Solutions to Selected Problems in a Course in Statistical Thermodynamics

71 Pages·1971·7.516 MB·English

by  Joseph Kestin (Eds.)

Solutions to Selected Problems
in
A COURSE IN STATISTICAL
THERMODYNAMICS
Joseph Kestin ). R. Dorfman
BROWN UNIVERSITY UNIVERSITY OF MARYLAND
COPYRIGHT © 1971, BY ACADEMIC PRESS, INC.
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PRINTED IN THE UNITED STATES OF AMERICA
1
Solution to Problem 1.2
First we note that for y = const we must have c = c„ = c . Indeed, from
p v
dh = Tds + vdP
we find that
(4 ■ /3φs\ ), *. ■/a v-A i%\—.
const ,
'T λ 'Τ
because
(ds\ _ 3(s,T) = 3(v,P) = /8v\
\3P/ " 3(Ρ,ϊ) 3(Ρ,Τ) ~\3τ/ '
T ρ
By integration
h - h = v(p - p ) + φ(τ)
o o
so that
u-u = h - h - (P - P )v = φ(Τ)
0 0 O
and
c (§) -$ = (!?) = cv = c <«y>
p v
P
Kence
h - h - v(p - p ) + c(T - T ) . (a)
o o o
Further
A = du + Pdv = dh -VdP = d£ = cdT
T T T T
and
T
s - SQ = cln — . (b)
The elimination of temperature from Equations (a) and (b) yields
h(s,P) = v(P - P ) + cT (e S/c - 1)
o o
with h = s = O a t P =P .
o oo
2
Solution to Problem 1.3
The solution to Problem 1.1 contains expressions for u(T,x) and s(T,x) .
Hence, the fundamental equation f = f(T,x) is found by direct substitution
into the definition f = u - Ts .
(a) f (T,v) = c (T - T ) - TÍcJto i- + R&i — + s }
v o V T s v o
o o
9
MZ - M2
(b) f (T,M) - c (T - T ) - T{c/n I _ 2.
M o + So>
o
(c) f (Τ,ε) = ^- Εε2 + αΕεΤ + c (T - T ) - T(c M £- + αΕε + s ) .
¿ o ε ο ε i o
o
Note that all expressions contain the troublesome term -Ts . This can be
properly normalized only with the aid of the Third Law of thermodynamics.
In the case of an incompressible fluid x = const, so that all derivatives
3.../3x become singular. This does not occur when derivatives with respect
to P are taken. Thus,
g(P,T) = h(P,T) - Ts(P,T) = v(P - PQ) + c(T - TQ) - T(c£n ~- + SQ) .
o
3
Solution to Problem 1.5
Each perfect-gas component in a mixture contributes the term
-n. R £nx. > 0
i = i
to the entropy of the mixture. Hence
S = ns = I n Si(P,T) - Σ n± R W
i i
with
(P,T) = c *n f - Si n f + soi
Si pi
* O O
Thus, the simplest fundamental equation is that where h = h(s,P) , with
H = nh = I n.h. = T n.c . (T - T )
J i i L i pi o
It follows that
h = T ■*)ï x.c . (a)
o\T i pi
1 \ o
and that
o ■ (to f ) l xiCpi " {δ ln^)}{ Xi - & pi 'nXi (b)
or
s¿ - is .(„ Í-) !f - j ¿ - *fc)
h t a
Note that
c . c . Ύ.
Pi Pi , L_
R m c . -c . Ύ. -1
pi vi i
and put
γ,χ.
r
1 i 1 S ιγ ι
Γ
Thus,
4
Problem 1.5 (continued)
S"s° , * y, V r . /ρ\1/Γ
- O 1 \ O'
= exp (s s )/£ n
t f 0\ *'* [ " ° ]
and
h(P,s) = c T °{(t)1/r(ni ^1 /Γ) txp(s ' s°}/=r) ■ 4
where
c = / x. c . .
P i i Pi
It is now easy to specialize for two components, with x- = x and x_ = 1
In this manner
c =xc- + (l-x)c
P Pi P20
*γ (ΐ-χ)γ
Ί 9
and
h(P ,x) = c τ {^)1/Γ(χ)[^Γ«(ΐ - Χ)(1-)/Γ««Φ((.-.) /8Γ(Χ))] -
