Modern Engineering Thermodynamics

60 CHAPTER 3: Thermodynamic Properties
EXAMPLE 3.1 (Continued )
Then, by multiplying this by (∂ρ/∂T) p and utilizing Eq. (3.2), we get the desired result:
∂p ∂ρ
= − ∂ρ
∂T ∂p ∂T
ρ
Exercises
1. If v =1/ρ is the specific volume of the material in Example 3.1, show that the result in this example can be written as
∂p ∂ν ∂ν
= −
∂T ∂p ∂T
ν
2. If a, b, and c are three independent intensive thermodynamic properties, use Eqs. (3.2) and (3.3) to show that they can
be related by
∂a
∂a
= −
∂c b
∂b c ∂b
∂c a
Τ
Τ
p
p
3.4 SOME EXCITING NEW THERMODYNAMIC PROPERTIES
In Chapter 2, we introduced the specific volume v, an intensive property, as
v = V/m (3.4)
where V is the total volume 1 and m is the total mass of the system. We are now free to establish v as a function of
any two other independent properties. For a single-phase (i.e., homogeneous) pure substance subjected to only one
work mode, the pressure and temperature are independent properties, and for such a system, we can then write
v = vp, ð TÞ
Differentiating this equation gives
dv =
∂v
dp +
∂v
dT
∂p
T
∂T p
The coefficients of dp and dT in the previous equation reflect the dependence of volume on pressure and temperature,
respectively. Because these terms have such important physical meaning, we introduce the following
notation:
and
1
v
− 1 v
∂v
∂T
= β = isobaric coefficient of volume expansion (3.5)
p
∂v
∂p
T
= κ = isothermal coefficient of compressibility (3.6)
where the thermodynamic term isobaric is from the Greek words iso meaning “constant” and baric meaning
“weight” or “pressure”; the term is to be taken to mean constant pressure in this text. Therefore, we can write
or
dv = −vκ dp + vβ dT
dv
= β dT − κ dp (3.7)
v
If κ and β are constant (or averaged) over small ranges of temperature and pressure, then Eq. (3.7) can be integrated
to give
ln v 2
= βðT 2 − T 1 Þ− κðp 2 − p 1 Þ
v 1
1 Remember that, in this text, volume is represented by the symbol V:. The symbol V is reserved for the magnitude of velocity.