Transcript First Law of Thermodynamics

First Law of Thermodynamics
Michael Moats
Met Eng 3620
Chapter 2
Thermodynamic Functions
• To adequately describe the energy state of a
system two terms are introduced
– Internal Energy, U
– Enthalpy, H
• Internal Energy related to the energy of a body.
– Example: Potential Energy
• Body has a potential energy related to its mass and height
(mgh)
• To move a body from one height to another takes work, w.
w = force * distance = mg * (h2-h1) = mgh2 – mgh1
= Potential Energy at height 2 – Potential Energy at Point 1
Work and Heat
• A system interacts with its environment
through work, w, or heat, q.
w = -(UB – UA) (if no heat is involved)
• If w < 0, work is done to the system
• If w > 0, work is done by the system
q = (UB – UA) (if no work is involved)
• If q < 0, work flows out of the body (exothermic)
• If q > 0, heat flows into the body (endothermic)
Change in Internal Energy
U  q  w
dU  q  w
Pressure
U is a state function. Therefore, the path between condition
1 and condition 2 do not affect the differential value. Heat
and work are not state functions. The path taken between
condition 1 and 2 does matter, hence the use of a partial
differential.
P1
1
b
a
w  PdV
c
2
P2
V1
V2
Volume
Fixing Internal Energy
• For a simple system consisting of a given
amount of substance of a fixed
composition, U is fixed once any two
properties (independent variables) are
fixed.
U  U (V , T )
U
U
dU  (
)T dV  (
) v dT
V
T
Convenient Processes
•
•
•
•
Isochore or Isometric - Constant Volume
Isobaric – Constant Pressure
Isothermal – Constant Temperature
Adiabatic – q = 0
Constant Volume Processes
• If during a process the system maintains a constant
volume, then no work is performed.
w  PdV
Recall
and
dU  q  w
Thus
dU  qv
where the subscript v means constant volume
• Hence in a constant volume process, the change in
internal energy is equal to heat absorbed or withdrawn
from the system.
Constant Pressure Processes
• Again starting with the first law and the definition
of work:
dU  q  w w  PdV
• Combining them and integrating gives
U 2  U1  q p  P(V2  V1 )
where the subscript p means constant pressure
• Solving for qp and rearranging a little gives
q p  (U 2  PV2 )  (U1  PV1 )
Enthalpy
• Since U, P and V are all state function, the
expression U+PV is also a state function.
• This state function is termed enthalpy, H
H = U + PV
• Therefore qp = H2 – H1 = ΔH
• In a constant pressure system, the heat
absorbed or withdrawn is equal to the
change in enthalpy.
Heat Capacity
• Before discussing isothermal or adiabatic
processes, a new term is needed to make the
calculations easier.
• Heat Capacity, C is equal to the ratio of the
heat absorbed or withdrawn from the system to
the resultant change in temperature.
q
C
T
• Note: This is only true when phase change does not occur.
Defining Thermal Process Path
• To state that the system has changed
temperature is not enough to define
change in the internal energy.
• It is most convenient to combine change in
temperature while holding volume or
pressure constant.
• Then calculations can be made as to the
work performed and/or heat generated.
Thermal Process at Constant V
• Define heat capacity
at a constant volume
• Recalling that at a
constant volume
• Leading to
dU
Cv  (
dT
q
Cv  ( ) v
dT
dU  qv
)v
or
dU  Cv dT
Thermal Process at Constant P
• Define heat capacity at
a constant pressure
• Recalling that at a
constant pressure
• Leading to
dH
Cp  (
dT
q
Cp  ( )p
dT
dH  q p
)p
or
dH  C p dT
Molar Heat Capacity
• Heat capacity is an extensive property
(e.g. dependent on size of system)
• Useful to define molar heat capacity
ncv  Cv
ncp  C p
where n is the number of moles and cv and cp are the molar heat
capacity at constant volume and pressure, respectively.
Molar Heat Capacity - II
• cp > cv
• cv – heat only needed to raise temperature
• cp – heat needed to raise temperature and
perform work
• Therefore the difference between cv and cp
is the work performed.
c p  cv  P (
V
)p
T
c p  cv  R
Long derivation and
further explanation
in section 2.6
Reversible Adiabatic Processes
•
•
•
•
•
•
•
Adiabatic; q = 0
Reversible; w  PdV
First law; dU  q  w
Substitution gives us; Cv dT  PdV
For one mole of ideal gas; c dT   RTdV
V
Recall that  dxx  ln x
Leading to
T2
V1
v
cv ln( )  R ln( )
T1
V2
Reversible Adiabatic – cont.
T2 Cv
V1 R
( ) ( )
T1
V2
• Rearranging gives
• Combining exponents and recalling that
( c c )
cp-cv=R gives T2 V1 c
p
T1
(
V2
)
v
v
T2
V1  1
• Defining a term,   c p / cv gives T  (V )
1
2
• From the ideal gas law
• Finally P V   PV  
2 2
1 1
T2
P2V2
(
)
T1
P1V1
constant
Reversible Isothermal
P or V Change
Recall dU  Cv dT
Isothermal means dT = 0, so dU = 0
Rearranging first law q  w
Substituting reversible work w  PdV and
ideal gas law gives q  w  PdV  RTdv/ V
V2
• Integration leaves q  w  RT ln(V )
1
• Isothermal process occurs at constant
internal energy and work done = heat
absorbed.
•
•
•
•
Example Calculation
• 10 liters of monatomic ideal gas at 25oC
and 10 atm are expanded to 1 atm. The cv
= 3/2R. Calculate work done, heat
absorbed and the change in internal
energy and enthalpy for both a reversible
isothermal process and an adiabatic and
reversible process.
First Determine Size of System
• Using the ideal gas law
PV
10atm*10l
n

 4.09moles
RT 0.08206liter * atm * 298K
K * mole
Isothermal Reversible Process
• Isothermal process; dT = 0, dU = 0
• To calculate work, first we need to know
the final volume. V  ( PV )  10atm *10liters  100liters
1 1
2
P2
1atm
• Then we integrate w  PdV
2
2
dV
w   PdV  nRT 
V
1
1
J
100liters
w  4.09m oles* 8.3144
* 298K * ln(
)  23.3kJ
m ole* K
10liters
Isothermal Reversible Process –
continued
• Since dU = 0, q = w = 23.3 kJ
• Recall definition of enthalpy
H = U + PV
H  U  ( P2V2  P1V1 )  P2V2  P1V1
H  nRT2  nRT1  nR(T2  T1 )  0
Isothermal = constant temperature
Reversible Adiabatic Expansion
• Adiabatic means q = 0
• Recall P3V3  P1V1  constant
  c p / cv c p  cv  R
• Since cv = 3/2R, then cp = 5/2R and   5 / 3
• Solve for V3
P1V15 / 3 3 / 5 10*105 / 3 3 / 5
V3  (
) [
]  39.8liters
P3
1
• Solve for T3
P3V3
1atm* 39.8liters
T3  (
) [
]  119K
liter * atm
nR
4.09moles* 0.08206
mole* K
Reversible Adiabatic Expansion continued
3
U   w   ncv dT  ncv (T3  T1 ) 
1
J
4.09m oles* 1.5 * 8.3144
* (119  298) K  9.13kJ
K * m ole
Text shows five examples of path does
not matter in determining U.
H  U  ( P3V3  P1V1 ) 
 9130J  [(1atm * 39.8liters)  (10atm * 10liters)] * 101.3
J
 15.2kJ
liter * atm