first law of thermodynamics michael moats met eng 3620 chapter 2
TRANSCRIPT
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First Law of Thermodynamics
Michael Moats
Met Eng 3620
Chapter 2
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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
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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)
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Change in Internal EnergywqU
wqdU 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.
Pre
ssur
e
Volume
P1
P2
V2V1
ab
c
1
2
PdVw
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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.
),( TVUU
dTT
UdV
V
UdU vT )()(
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Convenient Processes
• Isochore or Isometric - Constant Volume
• Isobaric – Constant Pressure
• Isothermal – Constant Temperature
• Adiabatic – q = 0
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• If during a process the system maintains a constant volume, then no work is performed.
Recalland
Thus
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• Hence in a constant volume process, the change in internal energy is equal to heat absorbed or withdrawn from the system.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 Volume Processes
PdVwwqdU
vqdU
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Constant Pressure Processes
• Again starting with the first law and the definition of work:
• Combining them and integrating gives
where the subscript p means constant pressure
• Solving for qp and rearranging a little gives
)( 1212 VVPqUU p
PdVwwqdU
)()( 1122 PVUPVUq p
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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.
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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.
• Note: This is only true when phase change does not occur.
T
qC
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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.
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Thermal Process at Constant V
• Define heat capacity at a constant volume
• Recalling that at a constant volume
• Leading to
vqdU
vv dT
qC )(
vv dT
dUC )( dTCdU vor
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Thermal Process at Constant P
• Define heat capacity at a constant pressure
• Recalling that at a constant pressure
• Leading to
pqdH
pp dT
qC )(
orpp dT
dHC )( dTCdH p
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Molar Heat Capacity
• Heat capacity is an extensive property (e.g. dependent on size of system)
• Useful to define molar heat capacity
vv Cnc pp Cnc
where n is the number of moles and cv and cp are the molar heat capacity at constant volume and pressure, respectively.
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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.
pvp T
VPcc )(
Long derivation and further explanation in section 2.6Rcc vp
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Reversible Adiabatic Processes
• Adiabatic; q = 0
• Reversible;
• First law;
• Substitution gives us;
• For one mole of ideal gas;
• Recall that
• Leading to
PdVwwqdU
PdVdTCv
V
RTdVdTcv
xx
dxln
)ln()ln(2
1
1
2
V
VR
T
Tcv
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Reversible Adiabatic – cont.
• Rearranging gives
• Combining exponents and recalling that cp-cv=R gives
• Defining a term,
• From the ideal gas law
• Finally
RC
V
V
T
Tv )()(
2
1
1
2
v
vp
c
cc
V
V
T
T)(
2
1
1
2 )(
1
2
1
1
2 )(
V
V
T
Tvp cc / gives
)(11
22
1
2
VP
VP
T
T
1122 VPVP constant
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Reversible Isothermal P or V Change
• Recall• Isothermal means dT = 0, so dU = 0• Rearranging first law• Substituting reversible work
ideal gas law gives• Integration leaves• Isothermal process occurs at constant
internal energy and work done = heat absorbed.
dTCdU v
wq PdVw and
VRTdvPdVwq /)ln(
1
2
V
VRTwq
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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.
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First Determine Size of System
• Using the ideal gas law
molesK
moleK
atmliterlatm
RT
PVn 09.4
298**
*08206.0
10*10
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Isothermal Reversible Process
• Isothermal process; dT = 0, dU = 0
• To calculate work, first we need to know the final volume.
• Then we integrate
litersatm
litersatm
P
VPV 100
1
10*10)(
2
112
PdVw
kJliters
litersK
Kmole
Jmolesw
V
dVnRTPdVw
3.23)10
100ln(*298*
*3144.8*09.4
2
1
2
1
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Isothermal Reversible Process –continued
• Since dU = 0, q = w = 23.3 kJ
• Recall definition of enthalpy
H = U + PV
0)(
)(
1212
11221122
TTnRnRTnRTH
VPVPVPVPUH
Isothermal = constant temperature
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Reversible Adiabatic Expansion• Adiabatic means q = 0
• Recall
• Since cv = 3/2R, then cp = 5/2R and
• Solve for V3
• Solve for T3
1133 VPVP constant
vp cc / Rcc vp
3/5
litersP
VPV 8.39]
1
10*10[)( 5/3
3/55/3
3
3/511
3
K
Kmole
atmlitermoles
litersatm
nR
VPT 119]
*
*08206.0*09.4
8.39*1[)( 33
3
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Reversible Adiabatic Expansion - continued
kJKmoleK
Jmoles
TTncdTncwU vv
13.9)298119(**
3144.8*5.1*09.4
)( 13
3
1
Text shows five examples of path does not matter in determining U.
kJatmliter
JlitersatmlitersatmJ
VPVPUH
2.15*
3.101*)]10*10()8.39*1[(9130
)( 1133