Chapter 11, The First Law of Thermodynamics
Mechanical Work
 
dW
  F.d s
Work in a volume change. dF = P.dA
d' W  PdAdS
or
d' W  PdV
An inexact differential
(path dependant)
Note that the integral is -ve for work done on the system, +ve
for work done by the system.
W 
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Vb
Va PdV
Phys235 : Thermodynamics 3
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P = const. (isobaric)
WP
Ideal gas at constant T
(Isothermal)
W 

Vb
Va dV  P(Vb
 Va )
Vb
Vb nRT
dV  nRTn
Va
Va
V
Van der Waals gas

a 
P  2 V  b  nRT
 V 
P
nRT
a
 2
V  b V
Vb  b  1
1 
W  nRTn
 a  
Va  b Va Vb 

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Work depends on the path
W  abd' W  
The total work around the loop is
Vb
Va PdV
W   PdV
Sometimes the configuration
changes without doing work. e.g
free expansion of gas
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The First Law of Thermodynamics
The total work is always the same in all adiabatic
processes between any two equilibrium states having
the same kinetic and potential energy.
d
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Internal Energy
The difference between the internal energy in state a and
b is equal to the work done by the system along any
adiabatic path from a to b.
dU = -d’Wab
Ua-Ub=Wab
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Example: An ideal gas, and a block of copper, have equal volumes
of 0.5 m3 at 300 K and atmospheric pressure. The pressure on
both is increased reversibly and isothermally to 5 atm. (a) Explain
with the aid of a P- V diagram why the work is not the same in the
two processes. (b) In which process is the work done greater? (c)
Find the work done on each if the compressibility of the copper is
0.7 x 10-6 atm-1. (d) Calculate the change in volume in each case.
(a) Vgas changes significantly
Vcopper almost no change
(b) Wgas > Wcopper
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(c) For gas
V2 
P1 
nRT
Wgas   PdV  
dV  nRTn  P1V1n 
V
V1 
P2 

 5 N 
1 
3
Wgas  10 2  0.5m n  8.15x10 4 J
 m 
5 


1 V 
    
V P T
For copper
W copper

V 
  PdV   P  dP  V  PdP
P T
6
1 2
0.7x10
 V P2  P12  
0.5 5 2 12 x1.013x10 5
2
2
 0.42J

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
Phys235 : Thermodynamics 3
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
7
(d) Volume Change
For gas
P1V1
V2 
 0.1m 3
P2
Vgas = (0.5-0.1)m3=0.4 m3
For copper
V =V0[1-(P2-P1)]
= 0.5[1-0.7x10-6(5-1)]
= 0.5[1-2.8x10-6]
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≈V0
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Example 2:
A volume of 10 m3 contains 8 kg of oxygen at a temperature of
300 K. Find the work necessary to decrease the volume to 5 m3,
(a) at a constant pressure and (b) at constant temperature. (c)
What is the temperature at the end of the process in (a)? (d) What
is the pressure at the end of the process in (b) ? (e) Show both
processes in the P- V plane.
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8x10 3
mass
n

 0.25x10 3 mole
molecular weight
32
nRT
P

V


0.25x10 3 8.314 300
10
 6.23x10 4 Nm2
(a) P = const
W = ∫PdV = P(V2-V1)=6.23x104(5-10)=-3.11x105 Joules
(b) T = const
V2 
1 
nRT
W   PdV  
dV nRTn  nRTn  4.32x10 5 J
2 
V
V1 

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(c) T2 = ? For (a)
T2 T1

V2 V1
T2 
T1
1
V2  300  150 K
V1
2
(d) P2 = ? For (b)
P1V1  P2V2
P1V1
4 2 
P2 
 6.23x10   12.46x10 4 Nm 2
1 
V2
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(e)
160x10
3
140
Pressure (Pa)
120
100
(b)
80
60
(a)
40
20
0
4
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6
8
3
Volume (m )
Phys235 : Thermodynamics 3
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12
12
Heat Flow
Q=Wtotal-Wadiabatic (i.e. non-adiabatic work)
Q -ve : heat flow out of the system
Q +ve : heat flow into the system
Using these quantities we can state the 1st law of thermodynamics
as
∆U=Ub-Ua=Q - W
Increase in internal energy = heat flow into the system - work done
by the system.
dU=d’Q-pdV
Note that d’Q depends on the path.
dU depends on the state of the system not the path taken.
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Mechanical Equivalent of Heat
1 Calorie = 1 g of H2O temperature increased by 1 K at
14.5 ˚C = 4.158 Joules
Heat Capacity
lim Q d'Q
Q
C
, C

T  0 T dT
T
Cp : P = const
Cv : V = cont
Cp and equation of state is sufficient to determine Cv
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Heat of transformation (latent heat) : 
A change of phase  change of volume  work is involved in
the process (1st order phase change).
At T=const. and P= const. w = P(v2- v1) (specific work).
From the 1st law
(u2- u1) =  - P(v2- v1)
 = (u2+ Pv2) - (u1+ Pv1).
If we define specific enthalpy h = u + Pv
then  = h2 - h1.
h is a state function whose value only depends on the state of the
system
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Heat flow in any reversible isobaric process is equal to the
change in enthalpy.
Terms in thermodynamics.
12 (solid-liquid)
Heat of Fusion
23 (liquid-vapour)
Heat of Vaporisation
13 (solid-vapour)
Heat of sublimation
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for H2O
23 = h'''- h''= 22.6x105 J kg-1
w = P(v'''- v'') = 1.8x105 J kg-1
(u'''- u'') = 23 - w = 20.8x105 J kg-1
So the enthalpy change is due to 92% change of internal energy
and 8% work done in expanding the H2O.
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Now
 dh  0
so if we have a cyclic process near the triple point in which the
substances goes from solid to vapor then back to liquid and
then to solid then
∆h1 + ∆h2 + ∆h3 = 0.
then 13 - 23 - 12 = 0
thus 13 = 23 + 12
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Phys235 : Thermodynamics 3
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General Form of the First Law
∆U + ∆Ek = Q - W
Where
∆U : Change of internal energy
∆Ek : Change of kinetic energy
and W =W* + Wc
where Wc is the work done by conservative force.
Wc = ∆Ep (potential energy)
So
∆U + ∆Ek + ∆Ep = Q - W*
∆E = Q - W*
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Phys235 : Thermodynamics 3
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