Isothermal Reversible expansion/compression of an ideal gas :
W = ? nRT ln (V
f
/V
i
)
Reversible and irreversible isochoric processes.
Since dV = 0
So dW = ? P
ext
 . dV = 0.
Reversible isobaric process :
W = P (V
f
 ? V
i
)
Adiabatic reversible expansion :
?
1
2 2
V T
? ?
= 
1
1 1
V T
? ?
Reversible  Work :
W = 
1
V P V P
1 1 2 2
? ?
?
 = 
1
) T T ( nR
1 2
? ?
?
Irreversible  Work :
W = 
1
V P V P
1 1 2 2
? ?
?
 = 
1
) T T ( nR
1 2
? ?
?
= nC
v
 (T
2
 ? T
1
) = ? P
ext
 (V
2
 ? V
1
)
and use 
2
2 2
1
1 1
T
V P
T
V P
?
Free expansion?Always going to be irrerversible and since P
ext
 = 0
so dW = ? P
ext 
. dV = 0
If no. heat is supplied q  = 0
then ?E = 0 so ?T = 0.
  
Application of Ist Law :
?U = ?Q + ?W ? ?W = ?P ?V
? ?U = ?Q ?P?V
Constant volume process
Heat given at constant volume = change in internal energy
? du = (dq)
v
du = nC
v
dT
C
v 
 =
dT
du
.
n
1
 = 
2
f
R
Constant pressure process :
H ? Enthalpy (state function and extensive property)
H = U + PV
? C
p
 ? C
v
 = R (only for ideal gas)
Second Law Of Thermodynamics :
?S 
universe
 = ?S 
system
  + ??S 
surrounding
  > 0 for a spontaneous process.
Entropy  (S) :
?S
system
 = 
?
B
A
rev
T
dq
Entropy calculation for an ideal gas undergoing a process :
State A
irr
S
irr
?
?? ?
State B
P
1
, V
1
, T
1
P
2
, V
2
, T
2
?S
system 
 = nc
v
 ln 
1
2
T
T
+ nR ln 
1
2
V
V
(only for an ideal gas)
Third Law Of Thermodynamics :
The entropy of perfect crystals of all pure elements & compounds is zero
at the absolute zero of temperature.
Gibb?s free energy ?G) :  (State function and an extensive property)
G
system
 =  H 
system
 ? TS 
system
Criteria of spontaneity :
(i) If ?G 
system 
 is (?ve) < 0 ? process is spontaneous
(ii) If ?G
 system
 is > 0 ? process is non spontaneous
(iii) If ?G
 system
    = 0 ? system is at equilibrium.
  
Physical interpretation of ?G :
??The maximum amount of non-expansional (compression) work which
can be performed.
?G = dw
non-exp
 = dH ? TdS.
Standard  Free  Energy  Change  (?Gº)  :
1. ?Gº = ?2.303 RT log
10
 K
2. At equilibrium ??G = 0.
3. The decrease in free energy (??G) is given as :
??G = W
net
 = 2.303 nRT log
10
1
2
V
V
4. ?
º
f
G for elemental state = 0
5. ?
º
f
G = 
º
products
G ?  
º
ts tan ac Re
G
Thermochemistry :
Change in standard enthalpy ?H° = 
0
2 , m
H ? 
0
1 , m
H
      = heat added at constant pressure.
      = C
P
?T.
If H
products
 > H
reactants
? Reaction should be endothermic as we have to give extra heat to reactants
to get these converted into products
and if H
products
 < H
reactants
? Reaction will be exothermic as extra heat content of reactants will be
released during the reaction.
Enthalpy change of a reaction :
?H
reaction
 = H
products
 ? H
reactants
?H°
reactions
 = H°
products
 ? H°
reactants
    = positive ? endothermic
    = negative ? exothermic
Temperature  Dependence  Of ?H : (Kirchoff's equation) :
For a constant pressure reaction
?H
2
° = ?H
1
° + ?C
P
 (T
2
 ? T
1
)
where  ?C
P
 = C
P
 (products) ? C
P
 (reactants).
For a constant volume reaction
?
? ? ? ? ? dT . C E E
V
0
1
0
2
Page 5


  
    
     
  
       
THERMODYNAMICS
Thermodynamic processes :
1. Isothermal process : T = constant
dT = 0
?T = 0
2. Isochoric process : V = constant
dV = 0
?V = 0
3. Isobaric process : P = constant
dP = 0
?P = 0
4. Adiabatic process : q = 0
or heat exchange with the surrounding = 0(zero)
IUPAC Sign convention about Heat and Work :
Work done on the system  = Positive
Work done by the system  = Negative
1
st
 Law of Thermodynamics
?U = (U
2 
? U
1
) = q + w
Law of equipartion of energy :
U = 
2
f
nRT (only for ideal gas)
?E  = 
2
f
 nR (?T)
where f = degrees of freedom for that gas. (Translational + Rotational)
f = 3 for monoatomic
  = 5 for diatomic or linear polyatmic
  = 6 for non - linear polyatmic
  
