KU Thermo notes - slide set 2

8/4/2019 KU Thermo notes - slide set 2

http://slidepdf.com/reader/full/ku-thermo-notes-slide-set-2 1/20

First Law (again)

We have seen two forms of the first law:

dU =∂U

∂T V dT −

∂U

∂V T dV

dU = ∂U

∂ S V

dS − ∂U

∂V S

dV

Where C is the heat capacity of the system.

More to come on C later.

So, the internal energy is a function of what two variables?

U = U(T,V)

U = U(S,V)

dU =C dT − P dV

dU =T dS − PdV



Gay-Lussac Experiment

Upon opening the valve, there is no change in T.

What is the significance of this fact? Which form of U dowe want?

dU =C V

dT − P dV

dU = ∂U

∂T V

dT − ∂U

∂V T

dV



Ideal Gas

For an IDEAL GAS:

∂U

∂V T

=0

Which means that for an ideal gas, U(T,V) = U(T).

Return to the first homework problem:

V

PA

B

What is the entropy changeduring isothermal expansion of an ideal gas piston from A to B?



Ideal Gas

V

PA

B

What is the entropy change during adiabatic expansion of an ideal gaspiston from A to B?



Ideal Gas

What is the entropy change during free expansion of an ideal gas from VA

to VA+B

?

Does no work, transfers no heat, therefore U is constant,which means T is constant.

Since the PV state variables are path independent, usethe isothermal expansion case to determine the entropychange.



What is entropy?

Why is it useful?

Thigh

Tlow

Consider two cubes in a box at differenttemperatures. A small amount of energymoves from the higher temperature cube to

the lower temperature cube.

S high=−Q

T high

S low=Q

T low

S total =Q 1

T low

−1

T high0



What is entropy?

Why is it useful?

Phigh

Plow

Consider two coupled pistons containing anideal gas at different pressures pushingagainst each other isothermally.

dS high=Qhigh

T

dS total =1

T P high− P low dV

= P high dV

T

dS low=− P low dV

T



Entropy?



Entropy

Entropy is the opposite of information.

The higher the entropy, the less is known aboutthe system.

HW: 2.4, 2.9, 2.10, 2.11, 2.13, 3.6



Mathematical background

∂∂ x ∂ z

∂ y x

y

= ∂∂ y ∂ z

∂ x y

x

1

∂ z

∂ y x

= ∂ y

∂ z x

∂ z

∂ y x

∂ y

∂ x z

∂ x

∂ z y=−1



Legendre Transformation

x,y

,

Every point on the curve can be described by aunique (x,y) coordinate or a ( , ) coordinate.

y dy

dx = y x −dy

dxx



Enthalpy

An example of a Legendre Transformation:

How can we write U(S,V) without the V?

U S , ∂U

∂V S

=U S , V − ∂U

∂V S

V

But . . .

From the first law we know that P =− ∂U

∂V S

So we call the new function Enthalpy and write

H S , P =U S , V P V

H =U P V Or simply



What good is Enthalpy?

Adiabatic throttling

Pressure and Volume both change as the system passes the throttle.

∆U = P1

V1

– P2

V2

Since ∆U = U2

– U1, we can rearrange to make

U2

+ P2

V2

= U1

+ P1

V1

H2

= H1



dH

H =U PV

dH =dU P dV VdP

dH =T dS − P dV P dV VdP

dH =T dS VdP

The term V dP can be thought of as non-mechanical work.So the first law still applies:

dH =δQ + V dP

But this time the work term is non-mechanical.



Isobaric Expansion

V

PA B

∆U = ∆Q - ∆W

U2

- U1

= ∆Q - (P2

- P1) V

H2

- H1

= ∆Q

Where do isobaric processes occur?



Specific Heat

We just determined that under isobaric conditions that

H2- H

1= ∆Q

We can also calculate the heat change as

∆Q = C∆T

Together these imply that

C P = H

T

1

n ∂ H

∂T P

The subscript P indicates constant pressure (isobaric),and the n normalizes the energy per mole.

Similarly, under constant volume,

C V =U

T

1

n ∂U

∂T V



Specific Heat notes

C V =Q

T =

1

n ∂U

∂T V

Note that ∆Q is not an exact differential, i.e. the magnitude of ∆Q is path dependent.However the heat capacity is an exact quantity. That is why heat capacities mustspecify a path, either constant pressure or volume.

C V =Q

T V

S =Q

T

Also, don't confuse calculations for entropy and heat capacity.

We had already decided that U was a function of T only (ideal gas). So we canassume:

C V =1

n ∂U

∂T V

dU =n C V dT



∆H during isothermal expansion

V

PA

B

What is the enthalpy change during adiabatic expansion of an ideal gas

piston from A to B?

H =U PV

H =U T n R T

Isothermal implies no change



Gas Constant: R

Lets assume a small temperature change, such that CV and CP can be consideredconstant.

dH =n C P dT dU =n C V dT and

From the first law we have:

Q=dU P dV Q=dH −V dP

Equate Qs, substitute dU and dH . . .

n C P dT −V dP =n C V dT P dV

n C P −C V dT = P dV V dP

C P −C V =1

n

d PV dT

= R



Homework 2.5, 3.8

Another constant which occurs frequently is

. =

C P

C V

You open a valve to fill an evacuated tank with air. The tank is insulated, the air isan ideal gas. What is the temperature of the gas in the tank after it is filled?

UF

- U0

= -∆W

T0, P

0, V

0

= P0

V0

UF

= U0

+ P0

V0

= H0

CVT

F= C

PT

0

TF

= T0

p2

p1

= T 2

T 1

−1

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KU Thermo notes - slide set 2