# Thermodynamic relations

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Thermodynamic relations
```PARTIAL DIFFERENTIATION
5.11 Thermodynamic relations
Thermodynamic relations provide a useful set of physical examples of partial
diﬀerentiation. The relations we will derive are called Maxwell’s thermodynamic
relations. They express relationships between four thermodynamic quantities describing a unit mass of a substance. The quantities are the pressure P , the volume
V , the thermodynamic temperature T and the entropy S of the substance. These
four quantities are not independent; any two of them can be varied independently,
but the other two are then determined.
The ﬁrst law of thermodynamics may be expressed as
dU = T dS − P dV ,
(5.44)
where U is the internal energy of the substance. Essentially this is a conservation
of energy equation, but we shall concern ourselves, not with the physics, but rather
with the use of partial diﬀerentials to relate the four basic quantities discussed
above. The method involves writing a total diﬀerential, dU say, in terms of the
diﬀerentials of two variables, say X and Y , thus
∂U
∂U
dX +
dY ,
(5.45)
dU =
∂X Y
∂Y X
and then using the relationship
∂2 U
∂2 U
=
∂X∂Y
∂Y ∂X
to obtain the required Maxwell relation. The variables X and Y are to be chosen
from P , V , T and S.
Show that (∂T /∂V )S = −(∂P /∂S)V .
Here the two variables that have to be held constant, in turn, happen to be those whose
diﬀerentials appear on the RHS of (5.44). And so, taking X as S and Y as V in (5.45), we
have
∂U
∂U
T dS − P dV = dU =
dS +
dV ,
∂S V
∂V S
and ﬁnd directly that
∂U
∂S
=T
and
V
∂U
∂V
= −P .
S
Diﬀerentiating the ﬁrst expression with respect to V and the second with respect to S, and
using
∂2 U
∂2 U
=
,
∂V ∂S
∂S∂V
we ﬁnd the Maxwell relation
∂T
∂P
=−
.
∂V S
∂S V
176
5.11 THERMODYNAMIC RELATIONS
Show that (∂S/∂V )T = (∂P /∂T )V .
Applying (5.45) to dS, with independent variables V and T , we ﬁnd
∂S
∂S
dU = T dS − P dV = T
dV +
dT − P dV .
∂V T
∂T V
Similarly applying (5.45) to dU, we ﬁnd
∂U
∂U
dU =
dV +
dT .
∂V T
∂T V
Thus, equating partial derivatives,
∂U
∂S
=T
−P
∂V T
∂V T
and
But, since
∂2 U
∂2 U
=
,
∂T ∂V
∂V ∂T
it follows that
∂S
∂V
+T
T
∂2 S
−
∂T ∂V
∂P
∂T
∂
∂T
i.e.
∂U
∂V
∂U
∂T
=T
V
=
T
∂
∂V
∂S
∂T
∂U
∂T
.
V
,
V
∂
∂2 S
∂S
T
=T
.
∂V
∂T V T
∂V ∂T
=
V
Thus ﬁnally we get the Maxwell relation
∂P
∂S
=
.
∂V T
∂T V
The above derivation is rather cumbersome, however, and a useful trick that
can simplify the working is to deﬁne a new function, called a potential. The
internal energy U discussed above is one example of a potential but three others
are commonly deﬁned and they are described below.
Show that (∂S/∂V )T = (∂P /∂T )V by considering the potential U − ST .
We ﬁrst consider the diﬀerential d(U − ST ). From (5.5), we obtain
d(U − ST ) = dU − SdT − T dS = −SdT − P dV
when use is made of (5.44). We rewrite U − ST as F for convenience of notation; F is
called the Helmholtz potential. Thus
dF = −SdT − P dV ,
and it follows that
∂F
∂T
= −S
and
V
∂F
∂V
Using these results together with
∂2 F
∂2 F
=
,
∂T ∂V
∂V ∂T
we can see immediately that
∂S
∂V
=
T
∂P
∂T
which is the same Maxwell relation as before. 177
,
V
= −P .
T
```
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