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MME 6701: Lecture 03
Thermodynamic Variables
and Relations
A. K. M. B. Rashid
Professor, Department of MME
BUET, Dhaka
Today’s Topics
 Review of thermodynamic relations
 The general procedure to obtain
thermodynamic relations
 The applications of thermodynamic relations
 Problem solving
1
Review of Thermodynamic Relations
1. Relations derived from the laws of
thermodynamics
The first law:
dU = dQ + dW + dW
Reversible heat absorbed by the system:
dQrev = TdS
Reversible mechanical work done on the system:
dWrev = - P dV
Combined statement of the first and second laws:
dU = TdS – PdV + dW
2. Definitions in Thermodynamics
Energy Functions
 Enthalpy, H
 Helmholtz Free Energy, F
 Gibbs Free Energy, G
Experimental Variables
 Coefficient of Thermal Expansion, a
 Coefficient of Compressibility, b
 Heat Capacity, CP
2
Energy Function
Definition
Combined Statement
Enthalpy, or
Heat Content
H  U + PV
dH = TdS + VdP + dW’
At const. P (and dW’ = 0)
dHP = TdSP = dQrev, P
Helmholtz Free
Energy or,
the Work Function
F  U - TS
dF = - SdT - PdV + dW’
At const. T,
dFT = - PdVT + dW’T
dFT = dWT + dW’T = dWtot
Gibbs Free Energy
G  H - TS
dG = - SdT + VdP + dW’
At const. P and T,
dG P, T = dW’ P, T
The Four Combined Statements
dU =
TdS – PdV + dW
dH = TdS + VdP + dW
dF = – SdT – PdV + dW
dG = – SdT + VdP + dW
3
The Experimental Variables
Coefficient of
Thermal Expansion, a
Coefficient of
Compressibility, b
Change in volume of material
with temperature at constant P
Change in volume of material
with pressure at constant T
a =
1
V
V
T
b =
K-1
1
V
P
V
P
atm-1
T
 These coefficients depend not only on material composition,
but also on temperature and pressure of the system.
Heat Capacity, C
 Heat required (Q) to bring about a certain temperature
change (T)
Qrev
C = T
and
C=
dQrev
dT
 The concept of heat capacity is only used the system
produces a finite temperature change (i.e. T0)
 The value of C changes with T, P, and Xk of the system.
 T dependence of C is usually expressed by the relation
C (T) = a + bT + cT–2
where a, b, c are constants
4
 Like heat, heat capacity is also a process variable.
CP =
CV =
dQrev
T
dQrev
T
or,
dQrev,P = CP dTP
or,
dQrev,V = CV dTV
P
V
 Heat capacity is an extensive variable.
Specific heat capacity (heat capacity per gram of substance at constant P)
and molar heat capacity (heat capacity per mole of substance at constant P
or V) are also used. Thus, for example, CP = n cP .
 Generally, for a given material, CP > CV. Why?
 If a, b, and CP are known for any simple system
(for which dW’=0)
 Changes of all the state functions can be computed
for any arbitrary process through which the system may be taken
 No additional information is required.
5
3. The Coefficient Relations
The function:
Coefficient Relations:
Z = Z (X, Y)
dZ = MdX + NdY
M (X,Y) = (Z/X)Y
N (X,Y) = (Z/Y)X
Combined Statements
Coefficient Relations
dU = TdS – PdV
T = (U/S)V ;
- P = (U/V)S
dH = TdS + VdP
T = (H/S)P ;
V = (H/P)S
dF = -SdT – PdV
- S = (F/T)V ; - P = (F/V)T
dG = -SdT + VdP
- S = (G/T)P ; V = (G/P)T
4. The Maxwell Relations
Z = Z (X, Y)
M = (Z/X)Y
Maxwell Relation:
dZ = MdX + NdY
N = (Z/Y)X
(M/Y)X = (N/X)Y
Combined Statements
Maxwell Relations
dU = TdS – PdV
(T/V)S = - (P/S)V
dH = TdS + VdP
(T/P)S = (V/S)P
dF = -SdT – PdV
(S/V)T = (P/T)V
dG = -SdT + VdP
- (S/P)T = (V/T)P
6
Example 4.1
Show that heat is not a state variable.
