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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
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:
AB
B
A
wwdww
B
A
dwwpath,
Instead we have to write:
Wrong!
We say dw is an inexact differential
Chapter 3 : Slide 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!
Chapter 3 : Slide 6
Preamble (little bit of a refresher):
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
Partial Derivatives and the Internal Energy
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
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
Chapter 3 : Slide 10
Partial Derivatives and the Internal Energy
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.
Chapter 3 : Slide 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 ).
dTCdVdU VT We had written:
Lets divide by dT VT CdT
dV
dT
dU
And impose constant p Vp
Tp
CT
V
T
U
Chapter 3 : Slide 12
Partial Derivatives and the Internal Energy
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
Chapter 3 : Slide 13
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
Chapter 3 : Slide 14
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
Chapter 3 : Slide 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 %.
Chapter 3 : Slide 16
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.
Chapter 3 : Slide 17
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
Chapter 3 : Slide 18
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
Chapter 3 : Slide 19
James Joule and the Gothic Era.
Chapter 3 : Slide 20
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.
Chapter 3 : Slide 21
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
Chapter 3 : Slide 22
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
Chapter 3 : Slide 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, , at 5o C, assuming it remains constant over this temperature range.
Chapter 3 : Slide 24
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.
Chapter 3 : Slide 25
The relation between CV and Cp
For any substance:
TVp
TVCC
2