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Chapter 3 : Slide 1

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We say dw is an inexact differential. Chapter 3 : Slide 5. Preamble (little bit of a refresher) ... U does not change with a change in V, i.e. pT = 0. ... – PowerPoint PPT presentation

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Title: Chapter 3 : Slide 1


1
Chapter 3 (7th Ed.) Rest of Chapter 2 (8th
Ed.) The First Law of Thermodynamics The
Machinery
2
  • OUTLINE
  • Partial Derivatives and the First Law
  • 1. Partial Derivatives and the Internal Energy,
    U
  • Internal Pressure, ?T, and the Expansion
    Coefficient, ?
  • Partial Derivatives and the Enthalpy, H
  • Isothermal Compressibility, ?T, and the
    Joule-Thomson Coefficient, ?
  • 3. The Relation Between Cp and CV.

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.
4
Preamble (little bit of a refresher)
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
Wrong!
Instead we have to write
We say dw is an inexact differential
5
Preamble (little bit of a refresher)
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!
6
Preamble (little bit of a refresher)
Homework from this lecture Exercises 3.4 3.7
3.4(a) part (a) Show that x2y3y2 has an exact
differential.
7
Partial Derivatives and the Internal Energy
For a closed system of constant composition
U f ( V, T )
8
Partial Derivatives and the Internal Energy
A small (infinitesimal) change in internal energy
is proportional to small changes in volume and
temperature.
We introduce a new term, ?T, the internal
pressure.
9
Partial Derivatives and the Internal Energy
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
10
Partial Derivatives and the Internal Energy
Consider an isothermal, closed system of constant
composition, i.e. T is constant i.e dT 0
?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.
11
Partial Derivatives and the Internal Energy
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 ).
We had written
Lets divide by dT
And impose constant p
12
Partial Derivatives and the Internal Energy
Exactly the same as on the previous slide
We introduce a new term, a, the expansion
coefficient.
A large a indicates that the samples volume
responds strongly to changes in temperature.
13
Partial Derivatives and the Enthalpy
Just like U, the enthalpy, H, is a state
function, we can write H f (p, T).
Recall
We may manipulate this to give
?T is the isothermal compressibility m is the
Joule-Thomson coefficient
14
Partial Derivatives and the Enthalpy
Exactly the same as on the previous slide
?T is the isothermal compressibility
m is the Joule-Thomson coefficient
15
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 .
16
The Perfect gas and pT, a, kT, m
Property
PD
Value for Perfect Gas
Info
17
The Perfect gas and pT, a, kT, m
Property
PD
Value for Perfect Gas
Info
18
The JOULE EXPERIMENT To measure pT 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
DU q w 0 (1st law). U does not change
with a change in V, i.e. pT 0. This is not a
very sensitive experiment because water has a
much higher Cp than air.
19
James Joule and the Gothic Era.
20
The Perfect gas and pT, a, kT, m
Property
PD
Value for Perfect Gas
Info
21
THE JOULE-THOMPSON EXPERIMENT A 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. Analysis w piVi - pfVf Since DU Uf
- Ui w (because q 0), Uf pfVf Ui
piVi i.e. DH 0 This is a constant
enthalpy (isenthalpic) process.
22
THE JOULE-THOMPSON EXPERIMENT
If m gt 0 then this indicates that the gas cools
when expanded (-ve Dp). If m lt 0 then this
indicates that the gas cools when compressed (ve
Dp). 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
23
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, m, at 5o
C, assuming it remains constant over this
temperature range.
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