Thermodynamic Property Relations

General

Relations, Part 1

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

• That’s not like algebra!

y xz

z x y1

x y z

y xz

z x y

x y z

= -1?

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

Problems?• To be solved in class!