)S p ο Ο
In order to obtain the form u = uis,v,x) , we write
s(T,v,x) = I c Än ~ + U n ~ - [ R Anx + S (a1)
x± vi X i ± Q
i o o i
with
u - Σ x c .(T - T ) , (b·)
± v o
and proceed in an analogous fashion as before.
The forms g(T,P,x) and f(Τ,ν,χ) are obtained directly from equations
(a), (b) or (a1), (bf), respectively
5
Solution to Problem 1.8
The Gibbs equation for this system is
dG = -SdT + VdP + Y μ. dn. .
L. i i
1
In a chemically active system
dn.
—~ = dξ (V. > 0 for products; V < 0 for reactants). ()
a
v± i
Hence
dG = -SdT + VdP + dξ £ μ.ν = -SdT + VdP - Αάζ, where A = -£ μ.ν. .
Since dG = -dW at T = const, P = const , we find dW° = Adξ . When
the process is performed irreversibly (dW = 0) , the rate of entropy
production is
A _ Αξ
Θ - —
Evidently Θ = 0 (or dG = 0) for
A = 0 or lvv = 0
± ±
Integrating the definition (a) from n, = n. to n. , we find
n. = n° + ν
±ξ
n = I n = i n° + ζ I v
ί x i 1 i
1 Κ1 X ξ [ν. (b)
+
i i
On the other hand
6
Problem 1.8 (continued)
μ. = g.(P,T) + R T An x. = g?(T) + R T £n £5- + R T An x.
1 & 1 = 1 &1 = * = 1
so that, either
I V± g±(P,T) + [ V. R T £n x. = 0
I v.g.(P*,T) +y V. RT An ^r + Y V.RT £n . = 0
L. i&i .1= * · 1= x1
1 1 P 1
The two forms yield:
Π x.1 = exp(-AG/RT)
■Δν
\Hi)
exp(-AG°/£T)
where
AG = I V g (Ρ,Τ); AG° = I V g°; Δν = £ V
i 11 i 11 i
Combining with (a) we obtain
v.
nfni + V )
= exp(-AG/RT)
i i
Δν
exp(-AG°/RT)
"(7)
7
Solution to Problem 1.9
The only additional complications consist in two facts:
First, we must write
dn. = J v d
tj ?j
for each of the r reactions j and integrate each term from 0 to ξ.
Secondly, there are now r equations leading to an equal number of
equilibrium conditions :
A. - -I v..^ = o
Here we note that the chemical potential y. for every species must be
computed from
n° + l ν..ξ.
We are led to r coupled equations of the type considered in Problem 1.8.
8
Solution to Problem 1.10
(a) The state in each compartment is described by two independent properties,
say, T , P and T, , P, . Since there is no interaction between the
J ' a' a b b
compartments, the total number of independent variables is f = 4 .
(b) Now, the total number has decreased to f = 2 .
The final state can be determined from the energy equation
m c T + m, c T = (m + m, )c T_ .
ava 1) v b a b v 2
m T +m, T,
a a D b
^2 m +
"b
independently of the pressures (perfect gas) P = m RT /V and
P, = m-BT./V . The final pressure
P2 = (ma + ν=Τ2/(νΐ + V2}
(maTa+Vb)S
V + V
VI V2
(c) Since the gas is in motion, its state is described by three independent
fields. We can choose these as
s(r,t), u(r,t), v(r,t)
Thus, at any instant
2(t) = j p(r,t) {u(r,t) + \ v2(r,t)}dV
S(t) = Pp((£r,,tt )s(r,t)dV .
JV
In a concrete case, the fields must be determined by solving the following
set of differential equations: