Question

8.3-2. Show that the fundamental equation of a monatomic ideal gas satisfies the criteria of intrinsic stability :-fe Aka oritoris

Answer 1

A. Single-Component Ideal Gas
We define a single-component ideal gas by saying that it has to have the
following two equations of state:
PV = NRT (1)
U = cNRT. (2)
Here P is the pressure, T is the temperature in K, U is the internal energy in
joules, N is the mole number, R is the gas constant (= 8.3144 J/(mole-K),
and c is a dimensionless constant. (c = 3/2 for a monatomic ideal gas, and
usually c = 5/2 for a diatomic ideal gas.)
From these two equations, we can get the “fundamental equation” for the
ideal gas, in the form S(U, V, N). First, we rewrite the equations of state as
1
T
=
cNR
U
=
Ã
∂S
∂U !
V,N
(3)
P
T
=
NR
V
=
Ã
∂S
∂V !
U,N
(4)
Eqs. (3) can be integrated to give
S(U,V,N) = cNR ln(U) + f(V,N) (5)
while eq. (4) can be similarly integrated to give
S(U,V,N) = NR ln(V ) + g(U,N). (6)
Eqs. (5) and (6) are consistent only if we can write S in the form
S(U,V,N) = cNR ln(U) + NR ln(V ) + k(N), (7)
where k(N) is a function of N alone.
Now, let us rearrange eq. (7) so that the expression for S satisfies certain
properties. Namely,• S must be extensive in U, V, and N; and
• the arguments of the logarithms must be dimensionless (otherwise, the
logarithm doesn’t make sense).
We can achieve both goals by rewriting S as
S = cNR ln µ
U
Nu0
¶
+ NR ln µ
V
Nv0
¶
+ Ns0. (8)
Here u0, v0, and s0 are constants with the dimensions of energy per mole, vol-
ume per mole, and entropy per mole. Eq. (8) thus implies that S(Nv0,Nu0) =
Ns0.
As a check, we can now get out the equations of state from this “funda-
mental equation.” Specifically, we have
1
T
=
Ã
∂S
∂U !
V,N
=
cNR
U
, (9)
and
P
T
=
Ã
∂S
∂V !
U,N
=
NR
V
, (10)
and
µ
T
= −
Ã
∂S
∂N !
U,V
= −
·
cR ln µ
U
Nu0
¶
+ R ln µ
V
Nv0
¶
+ s0
¸
+ (c + 1)R. (11)
Any one of these three equations can be expressed in terms of the other two.
Rather than writing S as a function of U, V, and N, we could also have
written it in terms of T, V, and N, using the relation (2) between U and T.
The result is
S(T,V,N) = NRc ln µ
T
T0
¶
+ NR ln µ
V
Nv0
¶
+ Ns′
0
, (12)
where T0 = u0/R and we have written s
′
0 = s0 + Rc ln c.

This is for single component