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Chapter 3
The Importance of State Functions:
Internal Energy and Enthalpy
Physical Chemistry 2nd Edition
Thomas Engel, Philip Reid
Objectives
• Express the infinitesimal quantities dU and dH
as exact differentials.
• Derive the change of U with T and V and the
change in H with T and P to experimentally
accessible quantities.
Chapter 3: The Importance of State Functions: Internal Energy and Enthalpy
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Outline
1. The Mathematical Properties of State Functions
2. The Dependence of U on V and T
3. Does the Internal Energy Depend More Strongly on
V or T?
4. The Variation of Enthalpy with Temperature at
Constant Pressure
5. How Are CP and CV Related?
6. The Variation of Enthalpy with Pressure at Constant
Temperature
7. The Joule-Thompson Experiment
8. Liquefying Gases Using an Isenthalpic Expansion
Chapter 3: The Importance of State Functions: Internal Energy and Enthalpy
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
3.1 The Mathematical Properties of State Functions
• Consider 1 mol of an ideal gas for which
P = f (V , T ) =
RT
V
• The change in P from a change in V or T is
proportional to the following partial
derivatives:
P (V + ∆V ,T ) − P (V ,T )
RT
 ∂P 
=
lim
=
−


∆V → 0
∆V
V2
 ∂V T
P (V ,T + ∆T ) − P (V , T ) R
 ∂P 
=
lim
=


∆T → 0
∆T
V
 ∂T V
Chapter 3: The Importance of State Functions: Internal Energy and Enthalpy
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
3.1 The Mathematical Properties of State Functions
• When P changes to P+dP,
 ∂P 
 ∂P 
dP =   dT + 
 dV
 ∂T V
 ∂V T
• When function f is a
state function,
 ∂  ∂f (V , T )  
 ∂  ∂f (V , T )  


  = 

 
 ∂T  ∂V T V  ∂V  ∂T V T
Chapter 3: The Importance of State Functions: Internal Energy and Enthalpy
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Example 3.1
a. Calculate
for the function
f ( x, y ) = ye x + xy + x ln y
b. Determine if f(x,y) is a state function of the variables
x and y.
c. If f(x,y) is a state function of the variables x and y,
what is the total differential df?
Chapter 3: The Importance of State Functions: Internal Energy and Enthalpy
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Solution
a.
 ∂f 
x
  = ye + y + ln y,
 ∂x  y
 ∂f 
 
 ∂y  x
f ( x, y) = ye x + xy + x ln y
x
= ex + x +
y
 ∂2 f

 ∂x 2

 ∂2 f

 ∂y 2


x
 =−
2

y
x

 = ye x ,

y
  ∂f 
 ∂ 
  ∂x  y
 ∂y



  ∂f  

 ∂  


1   ∂y  x 
1
x
x
=
e
+
1
+
,
=
e
+
1
+



y  ∂x 
y




x

y
Chapter 3: The Importance of State Functions: Internal Energy and Enthalpy
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Solution
b. Because we have shown that
  ∂f     ∂f  
 ∂    ∂  
  ∂x  y    ∂y  x 
 ∂y  =  ∂x 


 



x 
y
f(x,y) is a state function of the variables x and
y. Note that any well-behaved function that can
be expressed in analytical form is a state
function.
Chapter 3: The Importance of State Functions: Internal Energy and Enthalpy
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Solution
c. The total differential is given by
 ∂f 
 ∂f 
df =   dx +   dy
 ∂x  y
 ∂y  x
(
)
 x
x
= ye + y + ln y dx +  e + x +  dy
y

x
Chapter 3: The Importance of State Functions: Internal Energy and Enthalpy
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
3.1 The Mathematical Properties of State Functions
• Two differential calculus that used frequently:
a) A function of z=f(x,y),
 ∂x 
1
  =
 ∂y  z  ∂y 
 ∂x  z
b) The cyclic rule
 ∂x   ∂y   ∂z 
      = −1
 ∂y  z  ∂z  x  ∂x  y
Chapter 3: The Importance of State Functions: Internal Energy and Enthalpy
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
3.1 The Mathematical Properties of State Functions
• Combination of previous equations give the
coefficients.
1  ∂V 
β= 