Using the first law, for dW=0,
dU = dQ – PdV
dQ = dU + PdV
For the function U = U(T,V):
dU = (U/T)V dT + (U/V)T dV
dQ = (U/T)V dT + (U/V)T dV + PdV
dQ = (U/T)V dT + [(U/V)T + P] dV
If Q were to be a state function, Maxwell relation must be valid. Then
for the function Q = Q (T,V) the Maxwell relation is
(/V) [(U/T)V]T = (/T) [(U/V)T + P]V
2U
V.T
=
V, T
2U
T.V
+
T, V
P
T
V
Thus, heat is not a state variable.
But this requires that
(P/ T)V = 0
which is never true.
For example, for 1 mol of ideal gas,
PV = RT and (P/ T)V = R/V  0
How to Tackle a
Thermodynamic Problem ?
 To solve practical problems requires
finding or deriving the relationships between the sought (dependent)
variables and the given (independent) variables of the type:
Z = Z (X, Y, ....)
 Steps in solving practical problems:
 Identify which system properties are given (X, Y, ….)
 Identify which system property you want to find (Z)
 Establish the relationship that connect the sought and given properties
Z = Z (X, Y, …)
 Find required materials properties
 Perform calculation
7
General Strategy for Deriving
Thermodynamic Relations
 Identify the properties of the system about which the information is given.
 Independent variables
 Identify the property of the system about which one is seeking information.
 Dependent variable
 Find or derive a relationship between the sought and the given variables.
 The generic form of this relation Z = Z (X, Y)
 Contains quantities such as a, b, CP, etc.
 Obtain values for these quantities.
 Tabulated data (books, databases, experiments)
 Substitute values into relationship and carry out mathematical operations.
 Get a value for the dependent variable
Step three is critical
 To develop a general procedure for deriving such a relation
 Choose T and P as the independent variables
 Express all state variables as functions of T and P: Z = Z (T, P)
 Convert this relation to the required function: Z = Z (X, Y)
State Functions
Combined Law
Equation of State Variables (T, P, V)
dU = TdS – PdV
Energy Functions (U, H, F, G)
dH = TdS + VdP
Entropy Function, S
dF = - SdT - PdV
dG = - SdT + VdP
 Express V and S as functions of T and P:
V = V (T, P)
S = S (T, P)
8
Expressing V as a function of T and P
V = V (T, P)
dV =
V
dT +
T
P
a =
V
dP
P
T
V
( 1V )( T )P
b = -
( 1V )( V
)
P T
dV = Va dT – Vb dP
Expressing S as a function of T and P
S = S (T, P)
dS =
S
T
dT +
P
S
P
dP
T
Since, dQrev, P = TdSP = CP dTP
(S/T)P = CP/T
From Maxwell relation,
(S/P)T = - (V/T)P = -Va
dS = (CP/T) dT – Va dP
9
TABLE 4.2
Thermodynamic state functions expressed in terms of the independent
variables T and P
V = V (T, P):
dV = VadT – VbdP
S = S (T, P):
dS = (CP/T) dT – VadP
U = U (T, P): dU = TdS – PdV
dU = (CP – PVa)dT + V(Pb –Ta)dP
H = H (T, P): dH = TdS + VdP
dH = CPdT + V(1 – Ta)dP
F = F (T, P):
dF = – (S + PVa)dT + PVbdP
dF = – SdT – PdV
G = G (T, P): dG = – SdT + VdP
dG = – SdT + VdP
The coefficients in these differential equations contain the following factors:
 T and P (the independent variables specified in any application)
 a, b and CP (the experimental variables to be available in tables or data bases)
 S and V (can be evaluated as functions of T and P, given the value of a, b and CP)
General Procedure to Obtain
Thermodynamic Relations
1. Identify the variables: Z = Z (X, Y)
2. Write the differential form: dZ = M dX + N dY
3. Use Table 4.2 to express dX and dY in terms of the variables dT and dP:
dZ = M [XT dT + XP dP] + N [YT dT + YP dP]
4. Collecting terms:
dZ = [M XT + N YT ] dT + [M XP + N YP] dP
5. Obtain Z = Z (T, P) from Table 4.2: dZ = ZT dT + ZP dP
Check Units
6. Equating the like terms for the function Z = Z (T, P):
M XT + N YT = ZT
(4.37a)
M XP + N YP = ZP
(4.37b)
Energy (PV)
Entropy (PV/T)
Heat capacity (PV/T)
a (1/T)
b (1/P)
7. Solving Eq.(4.37) will result expressions for M and N
10
Example 4.3
Relate the entropy of a system to its temperature and volume.
