Unformatted text preview: 145 4.11. EQUILIBRIUM IN A TWOCOMPONENT SYSTEM
The pressure at equilibrium is
P2 =
=
= (nN2 + nN )RT
,
V
(0.888154 kmole + 0.223693 kmole) 8.314
498.84
111.185 kP a. (4.465)
kJ
kmole K (6000 K ) , (4.466)
(4.467) The pressure has increased because there are more molecules with the volume and temperature being
equal.
The molar concentrations ρi at equilibrium, are
ρN = kmole
0.223693 kmole
mole
= 4.484 × 10−4
= 4.484 × 10−7
,
3
3
498.84 m
m
cm3 (4.468) ρN 2 = kmole
0.888154 kmole
mole
= 1.78044 × 10−3
= 1.78044 × 10−6
.
3
3
498.84 m
m
cm3 (4.469) Now consider the heat transfer. One knows for the isochoric process that 1Q2 = U2 − U1 . The
initial energy is given by
U1 = nN2 uN2 , (4.470) = nN2 (hN2 − RT ), (4.471) = (1 kmole) 2.05848 × 105 = 1.555964 × 105 kJ. kJ
kJ
− 8.314
kmole
kmole K (6000 K ) , (4.472)
(4.473) The energy at the ﬁnal state is
U2 =
=
= = nN 2 u N 2 + nN u N ,
nN2 (hN2 − RT ) + nN (hN − RT ), (4.474)
(4.475) (0.888154 kmole) 2.05848 × 105 (4.476) kJ
kJ
(6000 K )
− 8.314
kmole
kmole K
kJ
kJ
+(0.223693 kmole) 5.9727 × 105
(6000 K ) ,
− 8.314
kmole
kmole K
2.60966 × 105 kJ. (4.477)
(4.478) So
1Q2 =
=
= U2 − U1 ,
2.60966 × 105 kJ − 1.555964 × 105 kJ,
1.05002 × 105 kJ. (4.479)
(4.480)
(4.481) Heat needed to be added to keep the system at the constant temperature. This is because the nitrogen
dissociation process is endothermic.
One can check for second law satisfaction in two ways. Fundamentally, one can demand that
Eq. (4.377), dS ≥ δQ/T , be satisﬁed for the process, giving
2 S2 − S1 ≥ 1 δQ
.
T (4.482) CC BYNCND. 18 November 2011, J. M. Powers. 146 CHAPTER 4. MATHEMATICAL FOUNDATIONS OF THERMODYNAMICS
For this isothermal process, this reduces to
S2 − S1 1Q2 ≥ T (nN2 sN2 + nN sN )2
− (nN2 sN2 + nN sN )1
yN2 P
Po
yN2 P
− nN2 so 2 − R ln
T,N
Po
PN2
nN2 so 2 − R ln
T,N
Po
PN2
− nN2 so 2 − R ln
T,N
Po
nN2 so 2 − R ln
T,N nN2 so 2 − R ln
T,N − nN2 so 2 − R ln
T,N nN2 RT
Po V
nN2 RT
Po V + nN
+ nN
+ nN
+ nN + nN yN P
Po
yN P
so − R ln
T,N
Po
PN
so − R ln
T,N
Po
PN
so − R ln
T,N
Po so − R ln
T,N + nN so − R ln
T,N nN RT
Po V so − R ln
T,N nN RT
Po V =0 1Q2 ≥ T , (4.483) , (4.484) , (4.485) , (4.486) . (4.487) 2 1 1Q2 ≥ T 2 1 2 1Q2 ≥ ≥ T 1Q2 T 1 Now at the initial state nN = 0 kmole, and RT /Po /V has a constant value of
kJ
8.314 kmole K (6000 K )
RT
1
=
=1
,
Po V
(100 kP a)(498.84 m3 )
kmole (4.488) so Eq. (4.487) reduces to
nN2 so 2 − R ln
T,N nN 2
1 kmole nN
1 kmole
nN 2
− R ln
1 kmole + nN so − R ln
T,N
− n N 2 so 2
T,N 2 1 ≥ 1Q2 T , (4.489)
((0.888143) (292.984 − 8.314 ln (0.88143)) + (0.223714) (216.926 − 8.314 ln (0.223714)))2
− ((1) (292.984 − 8.314 ln (1)))1
19.4181 kJ
kJ
≥ 17.5004
.
K
K ≥ 105002
,
6000
(4.490) Indeed, the second law is satisﬁed. Moreover the irreversible entropy production of the chemical reaction
is 19.4181 − 17.5004 = +1.91772 kJ/K.
For the isochoric, isothermal process, it is also appropriate to use Eq. (4.394), dAT,V ≤ 0, to check
for second law satisfaction. This turns out to give an identical result. Since by Eq. (4.307), A = U − T S ,
A2 − A1 = (U2 − T2 S2 ) − (U1 − T1 S1 ). Since the process is isothermal, A2 − A1 = U2 − U1 − T (S2 − S1 ).
For A2 − A1 ≤ 0, one must demand U2 − U1 − T (S2 − S1 ) ≤ 0, or U2 − U1 ≤ T (S2 − S1 ), or
S2 −S1 ≥ (U2 −U1 )/T . Since 1Q2 = U2 −U1 for this isochoric process, one then recovers S2 −S1 ≥ 1Q2 /T. CC BYNCND. 18 November 2011, J. M. Powers. ...
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 Fall '11
 ParkSou
 Dynamics, Trigraph, kmole, 1 19.4181 kJ, 1.91772 kJ

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