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Enthalpy is the sum of the internal energy and the product of pressure and volume and its general formula is H=U+PV.
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Chapter 2. The First Law.
P.
Justification 2.
The relation
∆
H = q
p
-^
For a general infinitesimal change in the state of the system,
U
changes to
U +
d U
,^ p
changes to
p +
d
p , and
V
changes to
V +
d
V , so
H
changes from
U + pV
to
Recognize
U
pV
=
H
on the right,
H
changes to
Now substitute d
U
= d
q^
w^
into this expression,
If the system is in mechanical equilibrium with its surroundings at a pressure
p
and does only expansion work, we can write d
w^
= −
p d
V^
and obtain
Now impose the condition that the heating occurs at constant pressure bywriting d
p^
= 0. Then
-^
dH=dq (at constant p, no additional work)
Example 2.
Relating
∆
H and
∆
U
-^
The internal energy change when 1.0 mol CaCO
in the form of calcite 3
converts to aragonite is +0.21 kJ. Calculate the difference between theenthalpy change and the change in internal energy when the pressure is 1.0bar given that the densities of the solids are 2.71 g cm
−
and 2.93 g cm
−3,
respectively.
-^
Answer:
-^
The volume of 1.0 mol CaCO
(100 g) as aragonite is 34 cm 3
3 , and that of 1.
mol CaCO
as calcite is 37 cm 3
(1 Pa m
3 = 1 J). Hence,
only 0.1% of
∆
U
. Usually ignore the difference between H and U of
condensed phases, except at very high p, when
pV
no longer negligible
-^
Self Test 2.
Calculate the difference between
∆
H
and
∆
U
when 1.0 mol Sn(s,
grey) of density 5.75 g cm
−
changes to Sn(s, white) of density 7.31 g cm
−
at
10.0 bar. At 298 K,
∆
H = +
2.1 kJ.
-^
Correct Answer:
∆
H −
∆
U = −
4.4 J
Enthalpy of perfect gas
Illustration 2.
The relation between
and
U for gas-phase reactions
(g) + O 2
(g) 2
O(l), 2
n
g^
3 mol. At 298 K, when
= 2.5 kJ
mol
−
, the enthalpy and internal energy
changes in the system
g)/(18.02 g mol
−
) = (0.798/18.02) mol H
the enthalpy of vaporization per mole ofH
O is 2
O(l) 2
O(g) , 2
n
g^
= +1 mol,
benzene at its boiling point (353.25 K) is30.8 kJ mol
−
. What is the molar internal
energy change? For how long would thesame 12 V source need to supply a 0.50 Acurrent in order to vaporize a 10 g sample?
-^
Correct Answer:
+27.9 kJ mol
−
, 660 s
Chapter 2. The First Law.
P.
2.5(c) The variation of enthalpy with temperature
Example 2.
Evaluating an increase in enthalpy with temperature
(p.45)
2
when it is heated from 25
°C to 100
( Table 2-2)
Chapter 2. The First Law.
P.
2.6 Adiabatic changes
Adiabatic processes
-^
Consider a stage in a reversible adiabaticexpansion when the pressure inside and out is
p
.
The work done when the gas expands by d
V
is
d
w
= −
p
d
V
; however, for a
perfect gas
, d
U
=
C
dV
T
.
-^
Therefore,because for an
adiabatic
change (d
q
= 0)
d
U
= d
w
q
= d
w
, we can equate these two
expressions for d
U
and write
We are dealing with a
perfect gas
, so we can
replace
p
by
nRT
/ V
and obtain
x
/ y
) = −ln(
y
/ x
), this expression
rearranges to
c
nR
we obtain (because ln
x
a^
a
ln
x
/f
)i c^
/i
) and,f
upon rearrangement,
Adiabatic processes
(4)
-^
The initial and final states of a perfect gas satisfy theperfect gas law regardless of how the change of statetakes place, so use
pV
=
nRT,
However, we have just shown that
where we use the definition of the heat capacity ratiowhere
γ
=
C
p ,m
/ C
V ,m
and the fact that, for a perfect
gas,
C
p ,m
C
V ,m
=
R
(
Sec.2.
). Then we combine the
two expressions, to obtain
which rearranges to
p
V i
γi
=
p
V f
γf
Illustration 2.
Work of adiabatic expansion
-^
Consider the adiabatic, reversible expansion of 0.020mol Ar, initially at 25
°C, from 0.50 dm
3
to 1.00 dm
The molar heat capacity of argon at constant volume is12.48 J K
−
mol
−
, so
c
= 1.501. From
It follows that
∆
T = −
110 K, from
Note that temperature change is independent of theamount of gas but the work is not.
-^
Test 2.
Calculate the final temperature, the work done,
and the change of internal energy when ammonia isused in a reversible adiabatic expansion from 0.50 dm
3
to 2.00 dm
3 , the other initial conditions being the same.
-^
Correct Answer:
195 K, −56 J, −56 J
Illustration 2.
The pressure change
accompanying adiabatic expansion
at 100 kPa expands reversibly andadiabatically to twice its initial volume thefinal pressure will be
final pressure would be 50 kPa.