V  ∂T  P
1
and κ = −
V
 ∂V 


 ∂P T
β = volumetric thermal expansion
coefficient
κ = isothermal compressibility
Chapter 3: The Importance of State Functions: Internal Energy and Enthalpy
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Example 3.2
You have accidentally arrived at the end of the range of
an ethanol in glass thermometer so that the entire
volume of the glass capillary is filled. By how much will
the pressure in the capillary increase if the temperature is
increased by another 10.0°C?
−1
−5
−1 ,
−5
,
κ
=
11
.
00
×
10
(
bar
)
β
=
11
.
2
×
10
(
°
C
)
ethanol
β glass = 2.00 × 10 (°C )
−4
−1
ethanol
Do you think that the thermometer will survive your
experiment?
Chapter 3: The Importance of State Functions: Internal Energy and Enthalpy
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Solution
β ethanol
β ethanol
1
1 Vf
dT
−
dV
≈
∆
T
−
ln
∫ κ
∫ κV
κ
κ Vi
(
β ethanol − β glass )
(
11 .2 − 0.200 ) × 10 − 4
=
∆T =
× 10 .0°C = 100 .bar
−5
κ
11.0 × 10
∆P =
In this calculation, we have used the relations:
V (T2 ) = V (T1 )(1 + β [T2 − T1 ])
ln (1 + x ) ≈ x if x << 1
The glass is unlikely to withstand such a large
increase in pressure.
Chapter 3: The Importance of State Functions: Internal Energy and Enthalpy
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
3.2 The Dependence of U on V and T
• As U is a state function, an infinitesimal
change in U can be written as
 ∂U 
 ∂U 
dU = 
 dT + 
 dV
 ∂T V
 ∂V  T
• The differential expression of the first law for
constant volume can be written as
dqV  ∂U 
=
 = CV
dT  ∂T V
• dqv/dT corresponds to a constant volume
path and is called the heat capacity at
constant volume, Cv
Chapter 3: The Importance of State Functions: Internal Energy and Enthalpy
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
3.2 The Dependence of U on V and T
• Atoms have only translational degrees of
freedom and low CV,m independent of
temperature.
• Molecules with vibrational
degrees of freedom
have higher CV,m.
Chapter 3: The Importance of State Functions: Internal Energy and Enthalpy
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
3.2 The Dependence of U on V and T
• After CV has been determined as a function of
T , the integral is numerically evaluated:
T2
T2
T1
T1
∆U V = ∫ CV dT = n ∫ CV , m dT
• Over a limited temperature range, CV,m can
be simplified into
qV = ∆U
Chapter 3: The Importance of State Functions: Internal Energy and Enthalpy
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
3.2 The Dependence of U on V and T
• The total differential of the internal energy
can be written as
  ∂P 

dU = dU V + dU T = CV dT + T 
 − P  dV
  ∂T 

• U is a state function, all
paths connecting Vi,Ti and
Vf,Tf are equally valid in
calculating ∆U.
Chapter 3: The Importance of State Functions: Internal Energy and Enthalpy
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
3.3 Does the Internal Energy Depend More Strongly on
V or T?
• U is a function of T alone for an ideal gas.
• Not true for real gases, liquids, and solids as
the change in U with V must be considered.
Chapter 3: The Importance of State Functions: Internal Energy and Enthalpy
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Example 3.3
Evaluate (∂U / ∂V )T for an ideal gas and modify
accordingly for the specific case of an ideal gas.
Solution:
nRT
 ∂P 
 ∂[nRT / V ] 
T
−P=0
 − P = T
 −P=
∂T
V
 ∂T V

