chapter 3 : slide 1 chapter 3 the first law of thermodynamics the machinery

Chapter 3 : Slide 1

Chapter 3

The First Law of Thermodynamics

The Machinery

Chapter 3 : Slide 2

OUTLINE

Partial Derivatives and the First Law

1. Partial Derivatives and the Internal Energy, U:Internal Pressure, T, and the Expansion Coefficient,

2. Partial Derivatives and the Enthalpy, H:Isothermal Compressibility, T, and the Joule-Thomson Coefficient,

3. The Relation Between Cp and CV.

Chapter 3 : Slide 3

Preamble (little bit of a refresher):

A state function is also known as a state property!

State property is an intuitive name for what we mean i.e. a state property is something like pressure, temperature, internal energy.

The “state” in state property reminds us that the properties we are talking about are only state dependent that is to say that the difference between a state property in one state and another state is independent of how we got from the old state to the new state.

AB

B

A

UUdUU

Chapter 3 : Slide 4


There is such a thing as a path property or path function.

An example of a path property is work, w.

The amount of work done in going from an initial state, i, to a final state, f, is totally dependent on the path.

This means we CANT write:

AB

B

A

wwdww

B

A

dwwpath,

Instead we have to write:

Wrong!

We say dw is an inexact differential

Chapter 3 : Slide 5


A state function is also known as a state property!

The term “state function” reminds us that the properties of a system are inter-related e.g. for a one component system:

p = f ( n, V, T ) recall pV = nRT

And also:

U = f ( n, V, T ) or U = f ( n, V, p ) or U = f ( n, p, T )

The multivariable nature of these state functions means that the use of partial derivatives is very common in thermodynamics!

Chapter 3 : Slide 6


Homework from this lecture: Exercises 3.4 – 3.7

3.4(a) part (a)

Show that x2y+3y2 has an exact differential.

Chapter 3 : Slide 7

Partial Derivatives and the Internal Energy

For a closed system of constant composition:

U = f ( V, T )

Constant T, change V to V+dV, then U changes to:

dVV

UUU

T

'

Constant V, change T to T+dT, then U changes to:

dTT

UUU

V

'

Change V to V+dV and T to T+dT, then U changes to:

dTT

UdV

V

UUU

VT

'

Because dU = U'-U, then:

dTT

UdV

V

UdU

VT

Chapter 3 : Slide 8


dTT

UdV

V

UdU

VT

A small (infinitesimal) change in internal energy is proportional to small changes in volume and temperature.

Recall:V

V T

UC

We introduce a new term, T, the internal pressure. T

T V

U

dTCdVdU VT

Chapter 3 : Slide 9


For NH3, T,m = 840 Pa at 300K and 1 bar and CV,m = 27.32 J K-1 mol-1. Calculate the change in molar internal energy as NH3 is heated from 298 K to 300 K while being compressed from 1 L to 0.9 L

Chapter 3 : Slide 10


Consider an isothermal, closed system of constant composition, i.e. T is constant i.e dT = 0

dVdU T

T gives an indication of how the internal energy changes with respect to changes in volume.

T gives a measure of the strength of the cohesive force in the sample.



Let us now consider the change in internal energy that accompanies a change in T when p is kept constant i.e. now U = f ( p, T ).

dTCdVdU VT We had written:

Lets divide by dT VT CdT

dV

dT

dU

And impose constant p Vp

Tp

CT

V

T

U



Vp

Tp

CT

V

T

U

We introduce a new term, , the expansion coefficient.pT

V

V

1

A large indicates that the samples volume responds strongly to changes in temperature.

VTp

CVT

U

Exactly the same as on the previous slide


Partial Derivatives and the Enthalpy

dTT

Hdp

p

HdH

pT

Just like U, the enthalpy, H, is a state function, we can write H= f (p, T).

Recall:p

p T

HC

dTCdpp

HdH p

T

We may manipulate this to give:

pTV

CT

H

1

T is the isothermal compressibility

is the Joule-Thomson coefficient


Partial Derivatives and the Enthalpy

pTV

CT

H

1

T is the isothermal compressibilityT

T p

V

V

1

is the Joule-Thomson coefficientH

p

T

Exactly the same as on the previous slide


3.16(b)

The isothermal compressibility of lead at 293 K is 2.21 10-6 atm-1. Calculate the pressure that must be applied in order to increase its density by 0.08 %.


The Perfect gas and T, , T,

Property PD Value for Perfect Gas Info:

TT V

U

Internal Pressure T=0 Strength/nature of

interactions between molecules

pT

V

V

1

Expansion Coefficient

=1 / TThe higher T, the less responsive is its volume to a change in temperature

T

T p

V

V

1Isothermal

CompressibilityT=1 / p The higher the p, the

lower its compressibility

Hp

T

Joule-Thomson Coefficient

=0 Another indication ofmolecular interactions.




TT V

U




The JOULE EXPERIMENTTo measure T i.e. test U as a function of V for a real gas.

Joule detected no temperature change i.e. q = 0.

Any work done? NO, so w = 0.

Interpretation U = q + w = 0 (1st law).

U does not change with a change in V, i.e. T = 0.

This is not a very sensitive experiment because water has a much higher Cp than air.

TT V

U


James Joule and the Gothic Era.




TT V

U



pT

V

V

1

Expansion Coefficient

=1 / TThe higher T, the less responsive is its volume to a change in temperature

T

T p

V

V

1Isothermal

CompressibilityT=1 / p The higher the p, the

lower its compressibility

Hp

T

Joule-Thomson Coefficient

=0 Another indication ofmolecular interactions.


THE JOULE-THOMPSON EXPERIMENTA further test of intermolecular forces in real gases.

Imagine a sample of gas pushed through a porous plug, in an isolated tube (adiabatic system). The temperature is measured on each side of the plug.

Analysisw = piVi - pfVf

Since U = Uf - Ui = w (because q = 0),

Uf + pfVf = Ui + piVi

i.e. H = 0

This is a constant enthalpy (isenthalpic) process.

Hp

T


If > 0 then this indicates that the gas cools when expanded (-ve p).If < 0 then this indicates that the gas cools when compressed (+ve p).

The "inversion temperature" indicates where on a T,p plot µ flips from +ve to –ve. For a real gas µ is non-zero (except at the inversion temperature) and thus H shows some variation with p.

The Linde process

THE JOULE-THOMPSON EXPERIMENT

Hp

T


3.13(b)

A vapor at 22 atm and 5o C was allowed to expand adiabatically to a final pressure of 1.00 atm; the temperature fell by 10 K. Calculate the Joule-Thomson Coefficient, , at 5o C, assuming it remains constant over this temperature range.


The relation between CV and Cp

pp T

HC

Recall:Vp

V T

UC

or

Therefore:Vp

Vp T

U

T

HCC

nRT

UnR

T

UCC

VpVp

For a perfect gas

Cp differs from CV by the work needed to change the volume of the system. This work is (a) the work required to push back the atmosphere and (b) the work of stretching the bonds of the material.


The relation between CV and Cp

For any substance:

TVp

TVCC

2

chapter 3 : slide 1 chapter 3 the first law of thermodynamics the machinery

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state property

t partial derivatives

new state

state dependent

state functions

old state

final state

initial state