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Internal Energy, Work, Enthalpy,
Ideal Gas
19 September 2013
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Processes 2 An isobaricprocess is one in which the pressure is constant.
An isochoricprocess is one in which the volume is constant. An isothermalprocess is one in which the temperature is
constant.
An adiabaticprocess is one in which no heat enters or leaves
the system; i.e. Q = 0. An isentropicprocess is one in which the entropy is constant.
It is a reversible adiabatic process.
If a system is left to itself after undergoing a non-quasistaticprocess, it will reach equilibrium after a time t much longerthan the longest relaxation time involved; i.e. t .
Metastable equilibrium occurs when one particularrelaxation time 0is much longer than the time t for which
the system is observed; i.e. 0 t .
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Three Types of Process
Adiabat
Isotherm
P
VHeat bath or reservoir
Isothermal process Adiabatic process
Adiabatic free expansionP
V
1
2
End points
System
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The Ideal Gas Law
The triple point of water is Ttr
= 273.16 K, Ptr
= 6.0 x 103atm.
Triple point of Water
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The Ideal Gas Law Fixed point (1954)
The triple point of water is Ttr
= 273.16 K, Ptr
= 6.0 x 103atm.
Ideal gas law PV = nRT or Pv = RT, where n is the no. of kmoles, v is the volume per kmole, T is the absolute temperature in K, and
the gas constantR = 8.314 x 103J/(K.kmol).
For a constant quantity of gas, P1V1/T1= P2V2/T2.
P P V
V
T increasing
T T
V increasing
P increasing
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Van der Waals Equation 1 The Van der Waals equation of state
(P + a/v2)(vb) = RT,
reproduces the behavior of a real gas more accurately thanthe ideal gas equation through the empirical parameters aand b, which represent the following phenomena.
i. The term a/v2
represents the attractive intermolecularforces, which reduce the pressure at the walls compared tothat within the body of the gas.
ii. The termb represents the volume occupied by a kilomoleof the gas, which is unavailable to other molecules.
As a and b become smaller, or as T becomes larger, theequation approaches ideal gas equation Pv = RT.
An inflection point, which occurs on the curve at the critical
temperatureTc, gives the critical point(Tc,Pc).
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Van der Waals Equation 2
Below the critical temperature Tc, the curves show maxima and minima. Aphysically reasonable result is obtained by replacing the portion xy, with a straightline chosen so that A1= A2. C is the critical point.
A vapor, which occurs below the critical temperature,differs from a gasin that itmay be liquefied by applying pressure at constant temperature.
P
V
P
V
C
C
Isotherms at
higher T
Tc
A1 A2
Tcxy
vapor
gas
lower T
Inflection point
Isotherms at
higher T
Isotherms at
higher T
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Thermodynamic Work 1 Sign convention
The work done by the system is defined to be positive.
With this definition, the work done on the systemtheexternal workof mechanicsis negative.
The work done in a reversible processthe configurationworkis given by the product of an intensive variable and itscomplementary external variable; e.g. dW = PdV.
Reversible isochoric process W = 0, since V = 0.
Reversible isobaric process W = P V = P(V2V1).
These results hold for all materials.
The work done is alwayspositive for expansion
andnegative for compression.
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Thermodynamic Work 2
Calculating the work done in a reversible isothermal process
requires the equation of state of the system to be known.
Reversible isothermal process for an ideal gas (PV = nRT)
W = PdV = nRT dV/V = nRT ln(V2/V1).
In both cases, the work done by the system equals the shadedarea under curve.
Isobaric process
W = P(V2V1)P
V1 V2
Isothermal
process
P
VV1 V2
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Thermodynamic Work 3 Reversible cyclic process
W = PdV equals thearea enclosed by the PVcurve.
W is positive if the area
is traversed in aclockwise sense (asshown), and negative iftraversed counter-clockwise.
P
V
The equality W = PdV applies to reversible processes only.The work done in an irreversible process is given by the
inequality W < PdV.
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Expansivity and Compressibiity An equation of statemay be written as
P = P(V,T), V = V(T,P) or T = T(P,V). Thus, for example,
dV = (V/T)PdT + (V/P)TdP.
In general (x/y)z (y/z)x (z/x)y=1,or (y/x)z =(y/z)x (z/x)y .
Two experimental quantities which may be used to
find the equation of state are the following:coefficient of volume expansion (1/V) (V/T)P;
isothermal compressibility (1/V) (V/P)T.
Thus dV = VdTV dP.
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Useful Theorem
Remember the
negative signs.
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Hysteresis Hysteresis curves are examples of processes which may be
quasistatic, but are not reversible.
Hysteresis is caused by internal friction, and is a well-knownfeature offerromagnetismandfirst-order phase transitions.
The specification of the state of a homogeneous system by asmall number of thermodynamic variables breaks down in the
presence of hysteresis, since the equilibrium state dependson the previous history of the system.