Calculation of heat (q) :
Total heat capacity :
C
T
 = 
dT
dq
T
q
?
?
?
= J/ºC
Molar heat capacity :
C = 
ndT
dq
T n
q
?
?
?
= J mole
?1 
K
?1
C
P
 = 
1 ?
R
?
?
C
V
 = 
1 ?
R
?
Specific heat capacity (s) :
S = 
mdT
dq
T m
q
?
?
?
= J gm
?1
 K
?1
WORK DONE  (w)  :
Isothermal Reversible expansion/compression of an ideal gas :
W = ? nRT ln (V
f
/V
i
)
Reversible and irreversible isochoric processes.
Since dV = 0
So dW = ? P
ext
 . dV = 0.
Reversible isobaric process :
W = P (V
f
 ? V
i
)
Adiabatic reversible expansion :
?
1
2 2
V T
? ?
= 
1
1 1
V T
? ?
Reversible  Work :
W = 
1
V P V P
1 1 2 2
? ?
?
 = 
1
) T T ( nR
1 2
? ?
?
Irreversible  Work :
W = 
1
V P V P
1 1 2 2
? ?
?
 = 
1
) T T ( nR
1 2
? ?
?
= nC
v
 (T
2
 ? T
1
) = ? P
ext
 (V
2
 ? V
1
)
and use 
2
2 2
1
1 1
T
V P
T
V P
?
Free expansion?Always going to be irrerversible and since P
ext
 = 0
so dW = ? P
ext 
. dV = 0
If no. heat is supplied q  = 0
then ?E = 0 so ?T = 0.
  
Application of Ist Law :
?U = ?Q + ?W ? ?W = ?P ?V
? ?U = ?Q ?P?V
Constant volume process
Heat given at constant volume = change in internal energy
? du = (dq)
v
du = nC
v
dT
C
v 
 =
dT
du
.
n
1
 = 
2
f
R
Constant pressure process :
H ? Enthalpy (state function and extensive property)
H = U + PV
? C
p
 ? C
v
 = R (only for ideal gas)
Second Law Of Thermodynamics :
?S 
universe
 = ?S 
system
  + ??S 
surrounding
  > 0 for a spontaneous process.
Entropy  (S) :
?S
system
 = 
?
B
A
rev
T
dq
Entropy calculation for an ideal gas undergoing a process :
State A
irr
S
irr
?
?? ?
State B
P
1
, V
1
, T
1
P
2
, V
2
, T
2
?S
system 
 = nc
v
 ln 
1
2
T
T
+ nR ln 
1
2
V
V
(only for an ideal gas)
Third Law Of Thermodynamics :
The entropy of perfect crystals of all pure elements & compounds is zero
at the absolute zero of temperature.
Gibb?s free energy ?G) :  (State function and an extensive property)
G
system
 =  H 
system
 ? TS 
system
Criteria of spontaneity :
(i) If ?G 
system 
 is (?ve) < 0 ? process is spontaneous
(ii) If ?G
 system
 is > 0 ? process is non spontaneous
(iii) If ?G
 system
    = 0 ? system is at equilibrium.
  
Physical interpretation of ?G :
??The maximum amount of non-expansional (compression) work which
can be performed.
?G = dw
non-exp
 = dH ? TdS.
Standard  Free  Energy  Change  (?Gº)  :
1. ?Gº = ?2.303 RT log
10
 K
2. At equilibrium ??G = 0.
3. The decrease in free energy (??G) is given as :
??G = W
net
 = 2.303 nRT log
10
1
2
V
V
4. ?
º
f
G for elemental state = 0
5. ?
º
f
G = 
º
products
G ?  
º
ts tan ac Re
G
Thermochemistry :
Change in standard enthalpy ?H° = 
0
2 , m
H ? 
0
1 , m
H
      = heat added at constant pressure.
      = C
P
?T.
If H
products
 > H
reactants
? Reaction should be endothermic as we have to give extra heat to reactants
to get these converted into products
and if H
products
 < H
reactants
? Reaction will be exothermic as extra heat content of reactants will be
released during the reaction.
Enthalpy change of a reaction :
?H
reaction
 = H
products
 ? H
reactants
?H°
reactions
 = H°
products
 ? H°
reactants
    = positive ? endothermic
    = negative ? exothermic
Temperature  Dependence  Of ?H : (Kirchoff's equation) :
For a constant pressure reaction
?H
2
° = ?H
1
° + ?C
P
 (T
2
 ? T
1
)
where  ?C
P
 = C
P
 (products) ? C
P
 (reactants).
For a constant volume reaction
?
? ? ? ? ? dT . C E E
V
0
1
0
2
  
Enthalpy of Reaction from Enthalpies of Formation :
The enthalpy of reaction can be calculated by
?H
r
° = ? ?
B
 ?H
f
°,
products
 ? ? ?
B
 ?H
f
°,
reactants
?
B
 is the stoichiometric coefficient.
Estimation of Enthalpy of a reaction from bond Enthalpies :
?H = 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
atoms  gasesous
the  from products  form
to  released  Enthalpy
atoms   gasesous
into  reactants  break
  to  required  Enthalpy
Resonance Energy :
?H°
resonance
 = ?H°
f, experimental
 ?  ?H°
f, calclulated
    = ?H°
c, calclulated
???  ?H°
c, experimental