V
Therefore, dU = CV dT , showing that for an
ideal gas, U is a function of T only.
Chapter 3: The Importance of State Functions: Internal Energy and Enthalpy
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
3.3 Does the Internal Energy Depend More Strongly on
V or T?
• U is a function of T alone for an ideal gas.
• Ideal gas molecules do not attract or repel
one another, no energy is required to change
their average distance of separation (increase
or decrease V).
Chapter 3: The Importance of State Functions: Internal Energy and Enthalpy
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Example 3.4
In Joule’s experiment to determine (∂U / ∂V )T ,
the heat capacities of the gas and the water
bath surroundings were related by
C surroundin g / Csystem ≈ 1000
If the precision with which the temperature of
the surroundings could be measured is ± 0 .006°C
, what is the minimum detectable change in the
temperature of the gas?
Chapter 3: The Importance of State Functions: Internal Energy and Enthalpy
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Solution
View the experimental apparatus as two
interacting systems in a rigid adiabatic
enclosure. The first is the volume within vessels
A and B, and the second is the water bath and
the vessels. Because the two interacting
systems are isolated from the rest of the
universe,
q = Cwater bath ∆Twater bath + Cgas ∆Tgas = 0
∆Tgas
Cwater bath
=−
∆Twater bath = −1000 (± 0.006 °C ) = ∓6°C
Cgas
Chapter 3: The Importance of State Functions: Internal Energy and Enthalpy
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
3.4 The Variation of Enthalpy with Temperature at Constant
Pressure
• The initial and final states for an undefined
process that takes place at constant pressure.
Chapter 3: The Importance of State Functions: Internal Energy and Enthalpy
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
3.4 The Variation of Enthalpy with Temperature at Constant
Pressure
• As P=Pf=Pi we have
∆H = q P
• Since H is a state function, dH is an exact
differential.
 ∂H 
 ∂H 
dH = 
 dT + 
 dP
 ∂T  P
 ∂P T
Chapter 3: The Importance of State Functions: Internal Energy and Enthalpy
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
3.4 The Variation of Enthalpy with Temperature at Constant
Pressure
• The heat capacity at constant pressure,
CP, is defined as
 ∂H 
Cp =
=

dT  ∂T  P
dq p
• In general, a constant pressure process with
no change in the phase of the system and no
chemical reactions,
Tf
Tf
Ti
Ti
∆ H P = ∫ C P (T )dT = n ∫ C P ,m (T )dT
Chapter 3: The Importance of State Functions: Internal Energy and Enthalpy
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Example 3.7
A 143.0-g sample of C(s) in the form of graphite is
heated from 300 to 600 K at a constant pressure. Over
this temperature range, CP,m has been determined to
be
Calculate ∆H and qP. How large is the relative error in
∆H if you neglect the temperature-dependent terms in
CP,m and assume that CP,m maintains its value at 300 K
throughout the temperature interval?
Chapter 3: The Importance of State Functions: Internal Energy and Enthalpy
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Solution
Answer:
∆H =
m
M
Tf
∫ C (T )dT
P, m
Ti
600
2
3
4
143.0 
T
−4 T
−7 T
−11 T
 − 12.19 + 0. 1126 − 1.947 × 10
=
+ 1.919 ×10
− 7.800 × 10
2
3