Signal amplitudeof high
temperature phase.
Signal amplitude
of low
temperature phase.
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First Law of Thermodynamics
Microscopic picture
The internal energy U is made up of the translational androtational KE, and intermolecular PE of the gas molecules ofthe system.
For anideal monatomic gas, U is the total translational KE,known as the thermal energy, since it is proportional to T.
Q
WU
U = QW ,whereU is the increase of internal energy of the system, Q is
the heat entering the system, and W is the work done by the
system.
U depends onW because gas molecules
rebound off the piston moving to right with a
lower speed, thus reducing the KE of the gas.
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Exact and Inexact Differentials The differential form of the 1stLaw is
dU = dQdW,where dU is an exact differential, because U is a statevariable,and both dQ and dW are inexact differentials,since Q and W are not state variables.
Exact differential dF(x,y)dF is an exact differential if F(x,y) is a function of thevariables x and y. Thus
dF = A(x,y) dx + B(x,y) dy,
whereA(x,y) = (F/x)yand B(x,y) = (F/y)x.
Inexact differential dF(x,y)
IfdF = A(x,y) dx + B(x,y) dy is an inexact differential, there
is no function F(x,y) from which dF can be derived.
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Tests for an Exact Differential
Note: for a state function F = F(V,T,N),
dF = (F/V)T,NdV + (F/T)N,VdT + (F/N)V,TdN,
and 2
F/VT = 2
F/TV, etc.
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Heat Capacities The heat capacity at constant parameter i is given by
Ci= (dQ/dT)i .Note that one cannot use the partial form (Q/T)i ,since dQ is an inexact differential.
Heat capacity at Constant Volume CV dQ = dU + PdV, so thatCV= (dQ/dT)V= (U/T)V.
Heat capacity at Constant Pressure CP
CP= (dQ/dT)P= (U/T)P + P(V/T)P. The enthalpy H is defined as H = U + PV, so that
(H/T)P= (U/T)P + P(V/T)P.
Thus, CP= (H/T)P .
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Adiabatic process in an Ideal Gas 1 Ratio of specific heats = cP/cV= CP/CV.
For a reversible process, dU = dQrPdV. For an adiabatic process, dQr= 0, so thatdU =P dV.
For an ideal gas, U = U(T), so thatCV= dU/dT.
Also, PV = nRTandH = U + PV, so thatH =H(T).Thus, H = H(T) andCP= dH/dT.
Thus,CPCV= dH/dTdU/dT = d(PV)/dT = nR.
CPCV= nR is known as Mayers Equation, which holdsfor an ideal gas only.
For 1 kmole,cPcV= R, where cPand cVare specificheats.
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Adiabatic process in an Ideal Gas 2
SincedQ= 0for an adiabatic process,
dU =P dV anddU = CVdT,so that dT =(P/CV) dV .
For an ideal gas,PV = nRT,
so thatP dV +V dP = nR dT = (nRP/CV) dV.
HenceV dP + P (1 +nR/CV) dV = 0.
Thus,CVdP/P + (CV+ nR) dV/V = 0.
For an ideal gas,CPCV= nR.
so that CVdP/P + CPdV/V = 0,or dP/P + dV/V = 0.
Integration gives ln P + ln V = constant, so thatPV = constant.
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Adiabatic process in an Ideal Gas 2 Work done in a reversible adiabatic process
Method 1: direct integration For a reversible adiabatic process,PV =K.
Since the process is reversible, W = PdV,
so thatW = K V dV = [K/( 1)] V(1) |
=[1/( 1)] PV |
W =[1/( 1)] [P2V2 P1V1].
For an ideal monatomic gas, = 5/3, so that
W =(3/2)] [P2
V2
P1
V1
].
V2
V1
P1V1
P2V2
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Adiabatic process in an Ideal Gas 3 Work done in a reversible adiabatic process
Method 2: from 1stLaw For a reversible process,W = QrU
so that W = U, since Qr= 0 for an adiabatic process.
For an ideal gas, U = CVT = ncVT = ncV(T2T1).Thus,W =ncV(T2T1).
For an ideal gas PV = nRT,
so thatW = (cV/R)[P2V2 P1V1]. ButR = cPcV (Mayers relationship for an ideal gas),
so that W = [cV/(cPcV)][P2V2 P1V1]
i.e. W = [1/( 1)] [P2
V2
P1
V1
].
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Reversible Processes for an Ideal Gas
Adiabaticprocess IsothermalprocessIsobaricprocess Isochoricprocess
PV= K
= CP
/CV
T constant P constant V constant
W = [1/( 1)]
.[P2V2 P1V1]
W = nRT ln(V2 /V1) W = P V W = 0
U = CVT U = 0 U = CVT U = CVT
PV = nRT, U = ncVT, cPcV = R, = cP/cV.
Monatomic ideal gascV= (3/2)R, = 5/3.