1.
2.
3.
4.
5.
6.
7.
S = S (T, V)
dS = M dT + N dV
Using Table 4.2: dS = M dT + N (VadT - VbdP)
dS = MdT + NVadT - NVbdP = (M + NVa) dT - NVbdP
From Table 4.2: dS = [CP/T]dT - VadP
Comparing coefficients:
M + NVa = CP/T ; - NVb = - Va
Solve this pair of equations for M and N:
M =
2
1
CP - TVa
T
b
S = S (T, V) :
dS =
and
a
N= b
1 C - TVa2
a
dT +
dV
P
T
b
b
Example 4.4
Find the relationship needed to compute the change in Gibbs
free energy when the initial and final states are specified by their
pressure and volume.
G = G (P, V)
Sb
dG = V - a dP - S dV
Va
Example 4.5
Derive an expression for the increase in temperature for process
in which the volume of the system is changed at constant entropy.
T = T (V, S)
dTS = - Ta dVS
CV b
11
The equation of state for an ideal gas:
PV = n RT
Universal gas constant
R
R
R
R
R
R
R
=
=
=
=
=
=
=
1.98717 cal/K-mol = 1.986 Btu/R-lb mol
10.73 ft3-psia/ R-lb mol = 1545 ft-lbf/R-lb mol
82.06 cm3 -atm/K-mol = 83.14 cm3 -bar/K-mol
62356 cm3 -torr/K-mol = 8314 cm3 -kPa/K-mol
0.082056 litre -atm/K-mol = 0.7302 ft3-atm/R-lb mol
0.082 m3 -bar/K-mol
8.31434 J/K-mol = 8.31434 m3 Pa/K-mol
PV = RT
a =
P
R
= (
=
( 1V )( V
(
)
)
T P
RT
P)
b = -
1
T
P
-RT
= -(
=
( 1V )( V
P )T
RT )( P2 )
1
P
2
CP - TVa = CP - TVP
= CP - R
T2
b
CP - R = CV
For monatomic gases:
CV = 3R/2
CP = 5R/2
12
dU = (CP - PVa) dT + V (Pb - Ta) dP
dU = (CP – PV/T) dT + V (P/P – T/T) dP = (CP – R) dT
dU = CV dT
U = CV (T2 – T1)
dH = CP dT + V(1 - Ta) dP
dH = CP dT + V(1 – T/T) dP
dH = CP dT
H = CP (T2 – T1)
Example 4.5
One mole of an ideal monatomic gas initially at temperature 298 K
and occupying volume 10 litres is compressed reversibly and
adiabatically to a final volume of 2 litres. Compute the final
temperature of the system.
Here, the function is: T = T (S, V)
For the function S = S (T, V):
dS = (CV/T) dT + (a/b) dV
Then, for the function T = T (S, V):
dT = (T/CV) dS - (Ta/bCV) dV
For adiabatic process, dS = 0. Thus,
Now, for ideal gas, a = 1/T, b=1/P.
Using PV = RT, P = RT/V, and
dTS = - (Ta/bCV) dVS
So,
dT = - (P/CV) dV
(dT/T) = - (R/CV) (dV/V)
13
(dT/T) = - (R/CV) (dV/V)
Integrating between limits:
ln (T2/T1) = -(R/CV) ln (V2/V1)
T2 = T1 (V1/V2) R/CV
Given data
V1 = 10 litre
V2 = 2 litre
T1 = 298 K
CV = 3R/2
T2 = (298 K) (10/2) 2/3 = 871.36 K
 a, b and CP are functions of T, P and X.
 For a given material, they are not strong functions of the state of
the system.
 When only an estimate of the thermodynamic functions is
sought, a, b and CP can be considered to be as constants.
 For precise calculations, the dependence upon P and T must
be obtained for these experimental variables a, b and CP.
14
Example 4.6
One mole of nickel initially at 300 K and 1 atm pressure is taken
through two separate processes:
(1) an isobaric change in temperature to 1000 K, and
(2) an isothermal compression to 1000 atm.
Compare the change in enthalpy of nickel for these two processes.
Given data:
V (300 K, 1 atm) = 6.57 cc/mol
a = 40x10-6 K-1
b = 1.5x10-6 atm-1
CP = 16.99 + 2.95x10-2T J/mol-K.
Answer:
H = H (T, P)
dH = CP dT + V (1 - Ta) dP
Process 1 (Isobaric)
dHP = CP dT
1000

H = dH =
300
1000
 (16.99 +
2.95x10-2
300
H = 11893.0 + 13422.5 J/mol
H = 25315.5 J/mol
T) dT
The second term on the right
hand side of the equation
arises from the temperature
dependent contribution to the
heat capacity.
If T dependency is ignored,
CP = 16.99, and then
H = 11893.0 J/mol
15
H = H (T, P)
dH = CP dT + V (1 - Ta) dP
Process 2 (Isothermal)
dHT = V (1 - Ta) dPT
Here V and a depend upon pressure
For an estimated calculation,
 V and a both can be treated as pressure independent.
H = V (1 - Ta) [P2 - P1]
H = 657.0 J/mol
For an exact calculation,
 the values of a are within the range of 10–6, so that a can be
considered as constant.
 but, pressure dependency of V cannot be ignored.
dV = Va dT – Vb dP
At constant T,
dVT = – Vb dPT
Integrating, ln (V / V0) = – b (P – P0)
V / V0 = exp – b (P – P0)
V  V0 1 + b (P – P0)
Then the enthalpy change
dH = V0 1 + b (P – P0) (1 – Ta) dP
16
dH = V0 1 + b (P – P0) (1 – Ta) dP
Integrating,
b (P – P0)2
H = V0 (1 – Ta) P +
2
1000
1
H = 657.0 + 0.07 J/mol
Only 0.07 J/mol amount of enthalpy is added because of
addition of the pressure variation of volume into the equation.
For solids and liquids, energy changes associated with
thermal influences tend to be much larger that those arises
from mechanical influences.
Problem Solving
4.9
Compute the change in entropy when one mole of an ideal
monatomic gas expands freely to twice its volume.
4.13
A copper rod is heated from 298 K to 650 K at constant pressure.
Calculate the changes in entropy, enthalpy and Gibbs free energy
for the process.
CP<Cu> = 22.6 + 5.6x10-3T J/mol-K
S298 = 33.14 J/mol-K.
4.16
Two moles of monatomic ideal gas are contained at 1 atm pressure
and 300 K temperature. 34166 joules of heat are transferred to the
gas, as a result of which the gas expands and does 1216 joules of
work against its surroundings. The process is reversible. Calculate
the final temperature of the gas.
17
4.17
A quantity of ideal gas occupies 5 litres at 10 atm and 150 K. The molar
heat capacity of the gas at constant volume, cV , is 3/2R. Calculate:
(1)
(2)
(3)
(4)
the final volume of the system,
the work done by the system,
the heat entering or leaving the system, and
the internal energy and enthalpy change in the system
if it undergoes
(a) a reversible isothermal expansion to 1 atm
(b) a reversible adiabatic expansion to 1 atm.
4.23
One mole of an ideal gas, initially at 273 K and 1 atm, is contained in a
chamber that permits programmed control of its state. Controlled
quantities of heat and work are supplied in the system so that its
pressure and volume change along a line given by the equation
V = 22.4 P litre/atm.
Assume the process is carried out reversibly. Compute the heat
required to be supplied to the system to take it to a final pressure of
0.5 atm. What is the final temperature of the gas?
Next Class
Lecture 04
Equilibrium in Thermodynamic
Systems
18

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