Download - Chapter 2-5
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Chapter 2
The First Law
Unit 5 state function and exact differentialsSpring 2009
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State function and Path function
State functiona property that is independent of how a sample is prepared.
example : T, P, U, H
Path functiona property that is dependent on the preparation of the state.
depends on the path between the initial and final states
example : W, q
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Example 2.7
Consider a perfect gas inside a cylinder fitted with a piston. Let the initial state be T, Vi and the final state be T,Vf. The change of state can be brought about in many ways, of which the two simplest are the following:
Calculating work, heat, and internal energyPath 1, in which there is free expansion against zero external pressure;
Path 2, in which there is reversible, isothermal expansion.
Calculate w, q, and U and DHfor each process.
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Example 2.7
Path 1isothermal free expansionIsothermal DU=0, DH=0
DU=q+w = 0q=-w
free expansion w = 0, q=0
Path 2isothermal reversible expansionIsothermal DU=0, DH=0
DU=q+w = 0q=-w
reversible expansion
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Self Test 2.8
Calculate the values of q, w, and U, DH for an irreversible isothermal expansion of a perfect gas against a constant nonzero externalIrreversible isothermal expansion
Isothermal DU=0, DH=0
DU=q+w = 0q=-w
Irreversible expansion w = - Pex DV , q= Pex DV
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Change in internal energy, DU
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Change in internal energy, DU
Internal pressure
Constant-pressure heat capacity
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Internal pressure
The variation of the internal energy of a substance as its volume is changed at constant temeperature.For a perfect gas pT = 0 For real gasesattractive forcepT > 0
repulsive forcepT < 0
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Internal pressure
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Joule experiment
Expands isothermally against vacuum (pex=0)w=0, q=0 so DU=0
and pT=0
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DU at constant pressure
Expansion coefficient (a):the fraction change in volume with a rise in temperature
Isothermal compressibility (kT):the fractional change in volumewhen the pressure increases in small amount
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E 2.32 b
The isothermal compressibility of lead at 293 K is 2.21 106 atm1. Calculate the pressure that must be applied in order to increase its density by 0.08 per cent. -
Example 2.8
Calculating the expansion coefficient of a gasDerive an expression for the expansion coefficient of a perfect gas.
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DU at constant pressure
For perfect gas pT = 0, -
Change in enthalpy, DH
(chain relation)
Joule-Thomson coefficient
m =
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Joule-Thomson coefficient, m
A vapour at 22 atm and 5C was allowed to expand adiabatically to a final pressure of 1.00 atm; the temperature fell by 10 K. Calculate the JouleThomson coefficient, , at 5C, assuming it remains constant over this temperature range. -
Joule-Thomson coefficient, m
For perfect gases m = 0For real gasesm > 0gas cools on expansion
m < 0gas heats on expansion
Inversion temperature -
Exercise 2.29a
When a certain freon used in refrigeration was expanded adiabatically from an initial pressure of 32 atm and 0C to a final pressure of 1.00 atm, the temperature fell by 22 K. Calculate the JouleThomson coefficient, , at 0C, assuming it remains constant over this temperature range. -
Joule-Thomson effect
Adiabatic process q=0, DU=wPi > PfOn the left
Cooling by isenthalpic expansionisothermal irreversible compression
Pi,Vi,Ti Pi,0,Ti
w1= -pi ( 0 - Vi )= pi Vi
On the rightisothermal irreversible expansion
Pf,0,Tf Pf,Vf,Tf
w2= -pf ( Vf - 0 )= -pf Vf
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Joule-Thomson effect
w = w1 + w2 = pi Vi - pf Vf w = DU=Uf -Ui = pi Vi - pf Vf Uf + pf Vf = Ui + pi Vi Hf = Hi
Cooling by isenthalpic expansionJoule-Thomson effect is an isenthalpic process
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Isothermal Joule-Thomson coefficient
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Liquefaction of gases
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Liquefaction of gases
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Review 1
Define internal pressure pTProve that, for ideal gas, pT = 0 -
Review 2
Define Expansion coefficient aDefine Isothermal compressibility kTProve that for ideal gasa= 1/T
kT= 1/p
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Review 3
Define Joule-Thomsom coefficientProve that Joule-Thomson experiment is an isentahlpic process. Explain the principle of using Joule-Thomson effect to liquefy gases.=
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