12. 01 300 
K
K
K
K4
∫
 T
d
 K

600
3
4
5
143.0 
T
T2
−5 T
−8 T
−11 T 
=
×  − 12.19 + 0.0563 2 − 6.49 × 10
+ 4. 798 × 10
− 1.56 ×10

3
4
12. 01 
K
K
K
K
K 5  300
= 46. 85kJ
Chapter 3: The Importance of State Functions: Internal Energy and Enthalpy
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Solution (cont’d)
If we had assumed CP,m=8.617 J mol-1 K-1,
which is the calculated value at 300 K,
∆H = 143 .0 / 12.01 × 8.617 × [600 . − 300 .] = 30 .81kJ
The relative error is (30.81 − 46.85) / 46.85 = −34%
. In this case, it is not reasonable to assume
that CP,m is independent of temperature.
Chapter 3: The Importance of State Functions: Internal Energy and Enthalpy
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
3.5 How Are CP and CV Related?
• The relationship between Cp and Cv is defined
as
β2
β2
CP = CV + TV
or CP ,m = CV , m + TVm
κ
κ
• When CP ≈ CV for a liquid or a solid,
 ∂U 
  ∂V 
CV >> 
 + P 

 ∂V T
  ∂T  P
Chapter 3: The Importance of State Functions: Internal Energy and Enthalpy
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
3.6 The Variation of Enthalpy with Pressure at Constant
Temperature
• When all systems containing pure substances
or mixtures at a fixed composition, provided
that no phase changes or chemical reactions
take place, we apply
 ∂H 
 ∂P   ∂V 
 ∂V 

 = T
 
 +V = V − T

 ∂P  T
 ∂T V  ∂P T
 ∂T  P
Chapter 3: The Importance of State Functions: Internal Energy and Enthalpy
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Example 3.8
Evaluate (∂H ∂P )T for an ideal gas.
Solution:
(∂P ∂T )V = (∂[nRT / V ]/ ∂T )V
= nR / V and (∂V ∂P )T = − nRT / P 2
for an ideal gas. Therefore,
nR  nRT 
nRT nRT
 ∂H 
 ∂P   ∂V 
+V = 0

 =T  
 +V = T
− 2  +V = −
V  P 
P nRT
 ∂P  T
 ∂T V  ∂P T
Chapter 3: The Importance of State Functions: Internal Energy and Enthalpy
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
3.7 The Joule-Thompson Experiment
• The Joule-Thompson experiment allows
to be measured with a much higher
sensitivity than in the Joule experiment.
Chapter 3: The Importance of State Functions: Internal Energy and Enthalpy
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
3.7 The Joule-Thompson Experiment
• The experimentally determined limiting ratio
of ∆T to ∆P at constant enthalpy is known as
the Joule-Thompson coefficient:
µ J −T
 ∆T 
 ∂T 
= lim 
=


∆P→ 0 ∆ P

 H  ∂P  H
Chapter 3: The Importance of State Functions: Internal Energy and Enthalpy
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
3.7 The Joule-Thompson Experiment
• From an isenthalpic process,
 ∂T 
 ∂H 
 ∂H 
CP 
 +
 = 0 giving 
 = − C p µ J −T
 ∂P  H  ∂P T
 ∂P T
• Values of µJ-T for selected gases are shown.
Chapter 3: The Importance of State Functions: Internal Energy and Enthalpy
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Example 3.11
 ∂U 
 ∂V 
 ∂H 
=
+
P



 +V


 ∂P T  ∂V  T
  ∂P  T
Show that
Solution:
µ J −T = −
µ J −T = 0
for an ideal gas.

1  ∂H 
1  ∂U   ∂V 
 ∂V 
+
P
+
V

 =−








C P  ∂P T
C P  ∂V T  ∂P T
 ∂P T



 ∂V 
0
+
P
+
V




∂
P


T



1   ∂[nRT / P ] 
1
=−
P
+
V
=
−




CP  
∂P
CP
T

=−
1
CP
 nRT

−
+
V
 P
=0


Chapter 3: The Importance of State Functions: Internal Energy and Enthalpy
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
3.8 Liquefying Gases Using an Isenthalpic
Expansion
• Heat is extracted from the gas exiting from
the compressor in Joule-Thompson expansion
• It is further cooled in the countercurrent heat
exchanger before expanding through a
nozzle.
• Because its temperature is
sufficiently low,
liquefaction occurs.
Chapter 3: The Importance of State Functions: Internal Energy and Enthalpy
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd