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First Law of
Thermodynamics
Created by: Marlon Flores Sacedon
Physics section, DMPS
June 2010
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The First Law of Thermodynamics
Thermodynamic system is a system that can
interact (and exchange energy) with its
surroundings, or environment, in at least two
ways, one of which is heat transfer.
Thermodynamic process is a process in which
there are changes in the state thermodynamic
system.
Work Done during volume changes
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Work Done by the system
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Work Done by the system
A
F
dS
dW  F  ds  pA  ds  p dV
dW  pdV
V2
W   pdV
V1
Where:
W = work done by the system [J]
p = pressure [pa]
dV = differential volume [m3]
V1 & V2 = initial and final volume [m3]
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Work Done by the system
V2
W   pdV
V1
Where:
W = work done by the system
p = pressure
dV = differential volume
V1 & V2 = initial and final volume
p
V2
W  Area   pdV
V
1
V
0
V1
pV-diagram
V2
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Work Done by the system
Signs of work done
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Work Done by the system
If the pressure is constant during thermodynamic process
pV  nRT
W  p [ V2  V1 ]
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Problem
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Work Done by the system
Paths Between Thermodynamics States
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Internal Energy (U)
Internal Energy of a system is the sum of
kinetic energies of all of its constituent
particles, plus the sum of all the potential
energies of interaction among these
particles.
U  U 2  U1
Where:
U = change in internal energy
U1 = initial internal energy
U2 = final internal energy
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The First Law of
Thermodynamics
Surroundings
(environment)
Q = 150J
System
W = 100J
U = Q-W = +50 J
U 2  U1  U  Q  W
Q  U  W
Where:
U = change in internal energy (J)
W = work done (J)
Q = heat quantity (J)
Surroundings
(environment)
Q = -150J
System
W = -100J
U = Q-W = -50 J
Surroundings
(environment)
Q = 150J
System
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W = 150J
U = Q-W = 0
The First Law of Thermodynamics
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The First Law of Thermodynamics
Ex. A gas in a cylinder is held at a constant pressure of 2.30x105 Pa and is
cooled and compressed from 1.70 m3 to 1.20 m3. The internal energy of the
gas decreases by 1.40x105 J. a) Find the work done by the gas. b) Find the
absolute value of the heat flow into or out of the gas, and state the direction
of heat flow. c) Does it matter whether or not the gas is ideal? J, b)
2.55x105J, out of gas, c) no
(Ans. a) -1.15x105
Ex. A gas in a cylinder is held at a constant pressure of 2.30 x 105 Pa and is
cooled and compressed from 1.70 m3 to 1.20 m3. The internal energy of the
gas decreases by 1.40 x 105 J. a) Find the work done by the gas, b) Find the
absolute value |Q| of the heat flow into or out of the gas, and state the
direction of heat flow, c) Does it matter whether or not the gas if ideal? Why
or who not?
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Kinds of Thermodynamic Process
1. Adiabatic Process (pronounced “ay-dee-ah-bat-ic”) is defined as
one with no heat transfer into or out of a system: Q = 0.
U 2  U1  U  W
(adiabatic process)
2. Isochoric Process (pronounced “eye-so-kor-ic”) is a constantvolume process. When the volume of thermodynamic system is constant
W=0.
U  U  U  Q (isochoric process)
2
1
3. Isobaric Process (pronounced “eye-so-bear-ic”) is a constant –
pressure process.
W  p(V2  V1 )
(Isobaric process)
4. Isothermal Process (pronounced “eye-so-bear-ic”) is a constant
–temperature process.
V2
W  nRT 
V1
dV
V
p
 nRT ln 2  nRT ln 1
V
V1
p2
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(Isothermal process)
Kinds of Thermodynamic Process
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Internal Energy of an Ideal Gas
Property of Ideal Gas: The
internal energy of an ideal gas
depends only on its temperature,
and not on its pressure and
volume.
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Heat Capacity of an Ideal Gas
Molar heat capacity at constant volume (CV)
Molar heat capacity at constant pressure (Cp)
Q  U  W
or
dQ  dU  dW
At constant volume
dQ  nCV dT
dW  0
dQ  dU (from First Law)
dU  nCV dT (because dQ=dU)
(First Law)
At constant pressure
dQ  nC p dT
dW  pdV
dW  nRdT ( from pV=nRT )
nC p dT  dU  nRdT
nC p dT  nCV dT  nRdT
C p  CV  R

Cp
CV
(Molar heat capacities (ratio of heat
of an ideal gas)
capacities)
Where:
Cp = molar specific at constant pressure (J/mol.K)
CV = molar specific at constant volume (J/mol.K)
R = ideal gas constant
 initial and final volume

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Molar Heat Capacities of Gases

CV
Cp
Cp-CV
(J/mol.K) (J/mol.K) (J/mol.K) (J/mol.K)
Type of
Gas
Gas
Monatomic
He
12.47
20.78
8.31
1.67
Ar
12.47
20.78
8.31
1.67
H2
20.42
28.74
8.32
1.41
N2
20.76
29.07
8.31
1.40
O2
20.85
29.17
8.31
1.40
CO
20.85
29.16
8.31
1.40
CO2
28.46
36.94
8.48
1.30
SO2
31.39
40.37
8.98
1.29
H2S
25.95
34.60
8.65
1.33
Diatomic
Polyatomic
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Heat Capacity of an Ideal Gas
Molar heat capacities for Monatomic ideal gas
3
5
C p  CV  R  R  R  R
2
2
Cp
5
2
3
2
Cp
7
2
5
2
R 5


  1.67
CV
R 3
Molar heat capacities for Diatomic ideal gas
5
7
C p  CV  R  R  R  R
2
2
R 7


  1.40
CV
R 5
Molar heat capacities for Polyatomic ideal gas
7
9
C p  CV  R  R  R  R
2
2
Cp
9
2
7
2
R 9


  1.29
CV
R 7
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Example. In an experiment to simulate conditions within an automobile
engine, 645J of heat is transferred to 0.185 mol of air-conditioned within a
cylinder of volume 40.0cm3. Initially the nitrogen is at a pressure of 3.00x106
Pa and a temperature of 780K. a) If the volume of the cylinder is held fixed,
what is the final temperature of the air? Assume that the air is essentially
nitrogen gas, use the Table. Draw a pV-diagram for this process. b) Find the
final temperature of the air if the pressure remains constant. Draw a pVdiagram for this process
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Adiabatic Process for an Ideal Gas
No heat transfer, Q = 0
U  W
dU  nCV dT
dW  pdV
dU  dW
nCV dT  pdV
nRT
nCV dT  
dV
V
dT R dV

0
T CV V
  Cp CV
C p C V C p
R


1   1
CV
Cp
CV
dT
dV
 (   1)
0
T
V
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Adiabatic Process for an Ideal Gas
ln T  (  1) ln V  cons tan t
ln T  ln V  1  cons tan t
ln(TV  1 )  cons tan t
TV  1  cons tan t
T1V1 1
T 2V2
 1 Adiabatic process,
ideal gas
pV  1
V
 cons tan t
nR
pV   cons tan t
p1V1 1  p 2 V2  Adiabatic process,
ideal gas
W  nCV (T1  T2 )
W
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CV
1
(p1V1  p 2 V2 ) 
(p1V1  p 2 V2 ) Adiabatic process,
ideal gas
R
 1