Transcript Chapter 2 The First Law Unit 5 state function and exact

Chapter 2
The First Law
Unit 5 state function and exact differentials
Spring 2009
State function and Path function
• State function
a property that is independent of how a sample is
prepared.
example : T, P, U, H …
• Path function
a property that is dependent on the preparation of
the state.
depends on the path between the initial and final
states
example : W, q …
Example 2.7
Calculating work, heat, and internal energy
• 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:
Path 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.
Example 2.7
• Path 1
isothermal free expansion
Isothermal DU=0, DH=0
DU=q+w = 0
q=-w
free expansion w = 0, q=0
• Path 2
isothermal reversible expansion
Isothermal DU=0, DH=0
DU=q+w = 0
q=-w
Vf 
reversible expansion w = nRT ln 
 Vi 
Vf 
q = nRT ln 
 Vi 
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 external
Irreversible isothermal expansion
Isothermal
DU=0, DH=0
DU=q+w = 0 q=-w
Irreversible expansion w =  Pex DV , q= Pex DV
Change in internal energy, DU
Change in internal energy, DU
U 

Internal pressure  T = 

 V T
Constant-pressure heat capacity
 U 
CV = 

 T V
dU =  T dV  CV dT
Internal pressure
• The variation of the internal energy of a substance as its
volume is changed at constant temeperature.
 U 
T = 

 V T
• For a perfect gas
• For real gases
attractive force
repulsive force
T = 0
T > 0
T < 0
Internal pressure
 U 
 P 

 = T   P
 V T
 T V
nRT
For ideal gas P =
V
 nRT 


nR
 U 
 P 
V


P =T
P=0

 = T
P =T
V
 T 
 V T
 T 




Joule experiment
• Expands isothermally against
vacuum (pex=0)
w=0, q=0 so DU=0
and T=0
DU at constant pressure
dU =  T dV  CV dT
 V 
 U 

=
  CV


T 
 T  P
 T  P
• Expansion coefficient (a):
the fraction change in volume with
a rise in temperature
1  V 
α= 

V  T  p
• Isothermal compressibility (kT):
the fractional change in
volumewhen the pressure
increases in small amount
1  V 
kT =   
V  P T
E 2.32 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 per cent.
kT = 
1  V

V  p

DV
  
VDp
T
1
DV
Dp = 

V kT
Example 2.8
Calculating the expansion coefficient of a gas
Derive an expression for the expansion coefficient
of a perfect gas.
DU at constant pressure
 U 
 V 

 = T 
  CV = a TV  CV
 T  P
 T  P
• For perfect gas T = 0,
 U 

 = CV
 T  P
Change in enthalpy, DH
(chain relation)
Joule-Thomson coefficient
 T 

 p  H
= 

dH = CP dp  C p dT
Joule-Thomson coefficient, 
 T 
 =  
 p  H
• A vapour at 22 atm and 5°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, µ, at 5°C,
assuming it remains constant over this temperature range.
 T 
DT
 10 K
 =   
=
= 0.48 K atm
 p  H Dp (1.00  22) atm
Joule-Thomson coefficient, 
• For perfect gases  = 0
• For real gases
 > 0 gas cools on expansion
 < 0 gas 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 0°C to a final pressure
of 1.00 atm, the temperature fell by 22 K.
Calculate the Joule–Thomson coefficient, µ, at
0°C, assuming it remains constant over this
temperature range.
 T 
DT
 22 K
 =   
=
= 0.71 K atm-1
 p  H Dp (1.00  32) atm
Joule-Thomson effect
Cooling by isenthalpic expansion
• Adiabatic process q=0, DU=w
• Pi > Pf
• On the left
isothermal irreversible compression
Pi,Vi,Ti → Pi,0,Ti
w1= -pi ( 0 - Vi )= pi Vi
• On the right
isothermal irreversible expansion
Pf,0,Tf → Pf,Vf,Tf
w2= -pf ( Vf - 0 )= -pf Vf
Joule-Thomson effect
Cooling by isenthalpic expansion
• w = w1 + w2 = pi Vi - pf Vf
• w = DU=Uf -Ui = pi Vi - pf Vf
• Uf + pf Vf = Ui + pi Vi
• Hf = Hi
Joule-Thomson effect is an isenthalpic process
Isothermal Joule-Thomson coefficient
 H 
 = C p
T = 
 p T
Liquefaction of gases
 T 
 =  
 p  H
Liquefaction of gases
 T 
 =  
 p  H
Review 1
• Define internal pressure T
• Prove that, for ideal gas, T = 0
Review 2
• Define Expansion coefficient a
• Define Isothermal compressibility kT
• Prove that for ideal gas
a= 1/T
kT= 1/p
Review 3
• Define Joule-Thomsom coefficient
• Prove that Joule-Thomson experiment is an
isentahlpic process.
• Explain the principle of using Joule-Thomson
effect to liquefy gases.