thermodynamic property relations - auburn university · thermodynamic property relations general...
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Objectives• Understand how thermodynamicists find
properties that can’t be directly measured
–What can be measured?
• Temperature, pressure, volume, mass
–Other properties?
• Internal Energy, enthalpy, . . .
• General relations for property changes
• Ideal Gases
Multivariable Calculus
• Functions of multiple variables:
• Partial derivatives:
• Total derivative:
z z(x,y)
y x
z zFirst derivatives : or
x y
y x
z zdz dx dy
x y
Eq. 1
Multivariable Calculus
• Functions of multiple variables:
• Partial derivatives:
• Total derivative:
x x(y,z)
yz
x xFirst derivatives : or
y z
yz
x xdx dy dz
y z
Eq. 2
Multivariable Calculus• Combine the two expressions for total
derivative: Put equation 2 into equation 1
y yz x
z x x zdz dy dz dy
x y z y
y y yz x
z x z z xdz dy dz
x y y x z
y y y z x
z x z x z1 dz dy
x z x y y
Multivariable Calculus
• This equation is true for all values of dy and dz
• It’s true if dy is zero and dz is not zero
• It’s true if dz is zero and dy is not zero
y y y z x
z x z x z1 dz dy
x z x y y
Multivariable Calculus
• The only way both of those can be true is:
y y
z x1 0
x z
y z x
z x z0
x y y
y y y z x
z x z x z1 dz dy
x z x y y
Eq. 3
Eq. 4
Reciprocity Relation• Equation 3 says
• Turns out just like algebra
y y
z x1 0
x z
y y
z x1
x z
y
y
z 1
xx
z
Cyclic Relation• Equation 4 says
• Using Reciprocity
y z x
z x z0
x y y
y z x
z x z
x y y
y xz
z x y1
x y z
Cyclic Relation - Example
• Let’s use a multivariable function we’re all familiar with:
RT RTv v(T,P) and P
P v
2
P P T v
v R T P v RT P R; ;
T P v R P P T v
2
P T v
T v P P RT R RT1
v P T R P v Pv
Let’s Apply These Rules• Find Enthalpy change in terms of things we
can measure
• From Calculus
• Using the definition of Cp
P T
h hh h(T,P) dh dT dP
T P
p
T
hdh C dT dP
P
Enthalpy Change• Hold that thought, we’ll be back
• From Calculus
• Put the above in 1st Law: dh = Tds + vdPP T
s ss s(T,P) ds dT dP
T P
P T
s sdh T dT dP vdP
T P
Enthalpy Change• Group like terms
• Recall:
P T
s sdh T dT T v dP
T P
p
T
hdh C dT dP
P
p
P T T
s h sC T ; T v
T P P
Enthalpy Change• Focusing on the change in h with respect to P
• One of the Maxwell relations (Eq. 12) is
• yielding
T T
h sT v
P P
T P
s v
P T
T P
h vv T
P T
Enthalpy Change• Putting it all together
• Everything on the right hand side can be measured . . . except Cp.
p
P
vdh C dT v T dP
T
Calorimetry• If two substances exchange heat with one
another . . .
• And we know the specific heat of one of the substances (1) but not the other (2) . . .
• We can measure everything we need to know to find the specific heat of the unknown in terms of the known.
• So, we count Cp as a measurable quantity.
1 p,known 1 2 p,unknown 2m C T m C T
Enthalpy Change• A finite enthalpy change:
• Note we’ve found an enthalpy CHANGE, not an enthalpy
2 2 2
p1 1 1P
vdh C dT v T dP
T
2 2
2 1 p1 1P
vh h C dT v T dP
T
Enthalpy Change
• This reminds us the property tables give us the enthalpy change from some arbitrarily chosen reference point.
2 2
2 1 p1 1P
vh h C dT v T dP
T
1 1
1 0 p0 0P
vh h C dT v T dP
T
Enthalpy Change• This reminds us the property tables give us the
enthalpy (for example) change from some arbitrarily chosen reference point.
• Not a problem:
2 2
2 0 p0 0P
vh h C dT v T dP
T
1 1
1 0 p0 0P
vh h C dT v T dP
T
2 0 1 0 2 1h h h h h h
Enthalpy Change – Ideal Gas
• What’s the enthalpy change for an ideal gas, Pv=RT? Let’s look at the integrand on the right
2 2
2 1 p1 1P
vh h C dT v T dP
T
P P
v RT R
T T P P
P
v RTv T v 0
T P
Enthalpy Change – Ideal Gas
• The integral with respect to P drops out, so
P
v RTv T v 0
T P
2
2 1 p1h h C dT
2 1 p,avg 2 1h h C T T
Internal Energy Change• Find Internal Energy change in terms of things
we can measure
• From Calculus
• Using the definition of Cv
v T
u uu u(T,v) du dT dv
T v
v
T
udu C dT dv
v
Internal Energy Change
• From Calculus
• Put the above in 1st Law: du = Tds - Pdv
v T
s ss s(T,v) ds dT dv
T v
v T
s sdu T dT dv Pdv
T v
Internal Energy Change• Group like terms
• Recall:
v T
s sdu T dT T P dv
T v
v
T
udu C dT dv
v
v
v T T
s u sC T ; T P
T v v
Internal Energy Change• Focusing on the change in u with respect to v
• One of the Maxwell relations (Eq. 11) is
• yielding
T T
u sT P
v v
T v
s P
v T
T v
u PT P
v T
Internal Energy Change• Putting it all together
v
v
Pdu C dT T P dv
T
2 2 2
v1 1 1v
Pdu C dT T P dv
T
2 2
2 1 v1 1v
Pu u C dT T P dv
T
Internal Energy Change, Pv=RT• What do you suppose happens to the last
integral if the substance is an Ideal gas?
2 2
2 1 v1 1v
Pu u C dT T P dv
T
v v
P RT R
T T v v
v
P RTT P P 0
T v
Collecting Results• General:
• Ideal Gas:
2 2
2 1 p1 1P
vh h C dT v T dP
T
2 2
2 1 v1 1v
Pu u C dT T P dv
T
2
2 1 p1h h C dT
2
2 1 v1u u C dT